Linear Algebra. Session 3 - Texas A&M Universityroquesol/Math_304_Spring_2020_Week… · Theorem...

Matrices and Matrix AlgebraDeterminants I

Linear Algebra. Session 3

Dr. Marco A Roque Sol

01/27/2020

Dr. Marco A Roque Sol Linear Algebra. Session 3


Matrices, matrix algebra

Inverting diagonal matrices

Theorem 3.1A diagonal matrix D = diag(d1, d2, · · · .dn) is invertible if and onlyif all diagonal entries are nonzero; di 6= 0 for 1 ≤ i < n

If D is invertible then D−1 = diag(d−11 , d−12 , · · · .d−1n )

d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n




proof

If all di 6= 0, then clearly,

diag(d1, d2, · · · .dn)diag(d−11 , d−12 , · · · .d−1n ) =

diag(1, 1, · · · .1) = In

diag(d−11 , d−12 , · · · , d−1n )diag(d1, d2, · · · .dn) =

diag(1, 1, · · · .1) = In

Now suppose that di = 0, for some i . Then for any n × n matrixB, the ith row of the matrix DB is zero. Hence DB 6= In. �




Inverting 2× 2 matrices

The determinant of a 2× 2 matrix

A =

(a bc d

)is denoted by det(A) and defined by det(A) = ad − bc

Theorem 3.2

A matrix

A =

(a bc d

)is invertible if and only if det(A) 6= 0 . If det(A) 6= 0, then

A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)Dr. Marco A Roque Sol Linear Algebra. Session 3



proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc

)In the case det(A) = ad − bc 6= 0 we have A−1 = [det(A)]−1B

In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ I2B = 0 ⇒ B = 0⇒ A = 0, but the zero matrix is singular !!!!! �




System of n linear equations in n variables:a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2

...an1x1 + an2x2 + · · ·+ annxn = bn

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn




Theorem 3.3

If the matrix A above, is invertible then the system has a uniquesolution, which is x = A−1b

General results on inverse matrices

Theorem 3.4

Given an n × n matrix A, the following conditions are equivalent:

(i) A is invertible.

(ii) x = 0 is the only solution of the matrix equation Ax = 0.

(iii) The matrix equation Ax = b has a unique solution for anyn-dimensional column vector b.




(iv) The row echelon form of A has no zero rows.

(v) The reduced row echelon form of A, is the identity matrix.

Theorem 3.5

Suppose that a sequence of elementary row operations converts amatrix A into the identity matrix. Then the same sequence ofoperations converts the identity matrix into the inverse matrix A−1




� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

Invertible case Noninvertible case

For any Invertible square matrix in row echelon form, the numberof columns with leading entries equals the number of rows withleading entries. For a Noninvertible square matrix, the number ofcolumns without leading entries (i.e., the number of free variablesin a related system of linear equations) equals the number of rowswithout leading entries (i.e., zero rows).




� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

Invertible case Noninvertible case

Hence the row echelon form of a square matrix A is either stricttriangular or else it has a zero row. In the former case, theequation Ax = b always has a unique solution. Also, in the formercase the reduced row echelon form of A is I .




Example 3.2

Check whether the matrix A, given by 3 −2 01 0 1−2 3 0

is invertible.

Solution

We convert it to row echelon form. Interchange the 1st row withthe 2nd row: 1 0 1

3 −2 0−2 3 0




Add −3 times the 1st row to the 2nd row: 1 0 10 −2 −3−2 3 0

Add 2 times the 1st row to the 3rd row : 1 0 1

0 −2 −30 3 2

Multiply the 2nd row by −1/2 : 1 0 1

0 1 3/20 3 2




Add −3 times the 2nd row to the 3rd row: 1 0 10 1 3/20 0 −5/2

Multiply the 3rd row by −2/5 : 1 0 1

0 1 3/20 0 1

We already know that the matrix A is invertible. Let’s proceedtowards reduced row echelon form .




Add −3/2 times the 3rd row to the 2nd row: 1 0 10 1 00 0 1

Add −1 times the 3rd row to the 1st row: 1 0 0

0 1 00 0 1

To obtain A−1, in this case, we need to apply the followingsequence of elementary row operations ( same applied to A before)to the identity matrix:




1) Interchange the 1st row with the 2nd row.

2) Add −3 times the 1st row to the 2nd row.

3) Add 2 times the 1st row to the 3rd row.

4) Multiply the 2nd row by −1/2 :

5) Add −3 times the 2nd row to the 3rd row.

6) Multiply the 3rd row by −2/5

7) Add −3/2 times the 3rd row to the 2nd row.

8) Add −1 times the 3rd row to the 1st row.




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (A|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1

Interchange the 1st row with the 2nd row: 1 0 1 0 1 0

3 −2 0 1 0 0−2 3 0 0 0 1

Add −3 times the 1st row to the 2nd row: 1 0 1 0 1 0

0 −2 −3 1 −3 0−2 3 0 0 0 1




1 0 1 0 1 00 −2 −3 1 −3 0−2 3 0 0 0 1

Add 2 times the 1st row to the 3rd row. 1 0 1 0 1 0

0 −2 −3 1 −3 00 3 2 0 2 1

Multiply the 2nd row by −1/2 : 1 0 1 0 1 0

0 1 3/2 −1/2 3/2 00 3 2 0 2 1




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 3 2 0 2 1

Add −3 times the 2nd row to the 3rd row. 1 0 1 0 1 0

0 1 3/2 −1/2 3/2 00 0 5/2 −3/5 1 0

Multiply the 3rd row by −2/5 1 0 1 0 1 0

0 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5

Add −3/2 times the 3rd row to the 2nd row. 1 0 1 0 1 0

0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

Add −1 times the 3rd row to the 1st row. 1 0 0 3/5 0 2/5

0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

= (I |A−1)




Thus

3 −2 01 0 1−2 3 0

3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=1

5

3 0 22 0 3−3 5 −2

That is 3 −2 0

1 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix A−1 is to merge the matrices A and I into onematrix (A|I ) and apply elementary row operations to this newmatrix.

Question: Why does it work?

Proposition

Any elementary row operation can be simulated as leftmultiplication by a certain ( elementary ) matrix




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying the ith row byr . (The matrix AE can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging the ith rowwith the jth row. (The matrix AE can be obtained from A,interchanging the ith column with the jth column)




Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A = I , where E1,E2, · · · ,Ek−1,Ek are elementarymatrices s imulating those operations.

Applying the same sequence of operations to the identity matrix,we obtain the matrix

B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover, B is invertible since elementary matricesare invertible. It follows that B−1(BA) = (B−1B)A = A, thenBA = AB, then A = B−1, so B = A−1.




Transpose of a matrix

Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

(1 22 0

)T

=

(1 22 0

)




Properties of transposes

(AT )T = A

(A + B)T = AT + BT

(rA)T = rAT

(AB)T = BTAT

(A1A2 · · ·Ak)T = ATk A

Tk−1 · · ·AT

2 AT1

(AT )−1 = (A−1)T




Definition

A square matrix A is said to be symmetric if A = AT

For example, any diagonal matrix is symmetric.

Proposition

For any square matrix A the matrices B = AAT and C = A + AT ,are symmetric.



Determinants I

The general definition of the determinant is quite complicated asthere is no simple explicit formula

Approach 1 (Axiomatic): We formulate properties that thedeterminant should have.

Approach 2 (Inductive): The determinant of an n × n matrix isdefined in terms of determinants of certain (n − 1)× (n − 1)matrices.

Approach 3 (Original): An explicit (but very complicated)formula is provided.



Determinants I

Approach 1 (Axiomatic)

Mn×n(R): The set of n × n matrices with real entries.

AXIOMS

There exists a unique function det :Mn×n(R) −→ R (called thedeterminant) with the following properties

A1 If a row of a matrix is multiplied by a scalar r , the determinantis also multiplied by r

A2 If we add a row of a matrix multiplied by a scalar to anotherrow, the determinant remains the same

A3 If we interchange two rows of a matrix, the determinantchanges its sign;

A4: det(I ) = 1



Determinants I

Theorem 3.6

Suppose A is a square matrix and B is obtained from A applyingelementary row operations. Then det(A) = 0 if and only ifdet(B) = 0.

Theorem 3.7

det(B) = 0 whenever the matrix B has a zero row

Theorem 3.8

det(A) = 0 if and only if the matrix is not invertible.



Determinants I

Idea of the proof of 3.8: Let B be the reduced row echelon form ofA. If A is invertible then B = I ; otherwise B has a zero row .

Remark. The same argument proves that properties (A1)-(A4) areenough to evaluate any determinant.

Theorem 3.9

If a matrix A has two proportional rows then det(A) = 0



Determinants I

Row echelon form of a square matrix

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

det(A) 6= 0 det(A) = 0



Determinants I

As an example, we have the matrix

A =

3 −2 01 0 1−2 3 0

Earlier we have transformed the matrix A into the identity matrixusing elementary row operations.

These included two row multiplications, by −0.5 and by −0.4, andone row exchange.

It follows that

det(A)(−0.5)(−0.4)(−1) = (−0.2)det(A) = det(I ) = 1

Hence det(A) = −5det(I ) = −5



Determinants I

Approach 2 (Inductive)

Let’s start by considering a 2× 2 system{ax + by = ecx + dy = f

whose solution is given by

x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice that the amount ad − bc playing an importantrole in the solution is a combination of the elements of the matrix(

a bc d

)Dr. Marco A Roque Sol Linear Algebra. Session 3


Determinants I

In a similar way, if we consider a 3× 3 systema11x + a12y + a13 = b1a21x + a22y + a23 = b2a31x + a32y + a33 = b3

we find, the solutions for the variables have in the denominator thequantity

a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which also plays an important role in the solution set. Thus, thiscould be a practical starting point of the concept of determinant.



Determinants I

Before introducing the concept of determinant, let’s start with acouple of definitions

Submatrices

Definition Given a matrix A, a k × k submatrix of A is a matrixobtained by specifying k columns and k rows of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition

Associated to every n × n matrix A there is a real number calledthe determinant of A denoted by |A| or det(A) and definedinductivly as follows

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

a12a23a31 + a13a21a32 − a31a22a13 − a32a23a11 − a33a21a12 =



Determinants I

a11a22a33 + a12a23a31+a13a21a32 − a31a22a13−a32a23a11 − a33a21a12

=

∣∣∣∣∣∣∣∣a11 a12 a13 a11 a12a21 a22 a23 a21 a22a31 a32 a33 a31 a32

− − − + + +

∣∣∣∣∣∣∣∣Now, for n ≥ 4, if let M1j be the corresponding minors to the firstrow, then we have

|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I

Theorem 3.10

For any 1 ≤ k ,m ≤ n we have that

det(A) =n∑

j=1

(−1)k+jakjMkj =n∑

j=1

akj(−1)k+jMkj =n∑

j=1

akjCkj

( Cofactor Ckj expansion, by the k − th row, )

det(A) =n∑

i=1

(−1)i+maimMim =n∑

j=1

aim(−1)i+mMim =n∑

j=1

aimCim

(Cofactor Cim expansion, by the m − th column, )



Determinants I

Theorem 3.11

Given an n × n matrix A, if B is an n × n matrix obtained from Aby

1) Additive law for rows

Suppose that matrices X ,Y ,Z are identical except for the ith row.The the ith row of Z is the sum of the ith rows of X and Y then

det(Z ) = det(X ) + det(Y )

∣∣∣∣∣∣a1 + a′1 a2 + a′2 a3 + a′3

b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣a′1 a′2 a′3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣Dr. Marco A Roque Sol Linear Algebra. Session 3


Determinants I

2) Adding a multiple of the ith row to the jth row then |B| = |A|∣∣∣∣∣∣a1 + rb1 a2 + rb2 a3 + rb3

b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣2) Interchanging two rows, then |B| = −|A|

3) Multiplying a row by a nonzero scalar α then |B| = α|A|



Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence, for every property of determinants involvingrows of a matrix there is an analogous property involving columnsof a matrix. )

2) |AB| = |A||B|

3) If A has a row (column) of zeros, then |A| = 0

4) If A has a two identical rows (columns), then |A| = 0



Determinants I

5) If two rows (columns) of A are proportional, then |A| = 0

6) If A is an upper (lower) triangular matrix, then|A| = a11a22a33 · · · ann

7) If A is an invertible matrix, then |A−1| = |A|−1

8) If A is an n × n matrix and r ∈ R, then |rA| = rn|A|


Linear Algebra. Session 3 - Texas A&M Universityroquesol/Math_304_Spring_2020_Week… · Theorem...

Documents

Transcript of Linear Algebra. Session 3 - Texas A&M Universityroquesol/Math_304_Spring_2020_Week… · Theorem...