Holtmann75 / Module Math 75 Holtmann Revised 2015-11-18.pdf

p. i

Math 75

Linear Algebra

Class Notes

Prof. Erich Holtmann

For use with

Elementary Linear Algebra,

7th

ed., Larson

Revised 21-Nov-2015

p. i

Contents

Chapter 1: Systems of Linear Equations 1

1.1 Introduction to Systems of Equations. 1

1.2 Gaussian Elimination and Gauss-Jordan Elimination. 9

1.3 Applications of Systems of Linear Equations. 15

Chapter 2: Matrices. 19

2.1 Operations with Matrices. 19

2.2 Properties of Matrix Operations. 23

2.3 The Inverse of a Matrix. 27

2.4 Elementary Matrices. 31

2.5 Applications of Matrix Operations. 37

Chapter 3: Determinants. 41

3.1 The Determinant of a Matrix. 41

3.2 Determinants and Elementary Operations. 45

3.3 Properties of Determinants. 51

3.4 Applications of Determinants. 55

Chapter 4: Vector Spaces. 61

8.1 Complex Numbers (Optional). 61

8.2 Conjugates and Division of Complex Numbers (Optional). 67

4.1 Vectors in Rn. 71

4.2 Vector Spaces. 77

4.3 Subspaces of Vector Spaces. 82

4.4 Spanning Sets and Linear Independence. 86

4.5 Basis and Dimension. 97

4.6 Rank of a Matrix and Systems of Linear Equations. 105

4.7 Coordinates and Change of Basis. 117

Chapter 5: Inner Product Spaces. 121

5.1 Length and Dot Product in Rn. 121

5.2 Inner Product Spaces. 129

5.3 Orthogonal Bases: Gram-Schmidt Process. 137

5.4 Mathematical Models and Least Squares Analysis (Optional). 145

5.5 Applications of Inner Product Spaces (Optional). 151

8.3 Polar Form and De Moivre's Theorem. (Optional) 157

8.4 Complex Vector Spaces and Inner Products. 163

p. ii

Chapter 6: Linear Transformations. 169

6.1 Introduction to Linear Transformations. 169

6.2 The Kernel and Range of a Linear Transformation. 175

6.3 Matrices for Linear Transformations. 183

6.4 Transition Matrices and Similarity. 191

6.5 Applications of Linear Transformations. 193

Chapter 7: Eigenvalues and Eigenvectors. 201

7.1 Eigenvalues and Eigenvectors. 201

7.2 Diagonalization. 209

7.3 Symmetric Matrices and Orthogonal Diagonalization. 215

7.4 Applications of Eigenvalues and Eigenvectors. 223

8.5 Unitary and Hermitian Spaces. 223

Chapter 1: Systems of Linear Equations

1.1 Introduction to Systems of Equations.

p. 1



Objective: Recognize and solve mn systems of linear equations by hand using Gaussian

elimination and back-substitution.

a1x1 + a2x2 + … + anxn = b is a linear equation in standard form in n variables xi.

The first nonzero coefficient ai is the leading coefficient. The constant term is b.

Compare to the familiar forms of linear equation s in two variables y = mx + b and x = a.

Example: Linear and Nonlinear Equations

(sin ) x1 – 4x2 = e2

sin x1 + 2x2 – 3x3 = 0

Linear Nonlinear

An mn system of linear equations is a set of m linear equations in n unknowns.

Example: Systems of Two Equations in Two Variables

Solve and graph each 22 system.

a. 2x – y = 1

x + y = 5

b. 2x – y = 1

–4x + 2y = –2

c. 2x – y = 1

–4x + 2y = 5

For a system of linear equations, exactly one of the following is true.

1) The system has exactly one solution (consistent, nonsingular system).

2) The system has infinitely many solutions (consistent, singular system).

Use a free parameter or free variable (or several free parameters) to represent the solution set.

3) The system has no solution (inconsistent, singular system).



p. 2

To solve mn systems of linear equations (when m and n are large) we use a procedure called

Gaussian elimination to find an equivalent system of equations in row-echelon form. Then we

use back-substitution to solve for each variable.

Row-echelon form means that the leading coefficients of 1 (called “pivots”) and the zero terms

below them form a stair-step pattern. You could walk downstairs from the top left. You might

have to move more than one column to the right to reach the next step, but you never have to

step down more than one row at a time.

Row-echelon form

91

841

11

731251

5

54

3

54321

x

xx

x

xxxxx

Row-echelon form

00

00

11

731251

3

54321

x

xxxxx

Not row-echelon form

931

8421

1231

731251

54

543

543

54321

xx

xxx

xxx

xxxxx

The goal of Gaussian elimination is to find an equivalent system that is in row-echelon form. The

three operations you can use during Gaussian elimination are

1) Swap the order of two equations.

2) Multiply an equation on both sides by a non-zero constant.

3) Add a multiple of one equation to another equation.

In Gaussian elimination, you start with Equation 1 (the first equation of your mn system).

1) Find the leading coefficient in the current equation. (Sometimes you need to swap

equations in this step.)

2) Eliminate the coefficients of the corresponding variable in all of the equations below the

current equation.

3) Move down to the next equation and go back to Step 1. Repeat until you run out of

equations or you run out of variables.

Solve using back-substitution: solve the last equation for the leading variable, then substitute into

the preceding (i.e. second-to-last) equation and solve for its leading variable, then substitute into

the preceding equation and solve for its leading variable, etc. Variables that are not leading

variables are free parameters, and we often set them equal to t, s, ….



p. 3

Examples: Gaussian Elimination and Back-Substitution on 33 Systems of Linear Equations.



p. 4



p. 5



p. 6



p. 7

Example: Chemistry Application

Write and solve a system of linear equations for the chemical reaction

(x1)CH4 + (x2)O2 (x3)CO2 + (x4)H2O

Solution: write a separate equation for each element, showing the balance of that element.

C: 1x1 + 0x2 = 1x3 + 0x4 so 1x1 + 0x2 – 1x3 + 0x4 = 0

H: 4x1 + 0x2 = 0x3 + 2x4 so 4x1 + 0x2 + 0x3 – 2x4 = 0

O: 0x1 + 2x2 = 2x3 + 1x4 so 0x1 + 2x2 – 2x3 – 1x4 = 0


1.2 Gaussian Elimination and Gauss-Jordan Elimination.

p. 9


Objective: Use matrices and Gauss-Jordan elimination to solve mn systems of linear equations

by hand and with software.

Objective: Use matrices and Gaussian elimination with back-substitution to solve mn systems

of linear equations by hand and with software.

Use row-echelon form or reduced row-echelon form to determine the number of solutions of a

homogeneous system of linear equations, and (if applicable) the number of free parameters.

A matrix is a rectangular array of numbers, called matrix

entries, arranged in horizontal rows and vertical columns.

Matrices are denoted by capital letters; matrix entries are

denoted by lowercase letters with two indices. In a given

matrix entry aij, the first index i is the row, and the second

index j is the column. The entries a11, a22, a33, … compose the

main diagonal. If m = n then A is called a square matrix.

A linear system

mnmnmmm

nn

nn

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

332211

22323222121

11313212111

can represented either by a coefficient matrix A and a column vector b

mnmmm

n

n

n

aaaa

aaaa

aaaa

aaaa

A

331

3333231

2232221

1131211

and

mb

b

b

b

3

2

1

b

or by an augmented matrix M, which I will sometimes write as [A | b]

mmnmmm

n

n

n

baaaa

baaaa

baaaa

baaaa

M

331

33333231

22232221

11131211

(The book, Mathematica, and the calculator do not display the dotted vertical line.)

To create M in Mathematica from A and b, type m=Join[a,b,2]

To create M on the TI-89 from A and b, type

Matrixaugment(A

B

M

mnmmm

n

n

n

aaaa

aaaa

aaaa

aaaa

A

321

3333231

2232221

1131211



p. 10

In a similar manner to that used for an mn system of linear equations, we can use a Gaussian

elimination on the coefficient side A of the augmented matrix [A | b] to find an equivalent

augmented matrix [U | c] in row-echelon form. Then we use back-substitution to solve for each

variable. U is called an upper triangular matrix because all non-zero entries are on or above the

main diagonal.

row-echelon form row-echelon form not row-echelon form

20000

41000

50100

31251

00000

00000

50100

31251

31000

42100

23100

31251

Example: Use Gaussian elimination and back-substitution to solve. The three elementary row

operations you can use during Gaussian elimination are

1) Swap the two rows.

2) Multiply a row by a non-zero constant.

3) Add a multiple of one row to another row.

2111

3123

8346

zyx

zyx

zyx

2111

3123

8346

2100

210

1

23

34

21

32

1)2()1( so

1)2(2 so 2

2

21

32

34

34

21

32

23

23

xzyx

yzy

z

Steps:



p. 11

Instead of using back-substitution, we can take the row-echelon form [U | c] and eliminate the

coefficients above the pivots by adding multiples of the pivot rows. The result [R | d] is called

reduced row-echelon form.

reduced row-echelon form reduced row-echelon form

20000

41000

50100

110051

30000

40000

23100

31051

Example: Use Gauss-Jordan elimination to solve.

2111

3123

8346

zyx

zyx

zyx

2111

3123

8346

2100

210

1

23

34

21

32

2100

1010

1001

2

1

1

z

y

x



p. 12

Example Using software to find the Reduced Row-Echelon Form (Exercise 1.2 #39)

Mathematica

Go to the Palettes Menu and open the Basic Math

Assistant. Under Basic Commands, open the Matrix

Tab.

Type a=

and click

Use the Add Row and Add Column buttons to expand

the matrix, so you can enter the augmented matrix

(If you make the matrix to large, use to remove rows and columns.) Press (or

on the number pad) when you are finished entering the augmented matrix.

Another way to enter the augmented matrix is to type (or download) a={{1,-1,2,2,6,6},{3,-2,4,4,12,14},{0,1,-1,-1,-3,-3},{2,-

2,4,5,15,10},{2,-2,4,4,13,13}}

MatrixForm[a]

You can download this matrix from http://holtmann75.pbworks.com. Click on Electronic

Data Sets and open 1133110878_323834.zip/DataSets/Mathematica/0102039.nb (Ch. 01,

Section 02, Problem 039). Notice that A and a are different variables! User-defined variables

should always begin with a lower-case letter, because Mathematica’s built-in fuctions and

commands begin with capital letters. (For example, N and C are already defined by

Mathematica.)

On the Basic Math Assistant palette, click on MatrixForm RowReduce and type a so you have

MatrixForm[RowReduce[a]]

Converting back to a system of linear equations, we have

x1 = 2 x4 = 5

x2 = 2 x5 = 1

x3 = 3

http://holtmann75.pbworks.com/

http://academic.cengage.com/resource_uploads/downloads/1133110878_323834.zip





p. 13

TI-89: Type

Data/Matrix Editor

New...

Type: Matrix

Folder: main

Variable: A

(A is above . If you type = instead of a, use .

If a already exists, then use Open... on the previous

screen instead of New....)

Use and the

arrow keys to type

in the coefficient matrix

(If you need to insert or delete a row or column, use )

When you are finished, press .

Another way to enter the matrix is to type (from the

home screen) 1,1,2,2,6,6

3,2,4,4,12,14

0,1,1,1,3,3

2,2,4,5,15,10

2,2,4,4,13,13

A

After the matrix is entered, type A

Matrix rref( A

Converting back to a system of linear equations, we have

x1 = 2

x2 = 2

x3 = 3

x4 = 5

x5 = 1



p. 14

Mathematica can help you perform row operations. If your augmented matrix is in a, then

a[[2]] is row 2 of the matrix. To view a in the usual format, type

MatrixForm[a]

To swap rows 2 and 3 of a, type a[[{2,3}]]=a[[{3,2}]]

To multiply row 2 of a by 7, type a[[2]]=7*a[[2]]

To add 7 times row 2 to row 1, type a[[1]]=a[[1]]+7*a[[2]]

Mathematica performs operations in the order that you type them in, not as they appear on the

screen. If you go back and edit your work, you can use Evaluate Notebook under the Evaluation

menu to recalculate the notebook in the order on the screen.

The TI-89 can also help you perform row operations. If your augmented matrix is in a, then …

To swap rows 2 and 3 of a and store the result in a1, type

Matrix Row opsrowSwap(A,2,3,)

A1

To multiply row 2 of a1 by 7 and store the result in a2, type

Matrix Row opsmRow(7,A1,2)

A2

To add 7 times row 2 of a2 to row 1 and store the result in a3, type

Matrix Row opsmRowAdd(7,A2,1,3)

A3

About notation:

43

21 and

43

21 are matrices, but

43

21 is a determinant (Ch. 3).

Theorem 1.1. Every homogeneous (constants on right-hand side are all zeroes) system of linear

equations is consistent. The number of free parameters in the solution set is the number of

variables minus the number of pivots (leading coefficients). If there are zero free parameters,

then there is exactly one solution.

Examples:

0112

0201

0511

0121428205

0559211

011150

042511

02131500

0377208

0454156

012352

Solutions:

0000

0310

0201


1.3 Applications of Systems of Linear Equations.

p. 15


Objective: Set up and solve a system of equations to fit a polynomial function to a set of data

points.

Objective: Set up and solve a system of equations to represent a network.

Polynomial Curve Fitting.

Given m data points (t1, y1), (t2, y2), …, (tm, ym). We want to find a polynomial of degree m–1 that

passes through these points.

p(t) = c0 + c1t + c2t2 + … + cmt

m Notice that yj = c0 + c1(tj)+ c2(tj)

2 + … + cm–1(tj)

m–1

This produces an mm linear system:

1

1

22

1

11

)(1

)(1

)(1

m

mm

m

m

tt

tt

tt

1

1

0

mc

c

c

=

my

y

y

2

1

that you can solve using Gauss-Jordan elimination.

Example 1.3#7

086028

21012

y

t



p. 16

Example 1.3#9

1275

200820072006

y

t

Network Analysis: write a system of linear equations using Kirchoff’s Laws.

1) Flow into each node (also called vertex) equals flow out.

2) In an electrical network, the sum of the products IR (I = current and R = resistance) around

any closed path of edges (lines) is equal to the total voltage in the loop from the batteries.

A resistor is represented by the symbol Resistance is measured in ohms ().

1 k = 1000

A battery is represented by the symbol If the current flows through the battery

from the short line (–) to the long line

(+), then the voltage is positive.

Current is measure in amps (A). 1 mA = 0.001 A.

Flow into a node is positive. Flow out of a node is negative.

To write a system of equations,

1) Pick a direction (at random) for each current I.

2) For each node, write an equation for the current input and output.

3) For each loop, write the V = IR equation.

I



p. 17

Example: 1.3.#32.

Solve for the currents.

Example: 1.3.29. The figure shows traffic flow in

vehicles/hour through a network of streets.

a) Solve for x1, x2, x3, and x4.

b) Find the traffic flow when x4 = 0.

c) Find the traffic flow when x4 = 100.

Chapter 2: Matrices.

2.1 Operations with Matrices.

p. 19



Objective: Determine whether or not two matrices are equal.

Objective: Add and subtract matrices and multiply a matrix by a scalar.

Objective: Multiply two matrices.

Objective: Write solutions to a system of linear equations in column vector notation.

Objective: Partition a matrix and perform block multiplication.

Three ways to represent the same matrix are A = [aij] =

mnmm

n

n

aaa

aaa

aaa

31

22221

11211

.

Two matrices A = [aij] and B = [bij] are equal if and only if they have the same dimensions or

order (mn) and aij = bij for all 1 i m and 1 j n.

Comment on logic:

An example of “P if Q” (also written P Q) is “It is cloudy if it is raining.”

An example of “Q only if P” (i.e. “if Q then P,” also written Q P) is “It is raining only

if it is cloudy.”

“R iff S” is shorthand for “R if and only if S” (also written R S). R iff S means that R is

equivalent to S.

To add (or subtract) two matrices A = [aij] and B = [bij] that have the same dimensions, add (or

subtract) corresponding matrix entries.

A + B = [aij + bij] A – B = [aij – bij]

The sum (and difference) of two matrices with different dimensions is undefined.

To multiply a matrix by a scalar (number), multiply each entry by that scalar.

cA = [caij]

Examples:

Let A =

103

421, B =

231

341, and D =

31

50. Find A + B, A – B, A + D, and 4A.



p. 20

In Mathematica, use a+b, a–b, and 4*a or 4a to add matrices, subtract matrices, and multiply

scalars c by matrices. Remember, you can enter D =

31

50 from the Palettes Menu::Basic

Math Assistant::Basic Commands::Matrix Tab.

On the TI-89, use ab, a b, and ca or 4a to add matrices A and B, subtract matrices,

and multiply scalars c by matrices. Remember, you can enter D =

31

50 from the home screen

as [0,5;-1,3]D or from Data/Matrix Editor. (Be sure to use

Type: Matrix)

Matrix multiplication is defined in a much more complex manner. For a 11 system, we can

write ax = b. We want a definition of matrix multiplication that allows us to write an mn system

mnmnmmm

nn

nn

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

332211

22323222121

11313212111

as Ax = b,

where A is the coefficient matrix

mnmm

n

n

aaa

aaa

aaa

A

21

22221

11211

,

and x and b are column matrices (or column vectors):

nx

x

x

2

1

x and

mb

b

b

2

1

b .

Observe that each row of A was multiplied by the column x to give the corresponding row of b.

If A = [aij] is an mn matrix and B = [bij] is an np matrix, then the product AB is an mp matrix

A = [cij] where

[cij] =

n

k

kjikba1

= ai1b1j + ai2b2j + ai3b3j + … ainbnj

If the column dimension of A does not match the row dimension of B, then the product is

undefined



p. 21

Examples

52

33

23

32

512

014

In Mathematica, use a.b to multiply matrices.(Do not use a*b)

On the TI-89, use ab multiply matrices.

Example: Write solutions to a system of linear equations in column vector notation.

Solve

2110

1011

3121

4

3

2

1

x

x

x

x

=

0

0

0

4

3

2

1

x

x

x

x

= t

1

1

1

2

Block multiplication on partitioned matrices works whenever the dimensions are OK.

Examples

2221

1211

1200

0001

0010

AA

AAA with

1200

00

00

01

10

2221

1211

AA

AA and

2221

1211

265

134

011

021

BB

BBB with

2

1

65

34

0

0

11

21

2221

1211

BB

BB

Then AB =

2222122121221121

2212121121121111

BABABABA

BABABABA =

2222212211

2211211111

000

000

BABAB

BABBA=

01213

021

011



p. 22

Ax=

mnmm

n

n

aaa

aaa

aaa

21

22221

11211

nx

x

x

2

1

=

nmnmm

nn

nn

xaxaxa

xaxaxa

xaxaxa

2211

2222121

1212111

=n

mn

n

n

mm

x

a

a

a

x

a

a

a

x

a

a

a

2

1

2

2

22

12

1

1

21

11

In block matrix notation, we write [ a1 | a2 | … | an ] =

mnmm

n

n

aaa

aaa

aaa

21

22221

11211

where the ai are

m1 column matrices. The Ax = [ a1 | a2 | … | an ]

nx

x

x

2

1

= a1x1 + a2x2 + … + anxn

Similarly, if B is an lm matrix (so BA is defined), then

BA = B[ a1 | a2 | … | an ] = [ Ba1 | Ba2 | … | Ban ], i.e. the columns of BA are Bai.

Notice that B is lm, ai is m1, Bai is l1, and BA is ln.

On the other hand, we could partition A into 1m row matrices ri

A =

mnmm

n

n

aaa

aaa

aaa

21

22221

11211

=

mr

r

r

2

1

. Then Ax =

mnmm

n

n

aaa

aaa

aaa

21

22221

11211

nx

x

x

2

1

=

mr

r

r

2

1

x =

xr

xr

xr

m

2

1

.

Notice that ri is 1n, x is n1, rix is 11 (i.e. a number), and Ax is m1.

If E =

lmll

m

m

eee

eee

eee

21

22221

11211

=

le

e

e

2

1

then EA =

le

e

e

2

1

A =

A

A

A

le

e

e

2

1

.

Notice that ei is 1m, A is mn, eiA is 1n, and EA is ln

If e = [ e1 | e2 | … | em], then eA = [ e1 | e2 | … | em]

mr

r

r

2

1

= e1r1 + e2r2 + … + emrm

Notice that e is 1m, A is mn, ei is a number, ri is 1n, and eA is 1n.


2.2 Properties of Matrix Operations.

p. 23


Objective: Know and use the properties of matrix operations (matrix addition and subtraction,

scalar multiplication, and matrix multiplication), and of the zero and identity matrices.

Objective: Know which properties of fields do not hold for matrices (commutativity of matrix

multiplication and existence of multiplicative inverses).

Objective: Find the transpose of a matrix and know properties of the transpose.

The real numbers R, together with the operations of addition and multiplication, is an example of

a mathematical field. (The complex numbers C with addition and multiplication is another

example.) Fields and their operations have all of the usual properties.

1) Closure under addition: if a and b R, then a + b R.

2) Addition is associative: (a + b) + c = a + (b + c)

3) Addition is commutative: a + b = b + a

4) Additive identity (zero): R contains 0, which has the property that a + 0 = a for all a.

5) Additive inverses (opposites): every a R. has an opposite –a, such that a + (–a) = 0.

We define subtraction a – b as a + (–b).

6) Closure under multiplication: if a and b R, then ab R.

7) Multiplication is associative: (ab)c = a(bc)

8) Multiplication is commutative: ab = ba

9) Multiplication distributes over addition: a(b + c) = ab +ac

10) Multiplicative identity (one): R contains 1, which has the property that 1a = a for all a.

11) Multiplicative inverses (reciprocals): every a R. has an inverse a–1

, such that aa–1

= 1.

We define division a b as ab–1

.

Matrix addition (and subtraction) has all of the usual properties: closure, associativity,

commutativity, zero matrices, and opposites.

A zero matrix has zero in all entries, but because the matrix can have any dimensions mn, we

have many zero matrices 0mn.

The opposite of a matrix [aij] is –[aij] = [–aij].

Multiplication of a scalar by a matrix has all of the usual properties: closure, associativity,

commutativity, multiplicative identity (the scalar 1), and distribution. For distribution, we have

both c(A + B) = cA + cB and (c + d)A = cA + dA.

Examples:

(–1)

97

98= –

97

98=

97

98,

356

295+

356

295=

000

000= 023



p. 24

Multiplication of matrices is closed, is associative, and distributes over matrix addition. The

multiplicative identity matrices are square matrices with ones on the main diagonal and zeros

everyplace else: Inn =

100

010

001

. If A is mn, then Imm Amn = Amn and Amn Inn = Amn

We define exponents for square matrices A and positive integers k: Ak =

timesk

AAA . Also, we

define A0 = I.

Multiplication of matrices is not commutative in general.

Many matrices do not have multiplicative inverses. Division of matrices is undefined.

Examples:

10

01

356

295=

356

295

100

010

001

=

97

98

356

295 =

356

295

97

98=

356

295

13

12

01

=

13

12

01

356

295=

Theorem 2.5: For a system of linear equations, exactly one of the following is true.

1) The system has no solution (inconsistent, singular system).

2) The system has exactly one solution (consistent, nonsingular system).

3) The system has infinitely many solutions (consistent, singular system).

Use a free parameter or free variable (or several free parameters) to represent the solution set.

Proof using matrix operations:

Given a system of linear equations Ax = b. Exactly one of the following is true: the system has

no solution, the system has exactly one solution, or the system has at least two solutions (call

them x1 and x2).



p. 25

*

If the system has two solutions, then Ax1 = b and Ax2 = b so

A(x1 – x2) = Ax1 – Ax2 = b – b = 0.

Let xh = x1 – x2, so xh is a nonzero solution to the homogenous equation Ax = 0.

Then for any scalar t, x1 + txh is a solution of Ax = b because

A(x1 + txh) = Ax1 + tAxh = b+ t 0 = b.

Thus, in the last case, the system has infinitely many solutions with a parameter t.

The transpose AT of a matrix A is formed by writing its columns as rows. For example,

if

mnmmm

n

n

n

aaaa

aaaa

aaaa

aaaa

A

321

3333231

2232221

1131211

then

mnnnn

m

m

m

T

aaaa

aaaa

aaaa

aaaa

A

321

3332313

2322212

1312111

.

Equivalently, the transpose of A is formed by writing rows as columns. The ij entry of AT is aji.

Example: Find the transpose of A =

8251

4803

0739

Mathematica: to take the transpose of a matrix, either type Transpose[a] (also available on

the Basic Math Assistant palette) or type a followed by tr (four separate keystrokes).

After the first three keystrokes, you will see atr. After the fourth keystroke, atr will

change to aT. Of course, at the end, type

TI-89: A

MatrixT

Properties of the transpose:

1) (AT)T = A

2) (A + B)T = A

T +B

T

3) (cA) T

=cAT

4) (AB) T

= BTA

T



p. 26

Example of Property #4:

Consider A = and B =

The 32 entry of (AB)T is the 23 entry of AB =

which is a21b13 + a22b23 + a23b33

If we look at AT and B

T, we much reverse the order of multiplication to obtain row column.

BTA

T =

The 32 entry BTA

T is b13a21 + b23a22 + b33a23. You can see that (AB)

T = B

TA

T

A matrix M is symmetric iff MT = M.

Example: M =

525

246

5610

is symmetric. Prove that AAT is symmetric for any matrix A.

Proof:


2.3 The Inverse of a Matrix.

p. 27


Objective: Find the inverse of a matrix (if it exists) by Gauss-Jordan elimination.

Objective: Use properties of inverse matrices.

Objective: Use an inverse matrix to solve a system of linear equations.

A square nn matrix is invertible (or nonsingular) when there exists an nn matrix A–1

such that

AA–1

= Inn and A–1

A = Inn

where Inn is the identity matrix. A–1

is called the (multiplicative) inverse of A. A matrix that does

not have an inverse is called singular (or noninvertible). Nonsquare matrices do not have

inverses.

Theorem 2.7: If A is an invertible matrix, then the inverse is unique.

Proof: let B and C be inverses of A. Then BA = I an AC = I. So

B = BI = B(AC) = (BA)C = IC =C. Therefore, B = C and the inverse of A is unique.

Example: Find the inverse of A =

21

53.

Solution: We need to solve the system AX = I, or

21

53

2221

1211

xx

xx =

10

01,

which gives four equations 121021

053153

22122111

22122111

xxxx

xxxx

To solve these four equations, we take the reduced row echelon forms

021

153

110

201 and

121

053

310

501, so A

–1 = X =

31

52

Since the row operations performed to find the reduced row echelon form depend only on the

coefficient part of the augmented matrix

21

53, we could solve all four equations

simultaneously by using a doubly augmented matrix

[ A | I ] =

1021

0153

3110

5201= [ I | A

–1 ]

To create the doubly augmented matrix in Mathematica from A, type

m=Join[a, IdentityMatrix[2],2]

To create the doubly augmented matrix on the TI-89 from A, type



p. 28

*

Matrix

augment(

A,

Matrix

identity(2))

M

We used IdentityMatrix[2] and identity(2) because we wanted a 22 matrix.

To find the inverse of an nn matrix A by Gauss-Jordan elimination, find the reduced row

echelon form of th n2n augmented matrix [ A | I ]. If the nn block on the left can be reduced to

I, then the nn block on the right is A–1

[ A | I ] [ I | A–1

]

If the nn block on the left cannnot be reduced to I, then A is not invertible.

Example: Invert A =

321

221

111

using Gauss-Jordan elimination.

Example: Invert A =

332

221

111

using Gauss-Jordan elimination.

Example: Invert M =

dc

ba using Gauss-Jordan elimination.

You can also use software. First clear the variables using Clear[a,b,c,d] or

Clear a-z, then type Inverse[m] or

M^1

(also on Basic Math Assistant palette)



p. 29

*

Theorem 2.8 Properties of Inverse Matrices

If A is an invertible matrix, k is a positive integer, and c is a nonzero scalar, then A–1

, Ak, cA, and

AT are invertible, and

1) (A–1

) –1

= A

2) (Ak)–1

= (A–1

)k

3) (cA) –1

= c1 A

–1

4) (AT)–1

= (A–1

)T

Proof:

1)

2) Proof by Induction (see the Appendix)

When k = 1,

If (Ak)–1

= (A–1

)k,

By mathematical induction, we conclude that (Ak)

–1 = (A

–1) for all positive integers k.

3)

4)



p. 30

*

*

*

Theorem 2.9 Inverse of a Product

If A and B are invertible nn matrices, then (AB)–1

= B–1

A–1

.

Proof:

Theorem 2.10 Cancellation Properties

Let C be an invertible matrix.

1) If AC = BC, then A = B. (Right cancellation property)

2) If CA = CB, then A = B. (Left cancellation property)

Theorem 2.11 Systems of Equations with Unique Solutions.

If A is an invertible matrix, then the system Ax = b has a unique solution x = A–1

b.

Review: What is wrong with the following “proof” that if AB = I and CA = I then B = C?

A

CA

A

AB

B = C

What is wrong with the following “proof” that if AB = I and CA = I and A is invertible then B =

C?

AB = CA

A–1

AB = CAA–1

IB = CI

B = C


2.4 Elementary Matrices.

p. 31


Objective: Factor a matrix into a product of elementary matrices.

Objective: Find the PA = LDU factorization of a matrix.

An elementary matrix is a square matrix of dimensions nn (or of order n) is a matrix that can be

obtained from the nn identity matrix by a single elementary row operation.

The three elementary row operations, with examples of corresponding elementary matrices, are

1) Swapping two rows, e.g. R2 R4 E1 =

0010

0100

1000

0001

2) Multiplying a single row by a nonzero constant, e.g. 3R3 R3 E2 =

300

010

001

3) Adding a multiple of one row to another row, e.g. –2R1 + R3 R3 E3 =

1000

0102

0010

0001

Theorem 2.12 Representing Elementary Row Operations

If we premultiply (multiply on the left) a matrix A by an elementary matrix, we obtain the same

result as if we had applied the corresponding elementary row operation to A.

Examples:

0010

0100

1000

0001

2514

9497

1913

0189

=

0010

0100

1000

0001

4

3

2

1

r

r

r

r

= R2 R4

300

010

001

497

913

189

= 3R3 R3



p. 32

1000

0102

0010

0001

2514

9497

1913

0189

= –2R1 + R3 R3

Gaussian elimination can be represented by a product of elementary matrices. For example,

A =

3963

5121

2310

R1 R2 E1 =

100

001

010

3963

2310

5121

3R1+ R3 R3 E2 =

103

010

001

12600

2310

5121

61 R3 R3 E3 =

6/100

010

001

2100

2310

5121

so

2100

2310

5121

= E3(E2(E1A)) =

6/100

010

001

103

010

001

100

001

010

3963

5121

2310

Two nn matrices A and B are row-equivalent when there exist a finite number of elementary

matrices such that B = EnEn–1…E2E1A.



p. 33

A square matrix L is lower triangular if all entries above the main diagonal are zero, i.e. lij = 0

whenever i < j. A square matrix U is upper triangular if all entries below the main diagonal are

zero, i.e. uij = 0 whenever i > j. A square matrix D is diagonal if all entries not on the main

diagonal are zero, i.e. dij = 0 whenever i j.

L =

0

00

U =

00

0 D =

00

00

00

Theorem 2.13+ Elementary Matrices are Invertible

If E is an elementary matrix, then E–1

exists and is an elementary matrix. Moreover, if E is lower

triangular, then E–1

is also lower triangular. And if E is diagonal, then E–1

is also diagonal.

Examples:

E1 =

100

001

010

R1 R2 E1–1

=

100

001

010

R1 R2

E2 =

103

010

001

3R1+ R3 R3 E2–1

=

103

010

001

–3R1+ R3 R3

E3 =

6/100

010

001

61 R3 R3 E3

–1 =

600

010

001

6R3 R3

You can check these using matrix multiplication.

E.g.

103

010

001

103

010

001

=

100

010

001

and

103

010

001

103

010

001

=

100

010

001

.

Theorem 2.14 Invertible Matrices are Row-Equivalent to the Identity Matrix

A square matrix A is invertible if and only if it is row equivalent to the identity matrix:

A = En…E2E1I = En…E2E1

if and only if A is a product of elementary matrices.

Proof of “If A is invertible, then A is a product of elementary matrices”:

Since A is invertible, Ax = b has a unique solution (namely, x = A–1

b). But this means that we

can use row operations to reduce [ A | b ] to [ I | c ] (where c = A–1

b, of course). If the



p. 34

corresponding elementary matrices are E1, E2, …, En, then I = EnEn–1…E2E1A so

A = E1–1

E2–1

… En–1–1

En–1

, which is a product of elementary matrices.

Proof of “If A is a product of elementary matrices, then A is invertible”:

If A = E1E2… En–1En, then A–1

exists and A–1

= En–1

En–1–1

…E2–1

E1–1

because every elementary

matrix is invertible.-

Theorem 2.15 Equivalent Conditions for Invertibility

If A is an n n matrix, then the following statements are equivalent.

1) A is invertible.

2) Ax = b has a unique solution for every n1 column matrix b (namely, x = A–1

b).

3) Ax = 0 has only the trivial solution.

4) A is row-equivalent to Inn.

5) A can be written as a product of elementary matrices.

Returning to our example of Gaussian elimination using elementary matrices, we found the

reduced echelon form of the augmented matrix

3963

5121

2310

2100

2310

5121

=

6/100

010

001

103

010

001

100

001

010

3963

5121

2310

Look just at the 33 coefficient matrix instead of the 34 augmented matrix. We have

U

formechelon -row

100

310

121

=

mRows

6/100

010

001

mRowAdds

103

010

001

swaps row

100

001

010

A

963

121

310

The row-echelon form is upper triangular. In general, we may have more than one mRow

(multiply a row by a nonzero constant) elementary matrix, more than one mRowAdd (add a

multiple of one row to another row) elementary matrix, and more than one row swap matrix. The

product of and arbitrary number of row swap matrices is called a permutation matrix P.

U=

mRows

12 FFFn

mRowAdds

12 EEEm PA



p. 35

ngularlower tria

mRowAdds

11

2

1

1

mEEE

diagonal

mRows

11

2

1

1

nFFF U= PA

Lemma The product of diagonal matrices is diagonal. The product of lower triangular matrices is

lower triangular.

LDU = PA

L

EEE m11

21

1

103

010

001

D

FFF n11

21

1

6/100

010

001

U

formechelon -row

100

310

121

=

P

swaps row

100

001

010

A

963

121

310

Theorem LU-Factorization

Every square matrix can be factored as PA = LDU, where P is a permutation matrix, L is lower

triangular with all ones on the main diagonal, D is diagonal, and U is upper triangular with all

ones on the main diagonal.

A variation on this is PA = LU, where this L equals the LD from above, and does not necessarily

have ones on the diagonal.

The PA = LU factorization is the usual method used by computers for solving systems of linear

equations, finding inverse matrices, and calculating determinants (Chapter 3). It is also useful in

proofs.



p. 36

*

Example: Find the PA = LDU factorization of A =

632

101

110

Solution: A =

632

101

110

21 RR

632

110

101

3312 RRR

430

110

101

3323 RRR

100

110

101

22 RR

100

110

101

= U

U

100

110

101

=

F

100

010

001

2

130

010

001

E

1

102

010

001

E

P

100

001

010

A

632

101

110

1

1

102

010

001

E

1

2

130

010

001

E

1

100

010

001

F

U

100

110

101

=

P

100

001

010

A

632

101

110

L

132

010

001

D

100

010

001

U

100

110

101

=

P

100

001

010

A

632

101

110


2.5 Applications of Matrix Operations.

p. 37

drop

A B

*


Objective: Write and use a stochastic (Markov) matrix.

Objective: Use matrix multiplication to encode and decode messages.

Objective: Use matrix algebra to analyze and economic system (Leontief input-output model).

Consider a situation in which members of a population occupy a finite number of states {S1, S2,

…, Sn}. For example, a multinational company has $4 trillion in assets (the population). Some of

the money is in the Americas, some in Asia, and the rest is in Europe (the three states). In a

Markov process, at each discrete step in time, members of the population may move from one

state to another, subject to the following rules:

1) The total number of individuals stays the same.

2) The numbers in each state never become negative.

3) The new state depends only on the current state (history is disregarded).

The behavior of a Markov process is described by a matrix

of transition probabilities (or stochastic matrix or Markov

matrix).

pij is the probability that a member of the population will

change from the j th

state to the i th

state. The rules above

become

1) Each column of the transition matrix adds up to one.

2) Every probability entry is 0 ≤ pij ≤ 1.

Example: Stochastic Matrix. A chemistry course is taught in two sections. Every week,41 of the

students in Section A and31 of the students in Section B drop, and

61 of each section transfer to

the other section. Write the transition matrix. At the beginning of the semester, each section has

144 students. Find the number of students in each section and the number of students who have

dropped after one week and after two weeks.

Solution:

P =

1

01

01

31

41

31

61

61

61

61

41

(d)

(B)

(A)

=

1

0

0

31

41

21

61

61

127

.

P

0

144

144

=

84

96

108

(d)

(B)

(A)

. P

84

96

108

=

143

66

79

(d)

(B)

(A)

to

from

2

1

21

22221

11211

21

nnnnn

n

n

n

S

S

S

ppp

ppp

ppp

P

SSS

61

61

31

004

1

127

21

1

61



p. 38

*--

Matrix multiplication can be used to encode and decode messages. The encoded messages are

called cryptograms.

To begin, assign a number to each letter of the alphabet (and assign 0 to a space).

0 1 2 3 4 5 6 7 8 9 10 11 12 13

_ A B C D E F G H I J K L M

14 15 16 17 18 19 20 21 22 23 24 25 26

N O P Q R S T U V W X Y Z

Use these numbers to convert a message in to a row matrix, including spaces but ignoring

punctuation. Then partition the row matrix into 13 uncoded row matrices.

Example:

M A K E _ I T _ S O _ _

[13 1 11] [5 0 9] [20 0 19] [15 0 0]

Example: Use the invertible matrix A =

310

672

141

to encode the message MAKE IT SO.

Solution:

[13 1 11]

310

672

141

= [11 –56 26] [20 0 19]

310

672

141

= [20 –99 37]

[5 0 9]

310

672

141

= [5 –29 22] [15 0 0]

310

672

141

= [15 –60 –15]

The sequence of encoded matrices is [11 –56 26] [5 –29 22] [20 –99 37] [15 –60 –15]

Removing the brackets yields the cryptogram 11 –56 26 5 –29 22 20 –99 37 15 –60 –15

In order to decrypt a message, we need to know the encryption matrix A.



p. 39

*-–-

Example: Use the invertible matrix A =

310

672

141

to decode

–8 26 31 13 –73 50 19 –97 44 16 –64 –16.

Solution:

A–1

=

112

436

171327

[–8 26 31]

112

436

171327

= [2 5 1] [19 –97 44]

112

436

171327

= [19 0 21]

[13 –73 50]

112

436

171327

= [13 0 21] [16 –64 –16]

112

436

171327

= [16 0 0]

[2 5 1] [13 0 21] [19 0 21] [16 0 0]

B E A M _ U S _ U P _ _

In economics, an input-output model (developed by

Leontief) consists of n different industries In, each of

which needs inputs (e.g. steel, food, labor, …) and has

and output. To produce a unit (e.g. $1 million) of output,

an industry may use the outputs of other industries and of

itself. For example, production of steel may use steel,

food, and labor.

Let dij be the amount of output the industry j needs from industry i to produce one unit of output

per year. (We assume the dij are constant, i.e. fixed prices.) The matrix of these coefficients is

called the input-output matrix or consumption matrix D. A column represents all of the inputs to

a given industry. For this model to work, 0 ≤ dij ≤ 1 and the sum of the entries in each column

must be less than or equal to 1. (Otherwise, it costs more than one unit to produce a unit in that

industry.)

Let xi be the total output matrix of industry i, and X = [xi]. If the economic system is closed (self-

sustaining: total output = “intermediate demand,” i.e. what is needed to produce it), then X = DX.

If the system is open with external demand matrix E (e.g. exports) then X = DX + E.

To find what output matrix is needed to produce a given external demand matrix, we solve

(Input)

Supplier

(Output)User

2

1

21

22221

11211

21

nnnnn

n

n

n

I

I

I

ddd

ddd

ddd

D

III



p. 40

*-–-

X = DX + E

X – DX = E

(I – D)X = E

X = (I – D)–1

E

Example: Input-Output Economic Model

Production of one unit of steel requires 0.4 units of steel, no food, and 0.5 units of labor.

Production of one unit of food requires no steel, 0.1 units of food, and 0.7 units of labor.

Production of one unit of labor requires 0.1 units of steel, 0.8 units of food, and 0.1 units of

labor. Find the output matrix when the external demands are 300 units of steel, 200 units of food,

and no labor.

Solution:

D =

1.07.05.0

8.01.00

1.004.0

. E =

0

200

300

. X =

1

1.07.05.0

8.01.00

1.004.0

100

010

001

0

200

300

2086

2076

848

We need 848 units of steel, 2076 units of food, and 2086 units of labor.

Chapter 3: Determinants.

3.1 The Determinant of a Matrix.

p. 41



Objective: Find the determinant of a 22 matrix.

Objective: Find the minors and cofactors of a matrix.

Objective: Use expansion by cofactors to find the determinant of a matrix.

Objective: Find the determinant of a triangular matrix.

Every square matrix can be associated with a scalar called its determinant. Historically,

determinants were recognized as a pattern of nn systems of linear equations. The system

2222121

1212111

bxaxa

bxaxa

has the solution

21122211

2122211

aaaa

baabx

and

21122211

2112112

aaaa

abbax

.

The determinant of a 11 matrix A = 11a is det(A) = |A| = a11. The |…| symbols mean

determinant, not absolute value.

The determinant of a 22 matrix A =

2221

1211

aa

aa is det(A) = |A| =

2221

1211

aa

aa = a11a22 – a12a21.

Geometrically, the signed area

of a parallelogram with

vertices at (0, 0), (x1, y1), (x2,

y2), and (x1 + x2, y1 + y2) is

A = 22

11

yx

yx

(The area is positive if the angle from

(x1, y1) to (x2, y2) is counterclockwise;

otherwise, the area is negative.)

Proof:

The area A of the parallelogram is

A = area of large rectangle areas of four triangles areas of two small rectangles

= (x1 + x2)(y1 + y2) 21 x1y1

21 x1y1

21 x2y2

21 x2y2 2x2y1

= x1y1 + x1y2 + x2y1 + x2y2 x1y1 x2y2 2x2y1

= x1y2 x2y1 = 22

11

yx

yx

(x2, y2)

(x1, y1)



p. 42

To define the determinant of a square matrix A of order (dimensions) higher than 2, we define

minors and cofactors. The minor Mij of the entry aij is the determinant of the matrix obtained by

deleting row i and column j of A. The cofactor Cij of the entry aij is Cij = (–1)i+j

Mij. Notice that

(–1)i+j

is a “checkerboard” pattern: (–1)i+j

=

Examples: Finding Cofactors. Let A =

Find C21. Solution: C21 = –3332

1312

aa

aa = –a12a33 + a13a32

Find C22. Solution: C22 = +3331

1311

aa

aa = a11a33 – a13a31

The determinant of an nn matrix A (n ≥ 2) is the sum of the entries in the first row of A

multiplied by their respective cofactors.

det(A) =

n

j

jjCa1

11 = a11C11 + a12C12 + … + a1nC1n

Example:

953

786

433

= 395

78

+ 3

93

76 + 4

53

86

= 3(–72 – 35) +3(54 – 21) + 4(30 + 24) = 3(–107) +3(33) +4(54) = –6

Theorem 3.1 Expansion by Cofactors

Let A be a square matrix of order n. Then the determinant of A is given by an expansion in any

row i

det(A) =

n

j

ijijCa1

= ai1Ci1 + ai2Ci2 + … + ainCin

and also by an expansion in any column j

det(A) =

n

i

ijijCa1

= a1jC1j + a2jC2j + … + anjCnj



p. 43

When expanding, you don’t need to find the cofactors of zero entries, because aijCij = (0)Cij = 0.

The definition of the determinant is inductive, because it uses the determinant of a matrix of

order n – 1 to define the determinant of a matrix of order n.

Example: Expanding by Cofactors to Find a Determinant

= = –1 + 3(– )

= –1(2 +7 ) +3(–2 ) = –1(2(3) + 7(–6)) + 3(–2)(0) = –(6 – 42) = 36

To find a determinant using Mathematica, type Det[a]

(also on Basic Math Assistant, More drop-down menu)

To find a determinant on the TI-89, type

Matrix

det(

A

To find a determinant of a 33 matrix, you can also use the following shortcut. Copy Columns 1

and 2 into Columns 4 and 5. To calculate the determinant, add and subtract the indicated

products.

333231

232221

131211

aaa

aaa

aaa

3231

2221

1211

333231

232221

131211

aa

aa

aa

aaa

aaa

aaa

333231

232221

131211

aaa

aaa

aaa

= a11a22a33 + a12a23a31 + a13a21a32 – a31a22a13 – a32a23a11 – a33a21a12

Example:

851

640

311

51

40

11

851

640

311

so

851

640

311

= 32 + 6 + 0 – 12 – 30 – 0 = –4

subtract

add

32 6 0

12 30 0



p. 44

Theorem 3.2 Determinant of a Triangular Matrix

The determinant of a triangular matrix A of order n is the product of the entries on the main

diagonal. det(A) = a11a22… ann

Proof by Induction for upper triangular matrices:

When k = 1, A = [a11] so |A| = a11

Assume that the theorem holds for all upper triangular matrices of order k. Let A be

an upper triangular matrix of order k + 1. Then expanding in the last row,

|A| =

1,1

1,

1,2222

1,111211

000

00

0

kk

kkkk

kk

kk

a

aa

aaa

aaaa

= ak+1,k+1

kk

k

k

a

aa

aaa

00

0 222

11211

= ak+1,k+1(a11a22…akk) = a11a22…akk ak+1,k+1

The proof for lower triangular matrices is similar.

Optional application to multivariable calculus:

Remember integration by substitution: 2

1

)(

u

u

duuf = 2

1

))((

x

x

dxdx

duxuf

For example, dxx)cot( = dxx

x

)sin(

)cos(

Let u = sin(x), dx

du = cos(x) so dx

x

x

)sin(

)cos( = dx

dx

du

u

1 = du

u

1 = ln|u| + C = ln|sin(x)| + C

In multivariable calculus,

V

dudvdwwvuf ),,( =

V

dxdydz

z

w

y

w

x

w

z

v

y

v

x

v

z

u

y

u

x

u

zyxwzyxvzyxuf )),,(),,,(),,,((

The determinant is called the Jacobian.

opti

onal


3.2 Determinants and Elementary Operations.

p. 45


Objective: Use elementary row operations to evaluate a determinant.

Objective: Use elementary column operations to evaluate a determinant.

Recognize conditions that yield zero determinants.

In practice, we rarely evaluate determinants using expansion by cofactors. The properties of

determinants under elementary operations provide a much quicker way to evaluate determinants.

Theorem 3.9 det(AT) = det(A). [Proof is in Section 3.4]

Theorem 3.3 Elementary Row (Column) Operations and Determinants.

Let A and B be nn square matrices.

a) When B is obtained from A by swapping two rows (two columns) of A, det(B) = –det(A).

b) When B is obtained from A by adding a multiple of one row of A to another row of A (or one

column of A to another column of A), det(B) = det(A).

c) When B is obtained from A by multiplying of a row (column) of A by a nonzero constant c,

det(B) = c det(A).

Theorem 3.4 Conditions that Yield a Zero Determinant.

If A is an nn square matrix and any one of the following conditions is true, then det(A) = 0

a) An entire row (or an entire column) consists of zeros.

b) Two rows (or two columns) are equal.

c) One row is a multiple of another row (or one column is a multiple of another column).

Proof by Induction of 3.3a (for rows):

When k = 2, A =

2221

1211

aa

aa and B =

1211

2221

aa

aa so

det(B) = a21a12 – a22a11 = –(a11a22 – a12a21) = –det(A)

Assume that the theorem holds for all matrices of order k. Let A be a matrix of order k

+ 1 and B be a matrix obtained by swapping two rows of A. To find det(A) and det(B),

expand in any row other than the swapped rows. The respective cofactors are

opposites, because they come from kk matrices that have two rows swapped. Thus,

det(B) = –det(A).

Proof of 3.4a (for rows): Suppose that row i of A is all zeroes. Expand by cofactors in row i.

det(A) =

n

j

ijijCa1

=

n

j

ijC1

0 = 0



p. 46

Proof of 3.4b: Let B be the matrix obtained from A by swapping the two identical rows

(columns) of A, so det(B) = –det(A). But B = A, so det(A) = –det(A) so det(A) = 0.

Proof of 3.3b (for rows): Suppose B is obtained from A by adding c times row k to row i. Expand

by cofactors in row i. Note that the cofactors of Cij are the same for matrices A and B,

because the matrices are the same everywhere except row i.

det(B) =

n

j

ijijCb1

=

n

j

ijijkj Caca1

)( =

n

j

ijkjCac1

+

n

j

ijijCa1

= c·0 + det(A) = det(A).

because

n

j

ijkjCa1

is the determinant of a matrix with two identical rows (row k and row i).

See Theorem 3.4b.

Another way of writing this is

nnn

inknik

knk

n

aa

acaaca

aa

aa

1

11

1

111

)()(

= c

nnn

knk

knk

n

aa

aa

aa

aa

1

1

1

111

+

nnn

ini

knk

n

aa

aa

aa

aa

1

1

1

111

= c·0 + det(A) = det(A).

Proof of 3.3c (for rows): Suppose B is obtained from A by multiplying row i by a nonzero scalar

c. Expand by cofactors in row i.

det(B) =

n

j

ijijCb1

=

n

j

ijijCca1

= c

n

j

ijijCa1

= c det(A)

Proof of 3.4c: Suppose B is a matrix with two equal rows (or two equal columns), and A is

obtained from B by multiplying one of those rows (or columns) by a nonzero scalar c. Using

3.3c on that row (or column), det(A) = c det(B). Using 3.4b, det(B) = 0. Thus, det(A) = 0.

Geometrically, the signed area of a parallelogram with edges from

(0, 0) to (x1, y1) and from (0, 0) to (x2, y2) has the same properties

as 22

11

yx

yx when you perform an elementary row operation. Also,

the signed area of a parallelepiped with edges from (0, 0) to (x1, y1,

z1), from (0, 0) to (x2, y2, z2), and from (0, 0) to (x3, y3, z3) has the

same properties as

333

222

111

zyx

zyx

zyx

when you perform an elementary

row operation.



p. 47

Row Swapping (Theorem 3.3a)

If we swap two rows, e.g. 22

11

yx

yx

11

22

yx

yx

or

333

222

111

zyx

zyx

zyx

333

111

222

zyx

zyx

zyx

then the sign of the area/volume changes because we change from a right-hand orientation

to a left-hand orientation.

Adding a multiple of one row to another row (Theorem 3.3b)

If we add a multiple of one row to another row, e.g. 22

11

yx

yx

1212

11

5.05.0 yyxx

yx

, then

the A = bh is unchanged.



p. 48

Adding a multiple of one row to another row (Theorem 3.3b)

If we multiply a row by a nonzero constant c, e.g. 22

11

yx

yx

22

11

yx

cycx, then the A = bh

(cb)h = cA is also multiplies by the constant c.

Example: Finding a Determinant Using Elementary Row Operations

432

145

234

=

432

145

01114

R1 + 2R2 R1

=

01322

145

01114

R3 – 4R2 R3

= –11322

1114

= –1[14(–13) – 11(–22)]

= –1(60)

= –60



p. 49

Example: Finding a Determinant Using Elementary Column Operations

2159

3157

5642

4331

=

______9

______7

______2

0001

44

33

212

____

____

3

CC

CC

CCC

= (__)

______9

______7

____12

0001

22___ CC

= (__)

______9

______7

0012

0001

4

3

__________

__________

C

C

= (__)(__) ______

______

001

= (__)(__)(__) ____

____

=


3.3 Properties of Determinants.

p. 51


Objective: Find the determinant of a matrix product and of a scalar multiple of a matrix.

Find the determinant of an inverse matrix and recognize equivalent conditions for a nonsingular

matrix.

Find the determinant of the transpose of a matrix.

Theorem 3.5 Determinant of a Matrix Product

If A and B are square matrices of the same order, then det(AB) = det(A) det(B).

Proof: To begin, let E be an elementary matrix. By Thm 2.12, EB is the matrix obtained from

applying the corresponding row operation to B. By Thm. 3.3,

det(EB) =

)det(

)det(

)det(

Bc

B

B

if the row operation is

cconstant nonzero aby row a gmultiplyin

another torow one of multiple a adding

rows twoexchanging

Also by Thm 3.3,

det(E) = det(EI) =

c

1

1

if the row operation is

cconstant nonzero aby row a gmultiplyin

another torow one of multiple a adding

rows twoexchanging

Thus, det(EB) = det(E) det(B). This can be generalized by induction to conclude that

|Ek…E2E1B| = |Ek| |…| |E2| |E1| |B| where the Ei are elementary matrices. If A is nonsingular,

then by Thm. 2.14, it can be written as the product A = Ek…E2E1 so |AB| = |A| |B|.

If A is singular, then A is row-equivalent to a matrix with an entire row of zeroes (for example,

the reduced row echelon form). From Thm 3.4, we know |A| = 0. Moreover, because A is

singular, it follows that AB must be singular. (Proof by contradiction: if AB were nonsingular,

then A[B(AB)-1

] = I would show that A is not singular, because A–1

= B(AB)-1

.) Therefore, |AB|

= 0 = |A| |B|.

Comment on Proof by Contradiction: “P implies Q” is equivalent to “not Q implies not P.”

Theorem 3.6 Determinant of a Scalar Multiple of a Matrix

If A is a square matrix of order n and c is a scalar, then det(cA) = cndet(A).

Proof: Apply Property (c) of Thm. 3.3 to each of the n rows of A to obtain n factors of c.

Theorem 3.7 Determinant of an Invertible Matrix

A square matrix A is invertible (nonsingular) if and only if det(A) 0.

Proof: On the one hand, if A is invertible, then AA–1

= I, so . |A| |A–1

| = | I | = 1. Therefore, |A| 0.

On the other hand, assume det(A) 0. Then use Gauss-Jordan elimination to find the reduced

row-echelon form R. Since R is in reduced row-echelon form, it is either the identity matrix or



p. 52

it must have at least one row of all zeroes. The second case is not possible: if R had a row of all

zeroes, then det(R) = 0, but then det(A) = 0 (which contradicts the assumption). Therefore, A is

row-equivalent to R = I, so A is invertible.

Theorem 3.8 Determinant of an Inverse Matrix

If A is an invertible matrix, then det (A–1

) = )det(

1

A

Proof: AA–1

= I, so . |A| |A–1

| = | I | = 1 and |A| 0, so |A–1

| =||

1

A.

Equivalent Conditions for a Nonsingular nn Matrix (Summary)

1) A is invertible.

2) Ax = b has a unique solution for every n1 column matrix b.

3) Ax = 0 has only the trivial solution for the n1 column matrix 0.

4) A is row-equivalent to I.

5) A can be written as a product of elementary matrices.

6) det(A) 0.

Theorem 3.9 Determinant of a the Transpose of a Matrix

If A is a square matrix, then det(AT) = det(A).

Proof: Let A be a square matrix of order n.

From Section 2.4, we know that A can be factored as PA = LDU, where P is a permutation

matrix, L is lower triangular with all ones on the main diagonal, D is diagonal, and U is upper

triangular with all ones on the main diagonal.

L is obtained from I by adding a multiple of the rows containing the diagonal ones to the rows

below the diagonal, so

|L| = |I | = 1

by Thm. 3.3b. Likewise, U is obtained from I by adding a multiple of the rows containing the

diagonal ones to the rows above the diagonal, so

|U| = |I | = 1

by Thm. 3.3b.

By Thm. 3.2,

|D| = d11d22… dnn

i.e. the product of its diagonal elements.

P is a product of elementary row-swap matrices, each of which has determinant –1. So |P| is

the product of some number of –1’s.

|P| = 1 if the number of row swaps is even;

|P| = –1 if the number of row swaps is odd.



p. 53

Let e1 =

0

0

1

, e2 =

0

1

0

, …, en =

1

0

0

be n1 matrices. Then P =

T

i

T

i

T

i

ne

e

e

2

1

where i1, i2, …, in is

some permutation of 1, 2, …, n. Now PT =

niii eee 21

, so

PPT =

nnnn

n

n

i

T

ii

T

ii

T

i

i

T

ii

T

ii

T

i

i

T

ii

T

ii

T

i

eeeeee

eeeeee

eeeeee

21

22212

12111

=

100

010

001

= I,

and by Thm. 3.5, det(P) det(PT

) = det(PPT

) = det(I) = 1.

Then either det(P) = 1 so det(PT) = 1, or det(P) = –1 so det(P

T) = –1. In both cases,

det(P) = det(PT).

So we have PA = LDU which gives us |P| |A| = |L| |D| |U| = (1) |D| (1) =|D|, so

|A| = ||

||

P

D.

Taking the transpose, we have ATP

T = U

TD

TL

T which gives us |A

T| |P

T| = |U

T| |D

T| |L

T|.

Now |PT| = |P|; |D

T| = |D| because

DT = D

since D is diagonal;

|LT| = 1

because LT is upper triangular with all ones on the main diagonal; and

|UT| = 1

because UT is lower triangular with all ones on the main diagonal.

Thus, |AT| |P

T| = |U

T| |D

T| |L

T| becomes |A

T| |P

T| = (1) |D| (1), so

|AT| =

||

||TP

D =

||

||

P

D = |A|.


3.4 Applications of Determinants.

p. 55


Objective: Find the adjoint of a matrix and use it to find the inverse of a matrix.

Objective: Use Cramer’s Rule to solve a system of n linear equations in n unknowns.

Objective: Use determinants to find area, volume, and the equations of lines and planes.

Using the adjoint of a matrix to calculate the inverse is time-consuming and inefficient. In

practice, Gauss-Jordan elimination is used for 33 matrices and larger. However, “adjoint” is

vocabulary you may be expected to know in future classes.

Recall from Section 3.1 that the cofactor Cij of a matrix A is (–1)i+j

times the determinant of the

matrix obtained by deleting row i and column j of A.

The matrix of cofactors of A is

nnnn

n

n

CCC

CCC

CCC

21

22221

11211

.

The adjoint of A is the transpose of matrix of cofactors: adj(A) =

nnnn

n

n

CCC

CCC

CCC

21

22212

12111

Theorem 3.10 The Inverse of a Matrix Given by Its Adjoint

If A is an invertible matrix, then A–1

=)det(

1

Aadj(A).

Proof: Consider A[adj(A)] =

nnnn

kiii

n

n

aaa

aaa

aaa

aaa

21

21

22221

11211

nnjnnn

nj

nj

CCCC

CCCC

CCCC

21

222212

112111

The ij entry of this product is ai1Cj1 + ai2Cj2 + … + ainCjn.

If i = j, this is det(A) (expanded by cofactors in row i). If i j, this is the determinant of the

matrix B, which is the same as A except that row j has been replaced with row i.

opti

onal



p. 56

A =

nnnn

jnjj

kiii

n

n

aaa

aaa

aaa

aaa

aaa

21

21

21

22221

11211

, B[adj(A)] =

nnnn

inii

kiii

n

n

aaa

aaa

aaa

aaa

aaa

21

21

21

22221

11211

nnjnnn

nj

nj

CCCC

CCCC

CCCC

21

222212

112111

The j column of the cofactor matrix is unchanged, because it does not depend on the j row of A

or B. Since two rows of B are the same, the cofactor expansion for i j is zero.

Thus, A[adj(A)] =

)det(00

0)det(0

00)det(

A

A

A

= det(A)I.

det(A) 1 because A is invertible, so we can write A[)det(

1

Aadj(A)] = I, so A

–1 =

)det(

1

Aadj(A).

For 33 matrices and larger, Gauss-Jordan elimination is much more efficient than the adjoint

method for finding the inverse of a matrix.

However, for a 22 matrix A =

dc

ba, adj(A) =

ac

bd so A

–1 =

bcad

1

dc

ba.

Cramer’s Rule to solve n linear equation in n variables is time-consuming and inefficient. In

practice, Gaussian elimination is used to solve linear systems. However, Cramer’s Rule is

vocabulary you may be expected to know in future classes.

Theorem 3.11 Cramer’s Rule

If an nn system Ax = b has a coefficient matrix with nonzero determinant |A| 0, then

)det(

)det( 11

A

Ax ,

)det(

)det( 22

A

Ax , …

)det(

)det(

A

Ax n

n

where Ai is the matrix A but with column i replace by b.



p. 57

Proof:

nx

x

x

2

1

x = A–1

b = )det(

1

Aadj(A)b =

)det(

1

A

nnnn

n

n

CCC

CCC

CCC

21

22212

12111

nb

b

b

2

1

so xi = )det(

1

A(b1C1i + b2C2i + … + bnCni). The sum in parentheses is the cofactor expansion of

det(Ai), so xi = )det(

)det(

A

Ai

Example: Solve 1543

1021

yx

yx using Cramer’s Rule.

Solution: x =

43

21

415

210

= 2

10

= –5; y =

43

21

153

101

= 2

15

=

2

15

Area, Volume and Equations of Lines and Planes:

We already know that the signed area of a parallelogram

is given by a 22 determinant (Section 3.1). A = 22

11

yx

yx

The signed area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is A =

1

1

1

33

22

11

yx

yx

yx

Proof:

The area of a triangle is half of the area of a

parallelogram. So the area of the triangle we want is

21

pos

33

22

yx

yx+

21

neg

11

33

yx

yx+

21

neg

22

11

yx

yx=

1

1

1

2

1

33

22

11

yx

yx

yx

For a triangle the signed area is A =

1

1

1

2

1

33

22

11

yx

yx

yx

If the vertices (x1, y1), (x2, y2), and (x3, y3) are ordered clockwise then the area is positive;

otherwise, it is negative. (The homework asks for the absolute value of the area.)

(x2, y2) (x1, y1)

(x3, y3)



p. 58

The area of the triangle is zero if and only if the three points are collinear.

(x1, y1), (x2, y2), and (x3, y3) are collinear if and only if

1

1

1

33

22

11

yx

yx

yx

= 0

The equation of a line through distinct points (x1, y1) and (x2, y2) is

1

1

1

22

11

yx

yx

yx

= 0

Similarly to the two-dimensional case of a parallelogram,

the signed volume of a parallelepiped is given

by a 33 determinant (Section 3.1). V =

333

222

111

zyx

zyx

zyx

Let’s find the volume of the tetrahedron (pyramid with four triangular faces) with vertices at

(x1, y1, z1), (x2, y2, z2), (x3, y3, z3), and (x4, y4, z4).

The volume of the tetrahedron with vertices at (0, 0, 0), (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) is 61

times the volume of the parallelepiped, i.e.

333

222

111

6

1

zyx

zyx

zyx

. For the triangle above

(21

1

1

1

33

22

11

yx

yx

yx

), we had 3 sides and the areas of 3 triangles to add/subtract. Now we have 4faces

and the volumes of 4 tetrahedrons to add/subtract. The signed volume of the tetrahedron with

vertices at (0, 0, 0), (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) is (imagine for the moment that (x4, y4,

z4) in front of the triangle (x1, y1, z1), (x2, y2, z2), (x3, y3, z3)

V =

333

444

111

6

1

zyx

zyx

zyx

+

222

211

444

6

1

zyx

zyx

zyx

+

444

333

222

6

1

zyx

zyx

zyx

+

111

222

333

6

1

zyx

zyx

zyx

For a tetrahedron, the signed volume is

V = –

1

1

1

1

6

1

444

333

222

111

zyx

zyx

zyx

zyx



p. 59

If when you wrap the fingers of your right from

(x1, y1, z1) to (x2, y2, z2) to (x3, y3, z3), your thumb points

toward (x4, y4, z4), then the signed volume is positive.

If when you wrap the fingers of your left from (x1, y1, z1)

to (x2, y2, z2) to (x3, y3, z3), your thumb points toward

(x4, y4, z4), then the signed volume is negative.

(The homework asks for the absolute value of the

volume.)

Four points (x1, y1, z1), (x2, y2, z2), (x3, y3, z3), and (x4, y4, z4) are coplanar if and only if

1

1

1

1

444

333

222

111

zyx

zyx

zyx

zyx

= 0 because that is when the tetrahedron has zero volume.

The equation of a plane through distinct points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) is

1

1

1

1

333

222

111

zyx

zyx

zyx

zyx

= 0

Chapter 4: Vector Spaces.

8.1 Complex Numbers (Optional).

p. 61



Objective: Use the Quadratic Formula to find all zeroes of a quadratic polynomial with real

coefficients.

Objective: Add and subtract two complex numbers and multiply a complex number by a real

scalar.

Objective: Graphically represent complex numbers in the complex plane as directed line

segments.

Objective: Multiply two complex numbers.

Objective: Multiply and find determinants of matrices with complex entries.

Objective: Perform Gaussian and Gauss-Jordan elimination on matrices with complex entries.

So far, the scalars we have been using have been real numbers R. However, any mathematical

field can be used as the scalars.

Properties of a Field F

Let a, b, and c be any elements of F. Then F is a field if it has two operations, addition and

multiplication, and it contains two distinct elements 0 and 1, such that the following properties

are true.

1) a + b is an element of F. Closure under addition

2) a + b = b + a Commutative property of addition

3) (a + b) + c = a + (b + c) Associative property of addition

4) a + 0 = a Additive identity property

5) There exists a –a such that a + (–a) = 0 Additive inverse property

6) ab is an element of F. Closure under multiplication

7) ab = ba Commutative property of multiplication

8) (ab)c = a(bc) Associative property of multiplication

9) 1a = a Multiplicative identity property

10) Except for a = 0, there exists an a–1

such that a(a–1

) = 1 Multiplicative inverse property

11) a(b + c) = ab + ac Distributive property

Three familiar examples of fields are the rational numbers Q = ,

the real numbers R, and the complex numbers C.

A familiar set that is not a field is the integers Z = {…, –2, –1, 0, 1, 2, …}. Why not?

We are going to need to use the field of complex numbers as our scalars, because we will need to

solve polynomial equations. All polynomials can be solved using complex numbers (complex

numbers are an “algebraically closed field”), but the same is not true of real numbers.



p. 62

The imaginary unit is 1def

i , so i2 = –1.

(Electrical engineers often write 1j because they use i for electric current.)

Example: Solve 5x2 – 6x + 5 = 0

Aside: the complex roots of a polynomial with real coefficients are complex conjugates (Section

8.2) of each other (a + bi and a – bi).

Notice that we have assumed a definition of multiplication by a real number:

101 (6 + 8i) = (

101 6) + (

101 8)i

Example: Solving the polynomial equation 2x3 +3x

2 +50x +75 = 0 using software.

Solution:

Mathematica: inputting Clear[x];Solve[2x^3+3x^2+50x+75==0,x]

yields output

You can also find Solve in the Palettes Menu::Basic Math Assistant::y = x menu.

TI-89: ComplexcSolve(2x^3+3x^2+50x+75=0,x)

A complex number is a number of the form a + bi, where a and b are real numbers. a is the real

part and bi is the imaginary part of a + bi. The form a + bi is the standard form of a complex

number, for example 1 + 2i, 0 + 3i, and –4 + 0i.

Geometrically, a complex number a + bi is represented in the

complex plane by a directed line segment from the origin to (a, b),

where a and b are Cartesian coordinates. In other words, the

horizontal axis is the real axis and the vertical axis is the imaginary

axis.

Operations in the Set of Complex Numbers C

Addition: (a + bi) + (c + di) def

(a + c) + (b + d)i

Multiplication by a real number: c(a + bi) def

ca + cbi

Negative: –(a + bi) def

–a + –bi. Notice that –(a + bi) = (–1)(a + bi).

Subtraction: (a + bi) + (c + di) def

(a + bi) + –(c + di) = (a – c) + (b – d)i



p. 63

Examples

Let z = 2 + 4i and w = –4 + 1i. Illustrate the following graphically.

z, w, z + w, 1.5z, – w, w – z



p. 64

Multiplication of complex numbers is defined using the distributive property and using i 2

= –1.

(a + bi)(c + di) def

(ac – bd) + (ad + bc)i

because (a + bi)(c + di) = ac + adi + bci + bdi 2

= ac + adi + bci + bdi 2

= ac + adi + bci – bd

Warning: When you multiply square roots of negative numbers, convert into standard form

a + bi (not using the square root of a negative number) before you multiply. For example,

1 1 = i·i = –1, not 1 1 = )1)(1( = 1 = 1.

Application (Electrical Engineering): V(t)= I(t)Z, I(t)= current = I0 cos(t) + i I0 sin(t)

V(t) = volatge, Z = impedance = R + Ci

1 + Li ,

R = resistance, C = capacitance, L = inductance

Example: Use software to check that i5

4

5

3 is a zero of the polynomial 5x

2 – 6x + 5.

Solution:

Mathematica: to input the imaginary unit i, denoted by i in Mathematica, type ii (four

separate keystrokes). After the first three keystrokes, you will see ii. After the fourth

keystroke, ii will change to i.

Type x=3/5+4ii/55x^2-6x+5

TI-89: (3/5+4/5)

5x^2-6x+5

Complex matrices


541

413

i

i by hand. Check your answer by multiplying.

Solution: A–1

=)det(

1

A

341

415

i

i.

det(A) = 5(3) – (–1 + 4i)(–1 – 4i) = 15 – (1 + 4i – 4i + 16) = –2

A–1

=2

1

341

415

i

i



p. 65

Check: AA–1

=

541

413

i

i

2

1

341

415

i

i

=

2

1

1516441205205

1231231644115

iiii

iiii =

2

1

20

02 =

10

01

Example: By hand, find the determinant of

Example: Perform Gauss-Jordan elimination

using row operations to solve

iziyixiw

iziyxiiw

ziyxiw

3)716()62()36(3

31)53(3)42(2

22)2(

392 Chapter 8 Complex Vector Spaces

8.1 Complex Numbers

Use the imaginary unit to write complex numbers.

Graphically represent complex numbers in the complex plane aspoints and as vectors.

Add and subtract two complex numbers, and multiply a complexnumber by a real scalar.

Multiply two complex numbers, and use the Quadratic Formula tofind all zeros of a quadratic polynomial.

Perform operations with complex matrices, and find the determinantof a complex matrix.

COMPLEX NUMBERS

So far in the text, the scalar quantities used have been real numbers. In this chapter, youwill expand the set of scalars to include complex numbers.

In algebra it is often necessary to solve quadratic equations such asThe general quadratic equation is and its

solutions are given by the Quadratic Formula

where the quantity under the radical, is called the discriminant. Ifthen the solutions are ordinary real numbers. But what can you conclude

about the solutions of a quadratic equation whose discriminant is negative? Forinstance, the equation has a discriminant of but there isno real number whose square is To overcome this deficiency, mathematiciansinvented the imaginary unit defined as

where In terms of this imaginary unit,With this single addition of the imaginary unit to the real number system, the

system of complex numbers can be developed.

Some examples of complex numbers written in standard form are and The set of real numbers is a subset of the set of complex

numbers. To see this, note that every real number can be written as a complex number using That is, for every real number,

A complex number is uniquely determined by its real and imaginary parts. So, twocomplex numbers are equal if and only if their real and imaginary parts are equal. Thatis, if and are two complex numbers written in standard form, then

if and only if and b � d.a � c

a � bi � c � di

c � dia � bi

a � a � 0i.b � 0.a

�6i � 0 � 6i.4 � 3i,2 � 2 � 0i,

i��16 � 4��1 � 4i.i2 � �1.

i � ��1

i,�16.

b2 � 4ac � �16,x2 � 4 � 0

b2 � 4ac � 0,b2 � 4ac,

x ��b � �b2 � 4ac

2a

ax2 � bx � c � 0,x2 � 3x � 2 � 0.

i

REMARK

When working with productsinvolving square roots of negative numbers, be sure toconvert to a multiple of beforemultiplying. For instance, consider the following operations.

Correct

Incorrect � 1

� �1

��1��1 � ��1��1�

� �1

� i2

��1��1 � i � i

i

Definition of a Complex Number

If and are real numbers, then the number

is a complex number, where is the real part and is the imaginary part ofthe number. The form is the standard form of a complex number.a � bi

bia

a � bi

ba

9781133110873_0801.qxp 3/10/12 6:52 AM Page 392

THE COMPLEX PLANE

Because a complex number is uniquely determined by its real and imaginary parts,it is natural to associate the number with the ordered pair With this association, complex numbers can be represented graphically as points in a coordinateplane called the complex plane. This plane is an adaptation of the rectangular coordinate plane. Specifically, the horizontal axis is the real axis and the vertical axisis the imaginary axis. The point that corresponds to the complex number is

as shown in Figure 8.1.

Plotting Numbers in the Complex Plane

Plot each number in the complex plane.

a. b. c. d. 5

SOLUTION

Figure 8.2 shows the numbers plotted in the complex plane.

a. b.

c. d.

Figure 8.2

Another way to represent the complex number is as a vector whose horizontal component is and whose vertical component is (See Figure 8.3.) (Notethat the use of the letter to represent the imaginary unit is unrelated to the use of torepresent a unit vector.)

Vector Representation of a Complex Number

Figure 8.3

1

−1

−2

−3

4 − 2iVertical

component

Horizontalcomponent

Realaxis

Imaginaryaxis

iib.a

a � bi

5 or (5, 0)Realaxis

Imaginaryaxis

−1 1 2 3 4 5

−2

1

2

3

4

−3i or (0, −3)

Realaxis

Imaginaryaxis

−1−2−3 1 2 3

−3

−2

−4

1

2

−2 − i

Realaxis

Imaginaryaxis

−1−2−3 1 2 3

−3

−4

1

2

or (−2, −1)

4 + 3ior (4, 3)

Realaxis

Imaginaryaxis

−1−2 1 2 3 4

−2

1

2

3

4

�3i�2 � i4 � 3i

�a, b�,a � bi

�a, b�.a � bi

8.1 Complex Numbers 393

(a, b) or a + bi

Realaxis

Imaginaryaxis

The Complex Plane

a

b

Figure 8.1

9781133110873_0801.qxp 3/10/12 6:52 AM Page 393


ADDITION, SUBTRACTION, AND SCALAR

MULTIPLICATION OF COMPLEX NUMBERS

Because a complex number consists of a real part added to a multiple of the operations of addition and multiplication are defined in a manner consistent with therules for operating with real numbers. For instance, to add (or subtract) two complexnumbers, add (or subtract) the real and imaginary parts separately.

Adding and Subtracting Complex Numbers

a.

b.

Using the vector representation of complex numbers, you can add or subtract two complex numbers geometrically using the parallelogram rule for vector addition, asshown in Figure 8.4.

Figure 8.4

Many of the properties of addition of real numbers are valid for complex numbersas well. For instance, addition of complex numbers is both associative and commutative.Moreover, to find the sum of three or more complex numbers, extend the definition ofaddition in the natural way. For example,

� 3 � 3i.

�2 � i� � �3 � 2i� � ��2 � 4i� � �2 � 3 � 2� � �1 � 2 � 4�i

Realaxis

1

1

2

−3

−3 2 3

Imaginaryaxis

w = 3 + i

z = 1 − 3i

z − w = −2 − 4iSubtraction of Complex Numbers

1

−2

−3

−4

−1 2 3 41 65

2

3

4

Addition of Complex Numbersw = 2 − 4i

z + w = 5

z = 3 + 4i

Imaginaryaxis

Realaxis

� �2 � 4i

�1 � 3i� � �3 � i� � �1 � 3� � ��3 � 1�i � 5

�2 � 4i� � �3 � 4i� � �2 � 3� � ��4 � 4�i

i,

Definition of Addition and Subtraction of Complex Numbers

The sum and difference of

and

are defined as follows.

Sum

Difference�a � bi� � �c � di� � �a � c� � �b � d �i

�a � bi� � �c � di� � �a � c� � �b � d �i

c � dia � bi

REMARK

Note in part (a) of Example 2that the sum of two complexnumbers can be a real number.

9781133110873_0801.qxp 3/10/12 6:52 AM Page 394

Another property of real numbers that is valid for complex numbers is the distributive property of scalar multiplication over addition. To multiply a complex number by a real scalar, use the definition below.

Scalar Multiplication with Complex Numbers

a.

b.

Geometrically, multiplication of a complex number by a real scalar corresponds tothe multiplication of a vector by a scalar, as shown in Figure 8.5.

Multiplication of a Complex Number by a Real Number

Figure 8.5

With addition and scalar multiplication, the set of complex numbers forms a vector space of dimension 2 (where the scalars are the real numbers). You are asked toverify this in Exercise 55.

Imaginaryaxis

Realaxis

1

1

2

−2

−3

3

2 3

−z = −3 − i

z = 3 + i

Realaxis

z = 3 + i

2z = 6 + 2i

1

1

2

−1

−2

3

4

2 3 4 65

Imaginaryaxis

� �1 � 6i

�4�1 � i� � 2�3 � i� � 3�1 � 4i� � �4 � 4i � 6 � 2i � 3 � 12i

� 38 � 17i

3�2 � 7i� � 4�8 � i� � 6 � 21i � 32 � 4i


LINEAR ALGEBRA APPLIED

Complex numbers have some useful applications in electronics. The state of a circuit element is described bytwo quantities: the voltage across it and the current flowing through it. To simplify computations, the circuit element’s state can be described by a single complex number of which the voltage and current aresimply the real and imaginary parts. A similar notation canbe used to express the circuit element’s capacitance andinductance.

When certain elements of a circuit are changing with time, electrical engineers often have to solve differentialequations. These can often be simpler to solve using complex numbers because the equations are less complicated.

z � V � li,

IV

Definition of Scalar Multiplication

If is a real number and is a complex number, then the scalar multipleof and is defined as

c�a � bi� � ca � cbi.

a � bica � bic

Adrio Communications Ltd/shutterstock.com

9781133110873_0801.qxp 3/10/12 6:52 AM Page 395


MULTIPLICATION OF COMPLEX NUMBERS

The operations of addition, subtraction, and scalar multiplication of complex numbershave exact counterparts with the corresponding vector operations. By contrast, there isno direct vector counterpart for the multiplication of two complex numbers.

Rather than try to memorize this definition of the product of two complex numbers,simply apply the distributive property, as follows.

Distributive property


Use

Commutative property


Multiplying Complex Numbers

a.

b.

Complex Zeros of a Polynomial

Use the Quadratic Formula to find the zeros of the polynomial and verify that for each zero.

SOLUTION

Using the Quadratic Formula,

Substitute each value of into the polynomial to verify that

In Example 5, the two complex numbers and are complex conjugates of each other (together they form a conjugate pair). More will be saidabout complex conjugates in Section 8.2.

3 � 2i3 � 2i

� 0 � 9 � 6i � 6i � 4 � 18 � 12i � 13

� �3 � 2i��3 � 2i� � 6�3 � 2i� � 13

p�3 � 2i� � �3 � 2i�2 � 6�3 � 2i� � 13

� 0 � 9 � 6i � 6i � 4 � 18 � 12i � 13

� �3 � 2i��3 � 2i� � 6�3 � 2i� � 13

p�3 � 2i� � �3 � 2i�2 � 6�3 � 2i� � 13

p�x� � 0.p�x�x

�6 � ��16

2�

6 � 4i

2� 3 � 2i. x �

�b � �b2 � 4ac

2a

p�x� � 0p�x� � x2 � 6x � 13

� 11 � 2i

� 8 � 3 � 6i � 4i

� 8 � 6i � 4i � 3��1� �2 � i��4 � 3i� � 8 � 6i � 4i � 3i2

��2��1 � 3i� � �2 � 6i

� �ac � bd � � �ad � bc�i � ac � bd � �ad �i � �bc�i

i2 � �1. � ac � �ad �i � �bc�i � �bd ��1� � ac � �ad �i � �bc�i � �bd �i2

�a � bi��c � di� � a�c � di� � bi�c � di�TECHNOLOGY

Many graphing utilities andsoftware programs can calculate with complex numbers. For example, onsome graphing utilities, youcan express a complex number

as an ordered pair Try verifying the result ofExample 4(b) by multiplying

and You shouldobtain the ordered pair �11, 2�.

�4, 3�.�2, �1�

�a, b�.a � bi

Definition of Multiplication of Complex Numbers

The product of the complex numbers and is defined as

�a � bi��c � di� � �ac � bd � � �ad � bc�i .

c � dia � bi

REMARK

A well-known result from algebra states that the complexzeros of a polynomial with realcoefficients must occur in conjugate pairs. (See ReviewExercise 81.)

9781133110873_0801.qxp 3/10/12 6:52 AM Page 396

COMPLEX MATRICES

Now that you are able to add, subtract, and multiply complex numbers, you can applythese operations to matrices whose entries are complex numbers. Such a matrix iscalled complex.

All of the ordinary operations with matrices also work with complex matrices, asdemonstrated in the next two examples.

Operations with Complex Matrices

Let and be the complex matrices

and

and determine each of the following.

a. b. c. d.

SOLUTION

a.

b.

c.

d.

Finding the Determinant of a Complex Matrix

Find the determinant of the matrix

SOLUTION

� �8 � 26i

� 10 � 20i � 6i � 12 � 6

� �2 � 4i��5 � 3i� � �2��3�

det�A� � �2 � 4i3

25 � 3i�

A � �2 � 4i3

25 � 3i�.

� � �27 � i

�2 � 2i3 � 9i�

� � �2 � 0�1 � 2 � 3i � 4i � 6

2i � 2 � 0i � 1 � 4 � 8i�

BA � �2ii

01 � 2i��

i2 � 3i

1 � i4�

A � B � � i2 � 3i

1 � i4� � �2i

i0

1 � 2i� � � 3i2 � 2i

1 � i5 � 2i�

�2 � i�B � �2 � i��2ii

01 � 2i� � �2 � 4i

1 � 2i0

4 � 3i�

3A � 3� i2 � 3i

1 � i4� � � 3i

6 � 9i3 � 3i

12�

BAA � B�2 � i�B3A

B � �2ii

01 � 2i�A � � i

2 � 3i1 � i

4�BA


Definition of a Complex Matrix

A matrix whose entries are complex numbers is called a complex matrix.

TECHNOLOGY

Many graphing utilities andsoftware programs can performmatrix operations on complexmatrices. Try verifying the calculation of the determinantof the matrix in Example 7. You should obtain the sameanswer, ��8, �26�.

9781133110873_0801.qxp 3/10/12 6:52 AM Page 397


8.1 Exercises

Simplifying an Expression In Exercises 1–6, determinethe value of the expression.

1. 2. 3.

4. 5. 6.

Equality of Complex Numbers In Exercises 7–10,determine such that the complex numbers in each pairare equal.

7.

8.

9.

10.

Plotting Complex Numbers In Exercises 11–16, plotthe number in the complex plane.

11. 12. 13.

14. 15. 16.

Adding and Subtracting Complex Numbers InExercises 17–24, find the sum or difference of the complex numbers. Use vectors to illustrate your answer.

17. 18.

19. 20.

21. 22.

23. 24.

Scalar Multiplication In Exercises 25 and 26, use vectors to illustrate the operations geometrically. Be sureto graph the original vector.

25. and , where

26. and , where

Multiplying Complex Numbers In Exercises 27–34,find the product.

27. 28.

29. 30.

31. 32.

33. 34.

Finding Zeros In Exercises 35–40, determine all thezeros of the polynomial function.

35. 36.

37. 38.

39. 40.

Finding Zeros In Exercises 41–44, use the given zeroto find all zeros of the polynomial function.

41. Zero:

42. Zero:

43. Zero:

44. Zero:

Operations with Complex Matrices In Exercises45–54, perform the indicated matrix operation using thecomplex matrices and

and

45. 46.

47. 48.

49. 50.

51. det 52. det

53. 54.

55. Proof Prove that the set of complex numbers, with theoperations of addition and scalar multiplication (withreal scalars), is a vector space of dimension 2.

57. (a) Evaluate for and 5.

(b) Calculate

(c) Find a general formula for for any positive integer

58. Let

(a) Calculate for and 5.

(b) Calculate

(c) Find a general formula for for any positive integer

True or False? In Exercises 59 and 60, determinewhether each statement is true or false. If a statement istrue, give a reason or cite an appropriate statement fromthe text. If a statement is false, provide an example thatshows the statement is not true in all cases or cite anappropriate statement from the text.

59. 60.

61. Proof Prove that if the product of two complex numbersis zero, then at least one of the numbers must be zero.

��10�2 � �100 � 10��2��2 � �4 � 2

n.An

A2010.

n � 1, 2, 3, 4,An

A � �0i

i0�.

n.in

i2010.

n � 1, 2, 3, 4,in

BA5AB

�B��A � B�

14iB2iA

12B2A

B � AA � B

B � �1 � i�3

3i�i�A � � 1 � i

2 � 2i1

�3i�B.A

x � 3ip�x� � x3 � x2 � 9x � 9

x � 5ip�x� � 2x3 � 3x2 � 50x � 75

x � �4p�x� � x3 � 2x2 � 11x � 52

x � 1p�x� � x3 � 3x2 � 4x � 2

p�x� � x4 � 10x2 � 9p�x� � x4 � 16

p�x� � x2 � 4x � 5p�x� � x2 � 5x � 6

p�x� � x2 � x � 1p�x� � 2x2 � 2x � 5

�2 � i��2 � 2i��4 � i��1 � i�3

�a � bi��a � bi��a � bi�2

�4 � �2 i��4 � �2 i ��7 � i��7 � i��3 � i�� 2

3 � i��5 � 5i��1 � 3i�

u � 2 � i�32u3u

u � 3 � i2u�u

�2 � i� � �2 � i��2 � i� � �2 � i��12 � 7i� � �3 � 4i�6 � ��2i�i � �3 � i��5 � i� � �5 � i��1 � �2 i� � �2 � �2 i��2 � 6i� � �3 � 3i�

z � 1 � 5iz � 1 � 5iz � 7

z � �5 � 5iz � 3iz � 6 � 2i

��x � 4� � �x � 1�i, x � 3i

�x2 � 6� � �2x�i, 15 � 6i

�2x � 8� � �x � 1�i, 2 � 4i

x � 3i, 6 � 3i

x

��i�7i 4i 3

��4��4�8��8��2��3

56. Consider the functions and

(a) Without graphing either function, determinewhether the graphs of and have -intercepts.Explain your reasoning.

(b) For which of the given functions is azero? Without using the Quadratic Formula, find theother zero of this function and verify your answer.

x � 3 � i

xqp

q�x� � x2 � 6x � 10.p�x� � x2 � 6x � 10

9781133110873_0801.qxp 3/10/12 6:52 AM Page 398

Section 8.1

1. 3. 5. 1 7. 9.11. 13.

15.

17. 19.

21. 23.

25.

27. 29. 8 31.33. 35. 37. 2, 3 39.

41. 43. 45.

47. 49.

51. 53. 55. Proof

57. (a) (b)

(c) , where is an integer.

59. False. See the Remark, page 392. 61. Proof

kin � �1i�1�i

n � 4kn � 4k � 1n � 4k � 2n � 4k � 3

i5 � ii4 � 1i3 � �ii2 � �1

i2010 � �1i1 � i

��525i

�15 � 10i15 � 30i��5 � 3i

��2 � 2i4 � 4i

2i6��2 � 2i

4 � 4i 2

�6i�� 2�1 � 2i

1 � 3i�4i��

32, ±5i1, 1 ± i

±2, ±2�12 ± 3

2i�2 � 2i�a2 � b2� � 2abi20 � 10i

4

Imaginary axis

−u = −3 + i

u = 3 − i

2u = 6 − 2i−2

−2

−4

−4 6

Realaxis

1

2

3

4

1 2 3 4

Realaxis

Imaginaryaxis

u = 2 + iv = 2 + i

u + v = 4 + 2i

4

2

−22 4

Realaxisu = 6

v = −2i

u − v = 6 + 2i

Imaginaryaxis

4 � 2i6 � 2i

Realaxis

Imaginaryaxis

−2

−4

2

4

6

8v = 5 − i

u − v = 2i

u = 5 + i

24

4

68

10

Imaginaryaxis

−2−4

6 8 10

u = 2 + 6i

v = 3 − 3i

u + v = 5 + 3i

Realaxis

2i5 � 3i

Realaxis

axis

1

2

3

4

5

1 2 3 4 5

z = 1 + 5i

Imaginary

Realaxis

axis

1

2

3

4

5

−1−2−3−4−5

z = −5 + 5i

Imaginary

2

−2

−4

axis

Realaxis

z = 6 − 2i

2 4 6

Imaginary

x � 3x � 6�4��6

Answer Key

9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A1


8.2 Conjugates and Division of Complex Numbers (Optional).

p. 67


Objective: Find the conjugate of a complex number.

Objective: Find the modulus of a complex number.

Objective: Divide complex numbers.

Objective: Perform Gaussian on and find the inverses of matrices with complex entries.

The conjugate of the complex number z = a + bi is denoted by 𝑧̅ or z* and is given by

z* = a – bi

Theorem 8.1 Properties of Complex Conjugates

For a complex numbers z = a + bi,

1) zz* = 𝑧𝑧̅ = a2 + b

2

2) zz* = 𝑧𝑧̅ 0

3) zz* = 𝑧𝑧̅ = 0 if and only if z = 0

4) (z*)* = (𝑧̅) = z

The modulus of the complex number z = a + bi is denoted by |z| and is given by

|z| = 22 ba

Theorem 8.2 The modulus of a complex number. |z| = *zz = √𝑧𝑧̅

Example: Find z* and |z| if z = 4 – i.

Solution: z* = 4 + i |z| = 22 )1(4 = 17

The quotient of two complex numbers z = a + bi and w = c + di, w 0, is

w

z =

w

z

*

*

w

w =

2||

*

w

zw =

22

))((

dc

dicbia

=

22

)(

dc

iadbcbdac

Example: Findi

i

34

72

.

Solution: i

i

34

72

=

)34)(34(

)34)(72(

ii

ii

=

)34(

21286822

ii = i

25

34

25

13



p. 68

Theorem 8.3 Properties of Complex Conjugates

For complex numbers z and w (w 0),

1) (z + w)* = z* + w* i.e. 𝑧 + 𝑤̅̅ ̅̅ ̅̅ ̅̅ = 𝑧̅ + �̅�

2) (z – w)* = z* – w* i.e. 𝑧 − 𝑤̅̅ ̅̅ ̅̅ ̅̅ = 𝑧̅ − �̅�

3) (zw)* = z*w* i.e. 𝑧𝑤̅̅ ̅̅ = 𝑧̅ �̅�

4) (z/w)* = z*/w* i.e. 𝑧/𝑤̅̅ ̅̅ ̅ = 𝑧̅/�̅�


ii

ii

3143

455 by hand.

Solution



p. 69

Example: Perform Gaussian elimination using row operations to solve

iziiyxi

iziiyix

iziyixi

10057)1615(20)55(

5767)1012(72

452)39()48()3(

Solution:

8.2 Conjugates and Division of Complex Numbers 399

8.2 Conjugates and Division of Complex Numbers

Find the conjugate of a complex number.

Find the modulus of a complex number.

Divide complex numbers, and find the inverse of a complex matrix.

COMPLEX CONJUGATES

In Section 8.1, it was mentioned that the complex zeros of a polynomial with real coefficients occur in conjugate pairs. For instance, in Example 5 you saw that the zerosof are and

In this section, you will examine some additional properties of complex conjugates.You will begin with the definition of the conjugate of a complex number.

Finding the Conjugate of a Complex Number

Complex Number Conjugate

a.

b.

c.

d.

Geometrically, two points in the complex plane are conjugates if and only if theyare reflections in the real (horizontal) axis, as shown in Figure 8.6. Complex conjugateshave many useful properties. Some of these are shown in Theorem 8.1.

PROOF

To prove the first property, let Then and

The second and third properties follow directly from the first. Finally, the fourth property follows from the definition of the complex conjugate. That is,

Finding the Product of Complex Conjugates

When you have zz � �1 � 2i��1 � 2i� � 12 � 22 � 1 � 4 � 5.z � 1 � 2i,

� a � bi � a � bi � z.� �a � bi��z�

zz � �a � bi��a � bi� � a2 � abi � abi � b2i2 � a2 � b2.

z � a � biz � a � bi.

z � 5z � 5

z � 2iz � �2i

z � 4 � 5iz � 4 � 5i

z � �2 � 3iz � �2 � 3i

3 � 2i.3 � 2ip�x� � x2 � 6x � 13

Definition of the Conjugate of a Complex Number

The conjugate of the complex number is denoted by and is given by

z � a � bi.

zz � a � bi

REMARK

In part (d) of Example 1, notethat 5 is its own complex conjugate. In general, it can be shown that a number is itsown complex conjugate if andonly if the number is real. (SeeExercise 39.)

Conjugate of a Complex Number

Figure 8.6

Realaxis2 3−2

−2−3−4−5

−3 5 6 7

12345

z = 4 − 5i

z = 4 + 5i

Imaginary axis

Realaxis

z = −2 − 3i

z = −2 + 3i

2

−2

−3

−3−4 −1 1 2

3

Imaginaryaxis

THEOREM 8.1 Properties of Complex Conjugates

For a complex number the following properties are true.

1. 2.3. if and only if 4. �z� � zz � 0.zz � 0

zz � 0zz � a2 � b2

z � a � bi,

9781133110873_0802.qxp 3/14/12 8:23 AM Page 399


THE MODULUS OF A COMPLEX NUMBER

Because a complex number can be represented by a vector in the complex plane, itmakes sense to talk about the length of a complex number. This length is called themodulus of the complex number.

Finding the Modulus of a Complex Number

For and determine the value of each modulus.

a. b. c.

SOLUTION

a.

b.

c. Because you have

Note that in Example 3, In Exercise 41, you are asked to prove that this multiplicative property of the modulus always holds. Theorem 8.2 states that themodulus of a complex number is related to its conjugate.

PROOF

Let then and zz � �a � bi��a � bi� � a2 � b2 � �z�2.z � a � biz � a � bi,

�zw� � �z� �w�.

�zw� � �152 � 162 � �481.

zw � �2 � 3i��6 � i� � 15 � 16i,�w� � �62 � ��1�2 � �37�z� � �22 � 32 � �13

�zw��w��z�w � 6 � i,z � 2 � 3i

Definition of the Modulus of a Complex Number

The modulus of the complex number is denoted by and is given by

�z� � �a2 � b2.

�z�z � a � bi


Fractals appear in almost every part of the universe. Theyhave been used to study a wide variety of applications suchas bacteria cultures, the human lungs, the economy, andgalaxies. The most famous fractal is called the MandelbrotSet, named after the Polish-born mathematician BenoitMandelbrot (1924–2010). The Mandelbrot Set is based onthe following sequence of complex numbers.

The behavior of this sequence depends on the value of thecomplex number For some values of the modulus ofeach term in the sequence is less than some fixed number

and the sequence is bounded. This means that is in theMandelbrot Set, and its point is colored black. For other values of the moduli of the terms of the sequence becomeinfinitely large, and the sequence is unbounded. This meansthat is not in the Mandelbrot Set, and its point is assigned a color based on “how quickly” the sequence diverges.

c

c,

cN,zn

c,c.

zn � �zn�1�2 � c, z1 � c

REMARK

The modulus of a complexnumber is also called theabsolute value of the number.In fact, when is a real number,

�z� � �a2 � 02 � �a�.z

THEOREM 8.2 The Modulus of a Complex Number

For a complex number �z�2 � zz.z,

Andrew Park/Shutterstock.com

9781133110873_0802.qxp 3/10/12 6:57 AM Page 400

8.2 Conjugates and Division of Complex Numbers 401

DIVISION OF COMPLEX NUMBERS

One of the most important uses of the conjugate of a complex number is in performingdivision in the complex number system. To define division of complex numbers,consider and and assume that and are not both 0. For the quotient

to make sense, it has to be true that

But, because you can form the linear system below.

Solving this system of linear equations for and yields

and

Now, because the following definition is obtained.

In practice, the quotient of two complex numbers can be found by multiplying thenumerator and the denominator by the conjugate of the denominator, as follows.

Division of Complex Numbers

a.

b.2 � i

3 � 4i�

2 � i

3 � 4i�3 � 4i

3 � 4i� �2 � 11i

9 � 16�

2

25�

11

25i

1

1 � i�

1

1 � i �1 � i

1 � i� �1 � i

12 � i2�

1 � i

2�

1

2�

1

2i

�ac � bd

c2 � d2�

bc � ad

c2 � d2i

��ac � bd� � �bc � ad�i

c2 � d 2

��a � bi��c � di��c � di��c � di�

a � bi

c � di�

a � bi

c � di �c � di

c � di�

zw � �a � bi��c � di� � �ac � bd� � �bc � ad�i,

y �bc � ad

ww.x �

ac � bd

ww

yx

dx � cy � b

cx � dy � a

z � a � bi,

z � w�x � yi� � �c � di��x � yi� � �cx � dy� � �dx � cy�i.

zw

� x � yi

dcw � c � diz � a � bi

Definition of Division of Complex Numbers

The quotient of the complex numbers and is defined as

provided c2 � d2 � 0.

�1

�w�2 �zw �

�ac � bd

c2 � d2�

bc � ad

c2 � d2i

z

w�

a � bi

c � di

w � c � diz � a � bi

REMARK

If then and In other words, asis the case with real numbers,division of complex numbersby zero is not defined.

w � 0.c � d � 0,c2 � d2 � 0,

9781133110873_0802.qxp 3/10/12 6:57 AM Page 401


Now that you can divide complex numbers, you can find the (multiplicative)inverse of a complex matrix, as demonstrated in Example 5.

Finding the Inverse of a Complex Matrix

Find the inverse of the matrix

and verify your solution by showing that

SOLUTION

Using the formula for the inverse of a matrix from Section 2.3,

Furthermore, because

it follows that

To verify your solution, multiply and as follows.

The last theorem in this section summarizes some useful properties of complexconjugates.

PROOF

To prove the first property, let and Then

The proof of the second property is similar. The proofs of the other two properties areleft to you.

� z � w.

� �a � bi� � �c � di� � �a � c� � �b � d �i

z � w � �a � c� � �b � d�i

w � c � di.z � a � bi

� �10

01�

110�

100

010 AA�1 � �2 � i

3 � i�5 � 2i�6 � 2i

110

��20�10

17 � i7 � i

A�1A

�110�

�20�10

17 � i7 � i.

�1

3 � i�1

3 � i��6 � 2i��3 � i��3 � i��3 � i�

�5 � 2i��3 � i��2 � i��3 � i�

A�1 �1

3 � i ��6 � 2i�3 � i

5 � 2i2 � i

� 3 � i

� ��12 � 6i � 4i � 2� � ��15 � 6i � 5i � 2� �A� � �2 � i��6 � 2i� � ��5 � 2i��3 � i�

A�1 �1

�A� ��6 � 2i�3 � i

5 � 2i2 � i.

2 � 2

AA�1 � I2.

A � �2 � i

3 � i

�5 � 2i

�6 � 2i

TECHNOLOGY

If your graphing utility or software program can performoperations with complex matrices, then you can verifythe result of Example 5. If youhave matrix stored on agraphing utility, evaluate A�1.

A

THEOREM 8.3 Properties of Complex Conjugates

For the complex numbers and the following properties are true.

1.2.3.4. zw � zw

zw � z wz � w � z � wz � w � z � w

w,z

9781133110873_0802.qxp 3/10/12 6:57 AM Page 402

8.2 Exercises 403

8.2 Exercises

40. Consider the quotient

(a) Without performing any calculations, describehow to find the quotient.

(b) Explain why the process described in part (a)results in a complex number of the form

(c) Find the quotient.

a � bi.

1 � i6 � 2i

.

Finding the Conjugate In Exercises 1–6, find the complex conjugate and geometrically represent both

and

1. 2.

3. 4.

5. 6.

Finding the Modulus In Exercises 7–12, find the indicated modulus, where and

7. 8.

9. 10.

11. 12.

13. Verify that where and

14. Verify that where and

Dividing Complex Numbers In Exercises 15–20,perform the indicated operations.

15. 16.

17. 18.

19. 20.

Operations with Complex Rational Expressions InExercises 21–24, perform the operation and write theresult in standard form.

21. 22.

23. 24.

Finding Zeros In Exercises 25–28, use the given zero tofind all zeros of the polynomial function.

25. Zero:

26. Zero:

27. Zero:

28. Zero:

Powers of Complex Numbers In Exercises 29 and 30,find each power of the complex number

(a) (b) (c) (d)

29. 30.

Finding the Inverse of a Complex Matrix In Exercises31–36, determine whether the complex matrix has aninverse. If is invertible, find its inverse and verify that

31. 32.

33. 34.

35. 36.

Singular Matrices In Exercises 37 and 38, determineall values of the complex number for which is singular. (Hint: Set and solve for )

37. 38.

39. Proof Prove that if and only if is real.

41. Proof Prove that for any two complex numbers andeach of the statements below is true.

(a)

(b) If then

42. Graphical Interpretation Describe the set ofpoints in the complex plane that satisfies each of thestatements below.

(a) (b)

(c) (d)

43. (a) Evaluate for 1, 2, 3, 4, and 5.

(b) Calculate and

(c) Find a general formula for for any positiveinteger

44. (a) Verify that

(b) Find the two square roots of

(c) Find all zeros of the polynomial x4 � 1.

i.

�1 � i�2 �2

� i.

n.�1i�n

�1i�2010.�1i�2000

n ��1i�n

2 � �z� � 5�z � i� � 2�z � 1 � i� � 5�z� � 3

�zw� � �z��w�.w � 0,�zw� � �z��w�

w,z

zz � z

A � �2

1 � i1

2i1 � i

0

1 � iz0A � � 5

3iz

2 � i

z.det�A� � 0Az

A � �100

01 � i

0

00

1 � iA � �i00

0i0

00i

A � �1 � i0

21 � iA � �1 � i

12

1 � i

A � �2i3

�2 � i3iA � � 6

2 � i3ii

AA�1 � I.A

A

z � 1 � iz � 2 � i

z�2z�1z3z2

z.

�1 � 3ip�x� � x3 � 4x2 � 14x � 20

�3 � �2 ip�x� � x4 � 3x3 � 5x2 � 21x � 22

�3 � ip�x� � 4x3 � 23x2 � 34x � 10

1 � �3 ip�x� � 3x3 � 4x2 � 8x � 8

1 � i

i�

3

4 � i

i

3 � i�

2i

3 � i

2i

2 � i�

5

2 � i

2

1 � i�

3

1 � i

3 � i

�2 � i��5 � 2i��2 � i��3 � i�

4 � 2i

5 � i

4 � i

3 � �2 i

3 � �2 i

1

6 � 3i

2 � i

i

v � �2 � 3i.z � 1 � 2i�zv2� � �z��v2� � �z��v�2,

w � �1 � 2i.z � 1 � i�wz� � �w��z� � �zw�,

�zv2��v��wz��zw��z2��z�

v � �5i.w � �3 � 2i,z � 2 � i,

z � �3z � 4

z � 2iz � �8i

z � 2 � 5iz � 6 � 3i

z.zz

9781133110873_0802.qxp 3/10/12 6:57 AM Page 403

Section 8.2 1. 3.

5. 4

7. 9. 11. 513.

15. 19.

21. 23. 25.

27.29. (a) (b) (c) (d)

31. 33. Not invertible

35. 37.

39. Proof 41. (a) and (b) Proofs43. (a)

(b)

where k is an integer(c) �1�i�n � �1

�i�1

i

n � 4k n � 4k � 1n � 4k � 2n � 4k � 3

,

�1�i�1 � 1�i � �i, �1�i�2 � �1, �1�i�3 � i,�1�i�4 � 1, �1�i�5 � �i�1�i�2000 � 1, �1�i�2010 � �1

53� �

103 iA�1 � �

�i00

0�i

0

00

� i�

A�1 � �13� i

�2 � i�3i

6�325 �

425i2

5 �15i2 � 11i3 � 4i

1, 2, �3 ± �2 i

23� , 1 ± �3i1

10 �910i�

12

52� i

13

10�

9

10i

71 � 2i 17.

11�

6�2

11 i

wz � �3 � i � �10

wz � �5�2 � �10

zw � �3 � i � �10

�65�5

1

2

3

−1

−2

31 2 4 5

z and z = 4Realaxis

Imaginaryaxis

4

8

84−4

−4

−8

−8

z = 8i

z = −8i

Imaginaryaxis

642

2

4

z = 6 − 3i

z = 6 + 3i

Realaxis

Imaginaryaxis

−2

−4

8i6 � 3i

Answer Key


4.1 Vectors in Rn.

p. 71

4.1 Vectors in Rn.

Objective: Represent a vector in the plane as a directed line segment.

Objective: Perform basic vector operations in R2 and represent them graphically.

Objective: Perform basic vector operations in Rn.

Prove basic properties about vectors and their operations in Rn.

In physics and engineering, a vector is an object with magnitude and direction and represented

graphically by a directed line segment. In mathematics we have a much more general definition

of a vector.

Geometrically, a vector in the plane is represented by a directed line segment with its initial point

at the origin and its terminal (final) point at (x1, x2). The same ordered pair used to represent the

terminal point is used to represent the vector. That is, x = (x1, x2). The coordinates x1 and x2 are

called the components of the vector x. Two vectors u = (u1, u2) and v = (v1, v2) are equal iff u1 =

v1 and u2 = v2.

Vector operations in R2

Vector Addition: u + v = (u1, u2) + (v1, v2)def

(u1 + v1, u2 + v2).

Scalar Multiplication: cu = c(u1, u2)def

(cu1, cu2).

Negative: –udef

(–u1, –u2). Notice that –u = (–1)u.

Subtraction: u – vdef

u + (–v) = (u1, u2) + (–v1, –v2) = (u1 – v1, u2 – v2).

The zero vector in R2 is 0 = (0, 0).


4.1 Vectors in Rn.

p. 72

Examples

Let u = (2, 4) and v = (–4, 1). Illustrate the following graphically.

u, v, u + v, 1.5u, – v, v – u


4.1 Vectors in Rn.

p. 73

Theorem 4.1 Properties of Vector Addition and Scalar Multiplication in the Plane (R2)

Let u, v, and w be vectors in R2, and let c and d be scalars.

1) u + v is a vector in R2. Closure under addition

2) u + v = v + u Commutative property of addition

3) (u + v) + w = u + (v + w) Associative property of addition

4) u + 0 = u Existence of additive identity

5) u + (–u) = 0 Existence of additive inverses

6) cv is a vector in R2. Closure under scalar multiplication

7) c(u + v) = cu + cv Distributive property over vector addition

8) (c + d)u = cu + du Distributive property over scalar addition

9) c(du) = (cd)u Associative property

10) 1u = u Multiplicative identity property

Proof of (3): Associative property of addition

(u + v) + w = [(u1, u2) + (v1, v2)] + (w1, w2)

=

=

= Assoc. prop. of addition of real numbers

=

=

= u + (v + w)


4.1 Vectors in Rn.

p. 74

Proof of (8): Distributive property of scalar multiplication over real number addition

(c + d)u = (c + d)(u1, u2)

=

= Distributive property of real numbers

=

=

= cu + du

To add (1, 4) + (2, –2) in Mathematica, type {1,4}+{2,-2}

You can also assign a variable by typing u={1,4}

You can perform scalar multiplication by 3u or 3*u

To add (1, 4) + (2, –2) on the TI-89, you can type 1,4

+

2,-2

or 1

4

+

2

-2

You can also assign a variable by typing 1,4

U

You can perform scalar multiplication by 3u or 3u

Vector operations in Rn

We can generalize from the 2-dimensional plane R2 to an n-space Rn

of ordered n-tuples. For

example, R1 = R = set of all real numbers; R2

= 2-space = set of all ordered pairs of real

numbers; R3 = 3-space = set of all ordered triples of real numbers.\

An n-tuple (x1, x2, …, xn) can be viewed as a point in Rn with the xi as its coordinates, or as a

vector with the xi as its components.

The standard vector operations in Rn are

Vector Addition: u + v = (u1, u2, …, un) + (v1, v2, …, vn) def

(u1 + v1, u2 + v2, …, un + vn).

Scalar Multiplication: cu = c(u1, u2, …, un) def

(cu1, cu2, …, cun).

Negative: –u def

(–u1, –u2, …, –un). Notice that –u = (–1)u.

Subtraction: u – v def

u + (–v) = (u1, u2, …, un) + (–v1, –v2, …, –vn)

= (u1 – v1, u2 – v2, …, un – vn).

The zero vector in Rn is 0 = (0, 0, …, 0).


4.1 Vectors in Rn.

p. 75

Theorem 4.2 Properties of Vector Addition and Scalar Multiplication in the Plane (Rn)

Let u, v, and w be vectors in Rn, and let c and d be scalars.

1) u + v is a vector in Rn. Closure under addition

2) u + v = v + u Commutative property of addition

3) (u + v) + w = u + (v + w) Associative property of addition

4) u + 0 = u Existence of additive identity

5) u + (–u) = 0 Existence of additive inverses

6) cv is a vector in Rn. Closure under scalar multiplication




10) 1u = u Multiplicative identity property

The vector 0 is called the additive identity in Rn and –v is the additive inverse of v.

Theorem 4.3 Properties of Vector Addition and Scalar Multiplication in Rn

Let v be a vector in Rn and let c be a scalar. Then

1) The additive identity is unique. That is, if v + u = v, then u = 0.

2) The additive inverse of v is unique. That is, if v + u = 0, then u = –v.

3) 0v = 0

4) c0 = 0

5) If cv = 0, then c = 0 or v = 0.

6) –(–v) = v

Proof of (1): Uniqueness of the additive identity

v + u = v Given

(v + u) + (–v) = v + (–v) Add –v to both sides

= v + (–v)

= v + (–v)

= 0

u = 0


4.1 Vectors in Rn.

p. 76

Proof of (2): Uniqueness of the additive inverse

v + u = 0 Given

(–v) + (v + u) = (–v) + 0

= (–v) + 0

= (–v) + 0

= (–v) + 0

u = –v


4.2 Vector Spaces.

p. 77

*

*

4.2 Vector Spaces.

Objective: Define a vector space and recognize some important examples of vector spaces.

Objective: Show that a given set is not a vector space. (Optional)

Theorem 4.2 listed ten properties of vector addition and scalar multiplication in Rn. However,

there are many other sets (Cn, sets of matrices, polynomials, functions) besides Rn

that can be

given suitable definitions of vector addition and scalar multiplication so that they too satisfy the

same ten properties. Hence, one branch of mathematics, linear algebra, can study all of these.

Definition of a Vector Space

Let V be a set on which two operations (vector addition and scalar multiplication) are defined. If

the axioms listed below are satisfied for every u, v, and w in V and every scalar c and d in a

given field F (usually, F = R or F = C), then V is called a vector space over F.

1) u + v is in V. Closure under addition

2) u + v = v + u Commutative property

3) (u + v) + w = u + (v + w) Associative property

4) V has a zero vector 0 such that Existence of additive identity

for every u in V, u + 0 = u

5) For every u in V, there is a vector Existence of additive inverses (opposites)

denoted by –u such that u + (–u) = 0

6) cv is a vector in V. Closure under scalar multiplication




10) 1u = u Scalar identity

Notice that a vector space actually consists of four entities: a set V of vectors, a field F of

scalars, and two defined operations (vector addition and scalar multiplication). Be sure all four

entities are clearly understood. (For example, I could keep the set V of vectors, the field F of

scalars, the same definition of scalar multiplication, but change the definition of how to add

vectors and end up with a different vector space, or end up with something that is no longer a

vector space.)

Examples of Vector Spaces. (Unless otherwise stated, assume the field is R.)

R2 with the standard operations

u + v = (u1, u2) + (v1, v2)def

(u1 + v1, u2 + v2). cu = c(u1, u2) def

(cu1, cu2).

0 = (0, 0) –u = (–u1, –u2)

Rn with the standard operations. Note that this includes R2

, which is just R with the usual

addition and multiplication.

Cn over the field C with the standard operations


4.2 Vector Spaces.

p. 78

.

More Examples of Vector Spaces. (Unless otherwise stated, assume the field is R.)

The vector space M2,3of all 23 real matrices with the standard operations

A + B =

232221

131211

aaa

aaa +

232221

131211

bbb

bbb =

232322222121

131312121111

bababa

bababa

and cA = c

232221

131211

aaa

aaa =

232221

131211

cacaca

cacaca

The vector space Mm,n of all mn real matrices with the standard operations.

The vector space P2 of all polynomials of degree 2 or less with the usual operations.

Let p(x) = a0 + a1x + a2x2 and q(x) = b0 + b1x + b2x

2.

Define the usual operations (p + q)(x) def

p(x) + q(x) and (cp)(x) def

c[p(x)]

We can verify closure under addition:

(p + q)(x) = p(x) + q(x) = a0 + a1x + a2x2 + b0 + b1x + b2x

2

= (a0 + b0) + (a1 + b1)x + (a2 + b2)x2 which is a polynomial of degree 2 or less

(less if a2 + b2 = 0). Notice that we have used the commutative and distributive properties of

real numbers. The other axioms can be verified in a similar manner.

Note that 0(x) = 0 + 0x + 0x2.

The vector space Pn of all polynomials of degree n or less with the usual operations.

The vector space P of all polynomials with the usual operations.

The vector space C(–,) of continuous real-

valued functions on the domain (–,) For

example, x2 + 1,

21

8

x, sin(x), and e

x are vectors

in this space.

Addition and scalar multiplication are defined in

the usual way.

(f + g)(x) def

f(x) + g(x) and (cf )(x) def

c[f (x)]

f, g, and f + g are vectors in C(–,), just as u, v, and u + v and are vectors in Rn. f (x) can be

thought of as a component of f, just as ui is a component of u. u has n components: u1, u2, …,

un. f has an infinite number of components: …, f (–2), …, f (32 ), …, f (0), …, f ( ), ….

The additive identity (zero function) is f0(x) = 0 (the x-axis), and given f (x), the additive

inverse of f is [–f ](x) = –[ f (x)].


4.2 Vector Spaces.

p. 79

.

Another Example of a Vector Spaces

The vector space C[a, b] of continuous real-valued functions on the domain [a, b] over the

field R.

The most important reason for defining an abstract vector space using the ten axioms above is

that we can make general statements about all vector spaces. I.e. the same proof can be used for

Rn and for C[a, b].

Theorem 4.4 Properties of Vector Addition and Scalar Multiplication

Let v be a vector in Vn and let c be a scalar. Then

1) 0v = 0

2) c0 = 0

3) If cv = 0, then c = 0 or v = 0.

4) –1v = –v

Proof of (2): c0 = 0

c0 = 0 Given

c0 = c(0 + 0) Additive identity

c0 =

c0 + –( c0)

c0 + –( c0) =

0 = c0 Additive inverse

Proof of (3): If cv = 0, then c = 0 or c 0. If c 0, then

cv = c0 Given

c–1

cv = c–1

0 Multiply both sides by Multiplicative

inverse in R


4.2 Vector Spaces.

p. 80

.() V =

c(c–1

)v = c–1

0 Commutative property of multiplication

1v = c–1

0 Multiplicative inverse in R

v c–1

0 Scalar identity

v = 0 Theorem 4.3(2) – just proved

Thus, either c = 0 or v = 0.

Examples that are not Vector Spaces

Z2 (ordered pair of integers) over the field R. Z2

is not closed under scalar multiplication, for

example 21 (1, 2) = (

21 , 1) Z2

.

Aside on notation: 1 Z means “1 is an element of (is a member) of the set of integers.

21 Z means “1 is not an element of the set of integers.

Although Z2 satisfied Axioms 1–5 and 10 of a vector space, it is not a vector space because not

all axioms are satisfied.

The set of second-degree polynomials is not a vector space because it is not closed under

addition. For example, let p(x) = x2 and q(x) = –x

2 + x + 1. Then p(x) + q(x) = x + 1 is a first

degree polynomial.

Let V = R2 with the standard vector addition but nonstandard scalar multiplication defined by

c(u1, u2) = (cu1, 0). Show that V is not a vector space.

It turns out that the only axiom that is not satisfied in this case is (10) Scalar identity. For

example, 1(2, 3) = (2, 0) (2, 3).

opti

onal


4.2 Vector Spaces.

p. 81

.

Another Example that is not a Vector Spaces

Rotations in three dimensions represented as arrows using the right-hand rule.

The direction of the arrow represents the direction of the rotation, via

the right-hand rule, while the length of the arrow represents the

magnitude of the direction in degrees. Scalar multiplication is the

standard operation (stretching the arrow, or reversing the direction if

the scalar is negative). “Vector addition” (e.g. ) is the first rotation followed by

the second. This is not a vector space because “vector addition” is not commutative.

(In Chapter 6, we will see that rotations can be represented not as vectors, but as matrices.)

opti

onal


4.3 Subspaces of Vector Spaces.

p. 82

.() V =

.() V =


Objective: Determine whether a subset W of a vector space V is a subspace of V.

Objective: Determine subspaces of Rn.

Many vector spaces are subspaces of larger spaces.

A nonempty subset W of a vector space V is a called a subspace of V when is a vector space

under the operations of vector addition and scalar multiplication defined in V.

Theorem 4.5 Test for a Subspace

If W is a nonempty subset of V, then W is a subspace of V if and only if the following

conditions hold.

0) W is not empty.

1) If u and v are in W, then u + v is in W. Closure under addition

2) If u is in W and c is a scalar, then cu is in W. Closure under scalar multiplication

Proof:

If W is a subspace of V, then W is a vector space satisfying the closure axioms, so

u + v is in W and cu is in W.

On the other hand, assume W is closed under vector addition and scalar multiplication. By

assumption, two axioms are satisfied.

1) u + v is in W. Closure under addition

6) cv is a vector in W. Closure under scalar multiplication

Then if u, v, and w are in W then they are also in V, so the following axioms are

automatically satisfied.

2) u + v = v + u Commutative property

3) (u + v) + w = u + (v + w) Associative property




10) 1u = u Scalar identity

Because W is closed under scalar multiplication, we know that for any v in W, 0v and (–1)v

are also in W. From Thm. 4.4, we know that 0v = 0 and (–1)v = –v so the remaining axioms

are also satisfied.

4) W contains the zero vector 0. Additive identity

5) For every v in W, W contains –v. Additive inverse (opposite)



p. 83

*

Examples of Subspaces and Sets that are not Subspaces.

Show that the set W = {(v1, 0, v3): v1 and v3 are real numbers} is a subspace

of R3 with the standard operations.

Graphically, W is the x-z plane in R3. W is nonempty, because it contains (0,

0, 0). W is closed under addition because if u, v W, then u + v = (u1, 0, u3)

+ (v1, 0, v3) = (u3 + v1, 0, u3 + v3) W. W is closed under scalar

multiplication because if c is a scalar and v W, then cv = c(v1, 0, v3) = (cv1,

0, cv3) W.

Is Z2 (ordered pair of integers) with the standard operations a subspace of

R2?

Z2 is closed under addition. But as we saw in 4.2, Z2

is not closed under

scalar multiplication, for example 21 (1, 2) = (

21 , 1) Z2

. So Z2 is not a

subspace of R2.

Show that the set W = {(v1, v2): v1 = 0 or v2 = 0} with the standard

operations is not a subspace of R2.

W is closed under scalar multiplication, but W is not closed under addition.

For example, (1, 0) + (0, 1) = (1, 1) W.

Is that the set W = {(v1, v2): v1 0 and v2 0} with the standard operations a

subspace of R2?

W is closed under addition, but W is not closed under scalar multiplication

when c < 0 and v 0. For example, (–1)(2, 3) = (–2, –3) W.

In general, a subspace is “straight” (line) or “flat” (plane or higher-dimensional object), is

“infinite” in all directions, has no “holes,” and contains the origin.

Let W be the set of all symmetric 22 matrices. Show that W is a subspace of M2,2 with the

standard operations.

“A is symmetric” means that A = AT. W is nonempty, because it contains

00

00. W is closed

under addition because if A, B W, then (A + B)T = A

T + B

T = A + B, so A + B is symmetric.

W is closed under scalar multiplication because if c is a scalar and A W, then (cA)T = c(A

T)

cA, so cA is symmetric.



p. 84

More Examples of Subspaces and Sets that are not Subspaces.

Let W be the set of all singular matrices of order 2. Show that W is not a subspace of M2,2 with

the standard operations.

00

01 and

10

00 are singular matrices (det = 0), but

00

01 +

10

00 =

10

01 is not

singular, so W is not closed under addition and is therefore not a subspace. (Notice that

W = {

dc

ba: ad – bc = 0}, and ad – bc = 0 is not a linear equation.)

Let W be the unit circle in R2, i. e. W = {(v1, v2): v1

2 + v2

2 = 1}, with the standard operations. Is

W a subspace of R2

No. W is not closed under addition: (1, 0) + (0, 1) = (1, 1) W. Another reason is that W is

not closed under scalar multiplication: 0(1, 0) = (0, 0) W.

Is W = {(0, 0)} with the standard operations a subspace of R2

Yes, the only addition to check is 0 + 0 = 0 W. Scalar multiplication is also easy to check:

c0 = 0 W. Is W = {0} and W = V are called the trivial subspaces of c.

Which of these two subsets is a subspace of R3 with the standard operations?

a) U = {(v1, v2, v3) R3: 2v1 + 3v2 + 4v3 = 12}

This is a the equation of a plane through the points (6, 0, 0),

(0, 4, 0), and (0, 0, 3). U is not closed under addition, because

(6, 0, 0) + (0, 4, 0) = (6, 4, 0), but 2(6) + 3(4) + 4(0) = 24 12.

Moreover, U is not closed under scalar multiplication, because

0(6, 0, 0) = (0, 0, 0), but 2(0) + 3(0) + 4(0) = 0 12. Either one of

these reasons is enough to show that U is not a subspace.

b) W= {(v1, v2, v3) R3: 2v1 –3 v2 + 4v3 = 0}

This is a the equation of a plane parallel to U, but W contains 0. If v and w W, then

v + w = (v1, v2, v3) + (w1, w2, w3) = (v1 + w1, v2 + w2, v3 + w3), and

2(v1 + w1) –3(v2 + w2) + 4(v3 + w3) = 2v1 + 2w1 –3v2 –3w2 + 4v3 + 4w3

= (2v1 –3v2 + 4v3) + (2w1 –3w2 + 4w3) = 0 + 0 = 0, so v + w W.

Also, if c is a scalar and v W, then cv = (cv1, cv2, cv3) and 2(cv1) –3(cv2) + 4(cv3)

= c(2v1 –3v2 + 4v3) = c·0 = 0, so cv W. Therefore, W is a subspace.



p. 85

.() V =

More Example of Subspaces.

P[0, 1] is the set of all real-valued polynomial funtions on the domain [0, 1].

C1[0, 1] is the set of all real-valued, continuously differentiable functions on the domain [0, 1].

C0[0, 1], also written as C[0, 1], is the set of all real-valued, continuous functions on the

domain [0, 1].

Let W be the set of all real-valued, integrable functions on the interval [0, 1].

Let V be the set of all real-valued functions on the interval [0, 1].

If we take the usual definitions of vector addition and scalar multiplication, and we use R as

the field of scalars, then V is a vector space, and the other four sets are subspaces of V. In fact,

P[0, 1] is a subspace of C1[0, 1], which is is a subspace of C

0[0, 1], which is is a subspace of

W, which is is a subspace of V.

Theorem 4.6 The Intersection of Two Subspaces is a Subspace.

If V and W are both subspaces of a vector space U, then the interaction of V and W (denoted

V W) is also a subspace of U. (Note: V W is the set of all vectors that are in both V W.)

Proof:

Because V and W are both subspaces of U, we know that they both

contain 0, so V W contains 0 and is not empty. To show that

V W is closed under addition, let u1 and u2 be two vectors in

V W. Because u1 and u2 are both in V, and—being a subspace—V

is closed, u1 + u2 is also in V. Likewise, u1 + u2 is also in W. Since

u1 + u2 is in both V and W, u1 + u2 is in V W, so V W is closed

under vector addition. A similar argument shows that V W is

closed under scalar multiplication, so V W is a subspace of U.

U

V

W

V W


4.4 Spanning Sets and Linear Independence.

p. 86


Objective: Write a vector as a linear combination of other vectors in a vector space V.

Objective: Determine whether a spanning set S of vectors in a vector space is a spanning set of

V.

Objective: Determine whether a set of vectors in a vector space V is linearly independent.

Objective: Prove results about spanning sets and linear independence.

A vector v in a vector space V is called a linear combination of the vectors u1, u2, …, uk in V if

and only if v can be written in the form

v = c1u1 + c2u2 + … + ckuk

where c1, c2, …, ck are scalars.

Example: Write the vector v = (–1, –2, 2) as a linear combination of the vectors in the set

S = {(2, –1, 3), (5, 0, 4)} (if possible).

Solution: We need to solve v = c1u1 + c2u2 for the scalars ci. Substituting in the given vectors, we

have

(–1, –2, 2) = c1(2, –1, 3) + c2(5, 0, 4)

(–1, –2, 2) = (2c1, –1c1, 3c1) + (5c2, 0c2, 4c2)

(–1, –2, 2) = (2c1 + 5c2, –1c1 + 0c2, 3c1 + 4c2)

which gives the system

243

201

152

21

21

21

cc

cc

cc

or

2

2

1

43

01

62

2

1

c

c

Solve this system by finding the reduced row-echelon form (using software)

243

201

162

000

110

201

or

0

1

2

00

10

01

2

1

c

c so c1 = 2 and c2 = –1.

Answer: (–1, –2, 2) = 2(2, –1, 3) – 1(5, 0, 4).

Example: Write the vectors v = (–3, 15, 18) and w = (31 ,

34 ,

21 ) as linear combinations of the

vectors in the set S = {(2, 0, 7), (2, 4, 5), (2, –12, 13)} (if possible).

Solution: We need to solve



p. 87

.

Example: Write the vector v =

2057

4824 as linear combinations of the vectors in the set

S = {

45

21,

26

72,

1211

94,

54

56} (if possible).

Solution:



p. 88

Let S = {v1, v2, …, vk} be a subset of a vector space V. Then S is called a spanning set of V if

and only if every vector in V can be written as a linear combination of vector in S. In such cases,

we say that S spans V.

Examples of spanning sets:

The set {(1, 0, 0), (0, 1, 0), (0, 0, 1)} spans R3 because any vector v = (v1, v2, v3) in R3

can be

written as v = v1(1, 0, 0) + v2(0, 1, 0) + v3(0, 0, 1).

The set {1, x, x2} spans P2 because any vector (polynomial) p(x) = ax

2 + bx + c in P2 can be

written as p(x) = c(1) + b(x) + a(x2).

Example: Determine whether the set S1 = {(5, 7, 6), (4, 2, 4), (1, –3, 2)} spans R3.

R3 consists of all the vectors of the form (v1, v2, v3), where v1, v2, and v3 are real numbers. S1

spans R3 if we can always solve for the scalars c1, c2, and c3 in the equation

(v1, v2, v3) = c1(5, 7, 6) + c2(4, 2, 4) + c3(1, –3, 2)

(v1, v2, v3) = (5c1, 7c1, 6c1) + (4c2, 2c2, 4c2) + (1c3, –3c3, 2c3)

(v1, v2, v3) = (5c1 + 4c2 + 1c3, 7c1 + 2c2 – 3c3, 6c1 4c2 + 2c3)

This vector equation is equivalent to the system

3213

3212

3211

246

327

145

cccv

cccv

cccv

or the matrix equation

3

2

1

3

2

1

246

327

145

v

v

v

c

c

c

.



p. 89

This equation can always be solved for c1, c2, and c3, because 0

246

327

145

, so the matrix is

invertible. (To be precise, the determinant is –32.) Therefore, S1 spans R3.

Example: Determine whether the set S2 = {(5, 7, 6), (3, 2, 4), (1, –3, 2)} spans R3.

only this number has changed.

Solution:

Geometrically, the vectors in S2 all lie in the same plane (a “2-dimensional” object), while the

vectors in S1 do not. We need a “3-dimensional” space to contain S1.

S1 S2.



p. 90

*

If S = {v1, v2, …, vk} is a set of vectors in a vector space V, then the span of S is the set of all

linear combinations of the vectors in S.

span(S) = {c1v1 + c2v2 + …+ ckvk: c1, c2, … ck, are scalars}

We sometimes write span{v1, v2, …, vk} instead of span(S).

Another notation for span(S)—which we will avoid because it will be confusing later—is

v1, v2, …, vk.

Theorem 4.7 Span(S) is a Subspace of V

If S = {v1, v2, …, vk} is a set of vectors in a vector space V, then span(S) is a subspace of V.

Moreover, span(S) is the smallest subspace of V that contains (S), in the sense that every

subspace that contains S must also contain span(S).

Proof

First, we want to show that span(S) is a subspace of V.

So we need to show that

Let c be a scalar and let u and w be any vectors in span(S). Then

Next, we want to show that every subspace that contains S must also contain span(S). This is

Lab Problem 4.4.55.



p. 91

Sometimes, one vector can be written “in terms of” other vectors. For example, in S2 = {(5, 7, 6),

(3, 2, 4), (1, –3, 2)} from above, (3, 2, 4) = 2

)2,3,1()6,7,5( . We could say that (3, 2, 4) “is

dependent” upon (5, 7, 6) and (1, –3, 2). But we could just as easily solve for (5, 7, 6); there is

no good reason to treat (3, 2, 4) as special. A more equitable equation is

1(5, 7, 6) – 2(3, 2, 4) + 1(1, –3, 2) = (0, 0, 0).

In fact, there are an infinite number of solutions to c1(5, 7, 6) + c2(3, 2, 4) + c3(1, –3, 2) = 0.

c1 = t, c2 = –2t, c3 = t. (This includes c1 = c2 = c3 = 0.)

On the other hand, for S1 = {(5, 7, 6), (4, 2, 4), (1, –3, 2)}, the only solution to

c1(5, 7, 6) + c2(4, 2, 4) + c3(1, –3, 2) = (0, 0, 0) or

0

0

0

246

327

145

3

2

1

c

c

c

is

0

0

0

0

0

0

246

327

1451

3

2

1

c

c

c

. This is called the trivial solution.

If S = {v1, v2, …, vk} is a set of vectors in a vector space V, then S is called linearly independent

if and only if the equation

c1v1 + c2v2 + …+ ckvk = 0

has only the trivial solution c1 = c2 = …= ck = 0. S is called linearly dependent if and only if

there are also nontrivial solutions.

Examples #1-4: Let w1 = (7, 3, –5), w2 = (1, 4, 6), w3 = (9, 11, 7), w4 = (12, –5, –4), and

w5 = (–6, 5, 8).

Example #1: Is {w1, w2} linearly independent? Does {w1, w2} span R3?

Solution:

To decide linear independence, we want to solve c1w1 + c2w2 = 0.

To decide spanning R3, we want to solve c1w1 + c2w2 = (v1, v2, v3) for arbitrary v1, v2, and v3.

Both equations look like c1(7, 3, –5) + c2(1, 4, 6) = (v1, v2, v3).

(For linear independence, the right-hand side is v1 = v2 = v3 = 0.)



p. 92

So we have (7c1, 3c1, –5c1) + (1c2, 4c2, 6c2) = (v1, v2, v3)

(7c1 + 1c2, 3c1 + 4c2, –5c1 + 6c2) = (v1, v2, v3)

3

2

1

2

1

65

43

17

v

v

v

c

c

65

43

17

is not square, so we cannot take the determinant as we did before.

Instead, find the reduced row-echelon form:

3

2

1

65

43

17

v

v

v

(Don’t worry about reproducing this result; I’ll explain soon why Mathematica and the TI-89

give a different answer.)

Writing this as a system of equations, we have

When the right-hand side is v1 = v2 = v3 = 0, we have only

the trivial solution c1 = c2 = 0, because each variable ci

corresponds to a pivot. Therefore, {w1, w2} is linearly

independent.

On the other hand, there are many choices of v1, v2, and v3

for which the last equation 0 = v1 – 3847 v2 +

3825 v3 can be

solved. This happens whenever the (reduced) row-

echelon form has a row of all zeroes. Therefore,{w1, w2}

does not span R3.

Notice that to answer these questions we didn’t need to pay attention to the coefficients on

the right-hand side of the line in the augmented matrix. All that we needed to know was the

coefficient matrix on the left-hand side of the line.

(Mathematica and the TI-89 both give

3

2

1

65

43

17

v

v

v

100

010

001

because they assume

you can divide Row 3 by v1 – 3847 v2 +

3825 v3, and they do not know that we intend the last



p. 93

column to be on the right-hand side of the equation. However, the right-hand column is not

important to deciding linear independence and spanning.)

Example #2: Is {w1, w2, w3} linearly independent? Does {w1, w2, w3} span R3?

Solution:

To decide linear independence, we want to solve c1w1 + c2w2 + c3w3 = 0.

To decide spanning R3, we want to solve c1w1 + c2w2 + c3w3 = v for arbitrary v.

Using block multiplication notation, we write [ w1 | w2 | w3 ]

3

2

1

c

c

c

=

3

2

1

v

v

v

so

765

1143

917

3

2

1

c

c

c

=

3

2

1

v

v

v

.

765

1143

917

is singular (not invertible) because its

determinant (found using software) is zero. This tells us that

sometimes we cannot solve c1w1 + c2w2 + c3w3 = v, so {w1,

w2, w3} does not span R3. This also tells us that c1w1 + c2w2 + c3w3 = 0 has an infinite number

of solutions, so {w1, w2, w3} is linearly dependent.

For another perspective, the reduced row-echelon form of

765

1143

917

is

000

210

101

.

To investigate linear independence, we set the right-hand side equal to zero. The third column

in the matrix, which doesn’t have a pivot, gives us a free parameter. (Fewer pivots than

variables.)

0000

0210

0101

321

321

321

ccc

ccc

ccc

has the solutions

tc

tc

tc

1

2

3

2 where t is a free parameter.

Since we have non-trivial solutions, {w1, w2, w3} is linearly dependent.

To investigate spanning, we set the right-hand side equal to arbitrary numbers. The third row of

all zeroes in the matrix gives us 0c1 + 0c2 + 0c3 = #, which does not have a solution when the

right-hand side is not zero. (Fewer pivots than equations.) Therefore, {w1, w2, w3} does not

span R3.

Example #3: Is {w1, w2, w4} linearly independent? Does {w1, w2, w4} span R3?

Solution:

To decide linear independence, we want to solve c1w1 + c2w2 + c4w4 = 0.

To decide spanning R3, we want to solve c1w1 + c2w2 + c4w4 = v for arbitrary v.



p. 94

Using block multiplication notation, we write [ w1 | w2 | w4 ]

4

2

1

c

c

c

=

3

2

1

v

v

v

so

465

543

1217

4

2

1

c

c

c

=

3

2

1

v

v

v

.

465

543

1217

is invertible (non-singular) because its

determinant (found using software) is 591. This tells us

that we can solve c1w1 + c2w2 + c4w4 = v, so {w1, w2,

w4} spans R3. This also tells us that c1w1 + c2w2 + c4w4

= 0 has a unique of solution, so {w1, w2, w4} is linearly

independent.

Example #4: Is {w1, w2, w4, w5} linearly independent? Does {w1, w2, w4, w5} span R3?

Solution:

To decide linear independence, we want to solve c1w1 + c2w2 + c4w4 + c5w5 = 0.

To decide spanning R3, we want to solve c1w1 + c2w2 + c4w4 + c5w5 = v for arbitrary v.

Using block multiplication notation, we write [ w1 | w2 | w4 | w5]

5

4

2

1

c

c

c

c

=

3

2

1

v

v

v

so

8465

5543

61217

5

4

2

1

c

c

c

c

=

3

2

1

v

v

v

.

We cannot take the determinant of a non-square matrix. The reduced row-echelon form of

8465

5543

61217

is

591263

591506

591128

100

010

001

.

To investigate linear independence, we see that the fourth column in the matrix (which doesn’t

have a pivot) gives us a free parameter, so {w1, w2, w4, w5} is linearly dependent.

To investigate spanning, we see that there is no row of all zeroes (every row has a pivot), so

we will never have an inconsistent equation 0 = #. Therefore, {w1, w2, w4, w5} spans R3.



p. 95

*

*

Summary:

{w1, …, wk} in Rn is linearly independent if and only if the reduced row-echelon form of the

matrix [w1 | … | wk ] has as many pivots as columns (variables).

{w1, …, wk} spans Rn if and only if the reduced row-echelon form of the matrix

[w1 | … | wk ] has as many pivots as rows (equations), i.e. no rows of all zeroes.

Theorem 4.8 A Property of Linearly Dependent Sets

A set S = {v1, v2, …, vk}, k 2, is linearly dependent if and only if at least one of the vectors vj

can be written as a linear combination of the other vectors in S.

Proof

To prove “only if (),” assume set S is linearly dependent and k 2.

Then

Because one of the coefficients must be nonzero, we can assume without loss of generality

(WOLOG) that c1 0. Then

Conversely (which means changing to ), suppose v1 is a linear combination of the other

vectors in S. Then

v1 =

0 =

Thus, {v1, v2, …, vk} is linearly dependent because

Example: Given that {w1, w2, w4, w5} is linearly dependent, and that

c1w1 + c2w2 + c4w4 + c5w5 = 0 has the solutions

tc

tc

tc

tc

509128

2

509506

2

509263

4

5

, write w2 as a linear

combination of the other vectors in the set.



p. 96

*

*

Solution:

509128 tw1 –

509506 tw2 +

509263 tw4 + tw5 = 0

so 128w1 + 263w4 + 509w5 = 506w2

so w2 = 506128 w1 +

506263 w4 +

506509 w5

Theorem 4.9 Corollary

Two vectors v1 and v2 in a vector space V are linearly dependent if and only if one is a scalar

multiple of the other.

Proof: From the proof of Theorem 4.8,

Theorem

Any set S = {0, v1, v2, …, vk}, containing the zero vector is linearly dependent.

Proof: 10 + 0v1 + 0v2 + … + 0vk = 0.

Since not all coefficients are zero, S is linearly dependent.


4.5 Basis and Dimension.

p. 97


Objective: Recognize bases in the vector spaces Rn, Pn, and Mm,n.

Objective: Find the dimension of a vector space.

Objective: Prove results about spanning sets and linear independence.

A set of vectors S = {v1, v2, …, vk} in a vector space V is called a basis for V when

1) S spans V, and

2) S is linearly independent.

“S spans V” says that the set S is not too small to be a basis; “S is linearly independent” says

that the set S is not too big to be a basis. (The plural of “basis” is “bases.”)

In the diagram on the left, you can see that S1 = {u1, u2} is too small to span R3. For example, v

is outside of span(S1). In the figure on the right, you can see that S1 = {u1, u2, u3} is large

enough to span R3. For example, you can see the linear combination that gives v. Finally, In the

figure on the right, you can see that S3 = {u1, u2, u3, v} is large to be linearly independent,

because v is a linear combination of u1, u2, and u3.



p. 98

*

*

Example. The standard basis for R3: Show that {(1, 0, 0), (0, 1, 0), (0 0, 1)} is a basis for R3

.

Solution:

To decide linear independence, we want to solve c1(1, 0, 0) + c2(0, 1, 0) + c3(0 0, 1) = 0.

To decide spanning R3, we want to solve c1(1, 0, 0) + c2(0, 1, 0) + c3(0 0, 1) = v for arbitrary v.

These equations are

100

010

001

3

2

1

c

c

c

=

0

0

0

and

100

010

001

3

2

1

c

c

c

=

3

2

1

v

v

v

The first equation has only the trivial solution, and the second equation always has a solution, so

{(1, 0, 0), (0, 1, 0), (0 0, 1)}is linearly independent and it spans R3. Therefore, {(1, 0, 0),

(0, 1, 0), (0 0, 1)} is a basis for R3.

The standard basis in Rn is

1,,0,0

0,,1,0

0,,0,1

2

1

ne

e

e

Example: Show that {(7, –2, 5), (–3, –9, –1), (1, 7, –7)} is a basis for R3.

Solution:

To decide linear independence, we want to solve

.

To decide spanning R3, we want to solve

As matrix equations, we obtain

The matrix is invertible (its determinant is 474). Because of this, the first equation has only the

trivial solution, and the second equation always has a solution, so {(7, –2, 5), (–3, –9, –1),

(1, 7, –7)} is linearly independent and it spans R3. Therefore, {(7, –2, 5), (–3, –9, –1),

(1, 7, –7)} is a basis for R3.



p. 99

*

The standard basis for Pn is {1, x, x2, …, x

n}.

The standard basis for M2,3 is {

00

00

01

,

00

00

10

,

00

01

00

,

00

10

00

,

01

00

00

,

10

00

00

}.

Theorem 4.9 Uniqueness of a Basis Representation

If S = {v1, v2, …, vk} is a basis for a vector space V, then every vector in V, can be written in

one and only one way as a linear combination of vectors in S.

Proof: Let u be an arbitrary vector in V,

u can be written in at least one way as a linear combination of vectors in S because

Now suppose that u can be written as a linear combination of vectors in S in two ways:

u = b1v1 + b2v2 + …+ bkvk and u = c1v1 + c2v2 + …+ ckvk

Subtracting these two equations gives

Since S is linearly independent,

Thus, u can be written as a linear combination of vectors in S in only one way.

Example: Using the basis {(7, –2, 5), (–3, –9, –1), (1, 7, –7)} for R3 from the previous example,

find the unique representation of u = (u1, u2, u3) using this basis. In other words, find the unique

solution for the constants c1, c2, c3 in the equation

u = c1(7, –2, 5), + c2(–3, –9, –1), + c3(1, 7, –7)}



p. 100

Solution: As a matrix equation, we obtain

Lemma 4.9½ Dimension of Spanning Sets and Linearly Independent Sets

If S1 = {v1, v2, …, vn} spans a vector space V and S2 = {u1, u2, …, um} is a set of m linearly

independent vectors in V, then m n.

Proof by contradiction:

Suppose m > n. To show that S2 is linearly dependent (a contradiction), we need to find scalars

k1, k1, …, km (not all zero) such that

k1u1 + k2u2 + … + kmum =

1

1

11

mm

mmkk

u

u

= 0

Because S1 spans V, each ui is a linear combination of vectors in S1:

1

1

1

1

nn

nm

mm

C

v

v

u

u

i.e.

nmnmm

nn

CC

CC

vvu

vvu

...

...

11

11111

Substituting into the first equation gives

1

1

11

nn

nmmm Ckk

v

v

= 0

Now consider

11

1

0

0

][

nmm

mn

T

k

k

C

This is n equations for m unknowns (ki), with n < m, so we have (infinitely many) nontrivial

(non-zero) solutions (k1, k2, …, km). Taking the transpose, we have a non-zero solution to



p. 101

*

*

*

n1

00 = nmmn Ckk 11

Therefore, we have a non-zero solution to

0 =

1

1

11

nn

nmmm Ckk

v

v

=

1

1

11

mm

mmkk

u

u

This says that S2 = {u1, u2, …, um} is linearly dependent, which contradicts the premise, thus

completing the proof by contradiction.

Theorem 4.10 Bases and Linear Dependence

If S = {v1, v2, …, vn} is a basis for a vector space V,

then every set containing more than n vectors in linearly dependent.

Proof:

S = {v1, v2, …, vn} spans V so by Lemma 4.9½, any set of m linearly independent vectors in V

has m n. Therefore, any set of m vectors in V where m > n must be linearly dependent.

Note: the last step is true because “P Q” is logically equivalent to its contrapositive “(not Q)

(not P)”.

Theorem 4.11 Bases and Linear Dependence

If a vector space V has one basis with n vectors, then every basis for V has n vectors.

Proof:

Let S1 = {u1, u2, …, un} be one basis for Vand let S2 = {v1, v2, …, vm} be another basis for V.

Because S1 is a basis, and S2 is linearly independent, Thm. 4.10 tells us that m < n. Similarly,

because S2 is a basis, and S1 is linearly independent, Thm. 4.10 tells us that n < m. Therefore,

m = n.

If a vector space V has basis consisting of n vectors, then the number n is called the dimension of

V, denoted by dim(V) = n. We define the dimension of the trivial vector space {0} to be n = 0.

The dimension of a vector space is well-defined (unambiguous) because of Thm. 4.11.



p. 102

.

Examples:

dim(Rn) = n

dim(Pn) = n + 1

dim(Mm,n) = mn

Example: Find the dimension of the subspace of R3 given by W= {(2a, a –3b, a + b): a and b are

real numbers}.

Solution:We can write (2a, a –3b, a + b) = a(2, 1, 1) + b(0, –3, 1), so W is spanned by S =

{(2, 1, 1), (0, –3, 1)}. Moreover, S is linearly independent, because the reduced row echelon

form of

11

31

02

is

00

10

01

(every column has a pivot). The only solution to

00

10

01

b

a =

0

0

0

is a = b = 0. So S is a basis, and dim(W) = 2.

Example: Find the dimension of the subspace W of P4 spanned by

S = {–3 – 5x + x2 + 2x

4,

–5x – 4x2 + 4x

3,

–12 – 10x + 11x2 – 12x

3 + 11x

4,

3 + 10x + 4x2 – 5x

4}

Solution: S spans W, but it might not be linearly independent. To decide linear independence, we

solve



p. 103

*

Theorem 4.12 Basis Tests in an n-Dimensional Space

Let V be a vector space of dimension n.

1) If S = {v1, v2, …, vn}is a linearly independent set of vectors in V, then S is a basis for V.

2) If S = {v1, v2, …, vn}spans V, then S is a basis for V.

Proof by Contradiction of Part(1)

Assume that S is not a basis for V. Since S is linearly independent, it must not span V. Choose a

vector u in V that is not in span(S) Then the set {v1, v2, …, vn, u} is also linearly independent.

To see this, note that c1v1 + c2v2 + …+ cnvn = u has no solution.

c1v1 + c2v2 + …+ cnvn = –cn+1u

has a solution only when cn+1 = 0, and in that case, c1 = c2 = … = cn = 0 because {v1, v2, …, vn}

is linearly independent.

{v1, v2, …, vn, u} being linearly independent contradicts Thm. 4.10—we have n + 1 linearly

independent vectors in an n-dimensional vector space. So S must be a basis for V

Proof by Contradiction of Part(2)

Assume that S is not a basis for V. Since S spans V, it must not be linearly independent. So

c1v1 + c2v2 + …+ cnvn = 0 has a solution where not all of the ci are zero. Without loss of

generality, we can assume that cn 0. Then the set {v1, v2, …, vn–1}also spans V.

To see this, first observe that vn = –nc

c1 v1 –nc

c2 v2 – … –n

n

c

c 1 vn–1.

Then


4.6 Rank of a Matrix and Systems of Linear Equations.

p. 105

.(

.(


Objective: Find a basis for the row space, a basis for the column space, and the rank of a matrix.

Find the nullspace of a matrix.

Find the solution of a consistent system Ax = b in the form xp + xh.

Objective: Prove results about subspaces associated with matrices.

Given an mn matrix A, in this section we will speak of its row vectors, which are in Rn

A =

mnmm

n

n

aaa

aaa

aaa

31

22221

11211

mnmm

n

n

aaa

aaa

aaa

,,

,,

,,

21

22221

11211

and its column vectors, which are in Rm

A =

mnmm

n

n

aaa

aaa

aaa

31

22221

11211

mn

n

n

mm a

a

a

a

a

a

a

a

a

2

1

2

22

12

1

21

11

Given an mn matrix A.

The row space of A is the subspace of Rn spanned by the row vectors of A.

The column space of A is the subspace of Rm spanned by the column vectors of A.

Theorem 4.13 Row-Equivalent Matrices Have the Same Row Spaces

If an mn matrix A is row-equivalent to an mn matrix U, then the row space of A is equal to the

row space of U.

Proof

Because the rows of U can be obtained from the rows of A by elementary row operations

(scalar multiplication and addition), it follows that the row vectors of U are linear combinations

of the row vectors of A. The row vectors of U lie in the row space of A, and the subspace

spanned by the row vectors of U is contained in the row space of A:

span(U) span(A). Since the rows of A can also be obtained from the rows of U by elementary

row operations, we also have that subspace spanned by the row vectors of A is contained in the

row space of U: span(A) span(U). Therefore, the two row spaces are equal: span(U) =

span(A).



p. 106

.(

Theorem 4.14 Basis for the Row Space of a Matrix

If a matrix A is row-equivalent to a matrix U in row-echelon form, the then the nonzero row

vectors of U form a basis for the row space of A.

Example: Show that the nonzero rows of

00000

10000

31000

26410

34291

(a matrix in row-echelon form)

are linearly independent.

Solution: We want to solve 00000 =

10000

31000

26410

34291

4

3

2

1

c

c

c

c

The first column of the right-hand side sum is c1, so c1 = 0. The second column of the right-

hand side sum is –9c1 + c2 = 0 + c2, so c2 = 0. The fourth column of the right-hand side sum is

0 + 0 + c3 (because c1 = c2 = 0), so c3 = 0. Finally, the fifth column of the right-hand side sum

is 0 + 0 + 0 + c4 (because c1 = c2 = c3 = 0), so c4 = 0. Therefore the nonzero rows are linearly

independent.

Alternate Solution: The transpose of 00000 =

10000

31000

26410

34291

4

3

2

1

c

c

c

c

is

0

0

0

0

0

= c1 1

3

4

2

9

1

+ c2

2

6

4

1

0

+ c3

3

1

0

0

0

+ c4

1

0

0

0

0

=

1323

0164

0042

0019

0001

4

3

2

1

c

c

c

c

.

Solving this by “forward substitution” (first row down to the last row) yields

4

3

2

1

c

c

c

c

=

0

0

0

0

.



p. 107

Example: Find a basis for the row space of

303240

52113769

4576374

2244334

4981

Solution: The row-echelon form is

0000

0000

1000

6810

4981

, so a basis for the row space is

{[1, –8, –9, 4], [0, 1, 8, –6], [0, 0, 0, 1]}.

You could also take the reduced row-echelon form

0000

0000

1000

0810

05501

to obtain the basis

{[1, 0, 55, 0], [0, 1, 8, 0], [0, 0, 0, 1]}.

Example: Find a basis for the subspace of R3 spanned by {[1, –6, 3], [1, –5, 7], [–7, 38, –37]}.

Solution:

Lemma: The pivot columns of a matrix U in row-echelon form (or reduced row-echelon form)

are linearly independent.



p. 108

.(

Example: U =

0000000

100000

1000

100

1

x

xxx

xxxx

xxxxxx

. If we look only at the pivot columns and solve

0000

1000

100

10

1

x

xx

xxx

4

3

2

1

c

c

c

c

=

0

0

0

0

0

, we see by back substitution that the unique solution is

4

3

2

1

c

c

c

c

=

0

0

0

0

, so the pivot columns of U are linearly independent.

Lemma: The pivot columns of a matrix U in row-echelon form (or reduced row-echelon form)

span the column space of U.

Example: U =

0000000

100000

1000

100

1

x

xxx

xxxx

xxxxxx

. The column space of U is the subset of R5 consisting

of vectors whose fifth component is zero (because we are taking linear combinations of vectors

whose fifth component is zero).

If we look only at the pivot columns, we can always solve

0000

1000

100

10

1

x

xx

xxx

4

3

2

1

c

c

c

c

=

0

4

3

2

1

v

v

v

v

, by back

substitution so the pivot columns of U span the column space of U.

Theorem Basis for the Column Space of a Matrix

Given an mn matrix A and its row-echelon (or reduced row-echelon) form U. Then the columns

of A that correspond to the pivot columns of U form a basis for the column space of A.

Proof: by combining the two preceding lemmas, we know that the pivot columns of U form a

basis for the column space of U. Let ai be the column vectors of A and ui be the column

vectors of U, so A = [a1| …| an ] and U = [u1| …| un ]. We can factor PA = LU, so Pai = Lui

for each column vector,

ui = L–1

Pai, and ai = P–1

Lui. (essential idea)



p. 109

.(

To show that the pivot columns of A are linearly independent, consider the equation

0 = pivotsi

iic a , where the terms in the sum include only the pivot columns. Then

L–1

P0 = L–1

P pivotsi

iic a , so 0 =

pivots

1

i

ii PLc a = pivotsi

iic u

Since the pivot columns of U are linearly independent, all of the ci (i pivots) are zero, so

the pivot columns of A are also linearly independent.

To show that the pivot columns of A span the column space of A, consider any vector v in the

column space of A. Then v can be written as v =

n

i

iic1

a so

L–1

Pv = L–1

P

n

i

iic1

a =

n

i

ii PLc1

1a =

n

i

iic1

u

Since L–1

Pv is a linear combination of the ui, it can be written as a linear combination of the

pivot columns (which are a basis for the column space of U):

L–1

Pv = pivotsi

iid u so v = P–1

L(L–1

Pv) = P–1

L pivotsi

iid u =

pivots

1

i

ii LPd u = pivotsi

iid a

Therefore, the pivot columns of A span the column space of A.

Taken together, the two parts of this proof show that the pivot columns of A form a basis for

the column space of A.

Example: Find a basis for the column space of

A =

303240

52113769

4576374

2244334

4981

.



p. 110

.(

.(

.(

Theorem 4.15 Row Space and Column Space have the Same Dimension.

If A is an mn matrix, then the row space and column space of A have the same dimension.

Proof: The dimension of the row space and the dimension of the column space are both equal to

the number of pivots in the row-echelon form (or in the reduced row echelon form) of A.

The rank r = rank(A) of a matrix is the dimension of its row space, the dimension of its column

space, and the number of pivots in the row-echelon form (or in the reduced row echelon form).

Theorem 4.16 Solutions to a Homogeneous System Ax = 0

If A is an mn matrix, then the set of all solutions to the homogeneous system of linear equations

Ax = 0 is a subspace of Rn.

Proof: Because A is mn, x is n1, so the set of all solutions to the homogeneous system of

linear equations Ax = 0 is a subset of Rn. We will call this subset N. N is not empty because it

contains 0: A0 = 0. Now we must show …

If A is an mn matrix, the subspace of Rn consisting of all solutions to the homogeneous system

Ax = 0 is a called the nullspace of A and is denoted by N(A). We sometimes call N(A) the

solution space of Ax = 0. The dimension of N(A) is called the nullity of A, and is sometimes

written as (the Greek letter nu).



p. 111

.(

Theorem 4.17+ Solutions of a Homogeneous System Ax = 0

If A is an mn matrix of rank r and nullity , then the dimension of the solution space of Ax = 0,

i.e. N(A), is = n – r. That is, n = rank(A) + nullity(A), or the number of variables = number of

pivot variables + number of free variables.

Moreover, the solution set to the homogeneous system Ax = 0 is the same as the solution set to

the reduced row-echelon form Rx = 0. This solution has the form

xh = t1x1 + t2x2 + … + tx

where the ti are free parameters, one for each free (non-pivot) variable, and {x1, x2, …, x} is a

basis for N(A).

Proof

Ax = 0 and Rx = 0 have the same solution set, i.e. N(A) = N(R), because A and R are row-

equivalent.

Ax = 0

iff En…E2E1Ax = En…E2E10

iff Rx = 0

Each column of R is multiplied by a component of x. R has r pivots, corresponding to r pivot

variables in x. The remaining n – r variables are free variables, which we can replace with free

parameters t1, t2, …, tn–r. Now we can solve r equations (on for each row of R that has a pivot,

and is therefore not all zeroes) for r pivot variables by back-substitution. The set of these

solutions xh is N(A). If we collect like terms in ti, and factor out the ti, we have

xh = t1x1 + t2x2 + … + tx

so N(A) is spanned by {x1, x2, …, x}. Moreover, {x1, x2, …, x} is linearly independent,

because the solution to 0 = t1x1 + t2x2 + … + tx = xh is 0 = xfree(i) = ti and for each free variable

and 0 = xpivot(j) = (some linear combination of the ti) for each pivot variable. Since all of the ti

are 0, {x1, x2, …, x} is linearly independent, so {x1, x2, …, x} is a basis for N(A), and N(A)

has dimension = nullity(A) = n – r. That is, the number of variables = number of pivot

variable. + number of free variables.

Example: Find a basis for nullspace of

A =

41394

22273

01152

00121

.



p. 112

Solution:

The reduced row-echelon form is [ R | 0 ] =

000000

021000

020110

040301

.

Now we need to solve the system

00000

21000

20110

40301

5

4

3

2

1

x

x

x

x

x

=

0

0

0

0

or the equivalent form

00

021

0211

0431

54

532

531

xx

xxx

xxx

The two free (non-pivot) variables are x3 and x5. We can choose parameters x3 = s and x5 = t,

and solve for the pivot variables x1, x2, and x4 using back substitution.

021

0211

0431

4

2

1

tx

tsx

tsx

x4 = 2t, x2 = s – 2t, and x1 = –3s + 4t

That means that N(A) is the set of all vectors of the form

5

4

3

2

1

x

x

x

x

x

=

t

t

s

ts

ts

2

2

43

= s

0

0

1

1

3

+ t

1

2

0

2

4

So {

0

0

1

1

3

,

1

2

0

2

4

} is a basis for N(A).

pivots



p. 113

S*

.(

Example: Find a basis for nullspace of

A =

2482

1110

1241

.

Solution:

Now that we have solved Ax = 0, let us consider Ax = b, where b 0. We have already seen (Ch.

1.2) that Ax = b has an infinite number of solutions if there are more variables than equations (n

> m). And since you can’t have more pivots than rows, m r, so n > r, or > 0.

Theorem 4.18 Solutions of a Nonhomogeneous System Ax = b

If xp is a particular solution to the nonhomogeneous equation Ax = b, then every solution of this

system can be written in the form x = xp + xh, where xh is a solution of the corresponding

homogeneous system Ax = 0, i.e. xh is in N(A). The general solution of Ax = b, parameterized by

the free parameters t1, t2, …, t is

x = xp + t1x1 + t2x2 + … + t x

Proof

Let x be any solution of Ax = b. Then (x – xp) is a solution of the homogeneous system Ax = 0,

because

A(x – xp) = Ax – Axp = b – b = 0

If we let xp = x – xp, then x = xp + xh.

(For ease of calculation, we usually find xp by solving Ax = b when all of the free variables are

set to zero.)

From Thm. 4.17+ we know that xh = t1x1 + t2x2 + … + tx.



p. 114

.(

Example: Find the general solution (the set of all solutions) of the system of linear system.

21521

145322

85533105

052

54321

54321

54321

54321

xxxxx

xxxxx

xxxxx

xxxxx

Solution:

2151121

1453221

85533105

051121

rref

000000

341000

260100

150021

To find xp, set the free variables x2 = 0, x5 = 0.

3)0(41

2)0(61

1)0(5)0(21

4

3

1

x

x

x

So x4 = –3, x3 = 2, x1 = 1. xp =

0

3

2

0

1

To find xh, set x2 = s, x5 = t. Ax = 0

041

061

0521

4

3

1

tx

tx

tsx

So x4 = –4t, x3 = –6t, x1 = –2s + 5t. xh =

t

t

t

s

ts

4

6

52

= s

0

0

0

1

2

+ t

1

4

6

0

5

x =

0

3

2

0

1

+ s

0

0

0

1

2

+ t

1

4

6

0

5



p. 115


82

12143

293272

18103

zyx

zyx

zyx

zyx

Solution:


822

29225

5426

19483

zyx

wzx

wzy

zyx

Solution:



p. 116

S*

Theorem 4.19 Consistency of a System of Linear Equations

The system Ax = b is consistent if and only if b is in the column space of A.

Proof

Suppose A is mn, so x is n1 and b is m1.

Then Ax =

mnmm

n

n

aaa

aaa

aaa

21

22221

11211

nx

x

x

2

1

= n

mn

n

n

mm

x

a

a

a

x

a

a

a

x

a

a

a

2

1

2

2

22

12

1

1

21

11

(see Section 2.1)

So Ax = b is if and only if b is a linear combination of the columns of A, i.e. b is in the column

space of A.

Summary of Equivalent Conditions for Square Matrices

If A is an nn matrix, then the following conditions are equivalent.

1) A is invertible.

2) Ax = b has a unique solution for any n1 matrix b.

3) Ax = 0 has only the trivial solution.

4) A is row-equivalent to In.

5) det(A) 0

6) rank(A) = n

7) The n row vectors of A are linearly independent.

8) The n column vectors of A are linearly independent.


4.7 Coordinates and Change of Basis.

p. 117

S*


Objective: Represent coordinates in general n-dimensional spaces.

Objective: Find a coordinate matrix relative to a basis in Rn.

Objective: Find the transition matrix from the basis B to the basis B in Rn.

Let B = {v1, v2, …, vn} be an ordered basis for a vector space V and let x be a vector in V such

that

x = c1v1 + c2v2 + …+ cnvn

The scalars c1, c2, …, cn are called the coordinates of the vector x relative to (or with respect to)

the basis B. The coordinate vector (or matrix) of x relative B is the column matrix in Rn whose

components are the coordinates of x.

[x]B =

nc

c

c

2

1

Example: The coordinate matrix of x = (x1, x2, x3) relative to the standard basis S = {e1, e2, e3} =

{(1, 0, 0), (0, 1, 0), (0 0, 1)} in R3 is simply

[x]S =

3

2

1

x

x

x

because x = x1e1 + x2e2 + x3e3

Example: Find the coordinate matrix of p(x) = 2x2 –5x + 6 relative to the standard ordered basis

S = {1, x, x2} in P2

Solution: p(x) = 6(1) – 5(x) + 2(x2) so [p]S =

6

5

6

Example: Find the coordinate matrix of A =

71

28 relative to the standard ordered basis

S = {

00

01,

00

10,

01

00,

10

00} in M2,2

Solution: A = 8

00

01 + 2

00

10 + 1

01

00 + 7

10

00 so [A]S =

7

1

2

8



p. 118

Example 1: Given that the coordinate matrix of x in R2 relative to the nonstandard ordered basis

B = {v1, v2} = {(1, 0), (1, 2)} is

[x]B =

2

3

Find the coordinate matrix of x relative to the standard basis B = {u1, u2} = {(1, 0), (0, 1)} =

S.

Solution: [x]B =

2

3 means that x = 3v1 + 2v2 = 3(1, 0) + 2(1, 2) = (5, 4).

Now (5, 4) = 5(1, 0) + 4(0, 1) = 5u1 + 4u2 so [x]B=S =

4

5.

Another method is to take the coordinate matrices relative to B of

x = 3v1 + 2v2 =

21 vv

2

3 to obtain [x]B=S =

'2'1 ][][ BB vv

2

3 =

20

11

2

3 =

4

5.

This procedure is called a change of basis from B to B. Given [x]B, we found [x]B using an

equation of the form

[x]B = P1

[x]B where P–1

=

'2'1 ][][ BB vv .

The matrix P–1

is called the transition matrix from B to B.

Example 2: Given that the coordinate matrix of x in R3 relative to the standard ordered basis

B = S = {e1, e2, e3} is

[x]B =

1

2

1

.

Find the coordinate matrix of x relative to the nonstandard ordered basis B = {u1, u2, u3} =

{(1, 0, 1), (0, –1, 2), (2, 3, –5)}.



p. 119

S*

Solution: Write x as a linear combination of the vectors in B:

x = c1u1 + c2u2 + c3u3=

321 uuu

3

2

1

c

c

c

= [x]B =

BBB ][][][ 321 uuu

3

2

1

c

c

c

Since B = S,

1

2

1

=

521

310

201

3

2

1

c

c

c

so

3

2

1

c

c

c

=

1

521

310

201

1

2

1

=

2

8

5

.

Thus we have [x]B=S =

3

2

1

c

c

c

=

2

8

5

.

Moreover, we found that [x]B =

BBB ][][][ 321 uuu [x]B

Comparing to [x]B = P1

[x]B after Example 1above, we must have

[x]B = P[x]B where P =

BBB ][][][ 321 uuu .

The matrix P is called the transition matrix from B to B.

Given nonstandard bases B = {v1, v2, …, vn}, B = {u1, u2, …, un}, and the standard basis S =

{e1, e2, …, en} of Rn. Define matrices

BBS =

SnSS ][][][ 21 vvv and (B)BS =

SnSS ][][][ 21 uuu .

(B and B are square, invertible matrices.) B is the transition matrix from B to S; B is the

transition matrix from B to S. To get P–1

, the transition matrix from B to B, first apply BBS,

then apply (B)–1

SB.

P–1

= (B)–1

B or (P–1

)BB = (B)–1

SB BBS

The inverse is

P = B–1

B or PBB = (B–1

)SB (B)BS



p. 120

Example: Given B = {v1, v2, v3} = {(1, 0, 1), (1, 1, 0), (0, 1, 1)} and [x]B =

1

3

2

. Find [x]S

Solution:

Example: Given B = {u1, u2, u3} = {(8, 11, 0), (7, 0, 10), (1, 4, 6)} and x =(3, 19, 2). Find [x]B

Solution:

Example: Given B = {v1, v2, v3} = {(1, 0, 2), (0, 1, 3), (1, 1, 1)},

B = {u1, u2, u3} = {(2, 1, 1), (1, 0, 0), (0, 2, 1)}, and [x]B =

1

2

1

.

Find the transition matrix P from B to B the transition matrix P–1

from B to B, and [x]B.

Solution:

Chapter 5: Inner Product Spaces.

5.1 Length and Dot Product in Rn.

p. 121

S*



Objective: Find the length of a vector and find a unit vector.

Objective: Find the distance between two vectors.

Objective: Find a dot product and the angle between two vectors, determine orthogonality, and

verify the Cauchy-Schwarz Inequality, the triangle inequality, and the Pythagorean Theorem.

Objective: Use a matrix product to represent a dot product.

From the Pythagorean Theorem, we know that the length, or norm, of a vector v in R2 is

||v|| =2

2

2

1 vv . Also from the Pythagorean Theorem, in R3 we have

||v|| = 2

3

22

2

2

1 vvv

=

2

3

2

2

2

1 vvv .

In Rn we define the length or norm of v to be ||v|| =

22

2

2

1 nvvv

Notice that

|v|| 0

||v|| = 0 if and only if v = 0.

A unit vector is a vector whose length is one. In R2 and R3

alternate notation is sometimes used

for the standard unit vectors

i = (1, 0), j = (0, 1) in R2

i = (1, 0, 0), j = (0, 1, 0) , k = (0, 0, 1) in R3

Two nonzero vectors u and v are parallel when one of them is a scalar multiple of the other:

u = cv. If c > 0, then u and v have the same direction. If c < 0, then u and v have opposite

directions.



p. 122

Theorem 5.1 Length of a scalar multiple

Let v be a vector in Rn and let c be a scalar. Then ||cv|| = |c| ||v||, where |c| is the absolute value of

c.

Proof: ||cv|| = 22

2

2

1 )()()( ncvcvcv = 22

2

2

1

2

nvvvc = |c| ||v||

Theorem 5.2 Unit Vector in the Direction of v

If v is a nonzero vector in Rn, then the vector u =

v

v has length one and has the same direction

as v. u is called the unit vector in the direction of v.

Proof: Because v 0, we know that ||v|| > 0 so v

1 > 0. Then u =

v

1v so u has the same

direction as v. Moreover, ||u|| = v

v =

v

1||v|| = 1, so u is a unit vector.

The process of finding the unit vector in the same direction as v is

called normalizing v.

The distance between two vectors u and v is d(u, v) = ||u – v||. In Rn,

d(u, v) = 22

22

2

11 )()()( nn vuvuvu

Notice that

d(u, v) 0

d(u, v) = 0 if and only if u = v.

d(u, v) = d(v, u)

Example:

Let u = (–2, 3, 4), v = (1, 2, –2). Find the length of u, normalize v,

and find the distance between u and v.

Solution by hand:

||u|| = 222 43)2( = 29 or 5.38516

v

v =

222 )2(21

)2,2,1(

=

3

2,

3

2,

3

1

d(u, v) = 222 ))2(4()23()12( =



p. 123

S*

222 61)3( = 46 = 6.78233

Solution using Mathematica:

u={-2,3,4} Out = {-2,3,4}

Norm[u] Out =

v={1,2,-2} Out = {1,2,-2}

Normalize[v] Out =

Norm[u-v] Out =

Solution using TI-89: [-2,3,4]

U

[1,2,-2]V

MatrixNorms norm(U)

MatrixVector ops unitV(V)

MatrixNorms norm(U-

V)

To find the angle between to vectors in Rn, we start with the Law of

Cosines in R2.

||u – v||2 = ||u||

2 + ||v||

2 – 2||u|| ||v|| cos( )

(u1 – v1)2 + (u2 – v2)

2 = u1

2 + u2

2 + v1

2 + v2

2 – 2||u|| ||v|| cos( )

u12 –2u1v1 + v1

2 + u2

2 –2u2v2 + v2

2 = u1

2 + u2

2 + v1

2 + v2

2 – 2||u|| ||v|| cos( )

–2u1v1 –2u2v2 = – 2||u|| ||v|| cos( )

u1v1 + u2v2 = ||u|| ||v|| cos( )

This equation leads us to the following definitions.

The dot product between two vectors u = (u1, u2, …, un) and v = (v1, v2, …, vn) in Rn

u•v = u1v1 + u2v2 + … + unvn

The angle between two nonzero vectors u and v in Rn is =

vu

vuarccos , 0 .

u

v

u – v



p. 124

Two vectors u and v in Rn are orthogonal (perpendicular) when u•v = 0. (Notice that the 0 vector

is orthogonal to every vector.)

We often represent a vectors u = (u1, u2, …, un) and v = (v1, v2, …, vn) in Rn by their coordinate

matrices relative to the standard basis {e1, e2, …, en}

u =

nu

u

u

2

1

and v =

nv

v

v

2

1

We then write uTv for u•v because u

Tv = [u1v1 + u2v2 + … + unvn].

Warning: In our formulas, we will be sloppy and pretend that uTv is equal to the scalar u•v.

However, Mathematica and the TI-89 know that uTv is a 11 matrix, and will not let you use it

as a scalar.

Example: Using u = (–2, 3, 4), v = (1, 2, –2) from the previous

example, find u•v and the angle between u and v.

Solution by hand:

u•v = (–2)(1) + (3)(2) + (4)(–2) = –4

||u|| = 222 43)2( = 29 or 5.38516

||v|| = = 222 )2(21 = 3

=

293

4arccos 1.821 or 104.3

Solution using Mathematica:

u.v Out = 4

ArcCos[u.v/Norm[u]/Norm[v]] Out =

Out = 1.82099

Solution using TI-89:

MatrixVector ops

dotP(U,

V)

MatrixVector ops dotP(u, v)

MatrixNorms norm(u)

MatrixNorms norm(v))

MatrixVector ops dotP(u, v)

period, not

asterisk



p. 125

MatrixNorms norm(u)

MatrixNorms norm(v))

To obtain a decimal approximation instead of an exact formula, use N[...] in Mathematica or

instead of on the TI-89. You can change modes between degrees and radians on

the TI-89 using Angle……… DEGREE

Theorem 5.3 Properties of the Dot Product

If u, v, and w are vectors in Rn and c is a scalar, then

1) u•v = v•u

2) u•(v + w) = u•v + u•w

3) c(u•v) = (cu)•v = u•(cv)

4) v•v = ||v||2

5) v•v 0 and v•v = 0 if and only if v = 0.

Proof of 5.3.2:

u•(v + w) =

Rn combined with the usual operations of vector addition, scalar multiplication, vector length,

and the dot product is called Euclidean n-space.

(Optional aside: non-Euclidean spaces include elliptic space, which is curved, cannot be

smoothly mapped Rn and a has a different definition of distance; hyperbolic space, which is

curved and has different definition of distance; and Minkowski space, used in Einstein’s special

relativity, which replaces the non-negative dot product with an “interval” that can be negative.)

opti

onal

opti

on

al



p. 126

S*

S*

Theorem 5.4 The Cauchy-Schwarz Inequality

If u and v are vectors in Rn, then |u•v| ||u|| ||v||, where |u•v| is the absolute value of u•v.

Proof

If u = 0, then equality holds, because

|u•v| = |0•v| = |0| = 0, and ||u|| ||v|

= ||0|| ||v|| = 0 ||v|| = 0.

If u 0, then let t be any real number and consider the vector tu + v. Then

0 || tu + v ||2 =

The left-hand side is a quadratic function f (t) = at2 + bt + c, where

a = , b = , c =

The quadratic formula is 𝑡 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎

Because f (t) 0, we have either one or zero roots, so 𝑏2 − 4𝑎𝑐 ≤ 0

Theorem 5.5 The Triangle Inequality

If u and v are vectors in Rn, then ||u + v|| ||u|| + ||v||.

Proof

||u + v||2 =



p. 127

S*

Theorem 5.6 The Pythagorean Theorem

If u and v are vectors in Rn, then u and v are orthogonal if and only if ||u + v||

2 = ||u||

2 + ||v||

2.

Proof

By definition, u and v are orthogonal if and only if

||u + v||2 =

The conclusion follows from these two equations.


5.2 Inner Product Spaces.

p. 129

*


Objective: Determine whether a function defines an inner product.

Objective: Find the inner product of two vectors in Rn, Mm,n, Pn, and C[a,b].

Objective: Use the inner product to find angles between two vectors and determine whether two

vectors are orthogonal.

Objective: Find an orthogonal projection of a vector onto another vector in an inner product

space.

The inner product is an extension of the concept of the dot product from Rn to a general vector

space. The standard dot product, also called the Euclidean inner product, on Rn is written as u•v,

while the general inner product on a vector space V is written as u, v.

Definition of Inner Product

Let u, v, and w be vectors in a vector space V, and let c be any scalar. An inner product on V is a

function that associates a real number u, v with each pair of vectors u and v. and satisfies the

following axioms.

1) u, v = v, u Symmetric

2) u, v + w = u, v + u, w

3) cu, v = cu, v

4) v, v 0 and v, v = 0 if and only if v = 0. Nonnegative

A vector space V with an inner product is called an inner product space.

Example: Consider R2. Let u =

2

1

u

u and v =

2

1

v

v. Instead of the Euclidean inner product

uTv, let u, v = u

T

31

12v = 2u1v1 – u1v2 – u2v1 + 3u2v2. Show that u, v is an inner

product.

Solution:

1) v, u = vT

31

12u = 2v1u1 – v1u2 – v2u1 + 3v2u2 = u, v

2) Let A =

31

12. We know that for matrix multiplication, u, v + w = u

TA(v + w) =

(uTA)(v + w) = (u

TA)v + (u

TA)w = u

TAv + u

TAw = u, v + u, w.

3) We know that for matrix multiplication, cu, v = c(uT

31

12v) = (cu

T)

31

12v =

= cu, v.



p. 130

4) v, v = 2v12 – 2v1v2 + 3v2

2 = v1

2 + 2v2

2 + v1

2 – 2v1v2 + v2

2 = v1

2 + 2v2

2 + (v1 – v2)

2 0,

and v, v = if and only if v1 = 0, v2 = 0, and v1 – v2 = 0 if and only if v = 0.

Example: Consider R4. Let u =

z

y

x

t

u

u

u

u

and v =

z

y

x

t

v

v

v

v

. Instead of the Euclidean inner product

uTv, let u, v = u

T

1000

0100

0010

0002c

v, where c is a positive constant (the speed of light), so

u, v = –c2utvt + uxvx + uyvy + uzvz. Show that u, v is not an inner product. (Optional aside: u,

v is called the spacetime interval.)

Solution:

4) v, v = –c2vt

2 + vx

2 + vy

2 + vz

2 which can be less than zero, for example when v =

0

0

0

1

.

Note: Axioms 2 and 3 are satisfied because of the rules of matrix multiplication. Axiom 1 is

satisfied because the matrix is symmetric (AT = A).

Theorem 5.7 Properties of Inner Products

If u, v, and w are vectors in an inner product space V “over the real numbers” and c is a scalar,

then

1) 0, v = v, 0 = 0

2) u + v, w = u, w + v, w

3) u, cv = cu, v

Proof of 5.7.1: 0, v = 0v, v = 0v, v = 0. By the symmetric property, v, 0 = 0, v = 0

Proof of 5.7.2: u + v, w = w, u + v Symmetry

= w, u + w, v Axiom 2

= u, w + v, w Symmetry

Proof of 5.7.3: u, cv = cv, u Symmetry

= cv, u Axiom 3

= cu, v Symmetry

opti

onal



p. 131

*

Together with the inner product axioms, Thm 5.7 says that the inner product is linear in each of

its arguments (we will study linear operators much more in Chapter 6.) Linearity is

essentially the same as distributivity:

Dot product: (2u + 3v)•(5x + 7y) = 10(u•x) + 14(u•y) + 15(v•x) + 21(v•y)

General inner product: 2u + 3v, 5x + 7y = 10u, x + 14u, y + 15v, x + 21v, y

Let u and v be vectors in an inner product space V.

The length, or norm, of u is ||u|| = uu, .

The distance between two vectors u and v is d(u, v) = ||u – v||.

The angle between two nonzero vectors u and v is given by =

vu

vu,arccos , 0 .

Two vectors u and v in Rn are orthogonal (perpendicular) when u, v = 0. (Notice that the 0

vector is orthogonal to every vector.)

If ||u|| = 1 then u is called a unit vector. If v is any nonzero vector, then u = v

v is the unit vector

in the direction of v. Notice that the definition of angle presumes that –1 vu

vu, 1. This is

guaranteed by the Cauchy-Schwarz Inequality (Theorem 5.8.1).

Properties of Length

1) ||u|| 0

2) ||u|| = 0 if and only if u = 0.

3) ||cu|| = |c| ||u||

These follow from the axioms in the definition of inner product and the definition of norm.

Properties of Distance

1) d(u, v) 0

2) d(u, v) = 0 if and only if u = v.

3) d(u, v) = d(v, u)

These follow from the axioms in the definition of inner product and the definition of distance.



p. 132

.(

Example: Let A =

2221

1211

aa

aa and B =

2221

1211

bb

bb be two matrices in M2,2 and define the inner

product A, B = a11b11 + a12b12 + a21b21 + a22b22.

Let I =

10

01, R =

01

10 and L =

45

43

43

45

.

a) Find the norm of I =

10

01.

Solution:

b) Find the inner product of and angle between R and L. Are they orthogonal?

Solution:

c) Find the inner product of and angle between I and L. Are they orthogonal?

Solution:



p. 133

.(

Example: Let p and q be two polynomials in P and define the inner product

p, q =

1

1

)()( dxxqxp

a) Find the norm of p(x) = 1. (Sometimes we will just say “Find the norm of 1.”)

Solution:

b) Find the inner product of p(x) = 1 and q(x) = x. Are they orthogonal?

Solution:

c) Find the angle between 1 and x2. Are they orthogonal?

Solution:

Theorem 5.8

Let u and v be vectors in an inner product space V.

1) Cauchy-Schwarz Inequality: | u, v | ||u|| ||v||

2) Triangle inequality: ||u + v|| ||u|| + ||v||

3) Pythagorean Theorem: u and v are orthogonal if and only if ||u + v||2 = ||u||

2 + ||v||

2.

The proofs are the same as the proofs of Theorems 5.4, 5.5, and 5.6, except that we replace u•v

with u, v. A result such as the triangle inequality, which seems intuitive in R2 and R3

, can be

generalized through linear algebra to some less obvious statements about functions in

C[a, b].



p. 134

Example: Consider f (x) = cos(x) and g(x) = 2

4

(/2 – x)(/2 + x) in C[–/2, /2] with the inner

product f , g =

2/

2/

)()(

dxxgxf . Verify the triangle inequality by direct calculation.

Solution: The triangle inequality is || f + g || || f || + || g ||.

|| f ||2 =

|| g ||2 =

|| f + g ||2 =

|| f + g || =

|| f || + || g || =

Orthogonal Projections: Let u and v be two vectors in R2. If v is nonzero,

then u can be orthogonally projected onto v. This projection is denoted by

projv u. Because projv u is a scalar multiple of v, we can write

projv u = av

v

(Recall that v

v is the unit vector in the direction of v.)

Then a = ||u|| cos = ||u|| vu

vu =

v

vu

projv u = vv

vu2

= v

vv

vu

projv u v

u



p. 135

*

.(

Let u and v be vectors in a general inner product space V. The orthogonal projection of u onto v

is

proj𝐯 𝐮 = ⟨𝐮, 𝐯⟩

⟨𝐯, 𝐯⟩ 𝐯

Example: Use the Euclidean inner product find the orthogonal projection of u = (1, 0, 0) onto

v = (1, 1, 1) and of v onto u .

Solution:

projv u =

projv u =



p. 136

.(

Theorem 5.9 Orthogonal Projection and Distance

Let u and v be two vectors in an inner product

space V such that v 0. Then

d(u, projv u) < d(u, cv) whenever c vv

vu

,

,

That is to say, the orthogonal projection projv u is the vector parallel to v that is closest to u.

Proof: Let b = vv

vu

,

,, c

vv

vu

,

,, p = cv – bv, q = u – bv, and t = u – cv.

The p is orthogonal to q, because

p, q = cv – bv, u – bv

= cv, u – cbv, v – bv, u + b2v, v

= cu, v – cvv

vu

,

,v, v –

vv

vu

,

,u, v +

2

2

,

,

vv

vuv, v

= cv, u – cv, u – vv

vu

,

,2

+ vv

vu

,

,2

= 0.

So the Pythagorean Theorem tells us ||t||2 = ||p||

2 + ||q||

2 > ||q||

2 because p 0 because c b.

Therefore, d(u, projv u) = ||q|| < ||t|| = d(u, cv).

Example: Consider C[–, ] with the inner product f , g =

dxxgxf )()( .

Find the projection of f (x) = x onto g(x) = sin(x).

Solution:

projg f = )(,

,xg

gg

gf.

cv v

u

d(u, cv)

projv u v

u

d(u, projv u)

u

t = u – cv q

bv

p = (c – b)v


5.3 Orthogonal Bases: Gram-Schmidt Process.

p. 137

S*

S*


Objective: Show that a set of vectors is orthogonal and forms an orthonormal basis.

Objective: Represent a vector relative to an orthonormal basis.

Objective: Apply the Gram-Schmidt orthonormalization process.

Although a vector space can have many different bases, some bases may be easier to work with

than others. For example, in R3, we often use the standard basis S = {e1, e2, e3} = {(1, 0, 0),

(0, 1, 0), (0 0, 1)}. Two properties that make the standard basis desirable are that

1) The vectors are normalized, i.e. each basis vector is a unit vector: ||ei|| = 1 for i = 1, 2, 3.

2) The vectors are mutually orthogonal: ei•ej = 0 whenever i j, i.e.

e1•e2 = 0,

e1•e3 = 0,

e2•e3 = 0

A set S of vectors in an inner product space V, is called orthogonal when every pair of vectors in

S is orthogonal. If, in addition, each vector in the set is a unit vector, then S is called

orthonormal.

If S = {v1, v2, …, vn} then these definitions can be written as

S is orthogonal if and only if vi•vj = 0 whenever i j.

S is orthonormal if and only if vi•vj = 0 whenever i j, and ||ei|| = 1 for i = 1, 2, …, n.

Example: Show that S = {(cos, sin, 0)}, (–sin, cos, 0), (0, 0, 1)} is an orthonormal set.

Solution: S is orthogonal, because

(cos, sin, 0) • (–sin, cos, 0) = – cos sin + sin cos = 0

(cos, sin, 0) • (0, 0, 1) = 0

and (–sin, cos, 0) • (0, 0, 1) = 0

S is normalized, because

||(cos, sin, 0)|| = 222 0sincos = 1

||(–sin, cos, 0)|| = 222 0cossin = 1



p. 138

||(0, 0, 1)|| = 222 100 = 1

Theorem 5.10 Orthogonal Sets are Linearly Independent

If S = {v1, v2, …, vn} is an orthogonal set of nonzero vectors in an inner product space V, then S

is linearly independent.

Proof: suppose c1v1 + c2v2 + …+ cnvn = 0

Then (c1v1 + c2v2 + …+ cnvn), vi = 0, vi for each of the vi

We can distribute the inner product on the left-hand side to obtain

Because S is orthogonal, when j i,

so equation becomes

Because vi 0,

Example: Consider C[0, 2] with the inner product f , g = 2

0

)()( dxxgxf .

Show that the set {1, cos(x), sin(x), cos(2x), sin(2x), …, cos(nx), sin(nx)} is a linearly

independent orthogonal set.

Solution: We use software to integrate the following expressions, and we use these identities to

evaluate the results:

1)2cos(

0)2sin(

k

k

whenever k is an integer

1, sin(nx) = 2

0

)sin(1 dxnx =

2

0

)cos(

n

nx =

nn

11 = 0

1, cos(nx) = 2

0

)cos(1 dxnx =

2

0

)sin(

n

nx = 0 – 0 = 0



p. 139

S*

S*

sin(mx), sin(nx) when m n is

2

0

)sin()sin( dxnxmx =

2

0)(2

))sin((

)(2

))sin((

nm

xnm

nm

xnm = (0 – 0) – (0 – 0) = 0

cos(mx), cos(nx) when m n is

2

0

)cos()cos( dxnxmx =

2

0)(2

))sin((

)(2

))sin((

nm

xnm

nm

xnm = (0 + 0) – (0 + 0) = 0

cos(mx), sin(nx) when m n is

2

0

)sin()cos( dxnxmx =

2

0)(2

))cos((

)(2

))cos((

nm

xnm

nm

xnm =

)(2

1

)(2

1

)(2

1

)(2

1

nmnmnmnm = 0

cos(nx), sin(nx) = 2

0

)sin()cos( dxnxnx =

2

0

2

2

)(cos

n

nx =

nn 2

1

2

1 = 0

Corollary Orthogonal Bases

If V is an inner product space of dimension n, then any orthogonal set of n nonzero vectors in V

is a basis for V.

Example: Show that S = {(0, 1, 0, 1), (–1, –1, 2, 1), (2, –1, 0, 1), (2, 2, 3, –2)} is an orthogonal

basis of R4.

Solution: S is orthogonal, because

(0, 1, 0, 1)•(–1, –1, 2, 1) = 0 – 1 + 0 + 1 = 0

(0, 1, 0, 1)•(2, –1, 0, 1) = 0 –1 + 0 – 1 = 0

(0, 1, 0, 1)•(2, 2, 3, –2) = 0 + 2 + 0 – 2 = 0

(–1, –1, 2, 1)•(2, –1, 0, 1) = –2 + 1 + 0 + 1 = 0

(–1, –1, 2, 1)•(2, 2, 3, –2) = –2 – 2 + 3 – 2 = 0

and (2, –1, 0, 1)•(2, 2, 3, –2) = 4 – 2+ 0 – 2 = 0

By the Corollary to Theorem 5.10, S is an orthogonal basis for R4.



p. 140

S*

Theorem 5.11 Coordinates Relative to an Orthonormal Basis

If B = {v1, v2, …, vn} is an orthonormal basis for an inner product space V, then the coordinate

representation of a vector w relative to B is

w = w, v1v1 + w, v2 v2 + … +w, vn vn

Proof: Because B is a basis for V, there are unique scalars c1, c2, …, cn such that

w = c1v1 + c2v2 + …+ cnvn

Taking the inner product with vi of both sides of the equation gives

w, vi = (c1v1 + c2v2 + …+ cnvn), vi

We can distribute the inner product on the left-hand side to obtain

Because S is orthonormal, when

so equation becomes

so

The coordinates of w relative to an orthonormal basis B are called the Fourier coefficients of w

relative to B. The corresponding coordinate matrix is

[w]B =

nc

c

c

2

1

=

nvw

vw

vw

,

,

,

2

1

Note: contrast Thm 5.11 with [w]B =

1

21 ][][][

SnSS vvv [w]S

from Section 4.7 for a general, non-orthonormal basis B.



p. 141

S*

S*

Example: Find the coordinate matrix of w = (–2, 6, 5) relative to the orthonormal basis

B = {(2

1,

2

1, 0), (–

6

2,

6

2,

3

22), (

3

2, –

3

2,

3

1)} of R3

.

Solution:

so the coordinate matrix is

Given any basis for a vector space, we can construct an orthonormal basis using a procedure

called the Gram-Schmidt Orthonormalization.

Theorem 5.12(Alt) (Alternative) Gram-Schmidt Orthonormalization Process

Let B = {v1, v2, …, vn} be a basis for an inner product space V.

Let B = {u1, u2, …, un}, where ui is given by

w1 = v1, ...........................................................................................................u1 = 1

1

w

w

w2 = v2 – v2, u1u1, ........................................................................................u2 = 2

2

w

w

w3 = v3 –v3, u1u1 – v3, u2u2,......................................................................u3 = 3

3

w

w

…

wn = vn – vn, u1u1 – vn, u2u2 – … – vn, u n–1un–1, ...................................un = n

n

w

w

Then B is an orthonormal basis for V. Moreover, span{ u1, u2, …, uk} = span{ v1, v2, …, vk} for

k = 1, 2, …, n.

Proof: First, B is normalized, since all of the ui are unit

vectors: ||ui||= 1. Second, observe that the terms of the form

vj, uiui are projections onto ui: vj, uiui = jivuproj

When we subtract jivuproj from vj, the resultant vector is

orthogonal to ui



p. 142

S*

Now we prove by induction that B is also an orthonormal set.

If n =1, then there are no pairs of vectors in B, so there is nothing to prove.

If {u1, u2, …, uk–1} is an orthonormal set, then consider uk = k

k

w

w where

wk = vk – vk, u1u1 – vk, u2u2 – … – vk, ujuj – … – vk, uk–1uk–1.

Let j = 1, 2, …, k – 1. Then

uk, uj = nw

1(vk – vk, u1u1 – vk, u2u2 – … – vk, ujuj – … vk, uk–1uk–1), uj

=nw

1 (vk, uj – vk, u1u1, uj – vk, u2u2, uj – … – vk, ujuj, uj – …

– vk, uk–1uk–1, uj)

=nw

1 (vk, uj – vk, u1(0) – vk, u2(0) – … – vk, uj(1) – … – vk, uk–1(0))

=nw

1 (vk, uj – vk, uj) = 0

So {u1, u2, …, uk} is also an orthonormal set. By mathematical induction, B is orthonormal, and

since it has n vectors in an n-dimensional space, B is an orthonormal basis.

Example: Use the Gram-Schmidt orthonormalization process to construct an orthonormal basis

from B = {(1, 2, 2), (–1, –1,0), (1, 0, 0)}.

Solution:

w1 = (1, 2, 2) u1 = )2,2,1(

)2,2,1( = (

31 ,

32 ,

32 )

w2 = (–1, –1,0) – ((–1, –1,0)•(31 ,

32 ,

32 ))(

31 ,

32 ,

32 )

= (–1, –1,0) – (–1)(1, 2, 2)

= (–32 , –

31 ,

32 ) u2 =

),,(

),,(

32

31

32

32

31

32

= (–

32 , –

31 ,

32 )

w3 =

u3 =



p. 143

Example: Use the Gram-Schmidt orthonormalization process to construct an orthonormal basis

from B = {1, x, x2} in P2 with the inner product p(x), q(x) =

1

1

)()( dxxqxp .

Solution: Let v1 = 1, v2 = x, and v3 = x2.

w1 = 1

u1 = 1

1

||1||2 =

1

1

11 dx = 1

1x = 2

u1 = 2

1

w2 =

u2 =


5.4 Mathematical Models and Least Squares Analysis (Optional).

p. 145


Objective: Find the orthogonal complement of a subspace.

Objective: Solve a least squares problem.

In this section, we consider inconsistent systems of linear equations, and we find the “best

approximation” to a solution.

Example: Given a table of data. We want to find the

coefficients c0 and c1 of the line y = c0 + c1t

that “best fits” these points.

Solution: The system of linear equations that we

have to solve comes from plugging the three points

(ti, yi) into the equation c0 + c1ti = yi

33

12

01

10

10

10

cc

cc

cc

or

3

1

0

31

21

11

1

0

c

c. Let A =

31

21

11

, b =

3

1

0

, and x =

1

0

c

c

The system is inconsistent: reduced row-echelon form is

100

010

001

. (We also knew that there

would be no solution because the three points in the graph are not collinear.) But what is the

“best approximation”?

Recall that Ax = 10

3

2

1

1

1

1

cc

is always in the column space of A. But since Ax = b has no

solution, b is not in the column space of A. We want to find an x that gives the Ax that is

closest to b.

Least Squares Problem: Given an mn

matrix A and a vector b in Rm, find x in

Rn, such that ||Ax – b||

2 is minimized.

This gives the Ax that is closest to b. It is

customary to minimize ||Ax – b||2 instead

of ||Ax – b|| because ||Ax – b|| involves a

square root. Intuitively, we see that

Ax – b is orthogonal to the column space

of A.

t y

1 0

2 1

3 3



p. 146

To solve the least squares problem, it helps to use the concept of orthogonal subspaces.

Let S1 and S2 be two subspaces of an n-dimensional vector space V.

S1 and S2 are orthogonal when v1, v2 = 0 for all v1 in S1 and v2 in S2.

The orthogonal complement of S1 is the set S1 = {u V: u, v= 0 for all vectors v S1}.

S1 is pronounced “S1 perp.”

If every vector x V can be written uniquely as a sum of a vector s1 from S1 and a vector s2

from S2, x = s1 + s2, then V is the direct sum of S1 and S2, and we write V = S1 S2.

Theorem 5.13 Properties of Orthogonal Complements

Let S be a subspace of an n-dimensional vector space V. Then

1) dim(S) + dim(S) = n

2) V = S S

3) (S) = S

Examples: Consider R3. {0}

= R3

. (R3)

= {0}. Let Sx = span(

0

0

1

), Sy = span(

0

1

0

),

Sz = span(

1

0

0

), Sxy = span(

0

0

1

,

0

1

0

), and Syz = span(

0

1

0

,

1

0

0

). Then Sx is orthogonal to Sy.

Sy is orthogonal to Sz. Sxy is not orthogonal to Syz. Sxy = Sx Sy. (Sz) = Sxy, and (Sxy)

= Sz.

Theorem 5.16 Fundamental Subspaces of a Matrix

Let A be an mn matrix. The notation R(A) means the column space of A (R for range). R(AT

) is

the row space of A. N(A) is the nullspace of A.

1) R(AT

) and N(A) are orthogonal complements. That is: all vectors in the row space of A

are orthogonal to all vectors in the nullspace of A; dim(R(AT

)) + dim(N(A)) = n, and

R(AT

) N(A) = Rn.

2) R(A) and N(AT

) are orthogonal complements. That is: all vectors in the column space of

A are orthogonal to all vectors in the nullspace of AT; dim(R(A)) + dim(N(A

T )) = m, and

R(A) N(AT

) = Rm.



p. 147

Sketch of the proof:

1) N(A) ={x Rn: Ax = 0} shows that the nullspace is orthogonal to the row space, because the

null vector x is orthogonal to each row of A.

Ax = 0 is equivalent to

0) of row(

0) of 2 row(

0) of 1 row(

x

x

x

Am

A

A

dim(R(AT

)) + dim(N(A)) = n is rank + nullity = n, which we already know.

The basis vectors of R(AT

)) are linear independent of the basis of N(A) because they are

orthogonal to the basis of N(A), so the union of a basis of R(AT

)) and a basis of N(A) is a basis

of Rn, so R(A

T ) N(A) = Rn

.

2) Replace A with AT in Part (1), and use (A

T )T = A. When we replace A with A

T, we must

interchange m and n.

Example: Find the orthogonal complement of S = span(

0

0

2

1

,

1

0

1

0

).

Solution: S is the column space R(A), where A =

10

00

12

01

, so S = N(A

T).

To find N(A), solve Ax = 0

01010

00021rref

01010

02001

so 1x1 + 0x2 + 0s – 2t = 0

and 0x1 + 1x2 + 0s + 1t = 0

so x2 = –t and x1 = 2t

x =

t

s

t

t2

= s

0

1

0

0

+ t

1

0

1

2

so S = N(A

T) = span(

0

1

0

0

,

1

0

1

2

)

If S is a subspace of an n-dimensional vector space V and v is a vector in V, then v can be

written uniquely as the sum of a vector from S and a vector from S

v = s1 + s2, where s1 S and s2 S

because V = S S. Then s1 S is the orthogonal projection of v onto S, written vSproj



p. 148

Theorem: v – vSproj is orthogonal to every vector in S.

Proof: v can be written uniquely as v = vSproj + s2, where s2 = v – vSproj S. Since

v – vSproj S, v – vSproj is orthogonal to every vector in S.

Theorem 5.15 Orthogonal Projection and Distance

Let S be a subspace of an n-dimensional vector space V and v be a vector in V. Then for all u

S, u vSproj , ||v – vSproj || < ||v – u||.

In other words, vSproj is the vector in S that is closest to v.

Proof: Let all u S, u vSproj so

v – u = (v – vSproj ) + ( vSproj – u)

Now v – vSproj is orthogonal to ( vSproj – u), so

by the Pythagorean Theorem,

||v – u||2 = ||v – vSproj ||

2 + || vSproj – u||

2

where || vSproj – u|| > 0 because u vSproj .

Therefore, ||v – u||2 > ||v – vSproj ||

2

||v – u|| > ||v – vSproj ||

To solve the least squares problem, we

need Ax – b (R(A)) = N(A

T ), so

AT(Ax – b) = 0, or

ATAx = A

Tb “Normal equations”

Solution to Least Squares Problem



p. 149

Example: Let’s finish the example from the

beginning of this section. Given a table of data. We

want to find the coefficients c0 and c1 of the line

y = c0 + c1t that “best fits” these points.

Solution:

33

12

01

10

10

10

cc

cc

cc

or

3

1

0

31

21

11

1

0

c

c. Let A =

31

21

11

, b =

3

1

0

, and x =

1

0

c

c

So ATAx = A

Tb , i.e.

146

63

1

0

c

c=

11

4, so

1

0

c

c =

1

146

63

11

4 =

2/3

3/5.

Therefore, the least-squares regression line for the data is y = 23 t –

35

Example: Given the table of data. Find the coefficients

c0, c1, and c2 of the quadratic y = c0 + c1t + c2t2 that

best fits these points.

Solution:

4331

5.2221

5.1111

2001

2

2

10

2

2

10

2

2

10

2

2

10

ccc

ccc

ccc

ccc

so

931

421

111

001

2

1

0

c

c

c

=

4

5.2

5.1

2

A =

931

421

111

001

, x =

2

1

0

c

c

c

, and b =

4

5.2

5.1

2

. ATAx = A

Tb so x = (A

TA)

–1A

Tb =

5.0

8.0

95.1

y = 1.95 – 0.8t + 0.5t2

t y

1 0

2 1

3 3

t y

0 2

1 1.5

2 2.5

3 4


5.5 Applications of Inner Product Spaces (Optional).

p. 151


Objective: Find the nth

-order Fourier approximation of a function.

Objective: Given a subspace with an orthogonal or orthonormal basis, find the projection of a

vector onto that subspace.

Objective: Find the cross product of two vectors in R3.

Recall from Thm. 5.15 that vSproj is the unique vector in S that is closest to v.

Theorems 5.14 & 5.19 Projection onto a Subspace & Least Squares Approximation

If {u1, u2, …, un} is an orthonormal basis for a subspace S of a vector space V, and v V, then

vSproj = v, u1u1 + v, u2u2 + … + v, unun

If V is a space of functions, e.g. V = C[a, b] (so the ui are functions), then

fSproj = f, u1u1 + f, u2u2 + … + f, unun

is the least-squares approximating function of f with respect to S.

Proof: we will show that vSproj = v, u1u1 + v, u2u2 + … + v, unun by showing that v can be

written uniquely as the sum of a vector from S and a vector from S

v = s1 + s2, where s1 S and s2 S. Then by definition, vSproj = s1.

Let s1 = v, u1u1 + v, u2u2 + … v, unun

and s2 = v – v, u1u1 – v, u2u2 – … – v, unun

Then s1 S because it is a linear combination of vectors in S.

For each basis vector ui of S

ui, s2 = ui, (v – v, u1u1 – v, u2u2 – … – v, uiui – … – v, unun)

= ui, v – v, u1ui, u1 – v, u2ui, u2 – … – v, uiui, ui – … – v, unui, un

= ui, v – 0 – 0 – … – v, ui (1) – … – 0

= 0

Since all vectors in S are linear combinations of the {ui}, s2 is orthogonal to all vectors in S,

so s2 S.

Example: we know from 5.3 that set {1, cos(x), sin(x), cos(2x), sin(2x), …, cos(nx), sin(nx)} is a

an orthogonal set using the inner product f , g = 2

0

)()( dxxgxf . Let’s normalize the set.



p. 152

||1||2 =

2

0

)1)(1( dx = 2.

For n 0, ||cos(nx)||2 =

2

0

)cos()cos( dxnxnx =

2

02

)cos()sin(

2

xn

nxnxx =

and ||sin(nx)||2 =

2

0

)sin()sin( dxnxnx =

2

02

)cos()sin(

2

xn

nxnxx =

So {2

1 ,

1 cos(x),

1 sin(x),

1 cos(2x),

1 sin(2x), …,

1 cos(nx),

1 sin(nx)} is an

orthonormal set.

The nth

-order Fourier approximation of a function f on the interval [0, 2] is the projection of f

onto span {2

1 ,

1 cos(x),

1 sin(x),

1 cos(2x),

1 sin(2x), …,

1 cos(nx),

1 sin(nx)}.

Fourier approximations are useful in modeling periodic functions such as sound waves, heart

rhythms, and electrical signals.

Example: Find the second-order Fourier approximation to

the periodic function

f (x) = – x for x [0, 2 ), and f (x + 2) = f (x)

Solution: g(x) = f, 2

1 2

1

+ f,

1 cos(x)

1 cos(x) + f,

1 sin(x)

1 sin(x)

+ f,

1 cos(2x)

1 cos(2x) + f,

1 sin(2x)

1 sin(2x)

Using software to evaluate the integrals, we find

f, 2

1 2

1 = 21

2

0

1 )( dxx = 0

f,

1 cos(x)

1 =

2

0

1 )cos()( dxxx = 0

f,

1 sin(x)

1 =

2

0

1 )sin()( dxxx = 2

f,

1 cos(2x)

1 =

2

0

1 )2cos()( dxxx = 0

f,

1 sin(2x)

1 =

2

0

1 )2sin()( dxxx = 1

So the 2nd

order Fourier approximation is g2(x) = 2sin(x) + sin(2x)



p. 153

See http://www.falstad.com/fourier/e-sawtooth.html

It is common notation with Fourier series to write the coefficients as

a0 =

2

0

1 )( dxxf ak =

2

0

1 )cos()( dxkxxf bk =

2

0

1 )sin()( dxkxxf

so that the nth

-order Fourier approximation is

gn(x) = 2

0a + a1cos(x) + b1sin(x) + a2cos(2x) + b2sin(2x) + … + ancos(nx) + bnsin(nx)

Example: Given that S = {(–6

2,

6

2,

3

22), (

3

2, –

3

2,

3

1)} is an orthonormal set in R3

. Find

the projection of v = (1, 0, 0) onto span{(–6

2,

6

2,

3

22), (

3

2, –

3

2,

3

1)}.

Solution:

vSproj = ((1, 0, 0)•(–6

2,

6

2,

3

22))(–

6

2,

6

2,

3

22)

+ ((1, 0, 0)• (3

2, –

3

2,

3

1))(

3

2, –

3

2,

3

1)

= –6

2(–

6

2,

6

2,

3

22) +

3

2(

3

2, –

3

2,

3

1) = (

2

1, –

2

1, 0)

In R3, the cross product of two vectors u = (u1, u2, u3) = u1i + u2j + u3k and v = (v1, v2, v3)

= v1i + v2j + v3k is

uv = (u2v3 – u3v2)i + (u3v1 – u1v3)j + (u1v2 – u2v1)k = =

321

321

vvv

uuu

kji

where the right-hand side is a “determinant” containing the vectors i, j, and k. The cross product

is undefined for vectors in vector spaces other than R3.

Theorem 5.17 Algebraic Properties of the Cross Product

If u, v, and w are vectors in R3 and c is a scalar, then

http://www.falstad.com/fourier/e-sawtooth.html



p. 154

1) uv = –vu

2) u(v + w) = uv + uw

3) c(uv) = (cu)v = u(cv)

4) u0 = 0u = 0

5) uu = 0

6) u•(vw) = (uv)•w =

321

321

321

www

vvv

uuu



p. 155

Theorem 5.18 Geometric Properties of the Cross Product

If u, v, and w are nonzero vectors in R3, then

1) uv is orthogonal both u and v

2) The angle between u and v is given by ||uv|| = ||u|| ||v|| sin( )

3) u and v are parallel if and only if uv = 0

4) The parallelogram having u and v as adjacent sides has an area of ||uv||

5) The parallelepiped having u, v, and w as edges has a volume of ||u•(vw)||

Example: Find the area of the parallelogram with vertices at (5, 2, 0), (3, –6, 7), (7, –2, 8), and

(5, –10, 15)

Solution: area = ||uv||. uv =

842

782

kji

= –36 + 30j + 24k

||–36i + 30j + 24k || = 222 )24()30()36( = 6 77 52.6498

Example: Find a vector orthogonal to u = (–4, 3, –4) and v = (4, 7, 1).

Solution: uv =

174

434

kji

= 31i – 12j – 40k

8.3 Polar Form and De Moivre's Theorem. (Optional)

p. 157


Objective: Determine the polar form of a complex number and convert between the polar form

and the standard form of a complex number.

Objective: Determine the exponential polar form of a complex number and convert between the

exponential polar form and the standard form of a complex number using Euler’s formula.

Objective: Multiply and divide complex numbers in polar form and in exponential polar form.

Objective: Use DeMoivre’s Theorem to find powers of numbers in polar form and in exponential

polar form.

Multiplication, division, and powers of complex number are tedious to calculate in standard

form. However, when complex numbers are written in polar form, multiplication, division, and

powers become easy to calculate and interpret.

The polar form of a nonzero complex number z = a + bi is given by

z = r (cos + i sin )

where a = r cos, b = r sin, r = 22 ba = |z|, and tan = b/a.

The number r is the modulus of z and is the argument of z.

There are infinitely many choices for the argument, because cosine and sine have a period of 2.

Usually, we choose the principal argument, which satisfies < 2.

The principal argument = Arg(z) of a nonzero complex number z = a + bi is given by

tan = b/a and – <

Example: Graph and find the polar form (using the

principal argument) of

a) –3 b) –4 – 3i c) 2 – 2 i

Solution: Real axis

Imag

inar

y a

xis

Arg(z) = tan–1(b/a) Arg(z) = tan–1(b/a) +

Arg(z) = tan–1(b/a) Arg(z) = tan–1(b/a) –

Arg(z) = Arg(z) = 0

Arg(z) = /2

Arg(z) = – /2

b)

c)

a)


p. 158

a) –3 = 3(cos + i sin)

b) r =22 )3()4( = 25 = 5

Arg(z) = + tan–1

(43

) 3.3785

–4 – 3i 5(cos(3.3785) + i sin(3.3785))

c) r =22 )2()2( = 4 = 2 Arg(z) = tan

–1(

2

2 ) = 4

2 – 2 i = 2(cos(4 ) + i sin(

4 ))

Example: Find the standard form of 10 (cos(6 ) + i sin(

6 ))

Solution: 10 (cos(6 ) + i sin(

6 )) = 10

i

2

1

2

3 = i535

Theorem. Euler’s Formula

cos + i sin = ei

where e is Euler’s number, e 2.71828

Proof: From calculus, we know that the Maclaurin series (the Taylor series expansion around

zero) of a function f is

f (x) = !0

1f (0) +

!1

1 f (0) x +

!2

1f (0) x

2 +

!3

1f (0) x

3 +

!4

1f

(4)(0) x

4 +

!5

1f

(5)(0) x

5 + …

Therefore, the Maclaurin series for ex is

...!7

1

!6

1

!5

1

!4

1

!3

1

!2

1

!1

11 765432 xxxxxxxe x

Substitute x = i. Note that i2 = –1, i

3 = –i, i

4 = 1, i

5 = i, etc. So

...!7

1

!6

1

!5

1

!4

1

!3

1

!2

1

!1

11 765432 iiiiei

Therefore, the Maclaurin series for cos and sin are

...0!6

10

!4

10

!2

101)cos( 642

...!7

10

!5

10

!3

10

!1

10)sin( 753


p. 159

Comparing the series for ei

, cos, sin, we see that ei

= cos + i sin

Example: Graph and find the polar form (using the


a) –i b) 1 – 3 i c) 3 + 2i

Solution

a) –i = e–i /2

b) r =22 )3(1 = 4 = 2

= tan–1

(– 3 /1) = – /3

1 – 3 i = 2 e–i /3

c) r =22 23 = 13 3.606

= tan–1

(2/3) 0.588

3 + 2i = 3.606 e0.588i

d) Example: Graph and find the polar form (using the


a) –i b) 1 – 3 i c) 3 + 2i

Example: Find the standard form of 3e5i/4

Solution: 3e5i/4

= 3 (cos

4

5+ i sin

4

5) = 3

i

2

2

2

2 = i

2

23

2

23

Theorem 8.4 Product and Quotient of Two Complex Nubmers

1

1

ier 2

2

ier = )(

2121 i

err )sin(cos)sin(cos 222111 irir = )]sin()[cos( 212121 irr

2

1

2

1

i

i

er

er =

)(

2

1 21 ie

r

r

)sin(cos

)sin(cos

222

111

ir

ir

= )]sin()[cos( 2121

2

1 ir

r, z2 0

Proof: The exponential formulas follow directly from the laws of exponents. The polar forms

follow from the exponential polar forms and Euler’s formula.

Example: Sketch and simplify.

a) 2ei/3

3ei/2

= 2(cos3 + i sin

3 )3(cos

2 + i sin

2 ) b)

)]sin()[cos(4

)]sin()[cos(3

4

3

43

434/3

i

i

e

ei

i

a)

b)

c)


p. 160

P

Solution:

a) 2ei/3

3ei/2

= (2)(3)ei(/3+/2)

= 6ei5/6

2(cos3 + i sin

3 )3(cos

2 + i sin

2 )

= 6[cos( 65 ) + i sin(

65 )]

/2 = 90, /3 = 60, 65 = 150

b) 4/34

3

i

i

e

e

= ))4/3((

4

3 ie = 4/7

4

3 ie

= )4/72(

4

3 ie = 4/

4

3 ie

)]sin()[cos(4

)]sin()[cos(3

43

43

i

i

)]sin()[cos(4

347

47 i

)]4/sin()4/[cos(4

3 i

Theorem 8.5 DeMoivre’s Theorem

(rei

)n = r

ne

in

[r (cos + i sin )]n = r

n [cos(n ) + i sin(n )]

Proof: The exponential formula follows directly from the laws of exponents. The polar forms

follow from the exponential polar forms and Euler’s formula.

– /4 = 2 – 7 /4

4

3

¾

–3 /4

7 /4 = –(–3 /4)

7 /4

5 /6 = /2 + /3

/2 /3

6

2

3

/3


p. 161

Example: Sketch and simplify.

a) ei2k/3

for k = –1, 0, 1, 2, 3, 4

b) eik/2

for k = –2, –1, 0, 1, 2, 3, 4, 5

Solution

a) ei(–1)/3

= –21 –

2

3 i = e–i/3

ei(0)/3

= 1 = e0

ei(1)/3

= –21 +

2

3 i = ei/3

ei(2)/3

= –21 –

2

3 i = e–i/3

ei(3)/3

= 1 = e0

ei(4)/3

= –21 +

2

3 i = ei/3

b) ei(–2)/2

= e–i

= –1 = ei

ei(–1)/2

= –i = e–i/2

ei(0)/2

= 1 = e0

ei(1)/2

= i = ei/2

ei(2)/2

= –1 = ei

ei(3)/2

= –i = e–i/2

ei(4)/2

= 1 = e0

ei(5)/2

= i = ei/2


8.3 Polar Form and DeMoivre’s Theorem

Determine the polar form of a complex number, convert betweenthe polar form and standard form of a complex number, and multiply and divide complex numbers in polar form.

Use DeMoivre’s Theorem to find powers and roots of complexnumbers in polar form.

POLAR FORM OF A COMPLEX NUMBER

At this point you can add, subtract, multiply, and divide complex numbers. However,there is still one basic procedure that is missing from the algebra of complex numbers.To see this, consider the problem of finding the square root of a complex number suchas When you use the four basic operations (addition, subtraction, multiplication, anddivision), there seems to be no reason to guess that

That is,

To work effectively with powers and roots of complex numbers, it is helpful to use apolar representation for complex numbers, as shown in Figure 8.7. Specifically, if

is a nonzero complex number, then let be the angle from the positive real axisto the radial line passing through the point and let be the modulus of This leads to the following.

So, from which the polar form of a complex number is obtained.

Because there are infinitely many choices for the argument, the polar form of a complex number is not unique. Normally, the values of that lie between and are used, although on occasion it is convenient to use other values. The value of thatsatisfies the inequality

Principal argument

is called the principal argument and is denoted by Arg( ). Two nonzero complex numbers in polar form are equal if and only if they have the same modulus and the same principal argument.

z

�� < � � �

��

a � bi � �r cos �� r sin ��i,

r � �a2 � b2

b � r sin �

a � r cos �

a � bi.r�a, b��a � bi

�1 � i�2 �2

� i.

�i �1 � i�2

.

i.

Definition of the Polar Form of a Complex Number

The polar form of the nonzero complex number is given by

where and The numberis the modulus of and is the argument of z.�z

rtan � � b�a.a � r cos �, b � r sin �, r � �a2 � b2,

z � r�cos � � i sin ��

z � a � bi

REMARK

The polar form of is expressed as

where is any angle.�

z � 0�cos � � i sin ��,

z � 0

Imaginaryaxis

Realaxis

br

a 0

θ

θ θ Standard Form: a + bi

Polar Form: r(cos + i sin )

(a, b)

Figure 8.7

9781133110873_0803.qxp 3/10/12 6:54 AM Page 404

Finding the Polar Form of a Complex Number

Find the polar form of each of the complex numbers. (Use the principal argument.)

a. b. c.

SOLUTION

a. Because and which implies thatFrom and

and

So, and

b. Because and then which implies that So,

and

and it follows that So, the polar form is

c. Because and it follows that and so

The polar forms derived in parts (a), (b), and (c) are depicted graphically in Figure 8.8.

a. b.

c.

Figure 8.8

z � 1�cos �

2� i sin

�

2�

Imaginaryaxis

Realaxis

θ

z = i1

1

z � �13 cos�0.98� � i sin�0.98�z � �2 �cos��

4� � i sin��

4��

Realaxis1 2

4

3

2

1θ

z = 2 + 3i

Imaginaryaxis

Imaginaryaxis

Realaxis

−1−1−2

1

2

2

−2

θ

z = 1 − i

z � 1�cos �

2� i sin

�

2�.

� � ��2,r � 1b � 1,a � 0

z � �13 �cos�0.98� � i sin�0.98��.

� � 0.98.

sin � �b

r�

3�13

cos � �a

r�

2�13

r � �13.r2 � 22 � 32 � 13,b � 3,a � 2

z � �2 �cos ��

4� � i sin��

4��.

� � ��4

sin � �b

r� �

1�2

� ��2

2 .cos � �

a

r�

1�2

��2

2

b � r sin �,a � r cos �r � �2.b � �1, then r2 � 12 � ��1�2 � 2,a � 1

z � iz � 2 � 3iz � 1 � i

8.3 Polar Form and DeMoivre’s Theorem 405

9781133110873_0803.qxp 3/10/12 6:54 AM Page 405


Converting from Polar to Standard Form

Express the complex number in standard form.

SOLUTION

Because and obtain the standard form

The polar form adapts nicely to multiplication and division of complex numbers.Suppose you have two complex numbers in polar form

and

Then the product of and is expressed as

Using the trigonometric identities and you have

This establishes the first part of the next theorem. The proof of the second part is left toyou. (See Exercise 75.)

z1z2 � r1r2 cos��1 � �2� � i sin��1 � �2�.

sin ��1 � �2� � sin �1 cos �2 � cos �1 sin �2,cos��1 � �2� � cos �1 cos �2 � sin �1 sin �2

� r1r2 �cos �1 cos �2 � sin �1 sin �2� � i �cos �1 sin �2 � sin �1 cos �2�. z1z2 � r1r2�cos �1 � i sin �1��cos �2 � i sin �2�

z2z1

z2 � r2�cos �2 � i sin �2�.z1 � r1�cos �1 � i sin �1�

z � 8�cos��

3� � i sin��

3�� 8�12

� i�32 � � 4 � 4�3i.

sin��3� � ��3�2,cos��3� � 1�2

z � 8�cos��

3� � i sin��

3��


Elliptic curves are the foundation for elliptic curve cryptography (ECC), a type of public key cryptography forsecure communications over the Internet. ECC has gainedpopularity due to its computational and bandwidth advantages over traditional public key algorithms.

One specific variety of elliptic curve is formed usingEisenstein integers. Eisenstein integers are complex

numbers of the form where

and and are integers. These numbers can be graphed as intersection points of a triangular lattice in the complexplane. Dividing the complex plane by the lattice of allEisenstein integers results in an elliptic curve.

ba

� � �12

��32

i,z � a � b�,

THEOREM 8.4 Product and Quotient of Two Complex Numbers

Given two complex numbers in polar form

and

the product and quotient of the numbers are as follows.

Product

Quotientz1z2

�r1r2

cos��1 � �2� � i sin��1 � �2�, z2 0

z1z2 � r1r2 cos��1 � �2� � i sin��1 � �2�

z2 � r2�cos �2 � i sin �2�z1 � r1�cos �1 � i sin �1�

thumb/Shutterstock.com

9781133110873_0803.qxp 3/10/12 6:54 AM Page 406

Theorem 8.4 says that to multiply two complex numbers in polar form, multiply moduli and add arguments. To divide two complex numbers, divide moduli and subtractarguments. (See Figure 8.9.)

Figure 8.9

Multiplying and Dividing in Polar Form

Find and for the complex numbers

and

SOLUTION

Because and are in polar form, apply Theorem 8.4, as follows.

multiply

add adddivide

subtract subtract

Use the standard forms of and to check the multiplication in Example 3. For instance,

To verify that this answer is equivalent to the result in Example 3, use the formulas forand to obtain

and sin�5�

12� � sin��

6�

�

4� ��6 ��2

4.cos�5�

12� � cos��

6�

�

4� ��6 ��2

4

sin �u � v�cos �u � v�

�53 �

�6 � �24

��6 � �2

4i�.

�5�6 � 5�2

12�

5�6 � 5�212

i

z1z2 � �5�22

�5�2

2i��3

6�

16

i� �5�612

�5�212

i �5�612

i �5�212

i2

z2z1

z1

z2

�5

1�3 �cos��

4�

�

6� � i sin��

4�

�

6�� 15�cos �

12� i sin

�

12�

z1z2 � �5��1

3� �cos��

4�

�

6� � i sin��

4�

�

6�� 5

3 �cos 5�

12� i sin

5�

12�

z2z1

z2 �13�cos

�

6� i sin

�

6�.z1 � 5�cos�

4� i sin

�

4�z1�z2z1z2

Imaginaryaxis

Realaxis

z1z2

r1r2

r2

r1

z1 z2

θ1θ1

θ 2

θ 2−

1 2To divide z and z :Divide moduli and subtract arguments.

Imaginaryaxis

Realaxis

z1z2

r1r2

r2r1

z1

z2

θ1

θ1

θ 2

θ 2+

1 2To multiply z and z :Multiply moduli and add arguments.


REMARK

Try verifying the division inExample 3 using the standardforms of and z2.z1

9781133110873_0803.qxp 3/10/12 6:54 AM Page 407

DEMOIVRE’S THEOREM

The final topic in this section involves procedures for finding powers and roots of complex numbers. Repeated use of multiplication in the polar form yields

and

Similarly,

and

This pattern leads to the next important theorem, named after the French mathematicianAbraham DeMoivre (1667–1754). You are asked to prove this theorem in Review Exercise 85.

Raising a Complex Number

to an Integer Power

Find and write the result in standard form.

SOLUTION

First convert to polar form. For

and

which implies that So,

By DeMoivre’s Theorem,

� 4096.

� 40961 � i�0�� 4096�cos 8� � i sin 8��

� 212�cos 12�2��

3� i sin

12�2��3 �

��1 � �3 i�12� �2�cos

2�

3� i sin

2�

3 ��12

�1 � �3 i � 2�cos 2�

3� i sin

2�

3 �.

� � 2��3.

tan � ��3

�1� ��3r � ��1�2 � ��3�2 � 2

�1 � �3 i,

��1 � �3 i�12

z5 � r 5�cos 5� � i sin 5��.

z4 � r4�cos 4� � i sin 4��

� r3�cos 3� � i sin 3��. z3 � r �cos � � i sin ��r2 �cos 2� � i sin 2��

� r2�cos 2� � i sin 2�� z2 � r �cos � � i sin ��r �cos � � i sin �� z � r �cos � � i sin ��


THEOREM 8.5 DeMoivre’s Theorem

If and is any positive integer, then

zn � rn�cos n� � i sin n��.

nz � r�cos � � i sin ��

9781133110873_0803.qxp 3/10/12 6:54 AM Page 408

Recall that a consequence of the Fundamental Theorem of Algebra is that a polynomial of degree has zeros in the complex number system. So, a polynomialsuch as has six zeros, and in this case you can find the six zeros by factoring and using the Quadratic Formula.

Consequently, the zeros are

and

Each of these numbers is called a sixth root of 1. In general, the th root of a complexnumber is defined as follows.

DeMoivre’s Theorem is useful in determining roots of complex numbers. To seehow this is done, let be an th root of where

and

Then, by DeMoivre’s Theorem

and because it follows that

Now, because the right and left sides of this equation represent equal complex numbers,equate moduli to obtain which implies that and equate principal arguments to conclude that and must differ by a multiple of Note that is apositive real number and so is also a positive real number. Consequently, forsome integer which implies that

Finally, substituting this value of into the polar form of produces the result statedin the next theorem.

w

�� 2�k

n .

k, n � � � 2�k,s � n�r

r2�.n�s � n�r ,sn � r,

sn �cos n � i sin n� � r�cos � � i sin ��.

wn � z,

wn � sn�cos n � i sin n�

z � r�cos � � i sin ��.w � s�cos � i sin �

z,nw

n

x �1 ± �3 i

2.x �

�1 ± �3i2

,x � ±1,

� �x � 1��x2 � x � 1��x � 1��x2 � x � 1� x6 � 1 � �x3 � 1��x3 � 1�

p�x� � x6 � 1nn


REMARK

Note that when exceedsthe roots begin to

repeat. For instance, whenthe angle is

which yields the same valuesfor the sine and cosine ask � 0.

� � 2�nn

��

n� 2�

k � n,

n � 1,k

Definition of the nth Root of a Complex Number

The complex number is an th root of the complex number when

z � wn � �a � bi�n.

znw � a � bi

THEOREM 8.6 The nth Roots of a Complex Number

For any positive integer the complex number has exactly distinct roots. These roots are given by

where k � 0, 1, 2, . . . , n � 1.

n�r �cos�� 2�kn � � i sin�� 2�k

n ��nn

z � r �cos � � i sin ��n,

9781133110873_0803.qxp 3/10/12 6:54 AM Page 409

The formula for the th roots of a complexnumber has a nice geometric interpretation, asshown in Figure 8.10. Because the th roots allhave the same modulus (length) they lie on acircle of radius with center at the origin.Furthermore, the roots are equally spacedaround the circle, because successive th rootshave arguments that differ by

You have already found the sixth roots of 1 byfactoring and the Quadratic Formula. Try solvingthe same problem using Theorem 8.6 to get theroots shown in Figure 8.11. When Theorem 8.6 isapplied to the real number 1, the th roots have aspecial name—the th roots of unity.

Finding the nth Roots of a Complex Number

Determine the fourth roots of

SOLUTION

In polar form,

so Then, by applying Theorem 8.6,

Setting and 3,

as shown in Figure 8.12.

Figure 8.12

Imaginaryaxis

Realaxis

13π

π8

π8

9π8

813π

8

9π8cos + i sin

5π8

5π8cos + i sin

cos + i sin

cos + i sin

z4 � cos 13�

8� i sin

13�

8

z3 � cos 9�

8� i sin

9�

8

z2 � cos 5�

8� i sin

5�

8

z1 � cos �

8� i sin

�

8

k � 0, 1, 2,

� cos��

8�

k�

2 � � i sin��

8�

k�

2 �.

i1�4 � 4�1 �cos��24

�2k�

4 � � i sin��24

�2k�

4 ��r � 1 and � � ��2.

i � 1�cos �

2� i sin

�

2�

i.

nn

2��n.n

n

n�r

n�r,n

n


rn

2π

2π

n

n

The nth Roots of aComplex Number

Realaxis

Imaginaryaxis

Figure 8.10

Imaginaryaxis

Realaxis

The Sixth Roots of Unity

1−1

1

1

1

1

2

2

2

2

2

2

2

2

3

3

3

3

+

−

+

−

i

i

i

i

−

−

Figure 8.11

REMARK

In Figure 8.12, note that wheneach of the four angles

and ismultiplied by 4, the result is ofthe form ��2� � 2k�.

13��8��8, 5��8, 9��8,

9781133110873_0803.qxp 3/10/12 6:54 AM Page 410

8.3 Exercises 411

8.3 Exercises

Converting to Polar Form In Exercises 1–4, expressthe complex number in polar form.

1. 2.

3. 4.

Graphing and Converting to Polar Form In Exercises5–16, represent the complex number graphically, andgive the polar form of the number. (Use the principalargument.)

5. 6.

7. 8.

9. 10.

11. 7 12. 4

13. 14.

15. 16.

Graphing and Converting to Standard Form InExercises 17–26, represent the complex number graphically, and give the standard form of the number.

17.

18.

19.

20.

21. 22.

23. 24.

25. 26.

Multiplying and Dividing in Polar Form In Exercises27–34, perform the indicated operation and leave theresult in polar form.

27.

28.

29.

30.

31.

32.

33.

34.

Finding Powers of Complex Numbers In Exercises35–44, use DeMoivre’s Theorem to find the indicatedpowers of the complex number. Express the result instandard form.

35. 36.

37. 38.

39.

40.

41.

42.

43.

44.

Finding Square Roots of a Complex Number InExercises 45–52, find the square roots of the complexnumber.

45. 46.

47. 48.

49. 50.

51. 52. 1 � �3 i1 � �3 i

2 � 2i2 � 2i

�6i�3i

5i2i

�5�cos 3�

2� i sin

3�

2 ��4

�2�cos �

2� i sin

�

2��8

�cos 5�

4� i sin

5�

4 �10

�3�cos 5�

6� i sin

5�

6 ��4

�5�cos �

9� i sin

�

9��3

�1 � �3i�3

��3 � i�7��1 � i�10

�2 � 2i�6�1 � i�4

9cos�3��4� � i sin�3��4�5cos��4� � i sin��4�

12cos��3� � i sin��3�3cos��6� � i sin��6�

cos�5��3� � i sin�5��3�cos� � i sin�

2cos�2��3� � i sin�2��3�4[cos�5��6� � i sin�5��6�]

�3�cos �

3� i sin

�

3��1

3 �cos

2�

3� i sin

2�

3 ��0.5�cos� � i sin�� 0.5�cos�� i sin��

�3

4 �cos �

2� i sin

�

2��6�cos �

4� i sin

�

4��3�cos

�

3� i sin

�

3��4�cos �

6� i sin

�

6��

9�cos � � i sin ��7�cos 0 � i sin 0�

6�cos 5�

6� i sin

5�

6 �4�cos 3�

2� i sin

3�

2 �

8�cos �

6� i sin

�

6�15

4 �cos �

4� i sin

�

4�

3

4 �cos

7�

4� i sin

7�

4 �

3

2 �cos 5�

3� i sin

5�

3 �

5�cos 3�

4� i sin

3�

4 �

2�cos �

2� i sin

�

2�

5 � 2i�1 � 2i

2�2 � i3 � �3 i

�2i6i

52��3 � i ��2�1 � �3i �2 � 2i�2 � 2i

Imaginaryaxis

Realaxis

3i

1

1−1

2

3

Imaginaryaxis

Realaxis

−6 1

2

−2

−2−3−4−5−6

−3

3

Imaginaryaxis

Realaxis

1 + 3i

1 2

1

2

3

−1

Imaginaryaxis

Realaxis

1

−1

−2

2

2 − 2i

9781133110873_0803.qxp 3/10/12 6:54 AM Page 411


Finding and Graphing nth Roots In Exercises 53–64,(a) use Theorem 8.6 to find the indicated roots, (b) represent each of the roots graphically, and (c) expresseach of the roots in standard form.

53. Square roots:

54. Square roots:

55. Fourth roots:

56. Fifth roots:

57. Square roots:

58. Fourth roots:

59. Cube roots:

60. Cube roots:

61. Cube roots: 8

62. Fourth roots:

63. Fourth roots: 1

64. Cube roots: 1000

Finding and Graphing Solutions In Exercises 65–72,find all the solutions of the equation and represent yoursolutions graphically.

65. 66.

67. 68.

69. 70.

71. 72.

73. Electrical Engineering In an electric circuit, theformula relates voltage drop current andimpedance where complex numbers can representeach of these quantities. Find the impedance when thevoltage drop is and the current is

75. Proof When provided with two complex numbersand

with prove that

76. Proof Show that the complex conjugate ofis

77. Use the polar forms of and in Exercise 76 to findeach of the following.

(a)

(b)

78. Proof Show that the negative of is

79. Writing

(a) Let

Sketch and in the complex plane.

(b) What is the geometric effect of multiplying a complex number by What is the geometriceffect of dividing by

80. Calculus Recall that the Maclaurin series for sin and cos are

(a) Substitute in the series for and show that

(b) Show that any complex number can beexpressed in polar form as

(c) Prove that if then

(d) Prove the formula


81. Although the square of the complex number is givenby the absolute value of the complex number is defined as

82. Geometrically, the th roots of any complex number are all equally spaced around the unit circle centered atthe origin.

zn a � bi � �a2 � b2.z � a � bi

�bi�2 � �b2,bi

ei� � �1.

z � re�i�.z � rei�,

z � rei�.z � a � bi

ei� � cos � � i sin �.exx � i�

cos x � 1 �x 2

2!�

x4

4!�

x6

6!� . . . .

sin x � x �x 3

3!�

x5

5!�

x7

7!� . . .

ex � 1 � x �x 2

2!�

x3

3!�

x4

4!� . . .

xx,ex,

i?zi?z

z�iiz,z,

z � r�cos � � i sin �� 2�cos �

6� i sin

�

6�.

�z � rcos�� i sin�� .

z � r�cos � � i sin ��z�z, z 0

zz

zz

z � rcos�� i sin��.z � r�cos � � i sin ��

z1z2

�r1r2

cos��1 � �2� � i sin��1 � �2�.

z2 0,z2 � r2�cos �2 � i sin �2�,z1 � r1�cos �1 � i sin �1�

2 � 4i.5 � 5i

Z,I,V,V � I � Z

x4 � i � 0x 3 � 64i � 0

x4 � 81 � 0x5 � 243 � 0

x3 � 27 � 0x 3 � 1 � 0

x4 � 16i � 0x4 � 256i � 0

81i

�4�2 �1 � i��

1252 �1 � �3i�

625i

�25i

32�cos 5�

6� i sin

5�

6 �

16�cos 4�

3� i sin

4�

3 �

9�cos 2�

3� i sin

2�

3 �

16�cos �

3� i sin

�

3�

74. Use the graph of the roots of acomplex number.

(a) Write each of the roots in trigonometric form.

(b) Identify the complex number whose roots aregiven. Use a graphing utility to verify your results.

(i) (ii)

Realaxis

Imaginaryaxis

45°45°

45°45°

3 3

3 3

30°30°

−11

Realaxis

Imaginaryaxis

2 2

2

9781133110873_0803.qxp 3/10/12 6:54 AM Page 412

Answer Key

Section 8.3

1. 3.

5.

7.

9.

11.

13.

15.

17.

19.

21.

15�2

8�

15�2

8i

123 θ = 4

r = 3.75

1 2 3−2−2−3

Realaxis

Imaginaryaxis

π

3

4�

3�3

4i

1

2

θ = 53

Realaxis

Imaginaryaxis

−2

−2 1 2

32r =

π

2i

1

θ = π2

−1−1

r = 2Realaxis

Imaginaryaxis

1

�5 cos��2.0344� � i sin��2.0344��

1 43

1

34

z = −1 − 2i

−1−3

−4

Realaxis

Imaginaryaxis

2�3�cos �

6� i sin

�

6

1 2 4

12

4z = 3 + 3 i

−1−4−2

−4

Realaxis

Imaginaryaxis

7�cos 0 � i sin 0�

3

6

z = 7

Realaxis

3 6−3−3

−6

Imaginaryaxis

6�cos �

2� i sin

�

2

2

4

Realaxis

Imaginaryaxis

z = 6i

−4

−4

−2 2 4

4�cos��2�

3 � i sin��2�

3 �

123

Realaxis

1 2 3−2−2−3

−3

z = −2 − 2 3 i

Imaginaryaxis

�8�cos��3�

4 � i sin��3�

4 �

1

2

1 2

Realaxis

z = −2 − 2i

Imaginaryaxis

6�cos � � i sin ��8�cos��

4 � i sin��

4 �

9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A2

23.

25.

7

27. 29.

31.

33. 35. 37.

39. 41. 43. 256

45.

Square roots:

47.

Square roots:

49.

Square roots:

51.

Square roots:

53. (a)

(b)

(c)

55. (a)

(b)

(c)

�3 � i�1 � �3 i��3 � i1 � �3 i

1r = 2

θθ

θ

θ

==

=

=

5

11

4

36

6

3

12

4

3

−1−1

1

Realaxis

Imaginaryaxis

π π

π

π

2�cos 11�

6� i sin

11�

6

2�cos 4�

3� i sin

4�

3

2�cos 5�

6� i sin

5�

6

2�cos �

3� i sin

�

3 �2�3 � 2i2�3 � 2i

2θ

θ

=

= 7

6

6

1

2

−2−2

r = 4Realaxis

Imaginaryaxis

π

π

4�cos 7�

6� i sin

7�

6 4�cos

�

6� i sin

�

6

�2�cos7�

6� i sin

7�

6 � ��62

��22

i

�2�cos�

6� i sin

�

6 ��62

��22

i

1 � �3i � 2�cos�

3� i sin

�

3

81�4�cos15�

8� i sin

15�

8 � 1.554 � 0.644i

81�4�cos7�

8� i sin

7�

8 � �1.554 � 0.644i

2�2i � 2�2�cos7�

4� i sin

7�

4

�3�cos7�

4� i sin

7�

4 ��62

��62

i

�3�cos3�

4� i sin

3�

4 � ��62

��62

i

�3i � 3�cos3�

2� i sin

3�

2

�2�cos5�

4� i sin

5�

4 � �1 � i

�2�cos�

4� i sin

�

4 � 1 � i

2i � 2�cos�

2� i sin

�

2 �

81

2�

81�3

2 i�8

�32i�44�cos �

6� i sin

�

6

12 �cos��

�

6 � i sin��

6 �

0.25�cos 0 � i sin 0�12�cos �

2� i sin

�

2

24 θ = 0

r = 7Realaxis

2 4 6−2−4−4−6

Imaginaryaxis

�4i

1 2 3

123

θ = 32

r = 4

−2−2−1−3

−3

Realaxis

Imaginaryaxis

π

Answer Key

9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A3

Answer Key

57. (a)

(b)

(c)

59. (a)

(b)

(c)

61. (a)

(b)

(c) 2

63. (a)

(b)

(c) 1

65.

67.

1

1

θ

θ

θ

=

=

=

π

5π

π

3

3

2

3

1

−1

Realaxis

Imaginaryaxis

r = 1

cos 5�

3� i sin

5�

3

cos � � i sin �

cos �

3� i sin

�

3

θ

θ

θ

θ

=

=

=

=

13

5

9

8

8

8

8

4

2

3

1

Realaxis

Imaginaryaxis

r = 42

2−2−2

π

π

π

π

4�cos13�

8� i sin

13�

8 4�cos

9�

8� i sin

9�

8 4�cos

5�

8� i sin

5�

8 4�cos

�

8� i sin

�

8 �i�1i

1

1

θ

θ

θ

θ

= 0

=

=

= π 32

2

1

4

2

3

r = 1

−1

Realaxis

Imaginaryaxis

π

π

cos 3�

2� i sin

3�

2

cos � � i sin �

cos �

2� i sin

�

2

cos 0 � i sin 0

�1 � �3 i�1 � �3 i

1θ

θ

θ

= 0

=

=

4

2

3

3

1

3

2

r = 2

−1−1 1

Realaxis

Imaginaryaxis

π

π

2�cos 4�

3� i sin

4�

3

2�cos 2�

3� i sin

2�

3 2�cos 0 � i sin 0�3.83 � 3.21i�4.70 � 1.71i0.868 � 4.92i

2

4

6 θ

θθ

1

32

=

==

4

1610

9

99

2 4 6−2

−4

−6

r = 5Realaxis

Imaginaryaxis

π

ππ

5�cos 16�

9� i sin

16�

9

5�cos 10�

9� i sin

10�

9

5�cos 4�

9� i sin

4�

9

5�2

2�

5�2

2 i

�5�2

2�

5�2

2 i

2

4

6θ

θ

1

2

=

=

3

7

4

4

2 4 6−2−6

−4

−6

r = 5Realaxis

Imaginaryaxis

π

π

5�cos 7�

4� i sin

7�

4

5�cos 3�

4� i sin

3�

4

9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A4

69.

71.

73. 75. Proof77. (a) (b)79. (a)

(b) Counterclockwise rotation of clockwise rotation of

81. True

�

2�

2;

−1−2

−2

−3

−3 1 2 3

Realaxis

Imaginaryaxis

z /i = 1 − 3 i

z = 3 + i

iz =−1 + 3 i

cos�2�� i sin�2��r2

32 �

12i

θ

θ

θ=

=

=

11

76

2

6

3

1

2

−2−2

2

2

r = 4Realaxis

Imaginaryaxis

π

π

π

4�cos 11�

6� i sin

11�

6

4�cos 7�

6� i sin

7�

6

4�cos �

2� i sin

�

2

1 2

2 θθ

θ

θθ

1

2

3

5

4

=

=

= π

==

3

9

7

5

5

5

5

−2

−2 −1

r = 3Realaxis

Imaginaryaxis

π

π

π

π

3�cos 9�

5� i sin

9�

5

3�cos 7�

5� i sin

7�

5 3�cos � � i sin ��

3�cos 3�

5� i sin

3�

5

3�cos �

5� i sin

�

5

Answer Key

9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A5

8.4 Complex Vector Spaces and Inner Products.

p. 163


Objective: Represent a vector in Cn by a basis.

Objective: Find the row space and null space of a matrix.

Objective: Solve a system of linear equations.

Objective: Find the Euclidean inner product, Euclidean norm, and Euclidean distance in Cn

A complex vector space is a vector space in which the scalars are complex numbers. The

complex version of Rn is the complex vector space Cn

consisting of ordered n-tuples of complex

numbers

v = (a1 + b1i, a2 + b2i, … an + bni)

We also represent vectors in Cn by their coordinate matrices relative to the standard basis

e1 = (1, 0, 0, …, 0), e2 = (0, 1, 0, …, 0), …, en = (0, 0, 0, …, 1),

v =

iba

iba

iba

nn

22

11

As in of Rn, the operations of vector addition and scalar multiplication in Cn

are performed

component by component.

The dimension of Cn is n. A basis of a subspace V of Cn

is any linearly independent set of vectors

that spans V. If V has dimension d, then any linearly independent set of d vectors in V is a basis

of V. Also, any set of d vectors that spans V is a basis of V.

Example: Is R2 a subspace of the complex vector space C2

over C?

Solution: R2 is a subset of C2

because R2 ={(a1 + b1i, a2 + b2i) where b1 = b2 = 0}. Moreover,


p. 164

Example: Show that S1 = {v1, v2, v3} = {(i, 2, 0), (2, i, 0), (1 + i, 0, 1 – i)} is a basis for C3.

Solution: Because the dimension of C3 is 3, the set S1 will be a basis if it is linearly

independent.

So we must check that

Example: Write u = (2 – i, 8 – 2i, 5 – i) as a linear combination of the vectors in the set S1 = {v1,

v2, v3} = {(i, 2, 0), (2, i, 0), (1 + i, 0, 1 – i)}, i.e. the same set as in the previous example.

Solution:

Example: Find the dimension of and a basis for

S2 = span{(–2 – 2i, –3i, –9 – 6i), (–1 – i, – 3, –3), (–3 – 2i, –1 + 3i, –6 – 5i)}.

Solution: S2 = the column space of


p. 165

Example: Find the rank, nullity, and bases for the row space and null space of

Solution:


p. 166

Example: Solve

Solution:

Let u and v be vectors in Cn. The Euclidean inner product of u and v is given by

u•v = u1v1* + u2v2* + … + unvn* = 𝑢1𝑣1̅̅ ̅ + 𝑢2𝑣2̅̅ ̅ + ⋯ + 𝑢𝑛𝑣𝑛̅̅ ̅

Theorem 8.7 Properties of the Euclidean inner product

Let u, v, and w be vectors in Cn and let k be a complex scalar. Then

1) u•v = (v•u)* = v∙u̅̅ ̅̅ Different from Rn!

2) (u + v)•w = u•w + v•w

3) (k u)•v = k (u•v)

4) u•(k v) = k*(u•v) = �̅�(u∙v) Different from Rn!

5) u•u 0 (This also says that u•u is real.)

6) u•u = 0 if and only if u = 0


p. 167

The Euclidean norm of u in Cn is ||u|| = uu

The Euclidean distance between u and v in Cn is d(u, v) = ||u – v||

To find a dot product u•v on the TI-89, use

MatrixVector opsdotP(u,v)

To find a dot product u•v in Mathematica, use u.vconj which displays as u.v*

To find a dot product u•v in PocketCAS, use dot(v,u) (PocketCAS takes the complex conjugate of the first vector, rather than the second.)

To find a norm ||v|| on the TI-89, use

MatrixNormsnorm(v)

To find a norm ||v|| in Mathematica, use Norm[v]

To find a norm ||v|| in PocketCAS, use l2norm(v) or norm(v)

Example: Let u = (i, 1, 0), v = (–i, 1, 0), and w = (–2 + 3i, –5 + i, –4 – 6i).

Find ||u||

Find ||w||

Find u•v

Find v•w

Find d(u, w)

8.4 Complex Vector Spaces and Inner Products 413

8.4 Complex Vector Spaces and Inner Products

Recognize and perform vector operations in complex vector spacesrepresent a vector in by a basis, and find the Euclidean

inner product, Euclidean norm, and Euclidean distance in

Recognize complex inner product spaces.

COMPLEX VECTOR SPACES

All the vector spaces studied so far in the text have been real vector spaces because thescalars have been real numbers. A complex vector space is one in which the scalars arecomplex numbers. So, if are vectors in a complex vector space, then alinear combination is of the form

where the scalars are complex numbers. The complex version of is the complex vector space consisting of ordered -tuples of complex numbers. So,a vector in has the form

It is also convenient to represent vectors in by column matrices of the form

As with the operations of addition and scalar multiplication in are performedcomponent by component.

Vector Operations in

Let

and

be vectors in the complex vector space Determine each vector.

a.

b.

c.

SOLUTION

a. In column matrix form, the sum is

b. Because and

c.

� �12 � i, �11 � i� � �3 � 6i, 9 � 3i� � ��9 � 7i, 20 � 4i�

3v � �5 � i�u � 3�1 � 2i, 3 � i� � �5 � i��2 � i, 4� � �5i, 7 � i �.

�2 � i�v � �2 � i��1 � 2i, 3 � i ��2 � i��3 � i� � 7 � i,�2 � i��1 � 2i� � 5i

v � u � �1 � 2i3 � i� � ��2 � i

4� � ��1 � 3i7 � i�.

v � u

3v � �5 � i�u�2 � i�vv � u

C2.

u � ��2 � i, 4�v � �1 � 2i, 3 � i�

Cn

CnRn,

v � �a1 � b1ia2 � b2i...an � bni

�.

Cn

v � �a1 � b1i, a2 � b2i, . . . , an � bni�.

CnnCn

Rnc1, c2, . . . , cm

c1v1 � c2v2 � � � � � cmvm

v1, v2, . . . , vm

Cn.CnCn,

9781133110873_0804.qxp 3/10/12 6:56 AM Page 413

Many of the properties of are shared by For instance, the scalar multiplicativeidentity is the scalar 1 and the additive identity in is The standard basis for is simply

which is the standard basis for Because this basis contains vectors, it follows thatthe dimension of is Other bases exist; in fact, any linearly independent set of vectors in can be used, as demonstrated in Example 2.

Verifying a Basis

Show that is a basis for

SOLUTION

Because the dimension of is 3, the set will be a basis if it is linearly independent. To check for linear independence, set a linear combination of the vectorsin equal to as follows.

This implies that

So, and is linearly independent.

Representing a Vector in by a Basis

Use the basis in Example 2 to represent the vector

SOLUTION

By writing

you can obtain

which implies that

and

So, v � ��1 � 2i�v1 � v2 � ��1 � 2i�v3.

c3 �2 � i

i� �1 � 2i .c1 �

2 � i

i� �1 � 2i,c2 � 1,

c3i � 2 � i

c2i � i

�c1 � c2�i � 2

� �2, i, 2 � i�� c1 � c2�i, c2i, c3i�

v � c1v1 � c2v2 � c3v3

v � �2, i, 2 � i�.S

Cn

�v1, v2, v3�c1 � c2 � c3 � 0,

c3i � 0.

c2i � 0

�c1 � c2�i � 0

��c1 � c2�i, c2i, c3i� � �0, 0, 0� �c1i , 0, 0� � �c2i, c2i, 0� � �0, 0, c3i� � �0, 0, 0�

c1v1 � c2v2 � c3v3 � �0, 0, 0�

0,S

�v1, v2, v3�C3

C 3.S � �v1, v2, v3� � ��i, 0, 0�, �i, i, 0�, �0, 0, i��

Cnnn.Cn

nRn.

en � �0, 0, 0, . . . , 1�

.

.

.e2 � �0, 1, 0, . . . , 0�e1 � �1, 0, 0, . . . , 0�

Cn0 � �0, 0, 0, . . . , 0�.Cn

Cn.Rn


REMARK

Try verifying that this linearcombination yields �2, i, 2 � i �.

9781133110873_0804.qxp 3/10/12 6:56 AM Page 414

Other than there are several additional examples of complex vector spaces. For instance, the set of complex matrices with matrix addition and scalar multiplication forms a complex vector space. Example 4 describes a complex vectorspace in which the vectors are functions.

The Space of Complex-Valued Functions

Consider the set of complex-valued functions of the form

where and are real-valued functions of a real variable. The set of complex numbersforms the scalars for and vector addition is defined by

It can be shown that scalar multiplication, and vector addition form a complex vector space. For instance, to show that is closed under scalar multiplication, let

be a complex number. Then

is in

The definition of the Euclidean inner product in is similar to the standard dotproduct in except that here the second factor in each term is a complex conjugate.

Finding the Euclidean Inner Product in

Determine the Euclidean inner product of the vectors

and

SOLUTION

Several properties of the Euclidean inner product are stated in the following theorem.Cn

� 3 � i

� �2 � i��1 � i� � 0�2 � i� � �4 � 5i��0� u � v � u1v1 � u2v2 � u3v3

v � �1 � i, 2 � i, 0�.u � �2 � i, 0, 4 � 5i�

C3

Rn,Cn

S.

� �af1�x� � bf2�x� � i�bf1�x� � af2�x� cf�x� � �a � bi��f1�x� � if2�x�

c � a � biS

S,

� �f1�x� � g1�x� � i�f2�x� � g2�x�. f�x� � g�x� � �f1�x� � if2�x� � �g1(x� � i g2�x�

S,f2f1

f�x� � f1�x� � if2�x�

S

m � nCn,


REMARK

Note that if and happen tobe “real,” then this definitionagrees with the standard inner(or dot) product in Rn.

vu

THEOREM 8.7 Properties of the Euclidean Inner Product

Let and be vectors in and let be a complex number. Then the following properties are true.

1. 2.3. 4.5. 6. if and only if u � 0.u � u � 0u � u � 0

u � �kv� � k�u � v��ku� � v � k�u � v��u � v� � w � u � w � v � wu � v � v � u

kCnwu, v,

Definition of the Euclidean Inner Product in

Let and be vectors in The Euclidean inner product of and is given by

u � v � u1v1 � u2v2 � � � � � unvn.

vuCn.vu

Cn

9781133110873_0804.qxp 3/10/12 6:56 AM Page 415


Definitions of the Euclidean Norm and Distance in

The Euclidean norm (or length) of in is denoted by and is

The Euclidean distance between and is

d�u, v� � u � v .

vu

u � �u � u�1�2.

uCnu

Cn

PROOF

The proof of the first property is shown below, and the proofs of the remaining properties have been left to you (see Exercises 59–63). Let

and

Then

The Euclidean inner product in is used to define the Euclidean norm (or length)of a vector in and the Euclidean distance between two vectors in .

The Euclidean norm and distance may be expressed in terms of components, asfollows (see Exercise 51).

Finding the Euclidean Norm and Distance in

Let and

a. Find the norms of and b. Find the distance between and

SOLUTION

a.

b.

� �47

� �1 � 5 � 41�1�2

� ��12 � 02� � ��2�2 � ��1�2� � �42 � ��5�2�1�2

� �1, �2 � i, 4 � 5i� d�u, v� � u � v

� �7

� �2 � 5 � 0�1�2

� ��12 � 12� � �22 � 12� � �02 � 02�1�2

v � � v1 2 � v2 2 � v3 2�1�2

� �46

� �5 � 0 � 41�1�2

� ��22 � 12� � �02 � 02� � �42 � ��5�2�1�2

u � � u1 2 � u2 2 � u3 2�1�2

v.uv.u

v � �1 � i, 2 � i, 0�.u � �2 � i, 0, 4 � 5i�

Cn

d�u, v� � � u1 � v1 2 � u2 � v2 2 � . . . � un � vn 2�1�2

u � � u1 2 � u2 2 � . . . � un 2�1�2

CnCnCn

� u � v.

� u1v1 � u2v2 � � � � � unvn

� v1u1 � v2u2 � . . . � vnun

� v1u1 � v2u2 � . . . � vnun

v � u � v1u1 � v2u2 � . . . � vnun

v � �v1, v2, . . . , vn �.u � �u1, u2, . . . , un �

9781133110873_0804.qxp 3/10/12 6:56 AM Page 416

COMPLEX INNER PRODUCT SPACES

The Euclidean inner product is the most commonly used inner product in . On occasion,however, it is useful to consider other inner products. To generalize the notion of aninner product, use the properties listed in Theorem 8.7.

A complex vector space with a complex inner product is called a complex inner product space or unitary space.

A Complex Inner Product Space

Let and be vectors in the complex space Show that thefunction defined by

is a complex inner product.

SOLUTION

Verify the four properties of a complex inner product, as follows.

1.

2.

3.

4.

Moreover, if and only if

Because all the properties hold, is a complex inner product.�u, v�

u1 � u2 � 0.�u, u� � 0

�u, u� � u1u1 � 2u2u2 � u1 2 � 2 u2 2 � 0

�ku, v� � �ku1�v1 � 2�ku2�v2 � k�u1v1 � 2u2v2 � � k �u, v� � �u, w� � �v, w�� u1w1 � 2u2w2 � � �v1w1 � 2v2w2 �

�u � v, w� � �u1 � v1�w1 � 2�u2 � v2�w2

� u1v1 � 2u2v2 � �u, v�v1u1 � 2v2u2�v, u� �

�u, v� � u1v1 � 2u2v2

C2.v � �v1, v2�u � �u1, u2�

Cn


Definition of a Complex Inner Product

Let and be vectors in a complex vector space. A function that associates and with the complex number is called a complex inner product whenit satisfies the following properties.

1.2.3.4. and if and only if u � 0.�u, u� � 0�u, u� � 0

�ku, v� � k�u, v��u � v, w� � �u, w� � �v, w��u, v� � �v, u�

�u, v�vuvu


Complex vector spaces and inner products have an importantapplication called the Fourier transform, which decomposes a function into a sum of orthogonal basis functions. The givenfunction is projected onto the standard basis functions for varying frequencies to get the Fourier amplitudes foreach frequency. Like Fourier coefficients and the Fourier approximation, this transform is named after the Frenchmathematician Jean-Baptiste Joseph Fourier (1768–1830).

The Fourier transform is integral to the study of signal processing. To understand the basic premise of this transform, imagine striking two piano keys simultaneously.Your ear receives only one signal, the mixed sound of thetwo notes, and yet your brain is able to separate the notes.The Fourier transform gives a mathematical way to take asignal and separate out its frequency components.

Eliks/Shutterstock.com

9781133110873_0804.qxp 3/10/12 6:56 AM Page 417


8.4 Exercises

Vector Operations In Exercises 1–8, perform the indicated operation using

and

1. 2.

3. 4.

5. 6.

7. 8.

Linear Dependence or Independence In Exercises9–12, determine whether the set of vectors is linearlyindependent or linearly dependent.

9.

10.

11.

12.

Verifying a Basis In Exercises 13–16, determinewhether is a basis for

13.

14.

15.

16.

Representing a Vector by a Basis In Exercises 17–20,express as a linear combination of each of the followingbasis vectors.

(a)

(b)

17. 18.

19. 20.

Finding Euclidean Inner Products In Exercises 21 and22, determine the Euclidean inner product

21. 22.

Properties of Euclidean Inner Products In Exercises23–26, let and Evaluate the expressions in parts (a) and (b)to verify that they are equal.

23. (a) 24. (a)

(b) (b)

25. (a) 26. (a)

(b) (b)

Finding the Euclidean Norm In Exercises 27–34,determine the Euclidean norm of

27. 28.

29. 30.

31. 32.

33.

34.

Finding the Euclidean Distance In Exercises 35–40,determine the Euclidean distance between and

35.

36.

37.

38.

39.

40.

Complex Inner Products In Exercises 41–44, determinewhether the function is a complex inner product, where

and

41.

42.

43.

44.

Finding Complex Inner Products In Exercises 45–48,use the inner product to find

45. and

46. and

47. and

48. and

Finding Complex Inner Products In Exercises 49 and50, use the inner product

where

and

to find

49.

50. v � � i3i

�2i�1�u � � 1

1 � i2i0�

v � �10

1 � 2ii�u � �0

1i

�2i��u, v�.

v � [v11

v21

v12

v22]u � [u11

u21

u12

u22]

�u, v� � u11v11 � u12v12 � u21v21 � u22v22

v � �2 � 3i, �2�u � �4 � 2i, 3�v � �3 � i, 3 � 2i�u � �2 � i, 2 � i�

v � �2 � i, 2i�u � �3 � i, i�v � �i, 4i�u � �2i, �i�

�u, v�.�u, v� � u1v1 � 2u2v2

�u, v� � u1v1 � u2v2

�u, v� � 4u1v1 � 6u2v2

�u, v� � �u1 � v1� � 2�u2 � v2��u, v� � u1 � u2v2

v � �v1, v2�.u � �u1, u2�

u � �1, 2, 1, �2i�, v � �i, 2i, i, 2�u � �1, 0�, v � �0, 1�u � ��2, 2i, �i�, v � �i, i, i�u � �i, 2i, 3i�, v � �0, 1, 0�u � �2 � i, 4, �i�, v � �2 � i, 4, �i�u � �1, 0�, v � �i, i�

v.u

v � �2, �1 � i, 2 � i, 4i�v � �1 � 2i, i, 3i, 1 � i�

v � �0, 0, 0�v � �1, 2 � i, �i�v � �2 � 3i, 2 � 3i�v � 3�6 � i, 2 � i�v � �1, 0�v � �i, �i�

v.

k�u � v�k�u � v�u � �kv��ku� � v

u � w � v � wv � u

�u � v� � wu � v

k � �i.w � �1 � i, 0�,v � �2i, 2 � i�,u � �1 � i, 3i�,

v � �1 � 3i, 2, 1 � i�v � �3i, 0, 1 � 2i�u � �4 � i, i, 0�u � ��i, 2i, 1 � i�

u � v.

v � �i, i, i�v � ��i, 2 � i, �1�v � �1 � i, 1 � i, �3�v � �1, 2, 0�

{�1, 0, 0�, �1, 1, 0�, �0, 0, 1 � i��{�i, 0, 0�, �i, i, 0�, �i, i, i�}

v

S � ��1 � i, 0, 1�, �2, i, 1 � i�, �1 � i, 1, 1��S � ��i, 0, 0�, �0, i, i�, �0, 0, 1��S � ��1, i�, �i, 1��S � ��1, �i�, �i, 1��

Cn.S

��1 � i, 1 � i, 0�, �1 � i, 0, 0�, �0, 1, 1��1, i, 1 � i�, �0, i, �i�, �0, 0, 1��1 � i, 1 � i, 1�, �i, 0, 1�, ��2, �1 � i, 0��1, i�, �i, �1��

2iv � �3 � i�w � uu � iv � 2iw

�6 � 3i�v � �2 � 2i�wu � �2 � i�viv � 3w�1 � 2i�w4iw3u

w � �4i, 6�.u � �i, 3 � i�, v � �2 � i, 3 � i�,

9781133110873_0804.qxp 3/10/12 6:56 AM Page 418

8.4 Exercises 419

51. Let

(a) Use the definitions of Euclidean norm and Euclideaninner product to show that

(b) Use the results of part (a) to show that

53. Let and If and the set is not a basis for what doesthis imply about and

54. Let and Determine a vectorsuch that is a basis for

Properties of Complex Inner Products In Exercises55–58, verify the statement using the properties of acomplex inner product.

55.

56.

57.

58.

Proof In Exercises 59–63, prove the property, whereand are vectors in and is a complex number.

59.

60.

61.

62.

63. if and only if

64. Writing Let be a complex inner product and letbe a complex number. How are and

related?

Finding a Linear Transformation In Exercises 65 and66, determine the linear transformation thathas the given characteristics.

65.

66.

Finding an Image and a Preimage In Exercises 67–70,the linear transformation is shown by

Find the image of and the preimage of

67.

68.

69.

70.

71. Find the kernel of the linear transformation in Exercise 68.

72. Find the kernel of the linear transformation in Exercise 69.

Finding an Image In Exercises 73 and 74, find theimage of for the indicated composition, where

and are the matrices below.

and

73. 74.

75. Determine which of the sets below are subspaces of thevector space of complex matrices.

(a) The set of symmetric matrices.

(b) The set of matrices satisfying

(c) The set of matrices in which all entries arereal.

(d) The set of diagonal matrices.

76. Determine which of the sets below are subspaces ofthe vector space of complex-valued functions (seeExample 4).

(a) The set of all functions satisfying

(b) The set of all functions satisfying

(c) The set of all functions satisfying


77. Using the Euclidean inner product of and in

78. The Euclidean norm of in denoted by is�u � u�2.

uCnu

u � v � u1v1 � u2v2 � . . . � unvn.Cn,vu

f�i� � f��i�.f

f �0� � 1.f

f�i� � 0.f

2 � 2

2 � 2

�A�T � A.A2 � 2

2 � 2

2 � 2

T1 � T2T2 � T1

T2 � [�ii

i�i]T1 � [0

ii0]

T2T1

v � �i, i�

w � �1 � i1 � i

i�v � �250�,A � �

0i0

1 i i

1�1 0�,

w � �2

2i3i�v � � 2 � i

3 � 2i�,A � �1ii

00i�,

w � �11�v � �

i0

1 � i�,A � �0i

i0

10�,

w � �00�v � �1 � i

1 � i�,A � �1i

0i�,

w.vT�v� � Av.T : Cm → Cn

T�0, i� � �0, �i�T�i, 0� � �2 � i, 1�T�0, 1� � �0, �i�T�1, 0� � �2 � i, 1�

T : Cm → Cn

�u, kv��u, v�k�u, v�

u � 0.u � u � 0

u � u � 0

u � �kv� � k�u � v��ku� � v � k�u � v��u � v� � w � u � w � v � w

kCnwu, v,

� k�u, v��u, kv��u, v� � �u, v� � �v, 2u��u, 0� � 0

�u, kv � w� � k�u, v� � �u, w�

C3.�v1, v2, v3�v3

v2 � �1, 0, 1�.v1 � �i, i, i�z3?z1, z2,

C3,�v1, v2, v3�v3 � �z1, z2, z3�v2 � �i, i, 0�.v1 � �i, 0, 0�

� . . . � un � vn 2�1�2.

d�u, v� � � u1 � v1 2 � u2 � v2 2

u � � u1 2 � u2 2 � . . . � un 2�1�2.

u � �a1 � b1i, a2 � b2i, . . . , an � bni�.

52. The complex Euclidean innerproduct of and is sometimes called the complex dot product. Compare the properties ofthe complex dot product in and those of the dotproduct in

(a) Which properties are the same? Which propertiesare different?

(b) Explain the reasons for the differences.

Rn.Cn

vu

9781133110873_0804.qxp 3/10/12 6:56 AM Page 419

Section 8.4 1. � 3.5. 7.9. Linearly dependent 11. Linearly independent

13. is not a basis for 15. is a basis for17. (a)

(b)

19. (a)

(b)

21.23. (a)

(b)The expressions are equal.

25. (a)(b)The expressions are equal.

27. 29. 31. 33. 35.37. 39. 41. Not a complex inner product43. A complex inner product 45. 47.49. 51. (a) and (b) Proofs53. and z2 can be any complex numbers.55.

57.

59–63. Proofs 65.

67. � , � 69.

71. where

73. 75. (a), (b), and (d) are subspaces.

77. False.u � v � u1v1 � u2v2 � . . . � unvn

�0

0� t � R�ker�T� � ��0, t, �ti�,

�2 � i

1 � 2i

�1 � 5i�, �2

1�1 � i

2i� �00

Tu�u1

2� � �2 � i

10

�i u��u1

2�

� �v, u � u� � �v, 2u�

�u, v� � �u, v� � �v, u� � �v, u�

� k�u, v� � �u, w� �u, kv � w� � �u, kv� � �u, w�z3 � 0, z1

�4 � i23 � 3i�6

�2�15�3�17�73�42�2

4 � i4 � i

1 � 4i1 � 4i

�4 � 3i

12

12 �� i��0, 0, 1 � i�

�2 � i��1, 1, 0� �

�1 � 3i��i, i, 0� � i�i, i, i� ��i, 2 � i, �1� � ��2 � 2i��1, 0, 0� �

��i, 2 � i, �1� � ��2 � 2i��i, 0, 0� �

�1, 2, 0� � i�i, 0, 0� � 2i�i, i, 0� � 0�i, i, i��1, 2, 0� � ��1, 0, 0� � 2�1, 1, 0� � 0�0, 0, 1 � i�

C 3.S C 2.S

��9 � 3i, 2 � 14i��5 � i, �4��8 � 4i, 6 � 12i��3i, 9 3i�

Answer Key

Chapter 6: Linear Transformations.

6.1 Introduction to Linear Transformations.

p. 169

S*

And T()*



Objective: Find the image and preimage of a function

Objective: Show that a function is a linear transformation.

Recognize and describe some common linear transformations.

A function T: V W maps a vector space V to a W. V is the domain and W is the codomain of

T. If v V and w W such that T (v) = w then w is the image of v under T. The set of all

images of vectors in V is called the range of T. The set of all vectors in V such that T (v) = w is

the preimage of w.

preimage domain v V

image range codomain w range W

Let V and W be vector spaces. Then the function (or map) T : V W is a linear transformation

when the following two properties are true for all v V and w W and for any scalar c.

1) T(u + v) = T(u) + T(v)

2) T(cu) = c T(u)

A linear transformation is “operation preserving” because applying an operation in V (before T )

gives the same result as applying the corresponding operation in W (after T ).

T(u + v) = T(u) + T(v) T(cu) = c T(u)

Example: Show that T : R2 R2

given by T(v1, v2) = (v1 – v2, v1 + v2) is a linear transformation,

find the image of (13, –4), and find the preimage of (2, 8). [Note: technically, we should write

T((v1, v2)) because v = (v1, v2).]

addition

in Vaddition

in W

scalar

multiplication

in W

scalar

multiplication

in V

T T T T



p. 170

S*

S*

Solution: T(u + v) = T(u1 + v1, u2 + v2)

= ((u1 + v1) – (u2 + v2), (u1 + v1) + (u2 + v2))

= (u1 – u2, u1 + u2) + (v1 – v2, v1 + v2)

= T(u) + T(v)

T(cu) = T(cv1, cv2)

(cv1 – cv2, cv1 + cv2)

(c(v1 – v2), c(v1 + v2))

c (v1 – v2, v1 + v2)

=c T(u)

Image of (13, –4):

T(13, –4) = (13 + (–4), 13 – (–4)) = (9, 17)

Preimage of (2, 8): T(v1, v2) = (v1 – v2, v1 + v2) = (2, 8)

v1 – v2 = 2 2 v1 = 10

v1 + v2 = 8 –2 v2 = –6

(v1, v2) = (5, 3)

Example: Show that f : R R given by f (x) = x2 is not a linear transformation.

Solution: f (x + y) = (x + y)2 = x

2 + 2xy + y

2.

Usually, this is not equal to f (x) + f (y) = x2 + y

2.

In particular, a specific counterexample is f (1 + 1) = 4, but f (1) + f (1) = 2.

Example: Show that f : R R given by f (x) = x + 1 is not a linear transformation.

Solution: f (x + y) = x + y + 1, which is never equal to f (x) + f (y) = x + 1 + y + 1.

[However, f (x) = x + 1 is a linear function, because its graph is a straight line.]

Two simple linear transformations are the zero transformation T : V W, T(v) = 0 for all v V

and the identity transformation T : V V, T(v) = v for all v V.

Theorem 6.1 Properties of Linear Transformations

Let T: V W be a linear transformation, and u and v be vectors in V. Then

1) T(0) = 0

2) T(–v) = –T(v)

3) T(u – v) = T(u) – T(v)

4) If v = c1v1 + c2v2 + …+ cnvn then

T(v) = T(c1v1 + c2v2 + …+ cnvn) = c1T(v1) + c2T(v2) + …+ cnT(vn)



p. 171

Proof:

1) T(0) =

2) T(–v) =

3) T(u – v) =

4) Prove T(c1v1 + c2v2 + …+ cnvn) = c1T(v1) + c2T(v2) + …+ cnT(vn) by induction:

T(c1v1) = c1T(v1) by the second property of linearity

Suppose T(c1v1 + c2v2 + …+ cnvn) = c1T(v1) + c2T(v2) + …+ cnT(vn).

Then T(c1v1 + c2v2 + …+ cnvn + cn+1vn+1)

= T(c1v1 + c2v2 + …+ cnvn) + T(cn+1vn+1) by the first property of linearity

= c1T(v1) + c2T(v2) + …+ cnT(vn) + T(cn+1vn+1) by supposition

= c1T(v1) + c2T(v2) + …+ cnT(vn) + cn+1T(vn+1) by the second property of linearity

Property 4 of Theorem 6.1 tells us that a linear transformation is completely determined by its

action on a basis for V. In other words, if {v1, v2, …, vn} is a basis for V and if T(v1), T(v2), …,

T(vn) are known, then T(v) is determined for any v V.

Example: Let T: R3 R3

be a linear transformation such that

T(1, 0, 0) = (3, 1, 4) T(0, 1, 0) = (1, 5, 9) T(0, 0, 1) = (2, 6, 5)

Find T(2, 7, 1).

Solution: T(2, 7, 1) = 2 T(1, 0, 0) + 7 T(0, 1, 0) + 1 T(0, 0, 1)

= 2(3, 1, 4) + 7(1, 5, 9) + 1(2, 6, 5)

= (17, 46, 51)



p. 172


the function such that

T(v) = Av =

76

34

57

2

1

v

v

Show that T is a linear function. Find T(1, 0) and T(0, 1).

Solution:

For any u and v in R2, T(u + v) = A(u + v) = Au + Av = T(u) + T(v).

Also, for any v in R2 and any scalar c, T(cv) = A(cv) = c(Av) = cT(v).

T(1, 0) =

76

34

57

0

1 =

6

4

7

= (7, –4, –6)

and

T(0, 1) =

76

34

57

1

0 =

7

3

5

= (5, –3, –7)

Notice that T(1, 0) and T(0, 1) are just the two columns of A.

Theorem 6.2: The Linear Transformation Given by a Matrix

Let A be an mn matrix. The function T: Rn Rm

the defined by T(v) = Av is a linear

transformation. Vectors in Rn are represented by n1 (column) matrices and vectors in Rm

are

represented by m1 (column) matrices.

T(v) =

nmmnmm

n

n

aaa

aaa

aaa

21

22221

11211

1

2

1

nnv

v

v

=

12211

2222121

1212111

mnmnmm

nn

nn

vavava

vavava

vavava

Example: Rotation. We show that the linear transformation T: R2 R2

represented by the

matrix

A =

)cos()sin(

)sin()cos(

rotates every vector in R2

counterclockwise by an angle about the

origin.



p. 173

T

)sin(

)cos(

r

r=

)cos()sin(

)sin()cos(

)sin(

)cos(

r

r

= r

)sin()cos()cos()sin(

)sin()sin()cos()cos(

=

)sin(

)cos(

r

r

Observe that T is a linear transformation. It

preserves vector addition and scalar

multiplication.

Example: projection. The linear transformation T: R3 R3

represented by the matrix

A =

000

010

001

projects a vector v = (x, y, z) to T(v) = (x, y, 0). In other words, T maps every

vector in R3 to its orthogonal projection in the x-y plane.

Example: Transpose. We show that the linear transformation T: Mm,n Mn,m given by T(A) = AT

is a linear transformation.

If A and B are mn matrices, then T(A + B) = (A + B)T = A

T + B

T = T(A) + T(B).

If A is an mn matrix and c is a scalar, then T(cA = (cA)T = c(A

T) = c T(A).

u

v

1.5v

u + v

T(u)

T(v)

T(1.5v) T(u + v)



p. 174

Example: The Differential Operator. Let C[a, b], also written as C1[a, b], be the set of all

functions whose derivatives are continuous on the interval [a, b]. We show that the differential

operator Dx defines a linear transformation from C[a, b] into C[a, b].

Using operator notation, we write Dx( f ) = dx

d[ f ], where f is in C[a, b].

From calculus, we know that for any functions f and g in C[a, b],

Dx( f + g) = dx

d[ f + g] =

dx

d[ f ] +

dx

d[ g] = Dx( f ) + Dx( g) C[a, b]

and for any function f in C[a, b] and for any scalar c,

Dx(c f ) = dx

d[c f ] = c

dx

d[ f ] = c Dx( f ) C[a, b]

Example: The Definite Integral Operator. Let P be the vector space of all polynomial functions.

We show that the definite integral operator T: P R defined by T(p) = b

a

dxxp )(

is a linear transformation.

From calculus, we know that for any polynomials p and q in P,

T(p + q) =

b

a

dxxqp )]([ =

b

a

dxxqxp )]()([ = b

a

dxxp )( + b

a

dxxq )( = T(p) + T(q)

and for any polynomial p in P and for any scalar c,

T(cp) = b

a

dxxcp )]([ = b

a

dxxcp )( = b

a

dxxpc )( = cT(p).


6.2 The Kernel and Range of a Linear Transformation.

p. 175

S*


Objective: Find the kernel of a linear transformation.

Objective: Find a basis for the range, the rank, and the nullity of a linear transformation.

Objective: Determine whether a linear transformation is one-to-one or onto.

Objective: Prove basic results about one-to-one and/or onto linear transformations.

Objective: Determine whether vector spaces are isomorphic.

Let T: V W be a linear transformation. Then the set of all vectors v in V such that T(v) = 0 is

the kernel of T and is denoted by ker(T).

Remember that for any linear transformation, T(0) = 0, so ker(T) always contains 0.

Example Find the kernel of the projection T: R3 R3

defined by T(x, y, z) = (x, y, 0).

Solution: The solution to T(x, y, z) = (0, 0, 0) is x = 0, y = 0. So ker(T) = {(0, 0, z): z R}

Example Find the kernel of the zero transformation T: V W defined T(v) = 0.

Solution: T(v) = 0 is true for all v V so ker(T) = V

Example Find the kernel of the identity transformation T: V V defined T(v) = v.

Solution: T(v) = 0 means v = 0 so ker(T) = {0}.

Example Find the kernel of the linear transformation T: R3 R2

defined by

T(x1, x2, x3) = (5x1 – 4x2 – 6x3, 4x2 –8x3).

Solution: T(x1, x2, x3) = (0, 0) yields the system

084

0645

32

321

xx

xxx

0840

0645rref

0210

0015

14

which has the solution x3 = t, x2 = 2t, x1 = t5

14 so ker(T) = {( t5

14 , 2t, t): t R}

Theorem 6.3: The Kernel of T: V W is a Subspace of V

The kernel of a linear transformation T: V W is a subspace of the domain V.

Proof: 0 ker(T) so ker(T) is not empty.

If u, v ker(T), then T(u) = 0 and T(v) = 0

so T(u + v) = T(u) + T(v) = 0 so u + v ker(T)



p. 176

S*

S*

If c is a scalar and v ker(T), then T(v) = 0

so T(cv) = cT(v) = 0 so cv ker(T)

Example Find a basis for the kernel of the linear transformation T: R5 R4

defined by

T(x) = Ax, where

Solution: ker(T) = the nullspace of A (i.e. the solution space of Ax = 0).

Let T: V W be a linear transformation. Then the set of all vectors w in W that are images of

vectors in V is the range of T, and is denoted by range(T). That is, range(T) = {T(v): v V}

kernel domain

0 range codomain

T T



p. 177

S*

S*

Theorem 6.4: The Range of T: V W is a Subspace of W

The range of a linear transformation T: V W is a subspace of the codomain W.

Proof: 0 range(T) because T(0) = 0, so range(T) is not empty.

If T(u) and T(v) are vectors in range(T), then u and v are vectors in V. V is closed under

addition, so u + v V, so T(u + v) range(T). T(u + v) = T(u) + T(v), so

T(u) and T(v) range(T) implies that T(u) + T(v) range(T).

Let c be a scalar and T(v) range(T) Then v V. Since V is closed under scalar

multiplication, cv V, so T(cv) range(T). T(cv) = cT(v), so

Let c be a scalar and T(v) range(T) implies that cT(v) range(T).

so T(cv) = cT(v) = 0 so cv ker(T)

Example Find a basis for the range of the linear transformation T: R5 R4

defined by

T(x) = Ax, where

Solution: range(T) = the column space of A.

Let T: V W be a linear transformation. The dimension of the kernel of T is called the nullity

of T and is denoted by nullity(T), the dimension of the range of T is called the rank of T and is

denoted by rank(T).



p. 178

S*

S*

S*

Theorem 6.5: The Sum of Rank and Nullity

Let T: V W be a linear transformation from the n-dimensional vector space V into a vector

space W. Then

rank(T ) + nullity(T ) = n

i.e. dim(range(T )) + dim(kernel(T )) = dim(domain(T ))

Example Find the rank and nullity of the linear transformation T: R8 R5

defined by

T(x) = Ax, where

Solution:

A function T: V W is one-to-one (injective) if and only if every vector in the range of T has a

single (unique) preimage. An equivalent definition is that T is one-to one if and only if T(u) =

T(v) implies that u = v.

A function T: V W is onto (surjective) if and only if the range of T is W (the codomain of T ).

And equivalent definition is that that T is one-to one if and only if every element in W has a

preimage in V.

Example: T: R2 R3

defined by T(x, y) = (x, y, x) is one-to-one

but not onto (points with z x have no preimage).

Example: The projection T: R3 R2

defined by T(x, y, z) = (x, y) is onto

but not one-to-one.



p. 179

S*

Example: The projection T: R2 R2

defined by T(x, y) = (x – 2y, 3x – 6y) = (x – 2y) (1, 3)

is neither one-to-one (because there are many different (x, y) pairs that give the same value of x –

2y ) nor onto (because points with y 3x are not in the range of T).

Example: The transformation T: Rn R n

defined by T(v) = Av, where A is an invertible

matrix, is one-to-one and onto.

T is one-to-one because T(u) = T(v) implies Au = Av. Since A is invertible, we can pre-multiply

by A–1

to obtain A–1

Au = A–1

Av or u = v.

T is onto because the rank of an invertible nn matrix (namely, A) is n. So we also have

rank(T ) = n. By Theorem 6.7 (below), T is onto because its rank equals the dimension of its

codomain.

Theorem 6.6: One-to-One Linear Transformations

Let T: V W be a linear transformation. Then T is one-to-one if and only if ker(T ) = {0}.

Proof

If T is one-to-one, then T(v) = 0 has only one solution, namely, v = 0. Thus, ker(T ) = {0}.

Conversely, suppose ker(T ) = {0} and T(u) = T(v). Because T is a linear transformation,

T(u – v) =

so u – v = because

Therefore,



p. 180

S*

S*

Theorem 6.7: Onto Linear Transformations

Let T: V W be a linear transformation, where W is finite dimensional. Then T is onto if and

only rank(T ) = dim(W).

Proof

Let T: V W be a linear transformation.

If T is onto, then W is equal to the range of T, so rank(T ) = dim(range(T)) = dim(W).

Let dim(W) = n. If rank(T ) = n, then there are n linearly independent vectors T(v1), T(v2), …,

T(vn) in the range of T. Since the range of T is in W, the vectors T(v1), T(v2), …, T(vn) are

linearly independent W. By Thm. 4.12 (n linearly independent vectors in an n-dimensional

space form a basis), the vectors T(v1), T(v2), …, T(vn) form a basis for W. So any vector w

W can be written as a linear combination

w = c1T(v1) + c2T(v2) + … + cnT(vn) = T(c1v1 + c2v2 + … + cnvn),

which is in the range of T. Therefore, T is onto.

Theorem 6.8: One-to-One and Onto Linear Transformations

Let T: V W be a linear transformation, where vector spaces V and W both have dimension n.

Then T is one-to-one if and only if it is onto.

Proof

If T is one-to-one, then by Thm 6.6, ker(T ) = {0}, and dim(ker(T )) = 0. By Thm 6.5,

dim(range(T)) =

Therefore, by Thm 6.7, T is onto.

Conversely, if T is onto, then dim(range(T)) = dim(W) = n

so by Thm 6.5, dim(ker(T)) =

Therefore, ker(T ) = {0}, and by Thm 6.6, T is one-to-one.



p. 181

S*

Example: Consider the linear transformations Let T: V W represented by T(x) = Ax. Find the

rank and nullity of T, and determine whether T is one-to-one or onto. (Notice that all of the

matrices are in row-echelon form.)

a) A =

100

410

531

b) A =

00

10

31

c) A =

410

531 d) A =

000

410

531

dim(codomain)

m

dim(domain)

n

dim(range)

r = rank(T)

dim(kernel)

= nullity(T)

One-to-one?

Onto?

a)

b)

c)

d)

A function that is both one-to-one (injective) and onto (surjective) is called bijective. A bijective

function is invertible.

A linear transformation T: V W that is one-to-one and onto is called an isomorphism. If V and

W are vector spaces such that there exists an isomorphism from V to W, then V and W are said

to be isomorphic to each other.



p. 182

S*

Theorem 6.9: Isomorphic Spaces and Dimension

Two finite-dimensional vector spaces V and W are isomorphic if and only if they are of the same

dimension.

Proof

Assume V is isomorphic to W, and V has dimension n. Then there exists a linear

transformation T: V W that is one-to-one and onto. Because T is one-to-one,

dim(ker(T)) = , so dim(range(T)) =

Moreover, because T is onto, dim(W) =

so V and W are of the same dimension.

Conversely, assume V and W both have dimension n. Let {v1, v2, … vn} be a basis for V, and

{w1, w2, … wn} be a basis for W. Then an arbitrary vector v V can be uniquely written as

v = c1v1 + c2v2 + … + cnvn. Define the linear transformation T: V W by

T(v) = c1w1 + c2w2 + … + cnwn.

Then T is one-to-one: If T(v) = 0,

so ker(T ) = {0}, and by Thm 6.6, T is one-to-one.

Also, T is onto: Since

the basis {w1, w2, … wn} is contained in the range of T, which is contained in W.

Thus, rank(T ) = n and Thm 6.7, T is onto.


6.3 Matrices for Linear Transformations.

p. 183

S*


Objective: Find the standard matrix for a linear transformation.

Objective: Find the standard matrix for the composition of linear transformations and find the

inverse on an invertible linear transformation.

Objective: Find the matrix for a linear transformation relative to a nonstandard basis.

Objective: Prove results in linear algebra using linearity.

Recall Theorem 6.1, Part 4: If v = c1v1 + c2v2 + …+ cnvn then

T(v) = T(c1v1 + c2v2 + …+ cnvn) = c1T(v1) + c2T(v2) + …+ cnT(vn)

This says that a linear transformation T: V W is completely determined by its action on a

basis of V. In other words, if {v1, v2, …, vn} is a basis of V and T(v1), T(v2), …, T(vn) are

given, then T(v) is determined for any v in V. This is the key to representing T by a matrix.

Recall that the standard basis for Rn in column vector notation is

S = {e1, e2, …, en} =

1

0

0

,,

0

1

0

,

0

0

1

Theorem 6.10: Standard Matrix for a Linear Transformation

Let T: Rn Rm

be a linear transformation such that for the standard basis vectors ei of Rn,

T(e1) =

1

21

11

ma

a

a

, T(e2) =

2

22

12

ma

a

a

, …, T(en) =

mn

n

n

a

a

a

2

1

.

Then the mn matrix whose n columns correspond to T(ei)

A =

mnmm

n

n

aaa

aaa

aaa

31

22221

11211

is such that T(v) = Av for every v in Rn. A is called the standard matrix for T.

Proof

Choose any v = v1e1 + v2e2 + …+ vnen in R n.

Then T(v) = T(v1e1 + v2e2 + …+ vnen) = v1 T(e1) + v2 T(e2) + …+ vn T(en)



p. 184

S

On the other hand, in column vector notation, v =

nv

v

v

2

1

Av =

mnmm

n

n

aaa

aaa

aaa

31

22221

11211

nv

v

v

2

1

=

nmnmm

nn

nn

vavava

vavava

vavava

2211

2222121

1212111

= v1

1

21

11

ma

a

a

+ v2

2

22

12

ma

a

a

+ … + vn

mn

n

n

a

a

a

2

1

= v1 T(e1) + v2 T(e2) + …+ vn T(en).

Example Find the standard matrix for the linear transformation T(x, y) = (2x – 3y, x – y, y – 4x).

Solution: . T: R2 R3

.

Example Find the standard matrix A for the linear transformation T that is the reflection in the

line y = x in R2, use A to find the image of the vector v = (3, 4) and sketch the graph of v and its

image T(v).

Solution:

A = [T(e1), T(e2)]

=

T(v) = T(3, 4)

=



p. 185

S*

S*

The composition of T1: Rn Rm

with T2: Rm Rp

is defined by T(v) = T2(T1(v)) where v is a

vector in Rn. This composition is denoted by T2 ◦ T1.

Theorem 6.11: Compostion of Linear Transformations

Let T1: Rn Rm

with T2: Rm Rp

be linear transformations with standard matrices A1 and A2,

respectively. The composition T: Rn Rm

, defined by T(v) = T2(T1(v)), is a linear

transformation. Moreover, the standard matrix A for T is given by the matrix product A = A2A1.

Proof

To show that T is a linear transformation, let u and v be vectors in R n and let c be a scalar.

Then T(u + v) = T2(T1(u + v)) =

T2(T1(u) + T1(v)) = T2(T1(u)) + T2(T1(v))

= T(u) + T(v)

Also, T(cv) = T2(T1(cv)) =

T2(cT1(v)) = cT2(T1(v))

= cT(v)

To show that A2A1 is the standard matrix for T,

T(v) = T2(T1(v)) = T2(A1v)

= A2(A1v)

= (A2A1)v using the associative property of matrix multiplication

Composition is not commutative because matrix multiplication is not commutative. We can

generalize to compositions of three or more linear transformations. For example, T = T3 ◦ T2 ◦ T1

is defined by T(v) = T3(T2(T1(v))) and is represented by A3A2A1.



p. 186

S

S

Example Given the linear transformations T1: R3 R2

with T2: R2 R3

defined by

T1

3

2

1

x

x

x

=

011

201

3

2

1

x

x

x

and T1

2

1

x

x =

01

20

11

2

1

x

x

Find the standard matrix for T = T2 ◦ T1.

Solution:

A linear transformation T1: Rn Rn

is invertible if and only if there exists a linear

transformation T2: Rn Rn

such that for every v in Rn,

T2(T1(v)) = v and T1(T2(v)) = v

T2 is the inverse of T1. When the inverse exists, it is unique, and we denote it by T1–1

.

Theorem 6.12 Existence of an Inverse Transformation

Let T: Rn Rn

be a linear transformation with a standard matrix A. Then the following

conditions are equivalent.

1) T is invertible.

2) T is an isomorphism.

3) A is invertible.

If T is invertible with standard matrix A, then the standard matrix for T–1

is A–1

.

Example Given the linear transformations T: R3 R3

defined by

T(x1, x2, x3) = (x1, 2x1 + x2, 3x1 + x2 + x3). Show that T is invertible, and find its inverse.

Solution:



p. 187

S0

Let V be an n-dimensional vector space with basis B = {v1, v2, …, vn}and W be an m-

dimensional vector space with basis B = {w1, w2, …, wn}. Let T: V W is a linear

transformation such that

[T(v1)]B =

1

21

11

ma

a

a

, [T(v2)]B =

2

22

12

ma

a

a

, …, [T(vn)]B =

mn

n

n

a

a

a

2

1

,

i.e. T(v1) = a11w1 + a21w2 + … + am1wm

T(v2) = a12w1 + a22w2 + … + am2wm

⁞ ⁞ ⁞ ⁞

T(vn) = a1nw1 + a2nw2 + … + amnwm

Then the mn matrix whose n columns correspond to [T(vi)]B

A =

mnmm

n

n

aaa

aaa

aaa

31

22221

11211

is such that [T(v)]B = A[v]B for every v in V. A is called the matrix of T relative to the bases

B and B.

Example Let Dx: P3 P2 be the differential operator that maps a polynomial p of degree 3 or

less to its derivative p. Find the matrix for Dx using the bases B = {1, x, x2, x

3} and B =

{1, x, x2}.

Solution: The derivatives of the basis vectors in B are

[Dx(1)]B = [0]B = [0(1) + 0x + 0x2]B =

0

0

0

[Dx(x)]B =



p. 188



p. 189

S0

Example Let B = B = {1, cos(x), sin(x), cos(2x), sin(2x)} and let Dx: span(B) span(B) be the

differential operator that maps a function f of to its derivative f . Find the matrix for Dx using

the bases B and B.

Solution: The derivatives of the basis vectors in B are

[Dx(1)]B = [0]B = [0(1) + 0cos(x) + 0sin(x) + 0cos(2x) + 0sin(2x)]B =

0

0

0

0

0

;


6.4 Transition Matrices and Similarity.

p. 191

S0

S0


Objective: Find and use a matrix for a linear transformation relative to a basis.

Objective: Use properties of similar matrices and diagonal matrices.

In 6.3 we saw that the matrix for a linear transformation T: V W depends on the choice of

basis B1 for V and B2 for W. Oftentimes, we can find bases that give a very simple matrix for a

given linear transformation. In this section, we will consider linear transformations T: V V

and we will use the same basis for the domain as for the codomain. We will find matrices

relative to different bases. (In Chapter 7, we will learn how to find the bases that give us simple

matrices.)

The matrix for T: V V relative to the basis B = {v1, v2, …, vn} is A, where [T(x)]B = A[x]B

The matrix for T: V V relative to the basis B = {v1, v2, …, vn} is A, where [T(x)]B = A[x]B

The transition matrix from B to B is PBB =

BnBB ][][][ 21 vvv , so P[x]B = [x]B

The transition matrix from B to B is P–1

, so P–1

[x]B = [x]B

So we have [T(x)]B = A[x]B

and we also have

[T(x)]B = P–1

BB A PBB [x]B

Therefore,

A = P–1

AP


which has the standard matrix A =

1360

25620

826

. Find the matrix A

for T relative to the basis B = {(1, 3, –2), (2, 7, –3), (3, 10, –4)}.

Solution: A is the matrix relative to the standard basis S.

PBS =

A [T(x)]B

[T(x)]B

[x]B

[x]B A

P P –1



p. 192

Diagonal matrices: Observe that A2 =

200

010

004

200

010

004

=

2

2

2

)2(00

0)1(0

00)4(

and Ak+1

=

k

k

k

)2(00

0)1(0

00)4(

200

010

004

=

1

1

1

)2(00

0)1(0

00)4(

k

k

k

.

Also, AT = A.

If D =

nd

d

d

000

000

000

000

2

1

is a diagonal matrix and none of the diagonal elements are zero,

then D–1

=

nd

d

d

/1000

000

00/10

000/1

2

1

Two square matrices A and B are similar if and only there is an invertible matrix P such that

B = P–1

AP. For example, the matrix representations of a linear transformation relative to two

different bases are similar.

Theorem 6:13: Properties of Similar Matrices

Let A, B, and C be square matrices. Then

1) A is similar to A.

2) If A is similar to B, then B is similar to A.

3) If A is similar to B and B is similar to C, then A is similar to C.


6.5 Applications of Linear Transformations.

p. 193


Objective: Identify linear transformations defined by reflections, expansions, contractions, or

shears in R2.

Objective: Use a linear transformation to rotate a figure in R3.

Reflections in R2.

Reflection in y-axis Reflection in x-axis

Horizonal reflection Vertical reflection

T(x, y) = (–x, y) T(x, y) = (x, –y)

The matrix A for Tv = Av is A = [ Te1 | Te2 ]



p. 194

Reflection in the line y = x T(x, y) = (y, x)

T(x, y) = (y, x)


Expansions and Contractions in R2.

Horizonal expansion Vertical expansion

T(x, y) = (kx, y) for k > 1 T(x, y) = (x, ky) for k > 1



p. 195

Horizonal contraction Vertical contraction

T(x, y) = (kx, y) for 0 < k < 1 T(x, y) = (x, ky) for 0 < k < 1




p. 196

Shears in R2.

Horizonal shear Vertical shear

T(x, y) = (–x, y) T(x, y) = (x, –y)




p. 197

Rotation in R2.

T(x, y) = (x cos( ) – y sin( ), x sin( ) + y cos( ))



p. 198

Rotation in R3.

The matrix A for Tv = Av is A = [ Te1 | Te2 | Te3 ]

Rotation about x-axis by /6 rad = 30 counterclockwise (right-hand rule).



p. 199

Rotation about y-axis by /6 rad = 30 Rotation about y-axis by /6 rad = 30

The matrix A for Tv = Av is A = [ Te1 | Te2 | Te3 ]

Chapter 7: Eigenvalues and Eigenvectors.

7.1 Eigenvalues and Eigenvectors.

p. 201

S*



Objective: Prove properties of eigenvalues and eigenvectors.

Objective: Verify eigenvalues and corresponding eigenvectors.

Objective: Find eigenvalues and corresponding eigenspaces.

Objective: Use the characteristic equation to find eigenvalues and eigenvectors.

Objective: Use software to find eigenvalues and eigenvectors.

Recall the example A =

01

10 from Section 6.5. Then Ae1 =

01

10

0

1 =

1

0 = e2

and Ae2 =

01

10

1

0 =

0

1 = e1

Let x1 =

5.1

5.1. Then Ax1 =

01

10

5.1

5.1 =

5.1

5.1 = 1x1

Let x2 =

5.1

5.1. Then Ax2 =

01

10

5.1

5.1 =

5.1

5.1 = –1x2

Given square matrix A, if we have a nonzero vector x and a scalar (“lambda”) such that

Ax = x

then x is an eigenvector of A with the corresponding eigenvalue .

Example:

01

10 has eigenvectors x1 =

5.1

5.1, x2 =

5.1

5.1 with eigenvalues 1= 1, 2 = –1.

“Eigen” is German

for “particular.”



p. 202

S*

S*

Example Show that

2

1 is an eigenvector of

60.030.0

05.065.0 , and find the corresponding

eigenvalue.

Solution:

Theorem 7.1: Eigenvectors of Form a Subspace

If A is an nn matrix with an eigenvalue , then the set of all eigenvectors of , together with the

zero vector

{x: x is an eigenvector of } {0} = {x: Ax = x}

is a subspace of Rn.

Proof:

{x: Ax = x} is not empty because it contains the zero vector.

{x: Ax = x} is closed under scalar multiplication because

for any vector x {x: Ax = x} and for any scalar c, A(cx) = c(Ax) = c(x) = (cx)

{x: Ax = x} is closed under addition because for any vectors x1, x2 {x: Ax = x},

A(x1 + x2) = Ax1 + Ax2 = x1 + x2 = (x1 + x2)

To find an eigenvalue of a matrix A, we start with the equation Ax = x.

Ax = x

Ax = Ix

0 = Ix – Ax

0 = (I – A)x

I – A is singular

det(I – A) = 0



p. 203

S*

Theorem 7.2: Eigenvalues and Eigenvectors of a Matrix

Let A be an nn matrix.

1) An eigenvalue of A is a scalar such det(I – A) = 0.

2) The eigenvectors of A corresponding to are the nonzero solutions of (I – A)x = 0

i.e. the nullspace of I – A.

Proof:

A has an eigenvector x 0 with eigenvalue if and only if

if and only if A – I is singular (so its determinant is zero) and x is in the nullspace of A – I.

det(I – A) = 0 is called the characteristic equation of A. The characteristic polynomial

det(I – A) is an nth

-degree polynomial in .

The Fundamental Theorem of Algebra states that an nth

-degree polynomial has exactly n roots

(i.e. an nth

-degree polynomial equation has exactly n solutions). In other words, we will always

have

det(I – A) = ( – 1) ( – 2) … ( – n)

where the n constants 1, 2, …, n are the eigenvalues of A. Note that some i may be complex

or some i may be repeated, for example 1 = 2.

Example Find the eigenvalues and corresponding eigenvectors of A =

13

21.

Solution by hand: 0 = det(



p. 204

so 1 =2 =

Eigenvector for 1 =

Eigenvector for 2 =


01

10.

Solution: 0 = det(

so 1 = 2 =



p. 205

Eigenvector for 1 =

Eigenvector for 2 =


2.18.04.0

8.02.14.0

8.08.06.1

.

Solution: 0 = det(

so 1 = 2 = 3 =

Eigenvector for 1 =



p. 206

Using rref we find which has a solution x =

E’vector for 2 = 3 =



p. 207

Using rref we find which has solution

Note that this eigenvalue, which has multiplicity two, has two linearly independent

eigenvectors.


a

ba

0 where b 0.

Solution: 0 = det(

so 1 = 2 =

Eigenvector for 1 =

Eigenvector for 2 =

Note that this eigenvalue has multiplicity two but has only one linearly independent

eigenvector.

Using software to find eigenvalues and eigenvectors. Let A =

23

41 and B =

01

10.

TI-89: Use

MatrixeigVl(A)

and

MatrixeigVc(A)



p. 208

S*

The eigenvectors will be the columns of the matrix returned by eigVc, corresponding in order

to the eigenvalues from eigVl. The eigenvectors are normalized (unit vectors).

However, eigVl and eigV will not work with complex eigenvalues and eigenvectors. In this

case you must find the characteristic polynomial using

Matrixdet(x

Matrixidentity(2)-b)

Then use the Algebra menu to solve: ComplexcSolve(x^2+1=0,x)

or factor : ComplexcFactor(x^2+1)

Use rref to find the eigenvectors:

Matrixrref(

Matrixidentity(2)-b)

Mathematica: On the Basic Math Assistant Palette, under

Basic Commands |

| More

use Eigenvalues[a] and Eigenvectors[a].

For example, the eigenvalue 5 corresponds to the eigenvector (1, 1).

Mathematica will handle real and complex eigenvalues and

eigenvectors. However, if you want the characteristic polynomial,

you can use Det[x*IdentityMatrix[2]-a]

Use the Factor[ ], Simplify[ ], and Solve[lhs ==rhs,var]

funtions as appropriate. To find the eigenvectors the long way, you can use MatrixForm[RowReduce[ ]]

Theorem 7.3: Eigenvalues of Triangular Matrices

If A is an nn triangular matrix, then its eigenvalues are the entries on the main diagonal.

Proof: If A is an nn triangular matrix, then I – A is a triangular matrix with – a11, – a22, …,

– ann on the diagonal. Since I – A is triangular, its determinant is the product of the entries

on the main diagonal ( – a11)( – a22)…( – ann) by Theorem 3.2. Therefore, the eigenvalues

of A are the entries on the main diagonal: a11, a22, …, ann.


7.2 Diagonalization.

p. 209

S*


Objective: Prove properties of eigenvalues and eigenvectors.

Objective: Find the eigenvalues of similar matrices, determine whether a matrix is

diagonalizable, and find a matrix P such that P–1

AP is diagonal.

Objective: Find, for a linear transformation T: V V a basis B for V such that the matrix for T

relative to B is diagonal.

An nn matrix A is diagonalizable when there exists an invertible matrix P such that D = P–1

AP

is a diagonal matrix.

Recall from Section 6.4 that two square matrices A and B are similar if and only there is an

invertible matrix P such that B = P–1

AP. So we can also say that A is diagonalizable when it it

similar to a diagonal matrix D.

Theorem 7.4: Similar Matrices Have the Same Eigenvalues

If A and B are similar nn matrices, then they have the same eigenvalues.

Proof: Because A and B are similar, there exists an invertible matrix P such that B = P–1

AP, so

|I – B| = | P–1

IP – P–1

AP | = |P–1

(I – A)P| = |P–1

| |(I – A)| |P| =||

1

P |(I – A)| |P| = |(I – A)|

Since A and B have the same characteristic polynomial, they must have the same eigenvalues.

Theorem 7.5: Condition for Diagonalization

An nn matrix A is diagonalizable if and only if it has n linearly independent eigenvectors.

Proof:

First, assume A is diagonalizable. Then there exists an invertible matrix P such that D = P–1

AP

is a diagonal matrix. Then PD = AP. Let the column vectors of P be p1, p2, …, pn and the main

diagonal entries of D be 1, 2, …, n.

Now PD = [ p1 | p2 | … | pn ]

n

000

00

00

2

1

= [ 1p1 | 2p2 | … | npn ]

and AP = A[ p1 | p2 | … | pn ] = [ Ap1 | Ap2 | … | Apn ]



p. 210

Since PD = AP, we have Api = ipi for each of the n column vectors of P. Since P is invertible,

the n column vectors are also linearly independent. So A has n linearly independent

eigenvectors.

Conversely, assume A has n linearly independent eigenvectors p1, p2, …, pn with

corresponding eigenvalues 1, 2, …, n. Let P be the matrix whose columns are the n

eigenvectors: P = [ p1 | p2 | … | pn ]. Then

AP = A[ p1 | p2 | … | pn ] = [ Ap1 | Ap2 | … | Apn ] = [ 1p1 | 2p2 | … | npn ]

= [ p1 | p2 | … | pn ]

n

000

00

00

2

1

Thus, AP = PD, where D =

n

000

00

00

2

1

is a diagonal matrix. Since the pi are linearly

independent, P is invertible, so D = P–1

AP and A is diagonalizable.

Example Diagonalize A =

020

200

123

by finding an invertible matrix P and a diagonal matrix D

such that D = P–1

AP. Use the characteristic polynomial to find the eigenvalues and the reduced

row-echechelon function of your software to find the eigenvectors.

Solution: 0 = det(

=

so 1 = 2 = 3 =



p. 211

Eigenvector for 1 =

Eigenvector for 2 =

Eigenvector for 3 =



p. 212

S*

Example: Diagonalize A =

366

252

221

by finding an invertible matrix P and a diagonal

matrix D such that D = P–1

AP. Use eigVl and EigVc on the TI-89, or Eigenvalues[ ]

and Eigenvectors[ ] in Mathematica, or eigenvalues and eigenvectors in PocketCAS.

Solution:

TI-89:

MatrixeigVl(A)

MatrixeigVc(A)P

P^-1*A*P

Mathematica:

Eigenvalues[a]

MatrixForm[p=Transpose[Eigenvectors[a]]]

(Note that Mathematica gives the transpose of P.)

MatrixForm[Inverse[p].a.p]

PocketCAS

eigenvalues(a) [3, 3, –3]

p:=eigenvectors(a)

333

130

103

p^-1*a*p

300

030

003



p. 213

Example: Try to diagonalize A =

201

121

001

Use eigVl and EigVc on the TI-89, or

Eigenvalues[ ] and Eigenvectors[ ] in Mathematica, or eigenvalues and

eigenvectors in PocketCAS.

Solution:

TI-89:

MatrixeigVl(A)

MatrixeigVc(A)P

P^(-1)*A*P Error: Singular Matrix

Mathematica:

Eigenvalues[a]

MatrixForm[p=Transpose[Eigenvectors[a]]]

(Note that Mathematica gives the transpose of P.)

MatrixForm[Inverse[p].a.p]

PocketCAS

eigenvalues(a) [2, 2, 1]

p:=eigenvectors(a) Not diagonalizable at eigenvalue 2


7.3 Symmetric Matrices and Orthogonal Diagonalization.

p. 215

S*

S*


Objective: Recognize, and apply properties of, symmetric matrices.

Objective: Recognize, and apply properties of, orthogonal matrices.

Objective: Find an orthogonal matrix P that orthogonally diagonalizes a symmetric matrix A.

A square matrix A is symmetric when A = AT.

Theorem 7.7: Properties of Symmetric Matrices

If A is a symmetric nn matrix, then

1) A is diagonalizable.

2) All eigenvalues of A are real.

3) If is an eigenvalue of A with multiplicity k, then has k linearly independent eigenvectors.

That is, the eigenspace of has dimension k.

The proof will be deferred until Ch. 8.5

Theorem 7.9: Property of Symmetric Matrices

Let A be an nn symmetric matrix. if 1 and 2 are distinct eigenvalues of A, then their

corresponding eignevectors x1 and x2 are orthogonal.

Proof:

Let 1 and 2 are distinct eigenvalues of A with corresponding eignevectors x1 and x2, so

Ax1 = 1x1 and Ax2 = 2x2. Recall that we can write x1•x2 = x1Tx2. Then

1 (x1•x2) = (1x1)•x2

= (Ax1)•x2

= 2(x1•x2)

Therefore, (1 – 2)(x1•x2) = 0. Since 1 – 2 0, we must have x1•x2 = 0 so x1 and x2 are

orthogonal.



p. 216

Example: Find the eigenvalues and eigenvectors of

011

101

110

. Are the eigenvalues real? What

are the dimensions of the eigenspaces? Find the angles between the eigenvectors.

Solution:

0 = det(

100

010

001

–

011

101

110

)

(using the calculator to take the determinant and factor it.)

1 = 2, 2 = –1, 3 = –1. All eigenvalues are real.

Solve (1I –

011

101

110

)x1 = 0 using rref.

Write x1=

3

2

1

. 3 = t is a free variable. We solve for 1 = t and 1 = t.

The solution is

t

t

t

= t

1

1

1

so x1 =

1

1

1

.

The eigenspace of = 2 (multiplicity = 1) is one-dimesional.

Solve (2I –

011

101

110

)x2 = 0 and

Solve (3I –

011

101

110

)x3 = 0 using rref. (Note 2 = 3.)

2 = s and 3 = t are free variables. We solve for 1 = –s – t.

The solution is

t

s

ts

= s

0

1

1

+ t

1

0

1

so x2 =

0

1

1

and x3 =

1

0

1

.

is the

Greek letter

xi



p. 217

S*

The eigenspace of = –1 (multiplicity = 2) is two-dimensional.

x1•x2 = (1, 1, 1)•(–1, 1, 0) = 0 so x1 is orthogonal to x2.

x1•x3 = (1, 1, 1)•(–1, 0, 1) = 0 so x1 is orthogonal to x3.

x2•x3 = )1,0,1()0,1,1(

)1,0,1()0,1,1(

=

22

1 =

2

1 so the angle between x2 and x3 is arccos(

2

1) =

3

.

Alternative method:

Use eigVl and eigVc on the calculator to obtain

1 = 3 = –1 and 2 = 2

with eigenvectors

p1 =

408248.

408248.

816497.

, p2 =

57735.

57735.

57735.

, p3 =

710756.

703401.

007355.

We can calculate ||p1|| = ||p2|| = ||p3|| = 1, p1•p2 = 10–14

, p1•p3 = 0.009008,

and p2•p3 = –6.810–15

,

so the eigenvectors are almost orthonormal.

A square matrix P is called orthogonal when it is invertible and P–1

= PT.

Theorem 7.8: Property of an Orthogonal Matrix

An nn matrix P is orthogonal if and only if its column vectors form an orthonormal set.

Proof:

Write P = [ p1 | p2 | … | pn ] in terms of its column vectors. Then

PT =

np

p

p

2

1

and PTP =

n

T

n

T

n

T

n

n

TTT

n

TTT

pppppp

pppppp

pppppp

21

22212

12111

.

Thus, PTP = I if and only if ||pi||

2 = pi•pi = 1 for each i and pi•pj = 0 whenever i j.

Therefore, PT = P

–1 if and only if its column vectors form an orthonormal set.



p. 218

S*

Because of this theorem, it would have been better if orthogonal matrices had been called

“orthonormal matrices.” but we are stuck with the name “orthogonal.”

Theorem 7.10: Fundamental Property of an Symmetric Matrices

Let A be a real nn matrix. Then A is orthogonally diagonalizable and has real eigenvalues if and

only if A is symmetric.

Proof:

Suppose A is orthogonally diagonalizable and real, so there exists a real, orthogonal matrix P

such that D = P –1

AP is diagonal and real. Then

= PDP –1

= PDPT

so

AT = (PDP

T)

T

= (PT)TD

TP

T

= PDPT

= A

Conversely, Theorems 7.7 and 7.9 together tell us that all symmetric matrices are orthogonally

diagonalizable with real eigenvalues.

An nn symmentric matrix has n eigenvalues, counting multiplicity. Because the dimension

of each eigenspace equals the multiplicity of the corresponding eigenvalue, we have n

linearly independent eigenvectors, so the matrix is diagonalizable (Theorem 7.5).

Eigenvectors with different eigenvalues are orthogonal. Within each eigenspace, we can use

the Gram-Schmidt process to find orthonormal eigenvectors. The set of all of the

orthonormal eigenvectors from all of the eigenspaces make up the columns of an orthogonal

matrix P, such that

D = P –1

AP

is diagonal.

Example (continued): Diagonalize

011

101

110

using…

1) the characteristic equation and rref

2) eigVl and eigVc on the TI-89

3) Eigenvalues[ ] and Eigenvectors[ ] in Mathematica

4) eigenvalues and eigenvectors in PocketCAS



p. 219

Solution:

1) From the first part of the example, 1 = 2, 2 = –1, 3 = –1.

Solve (1I –

011

101

110

)x1 = 0 using rref.

3 = t is a free variable. We solve for 1 = t and 1 = t.

The solution is

t

t

t

= t

1

1

1

so x1 =

1

1

1

.

Solve (2,3I –

011

101

110

)x2,3 = 0

2 = s and 3 = t are free variables. We solve for 1 = –s – t.

The solution is

t

s

ts

= s

0

1

1

+ t

1

0

1

so x2 =

0

1

1

and x3 =

1

0

1

.

Normalize p1 = 1

1

x

x =

1

1

1

3

1.

Gram-Schmidt: p2 = 2

2

x

x =

0

1

1

2

1

u3 = x3 – (x3•p2)p2 =

1

0

1

–

2

1

0

1

1

2

1 =

1

2/1

2/1

; p2 = 2

2

u

u =

2

1

1

6

1

so D =

100

010

002

and P =

6/203/1

6/12/13/1

6/12/13/1

Check that PTP = I and multiply P

–1AP = P

TAP to check that it gives you D.



p. 220

2) TI-89: Use eigVl and eigVc to obtain

D =

100

020

001

and P =

710756.57735.408248.

703401.57735.408248.

077355.57735.816497.

Check PTP =

.115E8.6009008.

15E8.6.114E.1

009008.14E.1.1

so P is almost orthogonal.

By looking at the ratios between components, it appears that

p1 = unitV([-2,1,1]) = 6/66/63/6

p2 = unitV([1,1,1]) = 3/33/33/3

p3 = unitV([0,-1,1]) = 2/22/20

Try a new P =

2/23/36/6

2/23/36/6

03/33/6

p gives

707107.57735.408248.

707107.57735.408248.

057735.816497.


–1AP = P


3) Mathematica: Eigenvalues[a] yields {2,-1,-1} so D =

100

010

002

Eigenvalues[a] gives {{1,1,1},{-1,0,1},{-1,1,0}} so

x1 =

1

1

1

, x2 =

1

0

1

, x3 =

0

1

1

Warning: MatrixForm[Eigenvalues[a]] gives

which has the eigenvectors in the rows (instead of in

the columns)!



p. 221

Orthogonalize[Eigenvalues[a]] gives the orthonormal basis

so P =

6/12/13/1

3/203/1

6/12/13/1

Warning: MatrixForm[Orthogonalize[Eigenvalues[a]]] gives PT, not P!


–1AP = P


4) PocketCAS: Computing M T

M shows in the off-diagonal

elements that the vectors are orthogonal,

and the diagonal elements give the norm

squared of each vector.

so D =

100

010

002

gramschmidt( ) will construct an eigenvectors( ) orthonormal basis from the rows of a matrix,

gives orthogonal, so use gramschmidt(transpose(p)) to

but unnormalized obtain PT.

eigenvectors.

so P =

6/12/13/1

6/203/1

6/12/13/1


–1AP = P




p. 222

Example: Diagonalize

444

422

422

.


8.5 Unitary and Hermitian Matrices.

p. 223


Objective: Find the conjugate transpose (Hermitian conjugate) of a complex matrix A.

Objective: Determine if a matrix A is unitary.

Objective: Find the eigenvalues and eigenvectors of a Hermitian matrix, and diagonalize a

Hemitian matrix.

In order to prove the theorems in Section 7.4 about real symmetric matrices, we must consider

complex “Hermitian” matices.

The conjugate transpose (or Hermitian conjugate) of a complex matrix A, denoted by A† or A

H or

A*, is given by A† = (𝐴 ̅)𝑇 = 𝐴𝑇̅̅̅̅ . That is, Take the complex conjugate of each of entry of A, and

transpose the matrix.

Notice that if A is real (all entries are real), then A† = A

T.

A is Hermitian if and only if A† = A. Notice that if A is real, then Hermitian is the same as

symmetric.

To take the Hermitian conjugate A† on the TI-89, use a

Matrix

T

To take the Hermitian conjugate A† in Mathematica, use act

To take the Hermitian conjugate A† in PocketCAS, use trn(a)

A is unitary if and only if A† = A

–1. Notice that if A is real, then unitary is the same as orthogonal.

Example: Classify the following matrices as Hermitian, unitary, both, or neither.

1) A =

0

0

i

i: A

† =

0

0

i

i. A

† = A so A is Hermitian. AA

† = I so A is also unitary.

2) B = : B† = . B

† B so B is not Hermitian.

BB† = I so B is unitary.

3) C =

523

234

7

1

i

i: C

† =

523

234

7

1

i

i. C

† = C so C is Hermitian.

CC† = I so C is not unitary.



p. 224

S*

4) D =

i0

01: D

† =

i0

01. D

† D so D is not Hermitian. D

†D =

10

01 I so D is not

unitary.

Theorem 8.8: Properties of the Hermitian Conjugate (Complex Transpose)

If A, B, and C complex matrices with dimensions mn, mn, and np respectively, and k is a

complex number, then the following properties are true.

1) (A†)† = A

2) (A + B)† = A

† + B

†

3) (kA)† = �̅�A

†

4) (AC)† = C

†A

†

Theorem 8.9: Unitary Matrices

An nn complex matrix U is unitary if and only if its row vectors form an orthonormal set in Cn

using the Euclidean inner product.

Proof:

Recall that the Euclidean inner product of u and v in Cn is given by

u•v = 𝑢1𝑣1̅̅ ̅ + 𝑢2𝑣2̅̅ ̅ + ⋯ + 𝑢𝑛𝑣𝑛̅̅ ̅

Note that if we write u and v as row vectors, then u•v = u v†

||u|| = uu and u is orthogonal to v if and only if u•v = 0.

Write A =

nu

u

u

2

1

in terms of its row vectors. Then

AT = [ u1

† | u 2

† | … | un

† ] and AA

† =

nnnn

n

n

uuuuuu

uuuuuu

uuuuuu

21

22212

12111

.

Thus, AA† = I if and only if ||ui||

2 = ui•ui = 1 for each i and ui•uj = 0 whenever i j.

Therefore, A† = A

–1 if and only if its row vectors form an orthonormal set.

Theorem 8.9.1: A is unitary if and only if A† is unitary.



p. 225

Corollary 8.9.2: Unitary Matrices

An nn complex matrix A is unitary if and only if its column vectors form an orthonormal set in

Cn using the Euclidean inner product.

Theorem 8.10: The Eigenvalues of a Hermitian Matrix

If H is a Hermitian matrix, then its eigenvalues are real.

Proof:

Let be an eigenvalue of H and let v 0 be its corresponding eigenvalue, so Hv = v.

Then (v†Hv)

† = v

†H

†(v

†)† = v

†Hv

Now v†Hv = v

†(Hv) = v

†(v) = (v

†v) = ||v||

2

So (v†Hv)

† = ‖𝐯‖2̅̅ ̅̅ ̅̅ ̅̅ = ̅||v||

2 because ||v|| is real.

Since ||v|| 0 and ||v||2 = ̅||v||

2, must be real.


8.5 Unitary and Hermitian Matrices

Find the conjugate transpose of a complex matrix

Determine if a matrix is unitary.

Find the eigenvalues and eigenvectors of a Hermitian matrix, anddiagonalize a Hermitian matrix.

CONJUGATE TRANSPOSE OF A MATRIX

Problems involving diagonalization of complex matrices and the associated eigenvalueproblems require the concepts of unitary and Hermitian matrices. These matricesroughly correspond to orthogonal and symmetric real matrices. In order to define unitary and Hermitian matrices, the concept of the conjugate transpose of a complexmatrix must first be introduced.

Note that if is a matrix with real entries, then * . To find the conjugatetranspose of a matrix, first calculate the complex conjugate of each entry and then takethe transpose of the matrix, as shown in the following example.

Finding the Conjugate Transpose

of a Complex Matrix

Determine * for the matrix

SOLUTION

Several properties of the conjugate transpose of a matrix are listed in the followingtheorem. The proofs of these properties are straightforward and are left for you to supply in Exercises 47–50.

A* � AT � �3 � 7i0

�2i4 � i�

A � �3 � 7i2i

04 � i� � �3 � 7i

�2i0

4 � i�

A � �3 � 7i2i

04 � i�.

A

� ATAA

A

A.A*

Definition of the Conjugate Transpose of a Complex Matrix

The conjugate transpose of a complex matrix denoted by *, is given by

*

where the entries of are the complex conjugates of the corresponding entriesof A.

A

� A TA

AA,

THEOREM 8.8 Properties of the Conjugate Transpose

If and are complex matrices and is a complex number, then the followingproperties are true.

1. 2.3. 4. �AB�* � B*A*�kA�* � kA*

�A � B�* � A* � B*�A*�* � A

kBA

9781133110873_0805.qxp 3/10/12 6:55 AM Page 420

UNITARY MATRICES

Recall that a real matrix is orthogonal if and only if In the complex system, matrices having the property that * are more useful, and such matricesare called unitary.

A Unitary Matrix

Show that the matrix is unitary.

SOLUTION

Begin by finding the product *.

Because

it follows that So, is a unitary matrix.

Recall from Section 7.3 that a real matrix is orthogonal if and only if its row (or column) vectors form an orthonormal set. For complex matrices, this property characterizes matrices that are unitary. Note that a set of vectors

in (a complex Euclidean space) is called orthonormal when the statements below are true.

1.

2.

The proof of the next theorem is similar to the proof of Theorem 7.8 presented inSection 7.3.

i � jvi � vj � 0,

�vi� � 1, i � 1, 2, . . . , m

Cn

�v1, v2, . . . , vm�

AA* � A�1.

AA* � �10

01� � I2

� �10

01�

�14�

40

04�

AA* �12�

1 � i1 � i

1 � i1 � i�

12 �

1 � i1 � i

1 � i1 � i�

AA

A �12 �

1 � i1 � i

1 � i1 � i�A

A�1 � AA�1 � AT.A

8.5 Unitary and Hermitian Matrices 421

Definition of Unitary Matrix

A complex matrix is unitary when

A�1 � A*.

A

THEOREM 8.9 Unitary Matrices

An complex matrix is unitary if and only if its row (or column) vectorsform an orthonormal set in Cn.

An � n

9781133110873_0805.qxp 3/10/12 6:55 AM Page 421

The Row Vectors of a Unitary Matrix

Show that the complex matrix is unitary by showing that its set of row vectors formsan orthonormal set in

SOLUTION

Let and be defined as follows.

Begin by showing that and are unit vectors.

Then show that all pairs of distinct vectors are orthogonal.

So, is an orthonormal set.�r1, r2, r3�

� 0

��5

65�

1 � 3i

65�

4 � 3i

65

r2 � r3 � �i

3�5i

215� � i3�

3 � i

215� � 13�

4 � 3i

215 � � 0

� �5i

415�

4 � 2i

415�

�4 � 3i

415

r1 � r3 � 12�

5i

215� � 1 � i2 � 3 � i

215� � �12�

4 � 3i

215 � � 0

�i

23�

i

23�

1

23�

1

23

r1 � r2 � 12��

i3� � 1 � i

2 � i3� � �

12�

13�

� 1

� �2560

�1060

�2560�

1�2

�r3� � � 5i

215�5i

215� � 3 � i

215�3 � i

215� � 4 � 3i

215 �4 � 3i

215 ��1�2

� 1

� �13

�13

�13�

1�2

�r2� � �� i3��

i3� � i

3�i

3� � 13�

13��

1�2

�r1� � �12�

12� � 1 � i

2 �1 � i2 � � �

12��

12��

1�2

� �14

�24

�14�

1�2� 1

r3r2,r1,

r3 � 5i

215,

3 � i

215,

4 � 3i

215 �

r2 � � i3

, i

3,

13�,r1 � 1

2,

1 � i

2, �

1

2�,

r3r1, r2,

A � �12

�i

35i

215

1 � i2i

33 � i

215

�12

13

4 � 3i

215

�C3.

A


REMARK

Try showing that the columnvectors of also form anorthonormal set in C3.

A

9781133110873_0805.qxp 3/10/12 6:55 AM Page 422

HERMITIAN MATRICES

A real matrix is symmetric when it is equal to its own transpose. In the complex system, the more useful type of matrix is one that is equal to its own conjugatetranspose. Such a matrix is called Hermitian after the French mathematician CharlesHermite (1822–1901).

As with symmetric matrices, you can recognize Hermitian matrices by inspection.To see this, consider the matrix

The conjugate transpose of has the form

If is Hermitian, then So, must be of the form

Similar results can be obtained for Hermitian matrices of order In other words,a square matrix is Hermitian if and only if the following two conditions are met.

1. The entries on the main diagonal of are real.

2. The entry in the th row and the th column is the complex conjugate of theentry in the th row and the th column.

Hermitian Matrices

Which matrices are Hermitian?

a. b.

c. d.

SOLUTION

a. This matrix is not Hermitian because it has an imaginary entry on its main diagonal.

b. This matrix is symmetric but not Hermitian because the entry in the first row andsecond column is not the complex conjugate of the entry in the second row and firstcolumn.

c. This matrix is Hermitian.

d. This matrix is Hermitian because all real symmetric matrices are Hermitian.

��1

23

20

�1

3�1

4��3

2 � i3i

2 � i0

1 � i

�3i1 � i

0�� 0

3 � 2i3 � 2i

4�� 1

3 � i

3 � i

i �

ijaji

jiaij

A

An � n.

A � � a1

b1 � b2ib1 � b2i

d1�.

AA � A*.A

� �a1 � a2ib1 � b2i

c1 � c2id1 � d2i

�.

� �a1 � a2i

b1 � b2i

c1 � c2i

d1 � d2i� A* � AT

A

A � �a1 � a2ic1 � c2i

b1 � b2id1 � d2i

�A.2 � 2


Definition of a Hermitian Matrix

A square matrix is Hermitian when

A � A*.

A

9781133110873_0805.qxp 3/10/12 6:55 AM Page 423

One of the most important characteristics of Hermitian matrices is that their eigenvalues are real. This is formally stated in the next theorem.

PROOF

Let be an eigenvalue of and let

be its corresponding eigenvector. If both sides of the equation are multipliedby the row vector then

Furthermore, because

it follows that * is a Hermitian matrix. This implies that * is a real number, so is real.

To find the eigenvalues of complex matrices, follow the same procedure as for realmatrices.

Finding the Eigenvalues of a Hermitian Matrix

Find the eigenvalues of the matrix

SOLUTION

The characteristic polynomial of is

So, the characteristic equation is and the eigenvalues of are and �2.�1, 6,

A�� 1�� 6�� 2� � 0,

� �� 1�� 6�� 2�. � �3 � 3�2 � 16� � 12

� ��3 � 3�2 � 2� � 6� � �5� � 9 � 3i� � �3i � 9 � 9�� 3i �1 � 3i� � 3�i�

� �� 3��2 � 2� � ��2 � i� ��2 � i�� 3i � 3��

��I � A� � � � � 3�2 � i

�3i

�2 � i�

�1 � i

3i�1 � i

� �A

A � �3

2 � i3i

2 � i0

1 � i

�3i1 � i

0�A.

�Avv1 � 1Avv

�v*Av�* � v*A*�v*�* � v*Av

v*Av � v*��v� � ��v*v� � ��a12 � b1

2 � a22 � b2

2 � . . . � an2 � bn

2�.

v*,Av � �v

v � �a1 � b1i

a2 � b2i

..

.

an � bni�

A�


THEOREM 8.10 The Eigenvalues of a Hermitian Matrix

If is a Hermitian matrix, then its eigenvalues are real numbers.A

REMARK

Note that this theorem impliesthat the eigenvalues of a realsymmetric matrix are real, asstated in Theorem 7.7.

9781133110873_0805.qxp 3/10/12 6:55 AM Page 424

To find the eigenvectors of a complex matrix, use a procedure similar to that usedfor a real matrix. For instance, in Example 5, to find the eigenvector corresponding tothe eigenvalue substitute the value for into the equation

to obtain

Solve this equation using Gauss-Jordan elimination, or a graphing utility or softwareprogram, to obtain the eigenvector corresponding to which is shown below.

Eigenvectors for and can be found in a similar manner. They are

and respectively.�1 � 3i

�2 � i

5�,�

1 � 21i

6 � 9i

13�

�3 � �2�2 � 6

v1 � ��1

1 � 2i

1�

�1 � �1,

��4

�2 � i�3i

�2 � i�1

�1 � i

3i�1 � i

�1��v1

v2

v3� � �

000�.

�� 3

�2 � i�3i

�2 � i�

�1 � i

3i�1 � i

��

v1

v2

v3� � �

000�

�� 1,



Quantum mechanics had its start in the early 20th centuryas scientists began to study subatomic particles and light.Collecting data on energy levels of atoms, and the rates oftransition between levels, they found that atoms could beinduced to more excited states by the absorption of light.

German physicist Werner Heisenburg (1901–1976) laid amathematical foundation for quantum mechanics usingmatrices. Studying the dispersion of light, he used vectors torepresent energy levels of states and Hermitian matrices torepresent “observables” such as momentum, position, andacceleration. He noticed that a measurement yields preciselyone real value and leaves the system in precisely one of a setof mutually exclusive (orthogonal) states. So, the eigenvaluesare the possible values that can result from a measurementof an observable, and the eigenvectors are the correspondingstates of the system following the measurement.

Let matrix be a diagonal Hermitian matrix that representsan observable. Then consider a physical system whosestate is represented by the column vector To measure thevalue of the observable in the system of state you canfind the product

Because is Hermitian and its values along the diagonalare real, is a real number. It represents the average of the values given by measuring the observable on asystem in the state a large number of times.u

Au*Au

A

� �u1u1�a11 � �u2u2�a22 � �u3u3�a33.

u*Au � u1 u2 u3��a11

00

0a22

0

00

a33��

u1

u2

u3�

u,Au.

A

TECHNOLOGY

Some graphing utilities andsoftware programs have built-inprograms for finding the eigenvalues and correspondingeigenvectors of complex matrices.

Jezper/Shutterstock.com

9781133110873_0805.qxp 3/10/12 6:55 AM Page 425


Just as real symmetric matrices are orthogonally diagonalizable, Hermitian matrices are unitarily diagonalizable. A square matrix is unitarily diagonalizablewhen there exists a unitary matrix such that

is a diagonal matrix. Because is unitary, *, so an equivalent statement is that is unitarily diagonalizable when there exists a unitary matrix such that *is a diagonal matrix. The next theorem states that Hermitian matrices are unitarily diagonalizable.

PROOF

To prove part 1, let and be two eigenvectors corresponding to the distinct (andreal) eigenvalues and Because and you have the equations shown below for the matrix product *

* *A* * * *

* * * *

So,

because

and this shows that and are orthogonal. Part 2 of Theorem 8.11 is often called theSpectral Theorem, and its proof is left to you.

The Eigenvectors of a Hermitian Matrix

The eigenvectors of the Hermitian matrix shown in Example 5 are mutually orthogonalbecause the eigenvalues are distinct. Verify this by calculating the Euclidean inner products and For example,

The other two inner products and can be shown to equal zero in a similar manner.

The three eigenvectors in Example 6 are mutually orthogonal because they correspond to distinct eigenvalues of the Hermitian matrix . Two or more eigenvectorscorresponding to the same eigenvalue may not be orthogonal. Once any set of linearlyindependent eigenvectors is obtained for an eigenvalue, however, the Gram-Schmidtorthonormalization process can be used to find an orthogonal set.

A

v2 � v3v1 � v3

� 0.

� �1 � 21i � 6 � 9i � 12i � 18 � 13

� ��1��1 � 21i� � �1 � 2i��6 � 9i� � 13

v1 � v2 � ��1��1 � 21i� � �1 � 2i��6 � 9i� � �1��13�

v2 � v3.v1 � v3,v1 � v2,

v2v1

�1� �2 v1*v2 � 0

��2 � �1�v1*v2 � 0

�2v1*v2 � �1v1*v2 � 0

v2�1v2 � �1v1v2 � v1v2 � ��1v1��Av1�v2�2v2 � �2v1Av2 � v1v2 � v1v2 � v1�Av1�

v2.�Av1�Av2 � �2v2,Av1 � �1v1�2.�1

v2v1

APPPAP�1 � PP

P�1AP

PA

THEOREM 8.11 Hermitian Matrices and Diagonalization

If is an Hermitian matrix, then

1. eigenvectors corresponding to distinct eigenvalues are orthogonal.2. is unitarily diagonalizable.A

n � nA

9781133110873_0805.qxp 3/10/12 6:55 AM Page 426

Diagonalization of a Hermitian Matrix

Find a unitary matrix such that * is a diagonal matrix where

SOLUTION

The eigenvectors of are shown after Example 5. Form the matrix by normalizingthese three eigenvectors and using the results to create the columns of

So,

Try computing the product * for the matrices and in Example 7 to see that

*

where and are the eigenvalues of You have seen that Hermitian matrices are unitarily diagonalizable. It turns out

that there is a larger class of matrices, called normal matrices, that are also unitarilydiagonalizable. A square complex matrix is normal when it commutes with its conjugate transpose: The main theorem of normal matrices states that a complex matrix is normal if and only if it is unitarily diagonalizable. You are askedto explore normal matrices further in Exercise 56.

The properties of complex matrices described in this section are comparable to theproperties of real matrices discussed in Chapter 7. The summary below indicates thecorrespondence between unitary and Hermitian complex matrices when compared withorthogonal and symmetric real matrices.

AAA* � A*A.

A

A.�2�1, 6,

AP � ��1

0

0

0

6

0

0

0

�2�P

PAAPP

P � ��

17

1 � 2i7

17

1 � 21i7286 � 9i728

13728

1 � 3i40

�2 � i40

540

� .

�v3� � ��1 � 3i, �2 � i, 5�� 10 � 5 � 25 � 40

�v2� � ��1 � 21i, 6 � 9i, 13�� 442 � 117 � 169 � 728

�v1� � ��1, 1 � 2i, 1�� 1 � 5 � 1 � 7

P.PA

A � �3

2 � i3i

2 � i0

1 � i

�3i1 � i

0�.

APPP


Comparison of Symmetric and Hermitian Matrices

A is a symmetric matrix A is a Hermitian matrix(real) (complex)

1. Eigenvalues of are real. 1. Eigenvalues of are real.2. Eigenvectors corresponding 2. Eigenvectors corresponding

to distinct eigenvalues are to distinct eigenvalues are orthogonal. orthogonal.

3. There exists an orthogonal 3. There exists a unitary matrix matrix such that such that

is diagonal. is diagonal.

P*APPTAP

PP

AA

9781133110873_0805.qxp 3/10/12 6:55 AM Page 427


8.5 Exercises

Finding the Conjugate Transpose In Exercises 1–4,determine the conjugate transpose of the matrix.

1. 2.

3. 4.

Finding the Conjugate Transpose In Exercises 5 and6, use a software program or graphing utility to find theconjugate transpose of the matrix.

5.

6.

Non-Unitary Matrices In Exercises 7–10, explain whythe matrix is not unitary.

7. 8.

9.

10.

Identifying Unitary Matrices In Exercises 11–16,determine whether is unitary by calculating *.

11. 12.

13. 14.

15.

16.

Row Vectors of a Unitary Matrix In Exercises 17–20,(a) verify that is unitary by showing that its rows areorthonormal, and (b) determine the inverse of

17. 18.

19.

20.

Identifying Hermitian Matrices In Exercises 21–26,determine whether the matrix is Hermitian.

21. 22.

23. 24.

25.

26.

Finding Eigenvalues of a Hermitian Matrix InExercises 27–32, determine the eigenvalues of the matrix

27. 28.

29. 30.

31. A � �100

4i0

1 � i3i

2 � i�A � � 0

2 � i2 � i

4�A � � 31 � i

1 � i2�

A � � 3�i

i3�A � � 0

�ii0�

A.

�1

2 � i5

2 � i2

3 � i

53 � i

6�� 1

2 � i2 � i

23 � i3 � i�

�0

2 � i0

ii1

100��

02 � i

1

2 � ii0

101�

� i0

0�i�� 0

�ii0�

A � �0

�1 � i6

26

1

0

0

01 � i3

13

�A �

1

22 �3 � i

3 � i

1 � 3 i

1 � 3 i�

A � �1 � i

2

1 2

�1 � i

2

1 2

�A � ��45

3 5

3 5

4 5

i

i�A.

A

A � ��

i2

i2

0

i3

i3

i3

i6

i6

�i

6

�

A � ��45

35

i

35

45

i�A � �

i2

i2

i2

�i

2�A � ��i

00i�

A � �1 � i1 � i

1 � i1 � i�A � �1 � i

1 � i1 � i1 � i�

AAA

A � �12

�i

3

�12

�12

13

12

1 � i2i

3

�1 � i

2�

A � �1 � i2

0

0

1

� i 2

0�A � �1

i

i

�1�A � � i

0

0

0�

�2 � i

0i

1 � 2i

12 � i2 � i

4

�12i

�i0

2i1 � i

1�2i

��1 � i2 � i1 � i

i

01i

2 � i

102

�1

�i2i4i0�

�250

i3i

6 � i��0

5 � i�2 i

5 � i64

2i43�

�1 � 2i1

2 � i1�� i

2

�i

3i�

9781133110873_0805.qxp 3/10/12 6:55 AM Page 428

8.5 Exercises 429

32.

Finding Eigenvectors of a Hermitian Matrix InExercises 33–36, determine the eigenvectors of thematrix in the indicated exercise.

33. Exercise 27 34. Exercise 30

35. Exercise 31 36. Exercise 28

Diagonalization of a Hermitian Matrix In Exercises37–41, find a unitary matrix that diagonalizes thematrix

37. 38.

39.

40.

41.

43. Show that is unitary by computing

44. Let be a complex number with modulus 1. Show thatthe matrix is unitary.

Unitary Matrices In Exercises 45 and 46, use the result ofExercise 44 to determine and such that is unitary.

45. 46.

Proof In Exercises 47–50, prove the formula, where and are complex matrices.

47. 48.

49. 50.

51. Proof Let be a matrix such that Provethat is Hermitian.

52. Show that det where is a matrix.

Determinants In Exercises 53 and 54, assume that theresult of Exercise 52 is true for matrices of any size.

53. Show that

54. Prove that if is unitary, then

55. (a) Prove that every Hermitian matrix can be written asthe sum where is a real symmetricmatrix and is real and skew-symmetric.

(b) Use part (a) to write the matrix

as the sum where is a real symmetricmatrix and is real and skew-symmetric.

(c) Prove that every complex matrix can bewritten as where and areHermitian.

(d) Use part (c) to write the complex matrix

as the sum where and areHermitian.

56. (a) Prove that every Hermitian matrix is normal.

(b) Prove that every unitary matrix is normal.

(c) Find a matrix that is Hermitian, but not unitary.

(d) Find a matrix that is unitary, but notHermitian.

(e) Find a matrix that is normal, but neitherHermitian nor unitary.

(f) Find the eigenvalues and corresponding eigenvectorsof your matrix in part (e).

(g) Show that the complex matrix

is not diagonalizable. Is this matrix normal?

� i0

1i�

2 � 2

2 � 2

2 � 2

CBA � B � iC,

A � � i2 � i

21 � 2i�

CBA � B � iC,An � n

CBA � B � iC,

A � � 21 � i

1 � i3�

CBA � B � iC,

A

�det�A�� 1.A

det�A*� � det�A�.

2 � 2A�A� � det�A�,iA

A* � A � O.A

�AB�* � B*A*�kA�* � kA*

�A � B�* � A* � B*�A*�* � A

n � nBA

A �12 �a

b

6 � 3i45

c�A �

12 �

�1b

ac�

Aca, b,

A �12�

ziz

z�iz�

Az

AA*.A � In

A � ��1

00

0�1

�1 � i

0�1 � i

0�A � � 4

2 � 2i2 � 2i

6�

A � �2

i2

�i

2

�i

2

2

0

i2

0

2�

A � � 02 � i

2 � i4�A � � 0

�ii0�

A.P

A � �2

i2

�i

2

�i

2

2

0

i2

0

2�

42. Consider the following matrix.

(a) Is unitary? Explain.

(b) Is Hermitian? Explain.

(c) Are the row vectors of orthonormal? Explain.

(d) The eigenvalues of are distinct. Is it possible to determine the inner products of the pairs ofeigenvectors by inspection? If so, state thevalue(s). If not, explain why not.

(e) Is unitarily diagonalizable? Explain.A

A

A

A

A

A � ��2

3 � i4 � i

3 � i1

1 � i

4 � i1 � i

3�

9781133110873_0805.qxp 3/10/12 6:55 AM Page 429

Answer Key

17. (a)

(b)

19. (a)

(b)

21. is Hermitian because 23. is not Hermitian because the entry on the main

diagonal, is not a real number. So,25. is not Hermitian because the matrix is not square.27. 29. 31.

33. 35.

37.

39.

41.

43.

Therefore and is unitary.

45. 47–53. Proofs

55. (a)

(b)

(c)

(d) � i �1

12

12

�2�� 0

2 �i2

2 �i2

1�A �

A � A*

2� i

A � A*

2i

�21

13� � i� 0

�110�

A �A � A

2� i

A � A

2i

A �1�2 ��1

�i

�1

i�InA* � A�1

AA* � �100...

010...

001...

. . .. . .. . .� � In

A* � �100...

010...

001...

. . .. . .. . .�

A � �100...

010...

001...

. . .. . .. . .�

P �1�6 �

�600

0�1 � i

2

02

1 � i�

P �12 �

�2�i

i

0�2�2

�2i

�i�

P �1�2�

1

�i

1

i�

v3 � �6 � 4i, 3i, 2�v2 � �2 � 2i, �1, 0�v2 � �1, i�v1 � �1, 0, 0�v1 � �1, �i�

3 � 2 � i2 � i2 � 42 � �11 � 11 � 11 � 1

AA � A*.

a22,AA � A*.A

A�1 �1

2�2 ��3 � i

1 � �3i�3 � i

1 � �3i��r1� � 1, �r2� � 1, r1 � r2 � 0

r2 �1

2�2��3 � i, 1 � �3 i�

r1 �1

2�2��3 � i, 1 � �3 i�,

A�1 � ��45

�35i

35

�45i�

�r1� � 1, �r2� � 1, r1 � r2 � 0

r1 � ��45, 35i�, r2 � �3

5, 45i�

9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A6

Section 8.5

1.

3.

5.

7. is not unitary because it is singular.9. is not unitary because it is not a square matrix.

11.

So, A is not unitary.

13.

So, A is unitary.

15.

So, A is unitary.

AA* � �10

01� � I2

AA* � I2 � �10

01�

AA* � � 4�4i

4i4� � I2

AA

�1 � i

01i

2 � i10

�2i

1 � i�i

2�4i

�i2 � i

�10�

�0

5 � i��2i

5 � i64

�2i43�

��ii

2�3i�


7.4 Applications of Eigenvalues and Eigenvectors.

p. 227


Objective: Model population growth using an age transion matrix and age distribution vector,

and find a stable age distribution vector.

Objective: Use a matrix equation to solve a system of first-order linear differential equations.

To model population growth of a population with a lifespan of L years, we partition the lifespan

into n equal-size classes

1st age class 2

nd age class … i

th age class … n

th age class

n

L,0

n

L

n

L 2,

…

n

iL

n

Li,

)1(

…

L

n

Ln,

)1(

The age distribution vector is x =

nx

x

x

2

1

where xi is the number of individuals in the ith

age class.

Let pi be the probability that after L/n years, a member of the ith

age class will survive to become

a member of the (i + 1)th

age class. (Note that 0 pi 1 for i = 0, 1, …, n – 1, and pn = 0.) Let bi

be the average number of offspring produced by a member of the ith

age class. (Note that 0 pi

for i = 0, 1, …, n.)

The age transition matrix or Leslie matrix is A =

0000

0000

0000

1

2

1

1321

n

nn

p

p

p

bbbbb

If xi is the age distribution vector at a specific time, then the age distribution vector L/n years

later is

xi+1 = Axi

Example: 80% of a population of mice survives the first year. Of that 80%, 25% survives the

second year. The maximum lifespan is three years.

The number of offspring for each member of the population is 3 in the first year, 6 in the second,

and 3 in the third year.

The population now consists of 120 members in each age class. How many members will there

be in each age class in one year? In two years?



p. 228

Solution: x1 = and A =

After one year, x2 = = for

3age2

2age1

1age0

After two years, x3 = = for

3age2

2age1

1age0

Example: A certain type of lizard has a maximum lifespan of two years. Only 8% of lizards

survive from their first year to their second year. The average number of offspring for each

member of the population is 1.5 in the first year, and 2 in the second year. Find a stable age

distribution vector for this population.

Solution: A = .

Using the TI-89, we find eigenvalues

with correspronding eigenvectors

A negative eigenvalue (and an eigenvector with some positive and some negative entries) does

not make sense. x and Ax = x must both have all non-negative entries, because the entries

represent the number of individuals in each age class.

We therefore use the eigenvector for = The number of individuals in each class

should be a whole number, so try x =

Now check: Ax =

The ratio of the age classes is stable at (first year) : (second year) =



p. 229

A system of first-order linear differential equations has the form

nnnnnn

nn

nn

yayayay

yayayay

yayayay

2211

22221212

12121111

with initial conditions

nn Cy

Cy

Cy

)0(

)0(

)0(

22

11

Where each iy is a function of t and dt

dyy i

i .

This is a linear system because the y′ =

ny

y

y

2

1

is a linear transformation of y =

ny

y

y

2

1

.

y′(t) = Ay(t), where A is a matrix of constants.

These are first-order differential equations because they contain first (and not higher)

derivatives.

Example: Solve the system

33213

32212

32111

00

00

00

yyyy

yyyy

yyyy


33

22

11

)0(

)0(

)0(

Cy

Cy

Cy

Solution:

From calculus, we know that the general solution to y′(t) = ky(t) is y(t) = Cekt.

So the solution to the system (with a diagonal matrix) is t

eCty 1

11 )(

t

eCty 2

22 )(

t

eCty 3

33 )(


313

212

3211 2

yyy

yyy

yyyy


2)0(

1)0(

4)0(

3

2

1

y

y

y



p. 230

Solution:

y′(t) = Ay(t), where A =

101

011

112

and y(0) =

2

1

4

Try a substitution y(t) = Pw(t), where P is some invertible constant matrix.

Then y′(t) = Pw′(t) and P–1

y′(t) = w′(t) so

y′(t) = A y(t)

P–1

y′(t) = P–1

A y(t)

w′(t) = P–1

A(Pw(t))

w′(t) = (P–1

AP)w(t)

If is P–1

AP diagonal, i.e. P–1

AP = D =

3

2

1

00

00

00

, then w(t) =

t

t

t

eB

eB

eB

3

2

1

3

2

1

for unknown

constants B1, B2, B3.

Using the TI-89 to find the eigenvalues and eigenvectors of A, we find

P =

707107.057735.0408248.0

707107.057735.0408248.0

057735.0816497.0

and D =

100

000

003

So w(t) =

t

t

eB

B

eB

3

2

3

1

, y(t) = Pw(t) = P

t

t

eB

B

eB

3

2

3

1

, and

y(0) =

2

1

4

= P

3

2

1

B

B

B

, so

3

2

1

B

B

B

= P–1

2

1

4

=

0.707107

0.57735

4.49073

Then y(t) = Pw(t) = P

t

t

e.

.

e.

7071070

577350

490734

=

t

t

t

e

e

e

3

3

3

833.15.03333.0

833.15.03333.0

667.33333.0

=

t

t

t

e

e

e

3

611

21

31

3

611

21

31

3

311

31

Other software may yield other P and D matrices, but w(t) will always be the same.



p. 231


33

3212

321

4

1044

53

yy

yyyy

yyy


0)0(

3)0(

2)0(

3

2

1

y

y

y

Solution: y′(t) = Ay(t), where A = and y(0) =

Try a substitution y(t) = Pw(t)

y′(t) = A y(t)

Pw′(t) = A Pw(t)

w′(t) = (P–1

AP) w(t)

We want diagonal P–1

AP =

3

2

1

00

00

00

, so w(t) = for unknowns B1, B2, B3.

Using the TI-89 to find the eigenvalues and eigenvectors of A, we find

P = and D =

So w(t) = , y(t) =

y(0) = = P

3

2

1

B

B

B

, so

3

2

1

B

B

B

=

Then y(t) = Pw(t) = P

Use Expand on the TI to obtain y(t) =



p. 232

An example from continuum mechanics of a

symmetric matrix is the stress tensor

T =

333231

232221

131211

.

The figure uses the notation that T: R3 R3

is the linear transformation iTi eTe

)(

ij is the force per unit area on the plan

perpendicular to ei, in the ej direction.

ii is is a pressure or normal stress. ij is a

shearing stress for i j. T is symmetric

because of conservation of angular

momentum.

Example: Let T =

103

014

341

.

Use the TI-89 to diagonalize T: P =

8042426404242640

6056568505656850

070710707071070

.. .

...

..

=

54

25

3

25

3

53

25

4

25

4

25

5

25

5 0

and D =

100

060

0011

The mutually orthogonal vectors p1 =

4242640

5656850

7071070

.

.

.

, p1 =

4242640

5656850

7071070

.

.

.

, and p3 =

80

6.0

0

.

are called the principal directions, where the stress vector )( ip

T is parallel to pi and where

there are no shear stresses dij for i j. The three stresses 1= 11, 2=6, and 3= 1, are called

principal stresses.



p. 233

Example (multivariable calculus): The Hessian matrix H of a function of several variables is

symmetric.

If f (x, y): R2 R is a (not necessarily linear) function of two variables, then

H =

2

22

2

2

2

y

f

xy

f

yx

f

x

f

has orthogonal eigenvectors. Near a critical point, the level curves of

f (x, y) = constant are ellipses if the eigenvalues of H have the same sign, a straight lines if

one eigenvalue is zero, or hyperbolas if the eigenvalues have opposite signs. See the text

for more discussion.

f (x, y) = 13x2 –8 xy + 7y

2 – 45

Eigenvalues = 5, 15

Eigenvectors =

1

2,

2

1

f (x, y) = 2𝑥2 + 4𝑥𝑦 + 2𝑦2 + 4√2𝑥 + 4√2𝑦 + 45

Eigenvalues = 0, 4

Eigenvectors =

1

1,

1

1

f (x, y) = 17x2 + 32xy – 7y

2 – 75

Eigenvalues = 25, –15

Eigenvectors =

1

2,

2

1

Holtmann75 / Module Math 75 Holtmann Revised 2015-11-18.pdf

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Transcript of Holtmann75 / Module Math 75 Holtmann Revised 2015-11-18.pdf