LINEAR ALGEBRA - cvut.czmath.feld.cvut.cz/vivi/ZbirkaLAL/LAL16f.pdfrational roots of the polynomial...

Czech Technical University in Prague

LINEAR ALGEBRA

A collection of solved and unsolved exercises

Paola Vivi

Department of Mathematics

Faculty of Electrical Engineering

Praha 2016

1

Chapter 1Polynomials and Gaussian elimination

1.1) Divide the polynomial p(x) = 2x5−x4 + 4x3 + 3x2−x+ 1 by the polynomial q(x) = x3 +x2−x+ 1 and findthe remainder.

1.2) Divide the polynomial p(x) = x5− 2x4−x3 + 3x2−x+ 2 by the polynomial q(x) = x3 + 2x2 +x+ 1 and findthe remainder.

1.3) Using Horner’s scheme, find p(a) and p(b), where p(x) = 2x4 − 3x3 − 5x2 − 9x+ 5 and a = 3, b = 12 .

1.4) Using Horner’s scheme, find p(a) and p(b), where p(x) = 2x4 − 3x3 + 5x2 − x+ 5 and a = −2, b = 1.

1.5) Given p(x) = x5 − 4x3 − 2x2 + 3x+ 2, find all roots of p(x) together with their multiplicity and express thegiven polynomial as a product of real irreducible polynomials.

1.6) Given p(x) = x5 + 8x4 + 7x3 − 52x2 − 44x + 80, find all roots of p(x) together with their multiplicity andexpress the given polynomial as a product of real irreducible polynomials.

1.7) Find all solutions of 3x3 − 7x2 − 7x+ 3 = 0.

1.8) Find all solutions of 2x4 − 3x3 − 4x2 + 3x+ 2 = 0.

1.9) Use Gaussian elimination to solve the following systems.

1.9.1)x+ 2y + z = 2

−3x− 4y − 5z = −52x− y + 7z = 3,

1.9.2)x+ 2y + z = 2

−3x− 5y − 5z = −5x+ 4y + z = 0.


1.10.1)x− 3y − 2z = 62x− 4y − 3z = 8−3x+ 6y + 8z = −5,

1.10.2)x+ 2y − 3z = 12x+ 5y − 8z = 4

3x+ 8y − 13z = 7.


1.11.1)x+ y + 2z = 4

2x+ 3y + 6z = 103x+ 6y + 10z = 17,

1.11.2)x1 + x2 − 2x3 + 4x4 = 52x1 + 2x2 − 3x3 + x4 = 3

3x1 + 3x2 − 4x3 − 2x4 = 1.


1.12.1)2x+ y − 3z = 2x− 3y − z = 3,

1.12.2)x+ 3y − 2z + 5t = 42x+ 8y − z + 9t = 9

3x+ 5y − 12z + 17t = 7.

1.13) Write down all solutions of the following system as p ∈ IR obtains all possible real values.

2x− y − z = p−x− y = p−x+ 2y + z = 1.


2x+ y − 4z = p2

−x− y + z = −3x+ 2y + z = 1.

2

Solutions of odd number problems - Polynomials and Gaussian elimination

1.1) Divide the polynomial p(x) = 2x5−x4 + 4x3 + 3x2−x+ 1 by the polynomial q(x) = x3 +x2−x+ 1 and findthe remainder.

Solution. Polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lowerdegree, it is a generalised version of the arithmetic technique called long division. Given any two polynomialsp(x) (the dividend) and q(x) (the divisor) so that q(x) is not zero and the degree of q(x) is lower or equal tothe degree of p(x), there exist and are uniquely determined a quotient s(x) and a remainder r(x) such that

p(x) = s(x)q(x) + r(x), or p(x)q(x) = s(x) + r(x)

q(x) . Moreover: either r(x) = 0 or the degree of r(x) is lower than

the degree of q(x). For the given polynomials, we have:

(2x5 − x4 + 4x3 + 3x2 − x+ 1) : (x3 + x2 − x+ 1)

−(2x5 + 2x4 − 2x3 + 2x2) 2x2 − 3x+ 9

− 3x4 + 6x3 + x2 − x+ 1

−(−3x4 − 3x3 + 3x2 − 3x)

9x3 − 2x2 + 2x+ 1

−(9x3 + 9x2 − 9x+ 9)

− 11x2 + 11x− 8

This means that r(x) = −11x2 + 11x − 8 is the reminder, s(x) = 2x2 − 3x + 9 is the quotient of thegiven polynomials, and

2x5 − x4 + 4x3 + 3x2 − x+ 1

x3 + x2 − x+ 1= 2x2 − 3x+ 9 +

−11x2 + 11x− 8

x3 + x2 − x+ 1.

1.3) Using Horner’s scheme, find p(a) and p(b), where p(x) = 2x4 − 3x3 − 5x2 − 9x+ 5 and a = 3, b = 12 .

Solution. The Horner’s scheme for p(3) reads:

2 − 3 − 5 − 9 53 6 9 12 9

2 3 4 3 14

It follows that p(3) = 14, and also that:

2x4 − 3x3 − 5x2 − 9x+ 5 = (x− 3)(2x3 + 3x2 + 4x+ 3) + 14,

or equivalently:2x4 − 3x3 − 5x2 − 9x+ 5

x− 3= 2x3 + 3x2 + 4x+ 3 +

14

x− 3.

The Horner’s scheme for p( 12 ) reads:

2 − 3 − 5 − 9 512 1 − 1 − 3 − 6

2 − 2 − 6 − 12 − 1

It follows that p( 12 ) = −1, and also that:

2x4 − 3x3 − 5x2 − 9x+ 5 = (x− 1

2)(2x3 − 2x2 − 6x− 12)− 1,

2x4 − 3x3 − 5x2 − 9x+ 5 = (2x− 1)(x3 − x2 − 3x− 6)− 1,

or equivalently:2x4 − 3x3 − 5x2 − 9x+ 5

x− 12

= 2x3 − 2x2 − 6x− 12 +−1

x− 12

,

2x4 − 3x3 − 5x2 − 9x+ 5

2x− 1= x3 − x2 − 3x− 6 +

−1

2x− 1.

3

1.5) Given p(x) = x5 − 4x3 − 2x2 + 3x+ 2, find all roots of p(x) together with their multiplicity and express thegiven polynomial as a product of real irreducible polynomials.

Solution. All possible rational roots of the given polynomial (that has 1 as coefficient of the term with highest degree)belong to the set {1,−1, 2,−2}, the possible integer divisors of the constant term, 2, of the polynomial. Thuswe use Horner’s scheme starting with a = 1:

1 0 − 4 − 2 3 21 1 1 − 3 − 5 − 2

1 1 − 3 − 5 − 2 0.

It follows that p(1) = 0, and

(x5 − 4x3 − 2x2 + 3x+ 2) : (x− 1) = (x4 + x3 − 3x2 − 5x− 2),

or(x5 − 4x3 − 2x2 + 3x+ 2) = (x− 1)(x4 + x3 − 3x2 − 5x− 2).

Now we continue using the Horner’s scheme with a = 1 for the polynomial q(x) = x4 + x3 − 3x2 − 5x− 2, tosee if 1 is a root of p(x) with multiplicity 2:

1 1 − 3 − 5 − 21 1 2 − 1 − 6

1 2 − 1 − 6 − 8

This implies that q(1) = −8, thus 1 is a root of p(x) with multiplicity only one. We now use Horner’s schemewith a = −1 for q(x), and we get:

1 1 − 3 − 5 − 2−1 − 1 0 3 2

1 0 − 3 − 2 0

This implies that q(−1) = p(−1) = 0, and

(x4 + x3 − 3x2 − 5x− 2) = (x+ 1)(x3 − 3x− 2),

and(x5 − 4x3 − 2x2 + 3x+ 2) = (x− 1)(x+ 1)(x3 − 3x− 2).

Now we continue using the Horner’s scheme with a = −1 for the polynomial s(x) = x3 − 3x− 2, to see if −1is a root of p(x) with multiplicity 2:

1 0 − 3 − 2−1 − 1 1 2

1 − 1 − 2 0

Since s(−1) = 0, the root −1 has multiplicity at least two for the given polynomial and:

(x5 − 4x3 − 2x2 + 3x+ 2) = (x− 1)(x+ 1)2(x2 − x− 2).

Even if we can find the roots of the polynomial r(x) = x2 − x− 2, using the quadratic formula, we prefer tocontinue checking the Horner’s scheme again with a = −1 for the polynomial r(x), to verify if −1 is a root ofp(x) with multiplicity three:

1 − 1 − 2−1 − 1 2

1 − 2 0

This implies that r(−1) = 0 and −1 is a root of p(x) with multiplicity three:

p(x) = (x5 − 4x3 − 2x2 + 3x+ 2) = (x− 1)(x+ 1)3(x− 2).

The given polynomial has simple roots 1 and 2, while −1 is a root with multiplicity three.

1.7) Find all solutions of 3x3 − 7x2 − 7x+ 3 = 0.

Solution. The given polynomial p(x) = 3x3 − 7x2 − 7x+ 3 has constant term 3 and 3 is also the coefficient of the termwith highest degree. Considering that the integer divisors of 3 are all the numbers {1,−1, 3,−3}, all possible

4

rational roots of the polynomial p(x) belong to the set {1,−1, 3,−3, 13 ,−13}. We use Horner’s scheme starting

with a = 1:3 − 7 − 7 3

1 3 − 4 − 113 − 4 − 11 − 8

This implies that p(1) = −8, thus 1 is not a root of p(x). We now use Horner’s scheme with a = −1 for p(x),and we get:

3 − 7 − 7 3−1 − 3 10 − 3

3 − 10 3 0

Thus −1 is a root of p(x) and:

3x3 − 7x2 − 7x+ 3 = (x+ 1)(3x2 − 10x+ 3).

The remaining two roots can be found using the formula for quadratic equations, or again with a Horner’sscheme, starting with the guess a = 3:

3 − 10 33 9 − 1

3 − 1 0

Thus 3 is also a root of p(x) and:

3x3 − 7x2 − 7x+ 3 = (x+ 1)(x− 3)(3x− 1).

The solutions of the given equation are the roots of the polynomial p(x), thus x = −1, x = 3, and x = 13 .


1.9.1)x+ 2y + z = 2

−3x− 4y − 5z = −52x− y + 7z = 3,

1.9.2)x+ 2y + z = 2

−3x− 5y − 5z = −5x+ 4y + z = 0.

Solution. Gaussian elimination, or Gaussian reduction, is an algorithm for determining the solutions of a system oflinear equations.

Given a linear system S, Ax = b, of m equations with n unknowns x1, x2, . . . , xn:a12x1 + a12x2 + . . .+ a1nxn = b1a21x1 + a22x2 + . . .+ a2nxn = b2

...am1x1 + am2x2 + . . .+ amnxn = bm.

we indicate with Ri the i− th equation of the system: Ri : ai1x1 + ai2x2 + . . .+ ainxn = biWe define the following elementary row operations:

(E1). (Row interchange) Interchange two rows, i.e. interchange Ri and Rj , Ri ↔ Rj .

(E2). (Row scaling) Multiply a non-zero constant throughout a row, i.e. multiply Rk by c 6= 0 thus replaceRk by cRk.

(E3). (Row addition) Replace a row by itself plus a constant multiple of another row, replace Rk byRk + cRj .

Theorem If a linear system S1 is obtained from a linear system S using elementary row operation, then S1

and S are equivalent, i.e. they have the same set of solutions.

Gaussian elimination is a method that consists in finding appropriate row operations to reduce the givensystem to subsequently easier ones whose solution can be directly found.

1.9.1) We will use the symbol ∼ to indicate the equivalence of systems.

x+ 2y + z = 2−3x− 4y − 5z = −5

2x− y + 7z = 3

R2 + 3R1

∼R3 − 2R1

x+ 2y + z = 22y − 2z = 1−5y + 5z = −1

5R2 + 2R3

∼

x+ 2y + z = 22y − 2z = 1

0 = 3.

5

The last row corresponds to the equation 0x + 0y + 0z = 3 that is inconsistent, we conclude that the givensystem has no solutions.

1.9.2) We now solve the following:

x+ 2y + z = 2−3x− 5y − 5z = −5x+ 4y + z = 0

R2 + 3R1

∼R3 −R1

x+ 2y + z = 2y − 2z = 1

2y = −2.

At this point, we can already get the solution of the system, starting from the last equation (y = −1) andsubstituting into the second and subsequently the first equation (we get z = −1, x = 5), but for the sake ofillustrating the Gaussian reduction method, we will continue to perform row operations till the result can beread directly from the final system:

R3 − 2R2

∼

x+ 2y + z = 2y − 2z = 1

4z = −4

14R3

∼

x+ 2y + z = 2y − 2z = 1

z = −1

R2 + 2R3

∼R1 −R3

x+ 2y = 3y = −1z = −1

R1 − 2R2

∼

x = 5y = −1z = −1.


1.11.1)x+ y + 2z = 4

2x+ 3y + 6z = 103x+ 6y + 10z = 17,

1.11.2)x1 + x2 − 2x3 + 4x4 = 52x1 + 2x2 − 3x3 + x4 = 3

3x1 + 3x2 − 4x3 − 2x4 = 1.

Solution. 1.11.1) We will use the symbol ∼ to indicate the equivalence of systems.

x+ y + 2z = 42x+ 3y + 6z = 103x+ 6y + 10z = 17

R− 2− 2R1

∼R3 − 3R− 1

x+ y + 2z = 4y + 2z = 23y + 4z = 5

−R− 3 + 3R2

∼

x+ y + 2z = 4y + 2z = 2

2z = 1.

At this point we see that the system has a solution, and we continue to perform row operations till the resultcan be read directly from the final system:

R2 −R3

R1 −R2

R3/2∼

x = 2y = 1z = 1

2 .

1.11.2) We now solve the system:

x1 + x2 − 2x3 + 4x4 = 52x1 + 2x2 − 3x3 + x4 = 33x1 + 3x2 − 4x3 − 2x4 = 1

R2 − 2R1

∼R3 − 3R1

x1 + x2 − 2x3 + 4x4 = 5x3 − 7x4 = −7

2x3 − 14x4 = −14

∼R3 − 2R2

x1 + x2 − 2x3 + 4x4 = 5x3 − 7x4 = −7

0 = 0

In this case we say that the pivot variables are x1 and x3, while x2 and x4 are free variables. We may continuethe Gaussian reduction with row operation R1 + 2R2, and get:

x1 + x2 − 10x4 = −9x3 − 7x4 = −7.

Now the infinitely many solutions of the given system can be expressed as:

x1 = −α− 10β − 9

x2 = α

x3 = −7β − 7

x4 = β,

where α, β ∈ IR are any possible real number.


2x− y − z = p−x− y = p−x+ 2y + z = 1.

6

Solution. We start a Gaussian elimination with row operations −R2 ↔ R1, R1 + 2R2 and R3 −R2, so to get:

x+ y = −p−3y − z = 3p3y + z = 1− p

−R2

∼R3 +R2

x+ y = −p3y + z = −3p

0 = 1 + 2p.

We now start a discussion: if 1+2p 6= 0, the system has no solutions because the last equation is inconsistent;while if 1 + 2p = 0, i.e. p = − 1

2 , then the last equation is 0 = 0, and can be disregarded (as it is satisfied forall possible values of x, y and z). Thus the given system is equivalent to:

x+ y =1

2

3y + z =3

2.

The infinitely many solutions of the system can be expressed with the use of a parameter: choosing y as afree variable, y = α, α ∈ IR, we have that for every possible α ∈ IR,

x =1

2− α

y = α

z =3

2− 3α

is a solution of the given system where p = − 12 .

7

Chapter 2Linear spaces

2.1) Given A =

(3 1 −24 0 2

)B =

(−2 0 16 4 −1

), calculate C = 2A− 3B.

2.2) Given v1 = (−1, 1, 1), v2 = (1, 0,−3), v3 = (2, 4, 5), calculate x = 2v1 − 3v2 + 4v3.

2.3) Determine if the vector v = (3,−5, 2) in IR3 is a linear combination of u1 = (1,−3, 2), u2 = (2,−4,−1) andu3 = (1,−5, 7).

2.4) Write x = (3,−1, 5) as a linear combination of v1 = (−1, 1, 1), v2 = (1,−1, 1), v3 = (1, 1,−1).

2.5) Determine if the matrix M =

(4 77 9

)is a linear combination of A =

(1 11 1

), B =

(1 23 4

), C =

(1 14 5

).

2.6) In M2,2 show that A =

(1 11 1

), B =

(1 00 1

)and C =

(1 10 0

)are Linearly Independent.


(1 23 1

), B =

(3 −12 2

)and C =

(1 −5−4 0

)are Linearly Dependent.

2.8) Prove that v1 = (5, 4, 3), v2 = (3, 3, 2), v3 = (8, 1, 3) are linearly dependent.

2.9) Prove that v1 = (2,−3, 1), v2 = (3,−1, 5), v3 = (1,−4, 3) are linearly independent.

2.10) Determine for what values of α ∈ IR the vectors v1 = (1, α, 1), v2 = (0, 1, α), v3 = (α, 1, 0) are linearlydependent.

2.11) For what values of λ ∈ IR are the polynomials p1(t) = 2 + 5t+ t2, p2(t) = 4 + 10t+ 8t2, p3(t) = 6 + 10t+ λt2

linearly dependent?

2.12) Given A =

(1 20 −1

), B =

(−1 23 6

), C =

(3 2−3 d

), find d such that A, B, C are linearly dependent.

2.13) Given A =

1 −20 31 −1

, B =

1 0−1 −8−4 2

, C =

−3 41 de f

, find d, e, f so that A, B, C are linearly

dependent.

2.14) In P2(t) prove that p1(t) = 1 + t2, p2(t) = 2 + t− t2, p3(t) = 1− t are linearly independent.

2.15) Given g = 5t − 7t2 in P(t), show that g ∈ span{g1, g2, g3} where g1 = 1 + t − 2t2, g2 = 7 − 8t + 7t2,g3 = 3− 2t+ t2.

2.16) In the linear space of polynomial functions P(t) are given p1(t) = 2 + t, p2(t) = 2 + t2, p3(t) = 2 + t3,q(t) = 1− 2t+ t2 − t3. Prove that q /∈ span{p1, p2, p3}.

2.17) Prove that in any linear space if x and y are linearly independent then x + y and x − y are two linearlyindependent vectors.

2.18) Prove that in any linear space if x + y + z = 0 then span{x,y} = span{y, z}.

2.19) Given u1 = (1, 1, 1), u2 = (1, 2, 3) and u3 = (1, 5, 8), show that span{u1,u2,u3} = IR3.

2.20) Prove that the set B = {(1, 0, 1), (1, 0, 0), (1, 1, 1)} is a basis of IR3. Find the coordinates of (2, 3,−1) withrespect to the ordered basis B.

2.21) Given p1(t) = 1, p2(t) = t− 3, p3(t) = (t− 1)(t+ 2), show that B = {p1, p2, p3} is a basis of the space P2(t)of all polynomial functions with degree ≤ 2. Find the coordinates of p(t) = 2t2 + t+ 2 with respect to B.

2.22) Show that v1 = (2, 1,−3), v2 = (3, 2,−5), v3 = (1,−1, 1) form a basis of IR3.Find coordinates of v = (7, 6,−14) with respect to this basis.

2.23) Given p1 = 1− x2, p2 = 1 + x, p3 = −1 + x+ x3, extend {p1, p2, p3} to a basis of P3. (Use coordinates)

8

2.24) Show that if {v1,v2,v3,v4} is a basis of a linear space L then {v1,v1]v2,v1]v2]v3,v1]v2]v3]v4} is also abasis of L. (Use coordinates)

2.25) Given the linear space P(t) of all polynomials, determine if the following are linear subspacesi) all polynomials with integer coefficientsii) all polynomials with even powers of t (i.e. containing only terms of the type t2n where n=0,1,2...)iii) all polynomials with degree greater or equal to six.

2.26) Is W = {(a, b, c) : a ≥ 0} a linear subspace of IR3?

2.27) In the linear space F of all real functions F = {f : IR → IR}, decide if the sets defined below form linearsubspaces or not.M1 = {f ∈ F : f(1) = 0},M2 = {f ∈ F : f(3) = 1}.

2.28) Consider the subsets of IRn defined below and for each of them prove if it is a linear subspace of IRn or not:M1 = {(x1, . . . , xn) ∈ IRn : x1 = xn},M2 = {(x1, . . . , xn) ∈ IRn : xi ∈ IZ, i = 1, . . . , n},M3 = {(x1, . . . , xn) ∈ IRn : x1 + x2 + · · ·+ xn = 0},M4 = {(x1, . . . , xn) ∈ IRn : x1 = 2},M5 = {(x1, . . . , xn) ∈ IRn : xk = 0 for k even},M6 = {(x1, . . . , xn) ∈ IRn : x1 ≤ 0}.

2.29) Given the following subsets of P3(t) determine if they are linear subspaces of P3(t) or not. In affirmativecase, find the dimension and a basis for each subspace:L1 = {p ∈ P3 : p(0) = 0}L2 = {p ∈ P3 : p(−t) = p(t)}.

2.30) Given the following subsets of IR3 determine if they are linear subspaces of IR3 or not. In affirmative case,find the dimension and a basis for the subspace:M1 = {(x, y, z) : 2x− y + z = 2}M2 = {(x, y, z) : x− 2y − z = 0}.

2.31) a) Prove that if M and N are subspaces of a vector space L then M ∩N is a subspace of L.b) Given M = span{(1, 2, 1, 0), (−1, 1, 1, 1)} and N = span{(1, 2, 1,−2), (2, 1, 0, 1)} (subspaces of IR4), findthe dimension and a basis for M ∩N .c) Consider the space U = span{M ∪N}, find the dimension and a basis of U .

2.32) In IR4 are given the vectors v1 = (0, 3,−2, 4), v2 = (0, 1,−1, 3), v3 = (0, 0, 1,−5), v4 = (0, 5,−4, 10). Find abasis of the space M = span{v1,v2,v3,v4}. Show that B = {(0, 2,−1, 1), (0, 1, 0,−2)} is also a basis of M .

2.33) Consider M = {p(t) ∈ P3(t) : p(t) is divisible by (t − 1)}. Prove that M a linear subspace of P3, find abasis and dimension of M .

2.34) Given the equation x+ 2y− z+ 5w = 0 prove that the set S of all its solutions is a subspace of IR4. Find thedimension and a basis of S.

2.35) Consider Q = {p(x) ∈ P4(x) : p(x) = x4 · p( 1x ) for any x ∈ IR, x 6= 0}. Prove that Q a linear subspace of P4,

find a basis and dimension of Q.

2.36) Consider the subset ofM2,2 formed by the symmetric matrices (aij = aji, for every i, j). Show that this is asubspace of M2,2 and find a basis and its dimension.

9

Solutions of odd number problems - Linear spaces

2.1) Given A =

(3 1 −24 0 2

)B =

(−2 0 16 4 −1

), calculate C = 2A− 3B.

Solution. We use the definition of sum introduced in the space of 2× 3 matrices:

C = 2

(3 1 −24 0 2

)− 3

(−2 0 16 4 −1

)=

(6 2 −48 0 4

)+

(6 0 −3−18 −12 3

)=

(12 2 −7−10 −12 7

).

2.3) Determine if the vector v = (3,−5, 2) in IR3 is a linear combination of u1 = (1,−3, 2), u2 = (2,−4,−1) andu3 = (1,−5, 7).

Solution. Using the definition of linear combination of vectors in a vector space, we must prove that there exist realnumbers a, b, c such that v = au1 + bu2 + cu3. Thus we look if the vector equation

(3,−5, 2) = a(1,−3, 2) + b(2,−4,−1) + c(1,−5, 7)

has solutions. Evaluating the sum on the right hand side we get

(3,−5, 2) = (a+ 2b+ c,−3a− 4b− 5c, 2a− b+ 7c)

that leads to the system of linear equations

a + 2b + c = 3−3a − 4b − 5c = −52a − b + 7c = 2.

We perform a Gaussian reduction and with operations R2 + 3R1 and R3 − 2R1 we transform the system into

a +2b +c = 32b −2c = 4−5b +5c = −4

a last row operation 2R3 + 5R2 leads to

a +2b +c = 32b −2c = 4

0 = 12

that is a system without solutions. We conclude that the vector v cannot be written as a linear combinationof u1,u2, and u3.

2.5) Determine if the matrix M =

(4 77 9

)is a linear combination of A =

(1 11 1

), B =

(1 23 4

), C =

(1 14 5

).

Solution. We must determine if there exist real numbers a, b, c such that M = aA + bB + cC, thus we must solve(4 77 9

)= a

(1 11 1

)+ b

(1 23 4

)+ c

(1 14 5

)equivalent to (

4 77 9

)=

(a aa a

)+

(b 2b3b 4b

)+

(c c4c 5c

)Performing the indicated operations(

4 77 9

)=

(a+ b+ c a+ 2b+ ca+ 3b+ 4c a+ 4b+ 5c

)that leads to the system of linear equations

a+ b+ c = 4

a+ 2b+ c = 7

a+ 3b+ 4c = 7

a+ 4b+ 5c = 9.

10

We perform a Gaussian reduction and with operations R2 − R1, R3 − R1 and R4 − R1 we transform thesystem into

a+ b+ c = 4b = 3

2b+ 3c = 33b+ 4c = 5

R3 − 2R1

∼R4 − 3R2

a+ b+ c = 4b = 3

3c = −34c = −4

.

From last two equations we deduce that c = −1, from the second b = 3, and, substituting into the first, a = 2.We conclude that M = 2A + 3B−C.


(1 23 1

), B =

(3 −12 2

)and C =

(1 −5−4 0

)are Linearly Dependent.

Solution. By definition, n vectors in a linear space are linearly dependent if at least one of them can be written as alinear combination of the others, but this condition is not simple to verify. Thus we use a basic theorem: nvectors in a linear space are linearly dependent if and only if there exists a non trivial combination of thevectors that equals the null vector of the space. This condition is easier to verify. We must prove that there

exist real numbers a, b, c, not all equal to zero, such that aA + bB + cC = O, where O =

(0 00 0

).

We solve

a

(1 23 1

)+ b

(3 −12 2

)+ c

(1 −5−4 0

)=

(0 00 0

)We perform the sum on the left hand side:(

a 2a3a a

)+

(3b −b2b 2b

)+

(c −5c−4c 0

)=

(0 00 0

)(

a+ 3b+ c 2a− b− 5c3a+ 2b− 4c a+ 2b

)=

(0 00 0

).

This leads to the homogeneous system of linear equations:

a+ 3b+ c = 0

2a− b− 5c = 0

3a+ 2b− 4c = 0

a+ 2b = 0.

We perform a Gaussian reduction and with operations R2 − 2R1, R3 − 3R1 and R4 − R1 we transform thesystem into:

a+ 3b+ c = 0−7b− 7c = 0−7b− 7c = 0−b− c = 0.

∼ a+ 3b+ c = 0b+ c = 0

.

This system has infinitely many solutions, for any possible real value of c, it is enough to choose b = −c anda = 2c to find a solution. To get a non trivial solution, we may choose c = 1, thus b = −1 and a = 2 and2A−B + C = O implies that A,B,C are linearly dependent.

2.9) Prove that v1 = (2,−3, 1), v2 = (3,−1, 5), v3 = (1,−4, 3) are linearly independent.

Solution. We must prove that the only linear combination of v1,v2,v3 that equals the zero vector of IR3 is the triviallinear combination. We prove that if av1 + bv2 + cv3 = 0 then a = b = c = 0, where 0 = (0, 0, 0). Supposethat av1 + bv2 + cv3 = 0, this means that

a(2,−3, 1) + b(3,−1, 5) + c(1,−4, 3) = (0, 0, 0) that implies (2a+ 3b+ c,−3a− b− 4c, a+ 5b+ 3c) = (0, 0, 0).

This is equivalent to the homogeneous system

2a+ 3b+ c = 0−3a− b− 4c = 0a+ 5b+ 3c = 0.

11

We perform a Gaussian reduction, with operations R3 ↔ R1, R1 − 2R3, R2 + 3R3 we get

a+ 5b+ 3c = 0−7b− 4c = 014b− 5c = 0

∼R3 + 2R2

a+ 5b+ 3c = 0−7b− c = 0−13c = 0

.

This system has a unique solution a = b = c = 0, thus the given vectors are linearly independent in IR3.Observation. If we solve the exercise after studying matrices and determinants, we may use a differentapproach. Three vectors in IR3 are linearly independent if and only if the matrix formed by the three vectorsas rows (or columns) has rank = 3, if and only if the determinant of such matrix is different from zero. Thus

we may calculate det

2 −3 13 −1 51 −4 3

= −6 − 15 − 12 + 1 + 40 + 27 = 36 6= 0 and conclude that the given

vectors are linearly independent.

2.11) For what values of λ ∈ IR are the polynomials p1(t) = 2 + 5t+ t2, p2(t) = 4 + 10t+ 8t2, p3(t) = 6 + 10t+ λt2

linearly dependent?

Solution. Three polynomials in P are linearly dependent if there exists a non trivial linear combination of the polyno-mials that equals the zero polynomial. We will look for the possible values of λ such that

ap1(t) + bp2(t) + cp3(t) = o

has a solution with a, b, c not all equal to zero. The previous equation is expanded into

a(2 + 5t+ t2) + b(4 + 10t+ 8t2) + c(6 + 10t+ λt2) = 0

and, after performing the operations on the left hand side

(2a+ 4b+ 6c) + (5a+ 10b+ 10c)t+ (a+ 8b+ λc)t2 = 0.

This leads to the following homogeneous system and its Gaussian reduction

2a+ 4b+ 6c = 05a+ 10b+ 10c = 0a+ 8b+ λc = 0

R1/2R2 − 5R1/2R3 −R1/2

a+ 2b+ 3c = 0−15c = 0

6b+ (λ− 3)c = 0.

The second equation implies that c = 0, substituting into the third equation we deduce b = 0, and finallysubstituting the obtained value into the first equation we get a = 0. The solution was deduced independentlyfrom λ, thus, for any possible value of λ the given polynomials are linearly independent.

Observation. If we solve the exercise after studying matrices and determinants, we may use a differentapproach. Three vectors in P2 are linearly independent if and only if their coordinates with respect to agiven basis of P2, seen as vectors of IR3, are linearly independent, if and only if the corresponding matrixformed by the three vectors as rows (or columns) has rank = 3, if and only if the determinant of suchmatrix is different from zero. Thus we may consider the canonical basis of P2, B = {1, t, t2} and thecoordinates of the given polynomials [p1]B = (2, 5, 1), [p2]B = (4, 10, 8), [p3]B = (6, 10, λ) and evaluate

det

2 5 14 10 86 10 λ

= 20λ+ 40 + 240− 60− 160− 20λ = 60 6= 0.

We conclude that the given polynomials are linearly independent for any possible value of λ.

2.13) Given A =

1 −20 31 −1

, B =

1 0−1 −8−4 2

, C =

−3 41 de f

, find d, e, f so that A, B, C are linearly

dependent.

Solution. The given matrices are vectors of the linear space M3,2, they are linearly dependent if and only if we canfind real numbers α, β, γ, not all equal to zero, such that αA + βB + γC = O where O is the zero vector inM3,2. We try to find values for the parameters d, e, f so that the equation αA + βB + γC = O is satisfiedfor some non trivial values of α, β, γ. We evaluate αA + βB + γC = O and we get

α

1 −20 31 −1

+ β

1 0−1 −8−4 2

+ γ

−3 41 de f

=

0 00 00 0

if and only if

12

α+ β − 3γ −2α+ 4γ−β + γ 3α− 8β + dγ

α− 4β + eγ −α+ 2β + fγ

=

0 00 00 0

.

This leads to the following homogeneous system and the indicated Gaussian reduction

α+ β − 3γ = 0−2α+ 4γ = 0−β + γ = 0

3α− 8β + dγ = 0α− 4β + eγ = 0−α+ 2β + fγ = 0

R2 + 2R1

R4 − 3R1

R5 −R1

R6 +R1

∼

α+ β − 3γ = 02β − 2γ = 0−β + γ = 0

−11β + (d+ 9)γ = 0−5β + (e+ 3)γ = 0

3β + (f − 3)γ = 0

R2/2R3 +R2/2R4/11−R3

R5/5−R3

R6/3 +R3

∼

α+ β − 3γ = 0β − γ = 0

0 = 0(d+9

11 − 1)γ = 0( e+3

5 − 1)γ = 0

( f−33 + 1)γ = 0

Only if the coefficients of γ in the last three equations are all equal to zero, the system has a non trivialsolution. Thus for d = 2, e = 2 and f = 0 the system reduces to

α+ β − 3γ = 0β − γ = 0

and we can find infinitely many non zero solutions for α, β and γ. For example, for β = γ = 1 we get α = 2,and 2A + B + C = O implies that A,B and C are linearly dependent for d = 2, e = 2 and f = 0.

2.15) Given g = 5t − 7t2 in P(t), show that g ∈ span{g1, g2, g3} where g1 = 1 + t − 2t2, g2 = 7 − 8t + 7t2,g3 = 3− 2t+ t2.

Solution. By definition, the span of a set of vectors is the set (we prove that it is a linear space) formed by all possiblelinear combinations of the vectors. In this exercise we must prove that g can be written as a linear combinationof g1, g2 and g3, i.e. we must find real values a, b, c such that g = ag1 + bg2 + cg3. We pose the condition:

5t− 7t2 = a(1 + t− 2t2) + b(7− 8t+ 7t2) + c(3− 2t+ t2).

This is equivalent to

5t− 7t2 = (a+ 7b+ 3c) + (a− 8b− 2c)t+ (−2a+ 7b+ c)t2.

Two polynomial functions are equal if and only if they have same corresponding coefficients, thus we look forvalues of a, b, c that solve the system:

a+ 7b+ 3c = 0a− 8b− 2c = 5−2a+ 7b+ c = −7.

We perform a Gaussian reduction with operations R2 −R1 and R3 + 2R1 to obtain

a+ 7b+ 3c = 0−15b− 5c = 5

21b+ 7c = −7

R2/− 5R3/7 +R2/5

∼a+ 7b+ 3c = 0

3b+ c = −10 = 0.

The system has infinitely many solutions: for any possible value α ∈ IR, a = 3 + 2α, b = α, c = −1 − 3α isa solution of the system. Thus g can be written as a linear combination of g1, g2, g3 in infinitely many ways,proving that g ∈ span{g1, g2, g3}.

2.17) Prove that in any linear space if x and y are linearly independent then x + y and x − y are two linearlyindependent vectors.

Solution. To prove that x + y and x − y are linearly independent, we will prove that the only linear combination ofx + y and x− y that equals the zero vector is the trivial linear combination. Suppose that

a(x + y) + b(x− y) = 0

thena(x + y) + b(x− y) = (a+ b)x + (a− b)y = 0.

By hypothesis, x and y are linearly independent, i.e. the only linear combination of x and y that equals thezero vector is the trivial linear combination, thus the last equation implies the following system of conditionson a and b

a+ b = 0a− b = 0

.

13

The only solution of this homogeneous system is a = b = 0, thus x + y and x − y are linearly independentvectors.

2.19) Given u1 = (1, 1, 1), u2 = (1, 2, 3) and u3 = (1, 5, 8), show that span{u1,u2,u3} = IR3.

Solution. In order to prove the equality span{u1,u2,u3} = IR3, we must prove two inclusions. The first,span{u1,u2,u3} ⊆ IR3, holds trivially. To prove the second, IR3 ⊆ span{u1,u2,u3}, we take an arbitraryvector v = (x, y, z) ∈ IR3 and we prove that it can be generated as a linear combination of u1,u2,u3, i.e thereexist real numbers a, b, c such that

v = au1 + bu2 + cu3.

Expanding the previous equality we get

(x, y, z) = a(1, 1, 1) + b(1, 2, 3) + c(1, 5, 8) = (a+ b+ c, a+ 2b+ 5c, a+ 3b+ 8c).

This leads to the following system:a+ b+ c = xa+ 2b+ 5c = ya+ 3b+ 8c = z

.

We start a Gaussian reduction with operations R2 −R1 and R3 −R1:

a+ b+ c = xb+ 4c = y − x2b+ 7c = z − x

R3 − 2R2 ∼a+ b+ c = xb+ 4c = y − x−c = z + x− 2y

.

Now we can conclude that, for any possible x, y, z the system has a unique solution for a, b, c, thus the vectorv belongs to the span of u1,u2,u3.

Observation. If we use the notion of basis, we can prove the statement just by proving that B = {u1,u2,u3}is a basis of IR3. Let’s recall that, by definition, a basis of a linear space is a set of linearly independent vectorsthat span the space. A theorem states that a basis is a maximal set of linearly independent vectors, thus,knowing that the dimension of IR3 is three, if we prove that B = {u1,u2,u3} is a linearly independent set, itfollows that it must be a basis, and thus span{u1,u2,u3} = IR3. In order to prove that the three vectors inIR3 are linearly independent, we then calculate the determinant of the matrix formed with the three vectorsas rows (or columns), and because

det

u1

u2

u3

= det

1 1 11 2 31 5 8

= −1 6= 0

we conclude that B = {u1,u2,u3} is a basis of IR3 and span{u1,u2,u3} = IR3.

2.21) Given p1(t) = 1, p2(t) = t− 3, p3(t) = (t− 1)(t+ 2), show that B = {p1, p2, p3} is a basis of the space P2(t)of all polynomial functions with degree ≤ 2. Find the coordinates of p(t) = 2t2 + t+ 2 with respect to B.

Solution. By definition, a basis of a linear space is a set of linearly independent vectors that span the space. A theoremstates that a basis is a maximal set of linearly independent vectors, thus, knowing that the dimension of P2(t)is three, if we prove that B = {p1, p2, p3} is a linearly independent set, it follows that it must be a basis.Let’s prove directly that p1, p2, p3 are linearly independent. Suppose that a linear combination of the threepolynomials equals the zero polynomial,

ap1(t) + bp2(t) + cp3(t) = 0

then we must prove that a = b = c = 0. Explicitly expressing the previous equality, we get

a(1) + b(t− 3) + c(t− 1)(t+ 2) = 0 that is (a− 3b− 2c) + (b+ c)t+ ct2 = 0.

The last equality leads to the linear system

a− 3b− 2c = 0

b+ c = 0

c = 0

14

that has the only trivial solution a = b = c = 0. Thus B = {p1, p2, p3} is a basis of the space P2(t). To findthe coordinates of p(t) = 2t2 + t+ 2 with respect to B, we look for real numbers a, b, c such that

ap1(t) + bp2(t) + cp3(t) = p(t).

This means

a(1) + b(t− 3) + c(t− 1)(t+ 2) = 2t2 + t+ 2 that is (a− 3b− 2c) + (b+ c)t+ ct2 = 2t2 + t+ 2.

The last equality leads to the linear system

a− 3b− 2c = 2

b+ c = 1

c = 2

that has the unique solution a = 3, b = −1, c = 2, thus [p(t)]B = (3,−1, 2).

Observation. A faster way to prove that the three polynomials p1, p2, p3 are linearly independent. Threepolynomials in P2(t) are linearly independent if and only if their coordinates with respect to a given basis ofP2, seen as vectors of IR3, are linearly independent, if and only if the corresponding matrix formed by thethree vectors as rows (or columns) has rank = 3, if and only if the determinant of such matrix is differentfrom zero. Thus we may consider the canonical basis of P2, C = {1, t, t2} and the coordinates of the given

polynomials [p1]C = (1, 0, 0), [p2]C = (−3, 1, 0), [p3]C = (−2, 1, 1) and evaluate det

1 0 10−3 1 0−2 1 1

= 1 6= 0.

We conclude that the given polynomials are linearly independent and thus they form a basis of P2(t).

2.23) Given p1 = 1− x2, p2 = 1 + x, p3 = −1 + x+ x3, extend {p1, p2, p3} to a basis of P3(t). (Use coordinates)

Solution. Given a linear space L with finite dimension n, we can always consider the isomorphism between L and IRn

that associate to any vector of L its coordinates with respect to a fixed ordered basis of L. Then the ”linear”property that we wish to prove for a given set of vectors in L can be proved for the corresponding vectors inIRn and viceversa. We will work with the coordinates of p1, p2, p3 with respect to the standard basis C of P3,C = {1, t, t2, t3}. Then [p1]C = (1, 0,−1, 0), [p2]C = (1, 1, 0, 0), [p3]C = (−1, 1, 0, 1). Let us use the symbol ∼to denote the coincidence of the corresponding spans. We have

( 1, 0,−1, 0)( 1, 1, 0, 0)(−1, 1, 0, 1)

∼R2 −R1

R3 +R1

(1, 0,−1, 0)(0, 1, 1, 0)(0, 1,−1, 1)

∼

R3 −R2

(1, 0,−1, 0)(0, 1, 1, 0)

(0, 0,−2, 1).

We see that adding the vector v = (0, 0, 0, 1) to the last set of three vectors, we will create a set of fourlinearly independent vectors of IR4, thus a basis of IR4. The polynomial g = t3 is such that [g]C = v, thusthe set {p1, p2, p3, g} is a basis of P3(t).Observation. Using the notion of matrix and the fact that four vectors in IR4 are linearly independent ifand only if the rank of the matrix formed with the vectors as rows is regular: we can try to guess and addany row of coordinates to

( 1, 0,−1, 0)( 1, 1, 0, 0)(−1, 1, 0, 1)

so that the resulting 4× 4 matrix has non zero determinant. For example, adding (1, 0, 0, 0) we get

det

1 0 −1 01 1 0 0−1 1 0 11 0 0 0

= −1 6= 0

and this implies that also the polynomial q(t) = 1 (for which [q]C = (1, 0, 0, 0)) is such that {p1, p2, p3, q} isa basis of P3(t).

2.25) Given the linear space P(t) of all polynomials, determine if the following are linear subspacesi) all polynomials with integer coefficientsii) all polynomials with even powers of t (i.e. containing only terms of the type t2n where n=0,1,2...)iii) all polynomials with degree greater or equal to six.

15

Solution. By definition, a non empty subset M of a linear space L is a subspace of L if:1) given any two vectors u,v ∈M , then u + v ∈M ,2) given any vector u ∈M and any real number α, then α · u ∈M .

For the given sets of polynomials we have:i) all polynomials with integer coefficients do not form a linear subspace of P(t). Indeed, condition 2) is notsatisfied, it is enough to consider any polynomial with integer coefficient, for example p(t) = t, and the realnumber α =

√2, to see that the polynomial α · p(t) =

√2t is no more a polynomial with integer coefficient.

ii) all polynomials containing only terms of the type t2n where n = 0, 1, 2... form a linear subspace of P(t).Indeed this subset of P(t) is not empty (for example p(t) = t2 belongs to it) and1) adding any two polynomial containing only terms with even powers of t we create a new polynomial withonly even powers of t;2) multiplying any polynomial containing only terms with even power of t by any non-zero number, we get anew polynomial containing exactly the same powers of t. While, if we multiply any polynomial by 0 ∈ IR weget the polynomial z(t) = 0 that belongs to the set by definition. We see how here it is of basic importancethat in the definition of the subset the values of n where specified as n = 0, 1, 2..., because if the value n = 0was not included, the given subset would not be a subspace.

iii) all polynomials with degree greater or equal to six do not form a linear subspace of P(t). Indeed, bothconditions are not satisfied. To see that condition 1) is not satisfied, let’s consider p1(t) = 5t − t6 andp2(t) = 1− 3t2 + t6, both polynomial have degree six, but

p1(t) + p2(t) = (5t− t6) + (1− 3t2 + t6) = 1 + 5t− 3t2

has degree three, less than six. For condition 2): if we multiply any polynomial with degree greater or equalto six by 0 ∈ IR we get the zero polynomial z(t) = 0, that is not in this subset of P(t) because it does nothave a degree equal or greater than six.

2.27) In the linear space F of all real functions F = {f : IR → IR}, decide if the sets defined below form linearsubspaces or not.M1 = {f ∈ F : f(1) = 0},M2 = {f ∈ F : f(3) = 1}.

Solution. At the beginning of the solution of exercise 25), we recall the definition of linear subspace. We use it toprove that M1 = {f ∈ F : f(1) = 0} is a linear subspace of F . M1 is clearly not empty as the zero functionf(x) = 0 belongs to it. In order to verify condition 1), let’s consider two functions f1, f2 belonging to M1, forsuch functions f1(1) = f2(1) = 0, therefore for their sum we have

(f1 + f2)(1) = f1(1) + f2(1) = 0 + 0 = 0

thus f1 + f2 belongs to M1. To verify condition 2), we consider f ∈ F and α ∈ IR, then for α · f we have

(α · f)(1) = α · f(1) = α · 0 = 0

thus αf belongs to M1.

M2 = {f ∈ F : f(3) = 1} is not a linear subspace of F . Indeed, both conditions are not satisfied. Consideringtwo functions f1, f2 belonging to M2, for such functions f1(3) = f2(3) = 1, therefore for their sum we have

(f1 + f2)(3) = f1(3) + f2(3) = 1 + 1 = 2 6= 1

thus f1 + f2 does not belong to M2. This is sufficient to say that M2 is not a linear subspace. Anyway, wemay also observe that also condition 2) is not satisfied: considering f ∈ M2 and 0 ∈ IR then 0 · f = 0 is theconstant zero function, but this does not belong to M2.

2.29) Given the following subsets of P3(t) determine if they are linear subspaces of P3(t) or not. In affirmativecase, find the dimension and a basis for each subspace:L1 = {p ∈ P3 : p(0) = 0}L2 = {p ∈ P3 : p(−t) = p(t)}.

Solution. For the definition of linear subspace we refer to the beginning of the solution of exercise 25).Let’s prove that L1 = {p ∈ P3 : p(0) = 0} is a linear subspace of P3. L1 is not empty as it contains forexample the zero polynomial. To verify condition 1), let’s consider two polynomials p1, p2 belonging to L1,for such polynomials of degree at most three, p1(0) = p2(0) = 0, therefore for their sum we have

(p1 + p2)(0) = p1(0) + p2(0) = 0 + 0 = 0

16

thus p1 + p2 belongs to L1. To verify condition 2), we consider p ∈ L1 and α ∈ IR, then for α · p we have

(α · p)(0) = α · p(0) = α · 0 = 0

thus αp belongs to L1. Now we need to find a basis for L1. Taken a general p(t) ∈ P3, p(t) = a+ bt+ ct2 +dt3

belongs to L1 if and only ifp(0) = a+ b0 + c0 + d0 = a = 0

thus any polynomial in L1 is of the form p(t) = bt+ ct2 +dt3 = b(t) + c(t2) +d(t3) and it can be generated bythe linearly independent polynomials t, t2, t3. We conclude that B = {t, t2, t3} is a basis of L1 and dimL1 = 3.

Let’s now prove that L2 = {p ∈ P3 : p(−t) = p(t)} is also a linear subspace of P3. L2 is not empty asit contains for example the zero polynomial. To verify condition 1), let’s consider two polynomials p1, p2belonging to L2, for such polynomials of degree at most three, p1(−t) = p1(t) and p2(−t) = p2(t), thereforefor their sum we have

(p1 + p2)(−t) = p1(−t) + p2(−t) = p1(t) + p2(t) = (p1 + p2)(t)

thus p1 + p2 belongs to L2. To verify condition 2), we consider p ∈ L2 and α ∈ IR, then for α · p we have

(α · p)(−t) = α · p(−t) = α · p(t) = (α· )(t)

thus αp belongs to L2. Now we need to find a basis for L2. Taken a general p(t) ∈ P3, p(t) = a+ bt+ ct2 +dt3

belongs to L2 if and only if p(−t) = p(t), that means

a+ b(−t) + c(−t)2 + d(−t)3 = a+ bt+ ct2 + dt3

ora− bt+ ct2 − dt3 = a+ bt+ ct2 + dt3.

This condition implies that b = d = 0. Thus any polynomial in L2 is of the form p(t) = a+ ct2 = a(1) + c(t2)and it can be generated by the linearly independent polynomials 1, t2. We conclude that B = {1, t2} is abasis of L2 and dimL2 = 2.

2.31) a) Prove that if M and N are subspaces of a vector space L then M ∩N is a subspace of L.b) Given M = span{(1, 2, 1, 0), (−1, 1, 1, 1)} and N = span{(1, 2, 1,−2), (2, 1, 0, 1)} (subspaces of IR4), findthe dimension and a basis for M ∩N .c) Consider the space U = span{M ∪N}, find the dimension and a basis of U .

Solution. For the definition of linear subspace we refer to the beginning of the solution of exercise 25).a) We prove that the intersection of two subspaces of a linear space is again a subspace. Being M and Nsubspaces of a linear space L, they both contain the zero vector of L, thus the intersection M∩N also containsthe zero vector of L and it is non-empty. Suppose now that we have two vectors u,v ∈ M ∩N , this impliesthat u,v ∈ M and u,v ∈ N , thus u + v ∈ M because M is a subspace, and u + v ∈ N because N is asubspace. This implies that u+v ∈M ∩N and condition 1) from the definition of linear subspace is satisfied.To prove that also condition 2) is satisfied, let’s consider u ∈M ∩N and α ∈ IR. Because u ∈M ∩N , u ∈Mand u ∈ N , thus α · u ∈ M and α · u ∈ N because both M and N are linear subspace and they satisfiedcondition 2). This implies that α · u ∈M ∩N .

b) Given M = span{(1, 2, 1, 0), (−1, 1, 1, 1)} and N = span{(1, 2, 1,−2), (2, 1, 0, 1)}, M ∩ N is composed byall vectors in IR4 that can be written at the same time as linear combination of (1, 2, 1, 0), (−1, 1, 1, 1) and aslinear combination of (1, 2, 1,−2), (2, 1, 0, 1). Given a vector v ∈M ∩N , there must be real numbers a, b, c, dsuch that

v = a(1, 2, 1, 0) + b(−1, 1, 1, 1) = c(1, 2, 1,−2) + d(2, 1, 0, 1).

This condition is equivalent to

(a− b, 2a+ b, a+ b, b) = (c+ 2d, 2c+ d, c,−2c+ d)

and leads to the system

a− b = c+ 2d

2a+ b = 2c+ d

a+ b = c

b = −2c+ d.

17

We rewrite the system as

a− b− c− 2d = 0

2a+ b− 2c− d = 0

a+ b− c = 0

b+ 2c− d = 0

and perform a Gaussian reduction. We start with operation R2 − 2R1 and R3 −R1 to get

a− b− c− 2d = 03b + 3d = 02b + 2d = 0b+ 2c− d = 0

R2/3R3/2−R2/3

∼R4 −R2/3

a− b− c− 2d = 0b + d = 0

0 = 02c− 2d = 0

∼a− b− c− 2d = 0

b + d = 0c− d = 0.

We now see that, taking d = α as a free variable, all possible solutions of the system can be written as a = 2α,b = −α, c = α, d = α, with α ∈ IR. Thus the vector v ∈M ∩N is such that

v = a(1, 2, 1, 0) + b(−1, 1, 1, 1) = 2α(1, 2, 1, 0)− α(−1, 1, 1, 1) = α(3, 3, 1,−1)

or equivalently

v = c(1, 2, 1,−2) + d(2, 1, 0, 1) = α(1, 2, 1,−2) + α(2, 1, 0, 1) = α(3, 3, 1,−1).

We conclude that M ∩N = span{(3, 3, 1,−1)}, and dimM ∩N = 1.

c) To find the dimension and a basis of U = span{M ∪ N}, we need to select a set of linearly independentvectors that generate the set

U = span{M ∪N} = span{(1, 2, 1, 0), (−1, 1, 1, 1), (1, 2, 1,−2), (2, 1, 0, 1)}

Let us use the symbol ∼ to denote the coincidence of the corresponding spans. We have

(1, 2, 1, 0)(−1, 1, 1, 1)(1, 2, 1,−2)(2, 1, 0, 1)

R2 +R1

∼R3 −R1

R4 − 2R1

(1, 2, 1, 0)(0, 3, 2, 1)

(0, 0, 0,−2)(0,−3,−2, 1)

R3/− 2∼

R4 +R2

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

Considering that the last two vectors are obviously dependent, we conclude that the first three span the givenspace:

U = span{M ∪N} = span{(1, 2, 1, 0), (0, 3, 2, 1), (0, 0, 0, 1)}

B = {(1, 2, 1, 0), (0, 3, 2, 1), (0, 0, 0, 1)} is a basis of U , and dimU = 3.

2.33) Consider M = {p(t) ∈ P3(t) : p(t) is divisible by (t − 1)}. Prove that M a linear subspace of P3, find abasis and dimension of M .

Solution. For the definition of linear subspace we refer to the beginning of the solution of exercise 25).The divisibility by a given polynomial means that the remainder after division is zero. Every polynomialp(t) ∈ M is either the zero polynomial (0 ∈ M 6= ∅) or can be written in the form p(t) = (t − 1)q(t), whereq(t) is a polynomial with degree at most two. Given two polynomials p1, p2 ∈M , we have p1(t) = (t− 1)q1(t)and p2(t) = (t− 1)q2(t), where q1, q2 ∈ P2(t), thus

p1(t) + p2(t) = (t− 1)q1(t) + (t− 1)q2(t) = (t− 1)[q1(t) + q2(t)].

This proves that p1(t) + p2(t) is divisible by (t− 1). Analogously, for every α ∈ IR and p(t) ∈M , there existsa q ∈ P2(t) so that p(t) = (t− 1)q(t), and we have

α · p(t) = α · (t− 1)q(t) = (t− 1)[α · q(t)].

This proves that α · p(t) is divisible by (t− 1). Thus M is a linear subspace of P3(t).Suppose now that p(t) = at3 + bt2 + ct + d is a polynomial in M , the fact that p(t) is divisible by (t − 1)implies that the remainder in the following division is zero:

18

(at3 + bt2 + ct+ d) : (t− 1) = at2 + (a+ b)t+ (a+ b+ c)

−(at3 − at2)

(a+ b)t2 + t+ d

−((a+ b)t2 − (a+ b)t)

(a+ b+ c)t+ d

−((a+ b+ c)t− (a+ b+ c))

(a+ b+ c+ d) = 0

We see that the coefficients of p(t) ∈M must be such that (a+ b+ c+ d) = 0. This means that a polynomialp(t) belongs to M if and only if there are coefficients a, b, c such that

p(t) = at3 + bt2 + ct− a− b− c = a(t3 − 1) + b(t2 − 1) + c(t− 1).

This proves that the set B = {t3 − 1, t2 − 1, t− 1} spans M and, because B is made of linearly independentpolynomials, B is a basis of M , and dimM = 3.

2.35) Consider Q = {p(x) ∈ P4(x) : p(x) = x4 · p( 1x ) for any x ∈ IR, x 6= 0}. Prove that Q a linear subspace of P4,

find a basis and dimension of Q.

Solution. For the definition of linear subspace we refer to the beginning of the solution of exercise 25). To see that Q isa non-empty set we may observe that the zero polynomial obviously satisfied the condition in the definitionof Q, but also for example the polynomial p(x) = x2 is such that p(x) = x2 = x4( 1

x2 ) = x4p( 1x ), thus also

x2 ∈ Q. In general, given p1, p2 ∈ Q, for every x ∈ IR, x 6= 0, we have p1(x) = x4p1( 1x ) and p2(x) = x4p2( 1

x ),thus:

p1(x) + p2(x) = x4p1( 1x ) + x4p2( 1

x ) = x4[p1( 1x ) + p2( 1

x )] = x4(p1 + p2)( 1x )

that means that p1 + p2 ∈ Q. Moreover for any α ∈ IR and p ∈ Q, we have

α · p(x) = α · x4p( 1

x) = x4(α · p( 1

x))

thus α · p ∈ Q, and this conclude the proof that Q is a linear subspace of P4(x).

Suppose now that p(x) = ax4 + bx3 + cx2 + dx + e is a polynomial in Q, the fact that p(x) = x4 · p( 1x ) is

equivalent to

ax4 + bx3 + cx2 + dx+ e = x4(a

1

x4+ b

1

x3+ c

1

x2+ d

1

x+ e)

= a+ bx+ cx2 + dx3 + ex4.

This condition is satisfied only if a = e and b = d, thus p(x) ∈ Q ⊂ P4 if and only if there exist coefficientsa, b, c ∈ IR such that

p(x) = ax4 + bx3 + cx2 + bx+ a = a(x4 + 1) + b(x3 + x) + c(x2).

This proves that the set B = {x4 + 1, x3 + x, x2} is a basis of Q, and dimQ = 3.

19

Chapter 3Matrices

3.1) Given A =

1 0 2 −10 2 1 11 −1 1 01 2 1 −1

, evaluate: AT , A2, rankA.

3.2) Given B =

2 0 0 −11 1 1 12 −1 2 0

, evaluate: BT , B ·BT , rankB.

3.3) a) Given a matrix A ∈Mn,n, prove that the set LA = {B ∈Mn,n : BA = AB} (the set of all matrices thatcommute with A) is a linear subspace of Mn,n.

b) For A =

(3 12 0

)find basis and dimension of LA.

3.4) Find all matrices B such that B ·A = A ·B where A =

(4 13 2

).

3.5) Solve the equation A + 3X = BC, where A =

(1 5−3 2

), B =

(1 23 4

)and C =

(−2 60 4

).

3.6) Solve the following system:

5X + 2Y = 16A−B

8X− 5Y = A− 18B

where A =

(1 2 32 −2 1

), B =

(0 1 −21 2 1

),X,Y ∈M2,3 .

3.7) Solve AX = B, where A =

(2 −1 −1−2 −1 1

), B =

(1 00 1

).

3.8) Find (all possible) X ∈M2,2 such that AX = X where A =

(2 4−3 −11

).

3.9) Find λ ∈ IR so that A =

1 −2 31 λ −11 1 −λ

has rank equal to two. For λ = 1, verify that rankA = rankAT .

3.10) Find all λ ∈ IR such that A =

1− λ 0 12 2 + λ 00 2 1

has rankA = 2.

3.11) Find the determinant of A a) reducing it to triangular form, b) using the theorem on expansion of determi-

nants: A =

1 0 −2 30 1 −3 43 2 0 −54 3 −5 0

. Then compute detA−1 and detA2.


nants: A =

3 2 42 0 41 −1 1

. Then compute detA−1 and detA−2.

3.13) Use the determinant to find the value of rankA for any possible α ∈ IR, where A =

16 0 4α0 7 4α−α 0 −1

.

20

3.14) Use the determinant of a matrix to find the values of λ ∈ IR for which the vectors v1 = (2, 5, 1), v2 = (4, 10, 8),v3 = (6, 10, λ) are linearly dependent.

3.15) For each of the following matrices find the inverse first using Gauss elimination method, then using the formulaA−1 = 1

detA adjA, where detA is the determinant of A and adjA is the classical adjoint matrix.

A =

1 1 2 −10 1 2 −10 0 1 20 0 1 0

, B =

1 2 3−1 1 22 1 −1

.



A =

1 0 11 −1 12 1 1

, P =

1 1 −11 −4 21 −1 1

.

3.17) a) Show that if A ∈M3,3 is upper triangular, i.e. A =

a11 a12 a130 a22 a230 0 a33

then A−1 is upper triangular, i.e.

A−1 =

b11 b12 b130 b22 b230 0 b33

. b) Find inverse of: B =

1 0 03 1 00 3 1

.

3.18) Given the matrix A =

1 a a2

1 b b2

1 c c2

, verify that detA = (c− b)(b− a)(c− a) and find A−1.

3.19) Given A =

(a bc d

). Find adjA. When does A = adjA? Show that adj(adjA) = A. Find A−1.

3.20) Given A =

1 1 01 1 10 2 1

. Find detA, adjA and A−1.


1 −2 10 2 0−1 0 1

and the vector b =

301

, find the inverse of the matrix A to solve

the system Ax = b.

3.22) Using an inverse matrix solve the matrix equation AX = B where

A =

1 −1 0−1 1 −11 0 1

and B =

1 2 32 3 43 4 5

.

3.23) Determine for what value of a ∈ IR the following matrix is regular and for those value find the inverse matrix:

A =

a+ 1 1 −a1 a 10 −1 a

.

3.24) Given A =

1 2−3 13 −1

, B =

1 03 24 5

, find (BTA)−1.

3.25) Use an inverse matrix to solve the matrix equation XA = (X + I)B where I is the identity matrix,

A =

6 9 14 2 4−1 0 −1

and B =

5 7 12 0 3−1 1 2

.

21

3.26) Use an inverse matrix to solve the matrix equation AX = B−X where

A =

2 2 14 2 0−1 0 −1

and B =

5 7 12 0 3−1 1 2

.

22

Solutions of odd number problems - Matrices

3.1) Given A =

1 0 2 −10 2 1 11 −1 1 01 2 1 −1

, evaluate: AT , A2, rankA.

Solution. By definition the transpose of a matrix is a new matrix whose rows are the columns of the original. In ourcase, A is a 4× 4 matrix, i.e. A ∈M4,4, thus AT ∈M4,4 and

AT =

1 0 1 10 2 −1 22 1 1 1−1 1 0 −1

.

In order to evaluate A2 = A ·A we need to perform a product of matrices. We recall that, in general, givenmatrices A = (aip) ∈ Mm,p, and B = (bpj) ∈ Mp,n, the product A ·B = C = (cij) is a matrix in the spaceMm,n, and

cij =

p∑k=1

aikbkj for 1 ≤ i ≤ m, 1 ≤ j ≤ n.

This means that, in the ij position, the product matrix C has the result of the ”scalar product” of the i-throw of A with the j-th column of B seen as a vectors of IRp.In our case n = m = p = 4, thus A ·A is a matrix in the space M4,4. Now we evaluate:

A2 =

1 0 1 10 2 −1 22 1 1 1−1 1 0 −1

·

1 0 1 10 2 −1 22 1 1 1−1 1 0 −1

=

=

1 + 0 + 2− 1 0 + 0 + 1 + 1 1 + 0 + 1 + 0 1 + 0 + 1− 10 + 0− 2− 2 0 + 4− 1 + 2 0− 2− 1 + 0 0 + 4− 1− 22 + 0 + 2− 1 0 + 2 + 1 + 1 2− 1 + 1 + 0 2 + 2 + 1− 1−1 + 0 + 0 + 1 0 + 2 + 0− 1 −1− 1 + 0 + 0 −1 + 2 + 0 + 1

=

2 2 2 10 5 −3 13 4 2 40 1 −2 2

By definition, the rank of a m × n matrix is the dimension of the linear space spanned by the m rows ofthe matrix, seen as vectors in IRn. Equivalently, we may say that rankA is the maximal number of linearlyindependent rows of A. Due to the fact (Theorem) that a Gaussian elimination does not change the rank,we use a Gaussian reduction:

A =

1 0 2 −10 2 1 11 −1 1 01 2 1 −1

R3 −R1

∼R4 −R1

1 0 2 −10 2 1 10 −1 −1 10 2 −1 0

2R3 +R2

∼R4 −R2

1 0 2 −10 2 1 10 0 −1 30 0 −2 −1

R4 − 2R3

∼

1 0 2 −10 2 1 10 0 −1 30 0 0 −7

.

This implies that rankA = 4.

3.3) a) Given a matrix A ∈Mn,n, prove that the set LA = {B ∈Mn,n : BA = AB} (the set of all matrices thatcommute with A) is a linear subspace of Mn,n.

b) For A =

(3 12 0

)find basis and dimension of LA.

Solution. a) For the definition of linear subspace we refer to the beginning of the solution of exercise 25).Suppose that a matrix A ∈ Mn,n is given, then the set LA = {B ∈ Mn,n : BA = AB} contains the zeromatrix because OA = AO = O, thus LA 6= ∅. Suppose now that two matrices B1,B2 ∈ LA are given, thenB1A = AB1 and B2A = AB2, thus for B1 + B2 we have

(B1 + B2)A = B1A + B2A = AB1 + AB2 = A(B1 + B2)

23

due to the (”left” and ”right”) distributivity property of the sum with respect to the product of matrices.This proves that B1 + B2 ∈ LA. In a similar way, using properties of matrix product, for any B ∈ LA andα ∈ IR we have

(α ·B)A = α ·B) = α(AB) = A(αB)

thus α ·B ∈ LA, and this concludes the proof that LA is a subspace of Mn,n.

b) Given A =

(3 12 0

), the subspace LA is formed by all matrices B =

(a bc d

), with a, b, c, d ∈ IR, such

that (a bc d

)(3 12 0

)=

(3 12 0

)(a bc d

).

Evaluating the indicated products, we have(3a+ b a3c+ 2d c

)=

(3a+ c 3b+ d

2a 2b

).

This matrix equation is equivalent to the following system

3a+ 2b = 3a+ c

a = 3b+ d

3c+ 2d = 2a

c = 2b.

Eliminating the first equation, that is obviously equivalent to the last, we rewrite the system and use aGaussian reduction:

a− 3b− d = 0−2a+ 3c+ 2d = 0−2b+ c = 0

2R1 +R2

∼

a− 3b− d = 0−6b+ 3c = 0−2b+ c = 0

3R3 −R2

∼

a− 3b− d = 0−2b+ c = 0

0 = 0.

We choose as free variables b and d, and write all solutions of the given system as a = 3α + β, b = α, d = βand c = 2α, where α, β ∈ IR. Then, any matrix B ∈ LA can be written as(

3α+ β α2α β

)= α

(3 12 0

)+ β

(1 00 1

)

thus LA = span{( 3 1

2 0

),

(1 00 1

)}, and dimLA = 2.

3.5) Solve the equation A + 3X = BC, where A =

(1 5−3 2

), B =

(1 23 4

)and C =

(−2 60 4

).

Solution. Using properties of matrix operations, we get 3X = BC −A and thus X = 13

[BC −A

]. We evaluate the

indicated operations:

X =1

3

[BC−A

]=

1

3

[(1 23 4

)·(−2 60 4

)−(

1 5−3 2

)]=

1

3

[(−2 + 0 6 + 8−6 + 0 18 + 16

)−(

1 5−3 2

)]=

=1

3

[(−2 14−6 34

)−(

1 5−3 2

)]=

1

3

(−3 9−3 32

)=

(−1 3−1 8

).

3.7) Solve AX = B, where A =

(2 −1 −1−2 −1 1

), B =

(1 00 1

).

Solution. We observe that A ∈M2,3 and B ∈M2,2. From this we deduce that X must be a matrix in the spaceM3,2.

Suppose that X =

a bc de f

, we are looking for a, b, c, d, e, f ∈ IR so that

(2 −1 −1−2 −1 1

)·

a bc de f

=

(1 00 1

).

24

After performing the product on the right hand side, we get(2a− c− e 2b− d− f−2a− c+ e −2b− d+ f

)=

(1 00 1

).

This is equivalent to the system of linear equations:

2a − c − e = 1

2b − d − f = 0

−2a − c + e = 0

−2b − d + f = 1.

With operations R2 ↔ R3, R3 +R1, R4 +R2, we start a Gaussian reduction and get the equivalent system:

2a − c − e = 1

−2c = 1

2b − d − f = 0

−2d = 1.

The system has infinitely many solutions: c = d = − 12 , e = 2a− 1

2 , f = 2b+ 12 , for any possible a, b ∈ IR. We

conclude that for any possible a, b ∈ IR a matrix X of the form

X =

a b− 1

2 − 12

2a− 12 2b+ 1

2

solves the given equation.

3.9) Find λ ∈ IR so that A =

1 −2 31 λ −11 1 −λ

has rank equal to two. For λ = 1, verify that rankA = rankAT .

Solution. We use the Theorem that guarantees that Gaussian elimination does not change the rank of a matrix, andthe usual symbol ∼ to denote the coincidence of rank, then we have 1 −2 3

1 λ −11 1 −λ

R2 −R1

∼R3 −R1

1 −2 30 λ+ 2 −40 3 −λ− 3

If λ = −2, the last matrix has rank equal to three. Supposing that λ 6= −2, we continue the Gaussianelimination with operation 3R2 − (λ+ 2)R3 and we get the matrix 1 −2 3

0 λ+ 2 −40 0 λ2 + 5λ− 6

.

Since λ2 + 5λ− 6 = 0 for λ = −6 and λ = 1, we conclude that if λ 6= 1, λ 6= −6, then rankA = 3, and if λ = 1or λ = −6, then rankA = 2.

For λ = 1 the given matrix is A =

1 −2 31 1 −11 1 −1

. Its rank is two (as we have already evaluated) obviously

due to the fact that the second and third row are equal, while the first and second are independent. To findrankAT , we perform a Gaussian reduction on AT :

AT =

1 1 1−2 1 13 −1 −1

R2 + 2R1

∼R3 − 3R1

1 1 10 3 30 −4 −4

R2/3∼

R3/4 +R2/3

1 1 10 1 10 0 0

.

Thus rankAT = 2 = rankA.


nants: A =

1 0 −2 30 1 −3 43 2 0 −54 3 −5 0

. Then compute detA−1 and detA2.

25

Solution. a) We recall that the determinant of a triangular matrix is equal to the product of the components in thediagonal. Moreover, if a matrix B is obtained from a matrix A using one of the following operations:i) exchanging two rows,ii) multiplying a row by a nonzero scalar λ ∈ IR,iii) substituting a row with the sum of that row with a linear combination of the others,then the new matrix is such thati) detB = − detA, ii) detB = λ detA, iii) detB = detA.Keeping this in mind, we proceed with the elimination method:

det

1 0 −2 30 1 −3 43 2 0 −54 3 −5 0

R3 − 3R1

=R4 − 4R1

det

1 0 −2 30 1 −3 40 2 6 −140 3 3 −12

R3/2=

R4/36 det

1 0 −2 30 1 −3 40 1 3 −70 1 1 −4

R3 −R2

=R4 −R2

= 6 det

1 0 −2 30 1 −3 40 0 6 −110 0 4 −8

R4/4↔ R3

=−24 det

1 0 −2 30 1 −3 40 0 1 −20 0 6 −11

R4 − 6R3

=−24 det

1 0 −2 30 1 −3 40 0 1 −20 0 0 1

from which we deduce that detA = −24.

b) We now use the formula detA =n∑j=1

aijAij , where Aij is the ij-th cofactor of A, and evaluate the

determinant of A along the second row, fixing i = 2:

detA = 0 + 1(−1)2+2 det

1 −2 33 0 −54 −5 0

+ (−3)(−1)2+3 det

1 0 33 2 −54 3 0

+ 4(−1)2+4 det

1 0 −23 2 04 3 −5

.

To evaluate the determinant of the 3× 3 matrices indicated, we now decide to use the formula (Sarruss’rule):

det

a b cd e fr s t

= aet+ bfr + cds− rec− sfa− dbt.

Therefore:

detA = (0+40−45−0−25−0)+3(0+0+27−24+15−0)+4(−10+0−18+16−0−0) = −30+54−48 = −24.

Knowing that in general detA−1 = 1detA and detA2 = det(A ·A) = (detA)2, we conclude that in our case

detA−1 = 124 and detA2 = 242.

3.13) Use the determinant to find the value of rankA for any possible α ∈ IR, where A =

16 0 4α0 7 4α−α 0 −1

.

Solution. For a square matrix A ∈ Mn,n we have detA 6= 0 ⇔ rankA = n ⇔ the rows are linearly independent asvectors in IRn. We start the discussion on the possible value of rankA for the given matrix evaluating itsdeterminant (using Sarruss’rule):

detA = det

16 0 4α0 7 4α−α 0 −1

= −16 · 7 + 7 · 4α2 = 7 · 4(α2 − 4).

Thus, if α 6= 2, α 6= −2, then A is regular and rankA = 3. For the values of α that make the determinantequal to zero, to determine the exact value of rankA (it could be 2 or 1), in general we substitute the value ofα into the matrix and perform a Gauss reduction. But in our case we may observe that the first and secondrow of A are linearly independent for any possible value of α, thus, if α = 2 or α = −2, then rankA = 2.



26

a) A =

1 1 2 −10 1 2 −10 0 1 20 0 1 0

, b) B =

1 2 3−1 1 22 1 −1

.

Solution. a) We will write A and I (the identity matrix) one beside the other and carry on a Gaussian elimination onA, simultaneously performing the same operations on I. The final stage is reached when A is reduced to theidentity matrix I, then I is reduced to A−1.

A =

1 1 2 −10 1 2 −10 0 1 20 0 1 0

1 0 0 00 1 0 00 0 1 00 0 0 1

R4 −R3

1 1 2 −10 1 2 −10 0 1 20 0 0 −2

1 0 0 00 1 0 00 0 1 00 0 −1 1

R1 −R2

R3 +R4

1 0 0 00 1 2 −10 0 1 00 0 0 −2

1 −1 0 00 1 0 00 0 0 10 0 −1 1

R2 − 2R3

R3 +R4/− 2

1 0 0 00 1 0 −10 0 1 00 0 0 1

1 −1 0 00 1 0 −20 0 0 10 0 1/2 −1/2

R2 +R4

1 0 0 00 1 0 00 0 1 00 0 0 1

1 −1 0 00 1 1/2 −5/20 0 0 10 0 1/2 −1/2

= A−1

We now evaluate the inverse of A using the formula A−1 = 1detA adjA. We start with calculating the

determinant of A along the fourth row that contains only one nonzero entry:

detA = 1(−1)3+4 det

1 1 −10 1 −10 0 2

= −2.

The matrix adjA is the adjugate, or classical adjoint of the square matrix A and it is defined as the transposeof its cofactor matrix. We will thus evaluate all cofactors of the given matrix.

A11 = (−1)1+1 · det

1 2 −10 1 20 1 0

= −2

A12 = (−1)1+2 · det

0 2 −10 1 20 1 0

= 0

A13 = (−1)1+3 · det

0 1 −10 0 20 0 0

= 0

A14 = (−1)1+4 · det

0 1 20 0 10 0 1

= 0

A21 = (−1)2+1 · det

1 2 −10 1 20 1 0

= 2

A22 = (−1)2+2 · det

1 2 −10 1 20 1 0

= −2

27

A23 = (−1)2+3 · det

1 1 −10 0 20 0 0

= 0

A24 = (−1)2+4 · det

1 1 20 0 10 0 1

= 0

A31 = (−1)3+1 · det

1 2 −11 2 −10 1 0

= 0

A32 = (−1)3+2 · det

1 2 −10 2 −10 1 0

= −1

A33 = (−1)3+3 · det

1 1 −10 1 −10 0 0

= 0

A34 = (−1)3+4 · det

1 1 20 1 20 0 1

= −1

A41 = (−1)4+1 · det

1 2 −11 2 −10 1 2

= 0

A42 = (−1)4+2 · det

1 2 −10 2 −10 1 2

= 5

A43 = (−1)4+3 · det

1 1 −10 1 −10 0 2

= −2

A44 = (−1)4+4 · det

1 1 20 1 20 0 1

= 1

Now we can write A−1 = − 12

−2 0 0 02 −2 0 00 −1 0 −10 5 −2 1

T

=

1 −1 0 00 1 1/2 −5/20 0 0 10 0 1/2 −1/2

.

b) For the given matrix B we follow the same approach as for the matrix A, we start with finding B−1 witha Gauss reduction:

B =

1 2 3−1 1 22 1 −1

1 0 00 1 00 0 1

R2 +R1

R3 − 2R1 1 2 30 3 50 −3 −7

1 0 01 1 0−2 0 1

R3 +R2

R2/3 1 2 30 1 5/30 0 −2

1 0 01/3 1/3 0−1 1 1

R3/− 2R2 + 5R3/6R1 + 3R3/2 1 2 0

0 1 00 0 1

−1/2 3/2 3/2−1/2 7/6 5/61/2 −1/2 −1/2

R1 − 2R2

1 0 00 1 00 0 1

1/2 −5/6 −1/6−1/2 7/6 5/61/2 −1/2 −1/2

= B−1.

28

We now evaluate the inverse of B using the formula B−1 = 1detB adjB. We start by calculating the determi-

nant of B using Sarruss’rule.

detB = det

1 2 3−1 1 22 1 −1

= −1 + 8− 3− 6− 2− 2 = −6.

We now proceed with the evaluation of adjB, calculating all cofactors of B:

B11 = (−1)1+1 · det

(1 21 −1

)= −3

B12 = (−1)1+2 · det

(−1 22 −1

)= 3

B13 = (−1)1+3 · det

(−1 12 1

)= −3

B21 = (−1)2+1 · det

(2 31 −1

)= 5

B22 = (−1)2+2 · det

(1 32 −1

)= −7

B23 = (−1)2+3 · det

(1 22 1

)= 3

B31 = (−1)3+1 · det

(2 31 2

)= 1

B32 = (−1)3+2 · det

(1 3−1 2

)= −5

B33 = (−1)3+3 · det

(1 2−1 1

)= 3

Now we can write B−1 = − 16

−3 3 −35 −7 31 −5 3

T

=

1/2 −5/6 −1/6−1/2 7/6 5/61/2 −1/2 −1/2

.

3.17) a) Show that if A ∈M3,3 is upper triangular, i.e. A =

a11 a12 a130 a22 a230 0 a33

then A−1 is upper triangular, i.e.

A−1 =

b11 b12 b130 b22 b230 0 b33

. b) Find inverse of: B =

1 0 03 1 02 3 1

.

Solution. a) Given a matrix that is upper (or lower) triangular, if ij, i 6= j is a position where the given triangularmatrix has a non-zero entry, then in the corresponding ij-th cofactor we get to evaluate the determinant of a

matrix that has a zero row or a zero column. For example, for the given matrix A =

a11 a12 a130 a22 a230 0 a33

, we

have

A12 = (−1)1+2 det

(0 a230 a33

)= 0, A13 = (−1)1+3 det

(0 a220 0

)= 0

and A23 = (−1)2+3

(a11 a120 0

)= 0.

Since the inverse matrix is a multiple of the adjoint matrix, and the latest is the transpose of the cofactormatrix, we get that the inverse matrix of an upper or lower triangular matrix is of the same form.

b) The given matrix B is lower triangular. From the above observation, we have B21 = B31 = B32 = 0. Wewill evaluate the remaining cofactors:

B11 = (−1)1+1 det

(1 03 1

)= 1

29

B12 = (−1)1+2 det

(3 02 1

)= −3

B13 = (−1)1+3 det

(3 12 3

)= 7

B22 = (−1)2+2 det

(1 02 1

)= 1

B23 = (−1)2+3 det

(1 02 3

)= −3

B33 = (−1)3+3 det

(1 03 1

)= 1.

To evaluate the determinant of a triangular matrix like B, it is enough to take the product of the diagonalelements, thus detB = 1. Now we use the formula B−1 = 1

detB adjB and write

B−1 = 1 ·

1 −3 70 1 −30 0 1

T

=

1 0 0−3 1 07 −3 1

.

3.19) Given A =

(a bc d

). Find adjA. When does A = adjA? Show that adj(adjA) = A. Find A−1.

Solution. The evaluation of the cofactor matrix of a 2× 2 matrix is very simple. Given A =

(a bc d

):

A11 = (−1)1+1d = d, A12 = (−1)1+2c = −c, A21 = (−1)2+1b = −b, A22 = (−1)2+2a = a.

Thus adjA =

(d −c−b a

)T=

(d −b−c a

). We deduce that A = adjA if and only if a = d and b = c = 0,

or, in other words, that A = adjA if and only if A ∈ span

{(1 00 1

)}.

We observe that adjA is a matrix that with respect to A has exchanged entries in the diagonal and opposite-sign entries in the remaining two positions. It is clear that repeating this pattern in finding adj adjA willlead to the original matrix A.To find the inverse of A we use the formula A−1 = 1

detA adjA, and since detA = ab− cd, we get

A−1 =1

ab− cd

(d −b−c a

).


1 −2 10 2 0−1 0 1

and the vector b =

301

, find the inverse of the matrix A to solve

the system Ax = b.

Solution. First of all we verify that A is invertible, we know that a matrix is invertible if and only if is regular if and onlyif has a non-zero determinant. Since detA = 2 + 0 + 0 + 2− 0− 0 = 4, A is invertible. Thus we may solve thegiven equation multiplying both sides of the equality by A−1 on the left, so that we get A−1 ·Ax = A−1 ·b,from which x = A−1 · b. We proceed evaluating the cofactor matrix:

A11 = (−1)1+1 det

(2 00 1

)= 2

A12 = (−1)1+2 det

(0 0−1 1

)= 0

A13 = (−1)1+3 det

(0 2−1 0

)= 2

A21 = (−1)2+1 det

(−2 10 1

)= 2

30

A22 = (−1)2+2 det

(1 1−1 1

)= 2

A23 = (−1)2+3 det

(1 −2−1 0

)= 2

A31 = (−1)3+1 det

(−2 12 0

)= −2

A32 = (−1)3+2 det

(1 10 0

)= 0

A33 = (−1)3+3 det

(1 −20 2

)= 2.

We now evaluate the inverse of A using the formula A−1 = 1detA adjA and we get

A−1 =1

4

2 0 22 2 2−2 0 2

T

=1

2

1 1 −10 1 01 1 1

.

To conclude, we evaluate the matrix product and get the result of the given equation:

x = A−1 · b =1

2

1 1 −10 1 01 1 1

· 3

01

=1

2

3 + 0− 10 + 0 + 03 + 0 + 1

=1

2

204

=

102

.

3.23) Determine for what value of a ∈ IR the following matrix is regular and for those value find the inverse matrix:

A =

a+ 1 1 −a1 a 10 −1 a

.

Solution. A square matrix is regular if and only if its determinant is non-zero, thus we evaluate the determinant of thegiven matrix:

detA = det

a+ 1 1 −a1 a 10 −1 a

= a2(a+ 1) + a+ (a+ 1)− a = (a+ 1)(a2 + 1).

Thus, if a 6= −1 the given matrix is regular. We will find the inverse of A using the formula A−1 = 1detA adjA.

We start evaluating the cofactors of A.

A11 = (−1)2 det

(a 1−1 a

)= a2 + 1

A12 = (−1)3 det

(1 10 a

)= −a

A13 = (−1)4 det

(1 a0 −1

)= −1

A21 = (−1)3 det

(1 −a−1 a

)= 0

A22 = (−1)4 det

(a+ 1 −a

0 a

)= a(a+ 1)

A23 = (−1)5 det

(a+ 1 1

0 −1

)= a+ 1

A31 = (−1)4 det

(1 −aa 1

)= a2 + 1

A32 = (−1)5 det

(a+ 1 −a

1 1

)= −2a− 1

31

A33 = (−1)6 det

(a+ 1 1

1 a

)= a2 + a− 1.

Thus A−1 = 1(a+1)(a2+1)

a2 + 1 −a −10 a(a+ 1) a+ 1

a2 + 1 −2a− 1 a2 + a− 1

T

= 1(a+1)(a2+1)

a2 + 1 0 a2 + 1−a a(a+ 1) −2a− 1−1 a+ 1 a2 + a− 1

.

3.25) Use an inverse matrix to solve the matrix equation XA = (X + I)B where I is the identity matrix,

A =

6 9 14 2 4−1 0 −1

and B =

5 7 12 0 3−1 1 2

.

Solution. The given matrix equation can be transformed as follows:

XA = (X + I)B ⇐⇒ XA = XB + B ⇐⇒ XA−XB = B ⇐⇒ X(A−B) = B

If the matrix A − B is invertible, then the latest equation can be solved by multiplying both sides of theequality by (A−B)−1 on the right, so to get X = B(A−B)−1. Let’s evaluate A−B and its determinant,to establish if this approach can be used.

A−B =

6 9 14 2 4−1 0 −1

− 5 7 1

2 0 3−1 1 2

=

1 2 02 2 10 −1 −3

.

Since det(A−B) = −6 + 1 + 12 = 7, we proceed to find its inverse. Let’s evaluate the cofactors:

(A−B)11 = (−1)2 det

(2 1−1 −3

)= −5

(A−B)12 = (−1)3 det

(2 10 −3

)= 6

(A−B)13 = (−1)4 det

(2 20 −1

)= −2

(A−B)21 = (−1)3 det

(2 0−1 −3

)= 6

(A−B)22 = (−1)4 det

(1 00 −3

)= −3

(A−B)23 = (−1)5 det

(1 20 −1

)= 1

(A−B)31 = (−1)4 det

(2 02 1

)= 2

(A−B)32 = (−1)5 det

(1 02 1

)= −1

(A−B)33 = (−1)6 det

(1 22 2

)= −2.

Now we use the formula (A−B)−1 = 1det(A−B) adj(A−B) and write

(A−B)−1 =1

7

−5 6 −26 −3 12 −1 −2

T

=1

7

−5 6 26 −3 −1−2 1 −2

.

To conclude, we evaluate the matrix product and get the result of the given matrix equation:

X = B(A−B)−1 =

5 7 12 0 3−1 1 2

1

7

−5 6 26 −3 −1−2 1 −2

=

=1

7

−25 + 42− 2 30− 21 + 1 10− 7− 2−10− 6 12 + 3 4− 65 + 6− 4 −6− 3 + 2 −21 − 4

=1

7

15 10 1−16 15 −2

7 −7 −5

.

32

Chapter 4Systems of linear equations

4.1) Write down all solutions of the following homogeneous system

x− 2y + 3z = 0x+ 2y − z = 0.


x− 2y + 4z = 0x+ y − 2z = 0.

4.3) Write down all solutions of the following homogeneous system for any possible value of a ∈ IR

x− 2y + z + 2w = 0−y + 2z + w = 0

−2x+ y + z − aw = 0ax− y + 2z + 7w = 0.


x+ 2y + z + 2w = 02x− 3y + 2w = 03x+ 5y − z + w = 0

4x+ 2y + 6z + aw = 0.

4.5) Use Gaussian elimination to find the general solution to the system:

x+ y + w = 12x+ 2y + z + w = 1

2x+ 2y + 2w = 2.


x+ 2y + 5z = 203x− y + 2z = 7x− 5y − 8w = −33.

4.7) Use Cramer’s rule to solve the system

x+ 2y + z = 32x+ 5y − z = −43x− 2y − z = 5.


2x− 5y + 2z = 2x+ 2y − 4z = 53x− 4y − 6z = 1.

4.9) Write down all solutions of the following system as the sum of a particular solution of the nonhomogeneoussystem with the linear space of solutions of the homogeneous system

x+ 2y + 4z = 52x− y + 3z = 5−x+ 3y + z = 0.


x+ y + z = 23x− y + z = 0.

33


x− 2y + z − w = 1x+ y − z + w = 22x− y + z − w = 1.


4x+ y + z + 4w = 335x− 4y + 2z − w = 182x− 3y + z − 2w = 1x+ 2y + 3w = 16.

4.13) Determine for what value of the parameter a the following system has a unique solution, infinitely manysolutions or no solution. Write down all solutions if any.

x+ ay − 3z = 5ax− 3y + z = 10x+ 9y − 10z = a+ 3.


x+ y + az = ax+ ay + z = 1ax+ y + z = 1.


x+ y − az = 1x− 2y + 3z = 2x+ ay − z = 1.


x+ 2y + 2w = 12x+ 3y + 2z + w = 2x+ 3y − 2z + aw = 1.

4.17) Find the solutions of the following system, depending on the value of the parameter λ ∈ IR:

x+ y + λz = λx+ λy + z = 1λx+ y + z = 1.

4.18) Find the solutions to the following system, depending on the value of the parameter λ ∈ IR:

λx+ y + z = 1x+ λy + z = λx+ y + λz = λ2.

4.19) Write down all solutions of the following system for any possible value of a, b ∈ IR

−2x+ y + z = 7ax+ 2y − z = 2−3x+ y + 2z = −b.

4.20) Write down all solutions of the following system for any possible value of a, b ∈ IR

x+ y − z = b−x+ y + z = 2−2y + az = 1.

34

Solutions of odd number problems - Linear Systems


x− 2y + 3z = 0x+ 2y − z = 0.

Solution. The augmented matrix associated to the system is

(1 −2 31 2 −1

∣∣∣∣∣∣∣∣ 00)

. We apply the matrix form of the

Gaussian elimination, we start with the operation R2 −R1 obtaining

(1 −2 30 4 −4

∣∣∣∣∣∣∣∣ 00)

.

We get the matrix to the echelon form with the operation R2/4 (dividing the last row by four):(1 −2 30 1 −1

∣∣∣∣∣∣∣∣ 00).

Since the rank of the matrix is two and the number of unknowns is three, we see that the dimension of thespace of all solutions in one. It is enough to guess one solution of the system associated to the last augmentedmatrix, this system is equivalent to the given one

x− 2y + 3z = 0y − z = 0.

Starting from the last equation, putting z = 1, we get y = 1, these values substituted into the first equationgive x = −1. Thus every solution x = (x, y, z) of the system can be expressed as a multiple of the one wehave found, i.e. x ∈ span{(−1, 1, 1)}.Observation. This method is equivalent to the usual one, where we introduce parameters for the freevariables, but it is often much more convenient. Anyway, in case we prefer, after we find the echelon formof the matrix and the associated system, we may conclude as follows: choosing z as a free variable, z = α,α ∈ IR, from the second equation we get y = α, that substituted into the first equation give x = −α. Thuswe see that every solution of the given system is of the form (x, y, z) = (−α, α, α) = α(−1, 1, 1), α ∈ IR.


x− 2y + z + 2w = 0−y + 2z + w = 0

−2x+ y + z − aw = 0ax− y + 2z + 7w = 0.

Solution. The augmented matrix of the given system is:1 −2 1 20 −1 2 1−2 1 1 −aa −1 2 7

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

Considering the presence of the parameter, we prefer to avoid a Gaussian elimination. In this case it ispossible because the given system has four equations and four unknowns, thus the matrix associated to thesystem is a square one.

A =

1 −2 1 20 −1 2 1−2 1 1 −aa −1 2 7

.

An n × n system of linear equations has a unique solution if and only if the determinant of the associatedmatrix is non-zero. Evaluating the determinant of A, we will be able to select the particular values of a forwhich the given system does not have a unique solution. We evaluate the determinant along the first column:

detA = (−1)1+1(1) det

−1 2 11 1 −a−1 2 7

+(−1)3+1(−2) det

−2 1 2−1 2 1−1 2 7

+(−1)4+1(a) det

−2 1 2−1 2 11 1 −a

=

= −18− 2(−18)− a(3a− 3) = −3a2 + 3a+ 18 = −3(a2 − a− 6).

35

Thus, the determinant of A is equal to zero only for a = −2, or a = 3. For all possible values of a ∈ IR, a 6= −2and a 6= 3, the given homogeneous system has a unique solution, the trivial one (x, y, z, w) = (0, 0, 0, 0). Let’snow analyse the case when a = −2. For this value of the parameter the augmented matrix is

1 −2 1 20 −1 2 1−2 1 1 2−2 −1 2 7

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

We proceed with the matrix form of Gaussian elimination, starting with the row operations R3+2R1, R4−R3,we get

1 −2 1 20 −1 2 10 −3 3 60 −2 1 5

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

−R2

−R3/3 +R2

∼−R4 + 2R2

1 −2 1 20 1 −2 −10 0 1 −10 0 3 −3

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

R4 − 3R3

∼

1 −2 1 20 1 −2 −10 0 1 −10 0 0 0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

Since the rank of the matrix is three and the number of unknowns is four, we see that the dimension of thespace of all solutions in one. It is enough to guess one solution of the system associated to the last augmentedmatrix, this system is equivalent to the given one

x− 2y + z + 2w = 0y − 2z − w = 0

z − w = 0.

Starting from the last equation, putting w = 1, we get z = 1, these values substituted into the second equationgive y = 3, and finally from the first equation we get x = 3. Thus every solution x = (x, y, z, w) of the systemcan be expressed as a multiple of the one we have found, i.e. x ∈ span{(3, 3, 1, 1)}.Let’s now analyse the case when a = 3. For this value of the parameter the augmented matrix is

1 −2 1 20 −1 2 1−2 1 1 −33 −1 2 7

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

We proceed with the matrix form of Gaussian elimination, starting with the row operations R3+2R1, R4−3R1,we get

1 −2 1 20 −1 2 10 −3 3 10 5 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

−R2

−R3 + 3R2

∼R4 + 5R2

1 −2 1 20 1 −2 −10 0 3 20 0 9 6

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

R4 − 3R3

∼

1 −2 1 20 1 −2 −10 0 3 20 0 0 0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

Since the rank of the matrix is three and the number of unknowns is four, we see that the dimension of thespace of all solutions in one. It is enough to guess one solution of the system associated to the last augmentedmatrix, this system is equivalent to the given one

x− 2y + z + 2w = 0y − 2z − w = 0

3z + 2w = 0.

Starting from the last equation, putting w = 3, we get z = −2, these values substituted into the secondequation give y = −1, and finally from the first equation we get x = −6. Thus every solution x = (x, y, z, w)of the system can be expressed as a multiple of the one we have found, i.e. x ∈ span{(6, 1, 2,−3)}.


x+ y + w = 12x+ 2y + z + w = 1

2x+ 2y + 2w = 2.

Solution. The augmented matrix associated to the system is 1 1 0 12 2 1 12 2 0 2

∣∣∣∣∣∣∣∣∣∣∣∣112

.

36

We apply the matrix form of the Gaussian elimination, we start with the operations R2 − 2R1, R3 − R2

obtaining 1 1 0 10 0 1 −10 0 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣

1−11

R3 +R2

∼

1 1 0 10 0 1 −10 0 0 0

∣∣∣∣∣∣∣∣∣∣∣∣

1−10

.

Since the rank of the augmented matrix is equal to the rank of the matrix associated to the homogeneoussystem, the system has solutions. Since the rank is two and the number of unknowns is four, we see that thedimension of the space of all solutions of the homogeneous system associated to the given one is two. Weconsider the system associated to the last augmented matrix, this is equivalent to the given one,

x+ y + w = 1z − w = −1.

Choosing as free variables w and y, and setting w = α and y = β, with α, β ∈ IR, we get from the lastequation z = −1 +α, and from the first x = 1−α− β. Thus any solution of the given system can be writtenas:

x = (x, y, z, w) = (1− α− β, β,−1 + α, α) = (1, 0,−1, 0) + α(−1, 0, 1, 1) + β(−1, 1, 0, 0), α, β ∈ IR,

or equivalentlyx = (1, 0,−1, 0) + span{(−1, 0, 1, 1), (−1, 1, 0, 0)},

where span{(−1, 0, 1, 1), (−1, 1, 0, 0)} is the space of all solutions of the homogeneous system associated tothe given one.


x+ 2y + z = 32x+ 5y − z = −43x− 2y − z = 5.

Solution. The matrix associated to the homogeneous system is

A =

1 2 12 5 −13 −2 −1

.

In order to use Cramer’s rule, we need to verify that detA 6= 0. Indeed detA = −5−6−4−15−2+4 = −28.Thus, Cramer’s rule tells us that the system has a unique solution that can be computed as follows:

x =detC1

detA=

det

3 2 1−4 5 −15 −2 −1

−28

=−15− 10 + 8− 25− 6− 8

−28=−56

−28= 2

y =detC2

detA=

det

1 3 12 −4 −13 5 −1

−28

=4− 9 + 10 + 12 + 5 + 6

−28=

28

−28= −1

z =detC3

detA=

det

1 2 32 5 −43 −2 5

−28

=25− 24− 12− 45− 8− 20

−28=−84

−28= 3.

The unique solution of the system is (x, y, z) = (2,−1, 3).


x+ 2y + 4z = 52x− y + 3z = 5−x+ 3y + z = 0.

37


A =

1 2 42 −1 3−1 3 1

.

Evaluating detA = −1 − 6 + 24 − 4 − 9 − 4 = 0, we verify that we cannot use Cramer’s rule to solve thesystem. The augmented matrix of the system is 1 2 4

2 −1 3−1 3 1

∣∣∣∣∣∣∣∣∣∣∣∣550

.

Let’s start a matrix form of Gaussian reduction: with operations R2 − 2R1, and R3 +R1 we get: 1 2 40 −5 −50 5 5

∣∣∣∣∣∣∣∣∣∣∣∣

5−55

−R− 2/5∼

R3 +R2

1 2 40 1 10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣510

.

Since the rank of the augmented matrix is equal to the rank of the matrix associated to the homogeneoussystem, the system has solutions. Considering that the rank is two and the number of unknowns is three,we see that the dimension of the space of all solutions of the homogeneous system is one. We consider thesystem associated to the last augmented matrix, this is equivalent to the given one,

x+ 2y + 4z = 5y + z = 1.

We now may proceed in two ways, a guessing method, or introducing a parameter.Guessing method. We know that the structure of the solution set of our system is: ”a particular solution plusthe space of solutions of the homogeneous system”. Knowing that the dimension of the space of solutions ofthe homogeneous system is one, we may guess one non-trivial solution xh of the homogeneous system and aparticular solution xp of the system. Looking for xh, we set z = 1 in the last (homogeneous) equation and gety = −1. These values, substituted into the first (homogeneous) equation, give x = −2. Thus xh = (−2,−1, 1).Looking for xp, we set y = 0 in the last (this time nonhomogeneous) equation and get z = 1, these values,substituted into the first equation, give x = 1. Thus xp = (1, 0, 1). All solutions of the given system can bewritten as

x = xp + span{xh} = (1, 0, 1) + span{(−2,−1, 1)}.

Introducing a parameter. We choose z as free variable, z = α, α ∈ IR. From the last equation we thus gety = 1 − α. Substituting into the first equation, we get x = 3 − 2α. Thus any solution of the system can bewritten as

x = (3− 2α, 1− α, α) = (3, 1, 0) + α(−2,−1, 1) = (3, 1, 0) + span{(−2,−1, 1)}.

Again we recovered the structure of the solution set of the system, as y = (3, 1, 0) is a particular solution ofthe system, and span{(−2,−1, 1)} is the space of solutions of the homogeneous system.


x− 2y + z − w = 1x+ y − z + w = 22x− y + z − w = 1.


A =

1 −2 1 −11 1 −1 12 −1 1 −1

and the augmented matrix is

B =

1 −2 1 −11 1 −1 12 −1 1 −1

∣∣∣∣∣∣∣∣∣∣∣∣121

.

38

Since A is not a square matrix, we proceed with a Gaussian reduction of B. We start with operations R2−R1,R3 − 2R1, and we get 1 −2 1 −1

0 3 −2 20 3 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣

11−1

R3 −R2

∼

1 −2 1 −10 3 −2 20 0 1 −1

∣∣∣∣∣∣∣∣∣∣∣∣

11−2

.

The rank of the augmented matrix is equal to the rank of the matrix associated to the homogeneous system,thus the system has solutions. Since the rank is three and the number of unknowns is four, we see that thedimension of the space of all solutions of the homogeneous system (associated to the given one) is one. Weknow that the structure of the solution set of our system is: xp + span{xh}, where xp is a particular solutionof the system, xh is a solution of the homogeneous system, and span{xh} is the space of solutions of thehomogeneous system. We consider the system associated to the last augmented matrix, this is equivalent tothe given one,

x− 2y + z − w = 13y − 2z + 2w = 1

z − w = −2.

We now may proceed in two ways, a guessing method, or introducing a parameter.Guessing method. We will now guess xh, a non trivial solution of the homogeneous system. Starting fromthe last (homogeneous) equation, we set w = 1 and get z = 1. These values, substituted into the second(homogeneous) equation, give y = 0 and finally from the first (homogeneous) equation we get x = 0. Thusxh = (0, 0, 1, 1). We now guess xp: starting again from the last equation, we set w = 1 and get z = −1. Thesevalues, substituted into the second equation, give y = −1 and finally from the first equation we get x = 1.Thus xp = (1,−1,−1, 1). Any solutions of the given system can be written as

x = xp + span{xh} = (1,−1,−1, 1) + span{(0, 0, 1, 1)}.

Introducing a parameter. We choose w as a free variable, w = α, α ∈ IR. From the last equation we getz = −2 + α. Substituting into the second and then into the first equation, we get y = −1 and then x = 1.Thus, for any possible α ∈ IR, we have solution of the system

x = (1,−1,−2 + α, α) = (1,−1,−2, 0) + α(0, 0, 1, 1) = (1,−1,−2, 0) + span{(0, 0, 1, 1)}.

It is important to recognise that we get the same solutions as in the previous case, because (1,−1,−2, 0)is only a different particular solution that can be obtain from the xp found before, by subtracting once thefound solution xh of the homogeneous system.


x+ ay − 3z = 5ax− 3y + z = 10x+ 9y − 10z = a+ 3.


A =

1 a −3a −3 11 9 −10

.

Since A is a square matrix, we will evaluate detA to distinguish the values of the parameter a for which wecan use Cramer’s rule (detA 6= 0).

detA = 30 + a− 27− 9− 9 + 10a2 = 2(5a2 − 13a+ 6).

Consequently, detA = 0 if and only if a = 2 or a = 35 . For any possible a ∈ IR, a 6= 2, a 6= 3

5 , the givensystem has a unique solution that can be evaluated using Cramer’s rule:

x =detC1

detA=

det

5 a −310 −3 1a+ 3 9 −10

2(5a2 − 13a+ 6)

=150 + a(a+ 3)− 270− 9(a+ 3)− 45 + 100a

2(5a2 − 13a+ 6)=

a2 + 94a− 192

2(5a2 − 13a+ 6)

39

y =detC2

detA=

det

1 5 −3a 10 11 a+ 3 −10

2(5a2 − 13a+ 6)

=−100 + 5− 3a(a+ 3) + 10− (a+ 3) + 50a

2(5a2 − 13a+ 6)=−3a2 + 40a− 88

2(5a2 − 13a+ 6)

z=detC3

detA=

det

1 a 5a −3 101 9 a+ 3

2(5a2 − 13a+ 6)

=−3(a+ 3) + 10a+ 45a+ 15− 90− a2(a+ 3)

2(5a2 − 13a+ 6)=−a3 − 3a2 + 52a− 84

2(5a2 − 13a+ 6).

For a = 2, we will perform a Gaussian reduction starting from the augmented matrix of the given system,where we substitute the value a = 2: 1 2 −3

2 −3 11 9 −10

∣∣∣∣∣∣∣∣∣∣∣∣

5105

R2 − 2R1

∼R3 −R1

1 2 −30 −7 70 7 −7

∣∣∣∣∣∣∣∣∣∣∣∣500

−R2/7∼

R3 +R2

1 2 −30 1 −10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣500

.

The rank of the augmented matrix is equal to the rank of the matrix associated to the homogeneous system,thus the system has solutions. Since the rank is two and the number of unknowns is three, we see that thedimension of the space of all solutions of the homogeneous system is one. We consider the system associatedto the last augmented matrix, this is equivalent to the given one,

x+ 2y − 3z = 5y − z = 0.

We introduce a parameter, choosing z as a free variable, z = α, α ∈ IR, from the last equation we get y = α.These values, substituted into the first equation, give x = 5 + α. Thus, any solution of the system can bewritten as

x = (x, y, z) = (5 + α, α, α) = (5, 0, 0) + α(1, 1, 1), α ∈ IR

or equivalentlyx = (5, 0, 0) + span{(5, 1, 1)},

where (5, 0, 0) is a particular solution of the system, and span{(5, 1, 1)} is the space of solutions of thehomogeneous system.

For a = 35 , we will perform a Gaussian reduction starting from the augmented matrix of the given system,

where we substitute the value a = 35 : 1 3

5 −335 −3 11 9 −10

∣∣∣∣∣∣∣∣∣∣∣∣

510185

R2 − 35R1

∼R3 −R1

1 35 −3

0 − 8425

145

0 425 −7

∣∣∣∣∣∣∣∣∣∣∣∣

57− 7

5

−25R2

∼2R3 + 5R2

1 35 −3

0 84 −700 0 0

∣∣∣∣∣∣∣∣∣∣∣∣

5−1751615

.

The rank of the augmented matrix is three, while the rank of the matrix associated to the homogeneoussystem is two, thus, by Frobenius’ theorem, the system has no solutions.


x+ y − az = 1x− 2y + 3z = 2x+ ay − z = 1.


A =

1 1 −a1 −2 31 a −1

.


detA = 2 + 3− a2 − 2a− 3a+ 1 = −a2 − 5a+ 6 = (a+ 6)(1− a)

40

Consequently, detA = 0 if and only if a = −6 or a = 1. For any possible a ∈ IR, a 6= −6, a 6= 1, the givensystem has a unique solution that can be evaluated using Cramer’s rule:

x =detC1

detA=

det

1 1 −a2 −2 31 a −1

(a+ 6)(1− a)

=2 + 3− 2a2 − 2a− 3a+ 2

(a+ 6)(1− a)=

(1− a)(2a+ 7)

(a+ 6)(1− a)=

2a+ 7

a+ 6

y =detC2

detA=

det

1 1 −a1 2 31 1 −1

(a+ 6)(1− a)

=−2 + 3− a+ 2a− 3 + 1

(a+ 6)(1− a)=

a− 1

(a+ 6)(1− a)=−1

a+ 6

z =detC3

detA=

det

1 1 11 −2 21 a 1

(a+ 6)(1− a)

=−2 + 2 + a+ 2− 2a− 1

(a+ 6)(1− a)=

−a+ 1

(a+ 6)(1− a)=

1

a+ 6.

For a = −6, we will perform a Gaussian reduction starting from the augmented matrix of the given system,where we substitute the value a = −6: 1 1 6

1 −2 31 −6 −1

∣∣∣∣∣∣∣∣∣∣∣∣121

R2 −R1

∼R3 −R1

1 1 60 −3 −30 −7 −7

∣∣∣∣∣∣∣∣∣∣∣∣110

−R2

∼3R3 − 7R2

1 1 60 3 30 0 0

∣∣∣∣∣∣∣∣∣∣∣∣

11−7

.


For a = 1, we will perform a Gaussian reduction starting from the augmented matrix of the given system,where we substitute the value a = 1: 1 1 −1

1 −2 31 1 −1

∣∣∣∣∣∣∣∣∣∣∣∣121

R2 −R1

∼R3 −R1

1 1 60 −3 20 0 0

∣∣∣∣∣∣∣∣∣∣∣∣110

.

The rank of the augmented matrix is equal to the rank of the matrix associated to the homogeneous system,thus the system has solutions. Since the rank is two and the number of unknowns is three, we see that thedimension of the space of all solutions of the homogeneous system is one. We consider the system associatedto the last augmented matrix, this is equivalent to the given one,

x+ y − z = 1−3y + 2z = 1.

We introduce a parameter, choosing z as a free variable, z = 3α, α ∈ IR, from the last equation we gety = 2α − 1

3 . These values, substituted into the first equation, give x = α + 43 . Thus, any solution of the

system can be written as

x = (x, y, z) = (α+4

3, 2α− 1

3, 3α) = (

4

3,−1

3, 0) + α(1, 2, 3), α ∈ IR

or equivalently

x = (4

3,−1

3, 0) + span{(1, 2, 3)},

where ( 43 ,−

13 , 0) is a particular solution of the system, and span{(1, 2, 3)} is the space of solutions of the

homogeneous system.

4.17) Find the solutions of the following system, depending on the value of the parameter λ ∈ IR:

x+ y + λz = λx+ λy + z = 1λx+ y + z = 1.


A =

1 1 λ1 λ 1λ 1 1

.

41

Since A is a square matrix, we will evaluate detA to distinguish the values of the parameter λ for which wecan use Cramer’s rule (detA 6= 0).

detA = λ+ λ+ λ− λ3 − 1− 1 = −(λ3 − 3λ+ 2).

By looking at the matrix A, we notice that, for λ = 1, the matrix has three identical rows, thus λ = 1 mustbe a value for which detA = 0. In order to find other possible values of λ that make the determinant equalto zero, we perform the following division of polynomials:

(λ3 − 3λ+ 2) : (λ− 1) = λ2 + λ− 2

−(λ3 − λ2)

λ2 − 3λ+ 2

−(λ2 − λ)

− 2λ+ 2

−(−2λ+ 2)

0

Consequently, detA = −(λ − 1)(λ2 + λ − 2) = −(λ − 1)2(λ + 2). Thus detA = 0 if and only if λ = 1 orλ = −2. For any possible λ ∈ IR, λ 6= 1, λ 6= −2, the given system has a unique solution that can be evaluatedusing Cramer’s rule:

x =detC1

detA=

det

λ 1 λ1 λ 11 1 1

−(λ− 1)2(λ+ 2)

=0

−(λ− 1)2(λ+ 2)= 0

y =detC2

detA=

det

1 λ λ1 1 1λ 1 1

−(λ− 1)2(λ+ 2)

=0

−(λ− 1)2(λ+ 2)= 0

z =detC3

detA=

det

1 1 λ1 λ 1λ 1 1

−(λ− 1)2(λ+ 2)

=detA

−(λ− 1)2(λ+ 2)= 1.

For λ = 1, we will perform a Gaussian reduction starting from the augmented matrix of the given system,where we substitute the value λ = 1: 1 1 1

1 1 11 1 1

∣∣∣∣∣∣∣∣∣∣∣∣111

R2 −R1

∼R3 −R2

1 1 10 0 00 0 0

∣∣∣∣∣∣∣∣∣∣∣∣100

.

The rank of the augmented matrix is equal to the rank of the matrix associated to the homogeneous system,thus the system has solutions. Since the rank is one and the number of unknowns is three, we see that thedimension of the space of all solutions of the homogeneous system is two. We consider the ”system” associatedto the last augmented matrix, this is equivalent to the given one, and it is composed by a single equation:

x+ y + z = 1.

This time we choose the guessing method. We know that the structure of the solution set of our systemis: xp + span{xh1,xh2}, where xp is a particular solution of the last equation, xh1 and xh2 are two linearlyindependent solutions of the homogeneous equation, and thus span{xh} is the space of all solutions of thehomogeneous equation. We easily guess xh1 = (1,−1, 0), xh2 = (0, 1,−1), and xp = (1, 0, 0). Thus, allsolutions of the system can be written as

x = (x, y, z) = (1, 0, 0) + span{(1,−1, 0), (0, 1,−1)}.

For λ = −2, we will perform a Gaussian reduction starting from the augmented matrix of the given system,where we substitute the value λ = −2: 1 1 −2

1 −2 1−2 1 1

∣∣∣∣∣∣∣∣∣∣∣∣−211

R2 −R1

∼R3 + 2R1

1 1 −20 −3 30 3 −3

∣∣∣∣∣∣∣∣∣∣∣∣−25−3

∼R3 +R2

1 1 −20 −3 30 0 0

∣∣∣∣∣∣∣∣∣∣∣∣−252

.

42


4.19) Write down all solutions of the following system for any possible value of a, b ∈ IR−2x+ y + z = 7ax+ 2y − z = 2−3x+ y + 2z = −b.


A =

−2 1 1a 2 −1−3 1 2

.


detA = −8 + 3 + a+ 6− 2− 2a = −a− 1.

Consequently, detA = 0 if and only if a = −1. For any possible a ∈ IR, a 6= −1, the given system has aunique solution that can be evaluated using Cramer’s rule:

x =detC1

detA=

det

7 1 12 2 −1−b 1 2

−a− 1

=28 + b+ 2 + 2b+ 7− 4

−a− 1=

3b+ 33

−a− 1

y =detC2

detA=

det

−2 7 1a 2 −1−3 −b 2

−a− 1

=−8 + 21− ab+ 6 + 2b− 14a

−a− 1=

2b− ab− 14a+ 19

−a− 1

z =detC3

detA=

det

−2 1 7a 2 2−3 1 −b

−a− 1

=4b− 6 + 7a+ 42 + 4 + ab

−a− 1=

4b+ ab+ 7a+ 40

−a− 1.

For a = −1, we will perform a Gaussian reduction starting from the augmented matrix of the given system,where we substitute the value a = −1:−2 1 1

−1 2 −1−3 1 2

∣∣∣∣∣∣∣∣∣∣∣∣

72−b

−R2 ↔ R1

R1 − 2R2

R3 − 3R2

∼

1 −2 10 −3 30 −5 5

∣∣∣∣∣∣∣∣∣∣∣∣−23

−b− 6

−R2/3∼

5R2 − 3R3

1 −2 10 1 −10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣−2−1

3b+ 33

.

We now see that the rank of the augmented matrix depends on the value of b. If b 6= −11, 3b+ 33 6= 0, thenthe rank of the augmented matrix is three, while the rank of the matrix associated to the homogeneous systemis two. By Frobenius’s theorem, the system has no solutions. If b = −11, then the system has infinitely manysolutions. The rank of the augmented matrix (equal to the rank of the matrix associated to the homogeneoussystem) is two and the number of unknowns is three, thus the dimension of the space of all solutions ofthe homogeneous system is one. We consider the system associated to the last augmented matrix, this isequivalent to the given one,

x− 2y + z = −2y − z = −1.

We introduce a parameter, choosing z as a free variable, z = α, α ∈ IR, from the last equation we gety = α− 1. These values, substituted into the first equation, give x = α− 4. Thus, any solution of the systemcan be written as

x = (x, y, z) = (α− 4, α− 1, α) = (−4,−1, 0) + α(1, 1, 1), α ∈ IRor equivalently

x = (−4,−1, 0) + span{(1, 1, 1)},where (−4,−1, 0) is a particular solution of the system, and span{(1, 1, 1)} is the space of solutions of thehomogeneous system.

Let’s summarise the final answer to the given exercise. If a 6= −1, for any possible b, the system has a uniquesolution x = ( 3b+33

−a−1 ,2b−ab−14a+19

−a−1 , 4b+ab+7a+40−a−1 ). If a = −1, and b 6= −11, the system has no solution. If

a = −1, and b = −11, the system has infinitely many solutions x = (−4,−1, 0) + α(1, 1, 1), α ∈ IR.

43

Chapter 5Linear transformations

5.1) Determine if the following are linear transformations:

a) f : IR2 → IR, f(x, y) = x+ y,b) f : IR2 → IR, f(x, y) = x+ 1,c) f : IR2 → IR, f(x, y) = xy,d) f : P2 → IR2, f(ax2 + bx+ c) = (a+ b, b+ c),e) f : IR2 → P1, f(a, b) = b+ a2x.

5.2) Determine if the following transformation f : IR3 → IR3 is linear:

a) f(x, y, z) = (2x− y, y − 1, x+ 2y),b) f(x, y, z) = (2x− y, y − x, xy),c) f(x, y, z) = (2x− y2 + z, y + 10z, x+ y),d) f(x, y, z) = (2x− y, y − x−

√3z, x+ 2y).

5.3) Given the linear transformation l : IR2 → IR2, l(x, y) = (2x − 2y,−x + y), write the matrix associated to lwith respect to the standard basis of IR2, find Ker l, Im l, their bases and dimensions. Find all vectors of IR2

that are mapped to (4,−2). Determine if l is invertible and, if possible, find its inverse.

5.4) Given the transformation l : IR3 → IR3, l(x, y, z) = (y + z, x− z, x+ y + z), write the matrix associated to lwith respect to the standard basis of IR3, find Ker l, Im l, their bases and dimensions. Find all vectors of IR3

that are mapped to (4,−2, 1). Determine if l is invertible and, if possible, find its inverse.

5.5) Given l : IR3 → IR3, l(x1, x2, x3) = (x1 + 2x2 + 3x3, 4x1 + 5x2 + 6x3, x1 + x2 + x3), find Ker(l), Im(l), theirbases and dimensions.

5.6) Given l : IR3 → IR2, l(x1, x2, x3) = (2x1−x2+3x3, x1+x2+x3), find Ker(l), Im(l), their bases and dimensions.

5.7) Determine the linear transformation l : IR3 → IR2 defined by:

l(1, 1, 0) = (2,−1), l(0, 1, 2) = (−1, 1), l(2, 0, 0) = (2,−3).

Find Ker(l), its basis and dimension. Calculate l(1, 2,−2).


l(2, 3) = (−1, 2, 1), l(1,−1) = (0, 3, 2).

Find Ker(l), Im(l) their basis and dimension. Calculate l(4, 11).

5.9) Given the linear transformation l : IR3 → IR3 defined by l(1, 2, 3) = (−3,−8,−3), l(1, 1, 0) = (1, 5, 2) andsuch that Ker l = span{(1, 1, 1)}, find the matrix associated with l with respect to the standard basis, findIm l, its basis and dimension. Find all v ∈ IR3 such that l(v) = (2, 3, 1).

5.10) Given l : IR4 → IR3 such that l(1, 1,−1, 0) = (0, 0, 0), l(1, 2,−1,−2) = (−1,−3, 1), l(1, 0, 0,−1) = (0, 0, 0),l(1, 1, 1, 1) = (5, 8, 2), find the matrix A associated with l with respect to the standard bases. Find all v ∈ IR4

such that Av = (13, 21, 5).

5.11) Given the linear transformation l : IR3 → P1, l(a, b, c) = b− c+ (2a− c)x, find the matrix associated with lwith respect to the standard bases, evaluate l(2, 1,−1), find Ker l its basis and dimension. Is l surjective?

5.12) Given l :M2,2 → P3 defined by l(A) = l

(a bc d

)= a + (2a − b)x + (b + c)x2 + (a − b + c + d)x3, find the

matrix associated with l with respect to the standard bases. Is l an isomorphism?

5.13) Given the matrix M =

(1 23 4

), we define the linear transformation l :M2,2 →M2,2 by l(X) = M ·X.

Is l an isomorphism? If possible, find its inverse.

44

5.14) Given the linear transformation l : IR4 → IR3 that has as associated matrix with respect to the standard

bases A(l) =

1 −1 1 11 0 2 −11 1 3 −3

, write down the general form of l(x, y, z, w), find Ker(l), Im(l), their bases

and dimensions.

5.15) Given the linear transformations

f : IR3 → IR2, f(x, y, z) = (2x− y, 2y + z), and g : IR2 → IR3, g(u, v) = (u, u+ v, u− v),

find the matrix associated to f ◦g and g◦f with respect to the standard basis. Find rank(f ◦g) and rank(g◦f),is one of the two compositions an isomorphism? If yes find its inverse, otherwise identify Kernel and Image.

5.16) Given the linear transformationsf : IR2 → IR2, f(x, y) = (4x− y, x− 2y),g : IR2 → IR2, g(1, 1) = (−1,−2), g(0, 1) = (2, 4)find the matrix associated to f ◦g and g◦f with respect to the standard basis. Find rank(f ◦g) and rank(g◦f),is one of the two compositions an isomorphism? If yes find its inverse, otherwise identify Kernel and Image.

5.17) For the space IR3 are given two ordered bases:

B = {(1, 0, 0), (1, 2, 0), (1, 2, 1)}, S = {(2,−2, 0), (2,−1, 0), (0, 2,−2)}.

Find PB→S the matrix of transition from basis B to basis S.(Hint: use C, the standard basis of IR3, and the fact that PB→S = P−1S→CPB→C).

5.18) In IR3 are given the standard basis C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} and the bases

B = {(1, 1, 0), (0, 1, 0), (0,−1, 1)}, D = {(1, 0, 1), (1, 1, 1), (0,−1, 1)}.

Find the transition matrices PB→C , PD→C , PB→D, PD→B .

5.19) Let A(l) =

(1 1 11 −1 2

)be the matrix associated to a linear transformation l : IR3 → IR2 with respect to

the standard bases of IR3 and IR2. Find the matrix associated to the given transformation with respect tothe bases S and B, where:

S = {(2,−2, 2), (2,−1,−2), (0, 2,−2)}, and B = {(1, 1), (1,−1)}.

5.20) Given l : IR2 → IR2 such that l(x, y) = (4x− 2y, 2x+ y), find the matrix associated to l with respect to thebasis F = {(1, 1), (−1, 0)}.

5.21) Given the linear transformation lh : IR3 → IR3, lh(x, y, z) = (2x+y, y−z, 2y+hz), write the matrix associatedto lh with respect to the standard basis of IR3, find Ker l, Im l, its bases and dimensions for every possiblevalue of h ∈ IR. Does there exist a value of h for which lh is an isomorphism?.

5.22) Given the transformation lh : IR3 → IR3 defined by lh(x, y, z) = (x− hz, x+ y − hz,−hx+ z), where h ∈ IRis a parameter. Find, for all possible values of h, Ker(lh), Im(lh), their bases and dimensions.Determine l−1h (1, 0, 1) = {(x, y, z) ∈ IR3 : lh(x, y, z) = (1, 0, 1)}.

45

Solutions of odd number problems - Linear transformations

5.1) Determine if the following are linear transformations:

a) f : IR2 → IR, f(x, y) = x+ y,b) f : IR2 → IR, f(x, y) = x+ 1,c) f : IR2 → IR, f(x, y) = xy,d) f : P2 → IR2, f(ax2 + bx+ c) = (a+ b, b+ c),e) f : IR2 → P1, f(a, b) = b+ a2x.

Solution. Given two linear spaces L and M a function f : L → M is called linear if the following two conditions aresatisfied:i) f(v + u) = f(v) + f(u), for every v,u ∈ L,ii) f(αv) = αf(v), for every v ∈ L, α ∈ IR.

a) Given f : IR2 → IR, defined as f(x, y) = x + y, we prove that f is linear. Indeed, given any u,v ∈ IR2,u = (x1, y1), v = (x2, y2), and α ∈ IR:

i) f(v + u) =f((x1, y1) + (x2, y2)) = f(x1 + x2, y1 + y2) = x1 + x2 + y1 + y2 =

=(x1 + y1) + (x2 + y2) = f(x1, y1) + f(x2, y2) = f(v) + f(u),

ii) f(αv) =f(α(x1, y1)) = f(αx1, αy1) = αx1 + αy1 = α(x1 + y1) = αf(x1, y1) = αf(v).

b) Given f : IR2 → IR, defined as f(x, y) = x + 1, we prove that f is not linear. Indeed, given u,v ∈ IR2,u = (x1, y1), v = (x2, y2), in general condition i) is not satisfied, i.e. we have

i) f(v + u) =f((x1, y1) + (x2, y2)) = f(x1 + x2, y1 + y2) = x1 + x2 + 1 6=6=(x1 + 1) + (x2 + 1) = f(x1, y1) + f(x2, y2) = f(v) + f(u),

it is enough to verify this inequality for v = u = (1, 2), when we get

f(v + u) = f(2, 4) = 2 + 1 = 3, while f(v) + f(u) = f(1, 2) + f(1, 2) = 1 + 1 + 1 + 1 = 4.

c) Given f : IR2 → IR, defined as f(x, y) = xy, we prove that f is not linear. Indeed, it is easy to disprovecondition i): we need to find u,v ∈ IR2, u = (x1, y1), v = (x2, y2), and prove that

i) f(v + u) =f((x1, y1) + (x2, y2)) = f(x1 + x2, y1 + y2) = (x1 + x2)(y1 + y2) 6=6=(x1y1) + (x2y2) = f(x1, y1) + f(x2, y2) = f(v) + f(u),

it is enough to choose v = u = (1, 2), then we get

f(v + u) = f(2, 4) = 2 · 4 = 8, while f(v) + f(u) = f(1, 2) + f(1, 2) = 1 · 2 + 1 · 2 = 4.

d) Given f : P2 → IR2, defined as f(ax2 + bx + c) = (a + b, b + c), we prove that f is linear. Indeed, givenany u,v ∈ P2, u = a1x

2 + b1x+ c1, v = a2x2 + b2x+ c2, and α ∈ IR:

i) f(v + u) =f((a1x2 + b1x+ c1) + (a2x

2 + b2x+ c2)) = f((a1 + a2)x2 + (b1 + b2)x+ (c1 + c2)

)=

=(a1 + a2 + b1 + b2, b1 + b2 + c1 + c2 =((a1 + b1) + (a2 + b2), (b1 + c1) + (b2 + c2)

)=

=(a1 + b1, b1 + c1) + (a2 + b2, b2 + c2) = f(a1x2 + b1x+ c1)+f(a2x

2 + b2x+ c2) = f(v) + f(u),

ii) f(αv) =f(α(a1x2 + b1x+ c1)) = f(αa1x

2 + αb1x+ αc1) = (αa1 + αb1, αb1 + αc1) =

=α(a1 + b1, b1 + c1) = αf(a1x2 + b1x+ c1) = αf(v).

e) Given f : IR2 → P1, defined as f(a, b) = b + a2x, we prove that f is not linear. Indeed, we will disprovecondition ii), we need to find v = (a, b) ∈ IR2 and α ∈ IR such that

ii) f(αv) = f(α(a, b)) = f(αa, αb) = αb+ (αa)2x 6= α(b+ a2x) = αf(a, b) = αf(v).

It is enough to choose v = (1, 2), and α = 3, then we get

f(αv) = f(3, 6) = 6 + 9x2, while αf(v) = 3f(1, 2) = 3(2 + x2) = 6 + 3x2.

46

5.3) Given the linear transformation l : IR2 → IR2, l(x, y) = (2x − 2y,−x + y), write the matrix associated to lwith respect to the standard basis of IR2, find Ker l, Im l, their bases and dimensions. Find all vectors of IR2

that are mapped to (4,−2). Determine if l is invertible and, if possible, find its inverse.

Solution. Given two linear spaces L, M with finite dimension, and two respective bases BL = {b1,b2, . . . ,bn}, BM ={c1, c2, . . . , cm}, every linear transformation l : L→M is uniquely determined by a matrixA = A(l, BL, BM ) = (aij) ∈Mm,n, where aij are as follows:

l(b1) =a11c1 + a21c2 + · · ·+ am1cm

l(b2) =a12c1 + a22c2 + · · ·+ am2cm

...

l(bn) =a1nc1 + a2nc2 + · · ·+ amncm.

Such matrix A has the property that, for every v ∈ L,

[l(v)]BM= A · [v]BL

where [v]B are the coordinates of the vector v with respect to the basis B seen as a column.

Following this definition, we are looking for the matrix A that represents the given linear transformation withrespect to the canonical basis B = {(1, 0), (0, 1)} of IR2 (both as starting and landing space: L = M = IR2).We have:

l(1, 0) = (2,−1) and l(0, 1) = (−2, 1).

Thus A =

(2 −2−1 1

). In order to find Ker l and Im l, we recall this two fundamental definitions. Given a

linear map l : L→M :

Ker l = {v ∈ L : l(v) = 0M}, (Ker l is a subspace of L),

Im l = {u ∈M : there exists v ∈ L, l(v) = u}, (Im l is a subspace of M).

In our case, Ker l =

{(x, y) ∈ IR2 : A

(xy

)=

(00

)}is the space of solutions of the homogeneous system

associated to A. We use the matrix form of Gaussian reduction and get:(2 −2−1 1

∣∣∣∣∣∣∣∣ 00)

R1/2R1/2 +R2

∼(

1 −10 0

∣∣∣∣∣∣∣∣ 00).

Thus, Ker l = span{(1, 1)}, dim Ker l = 1, and a basis of Ker l is BK = {(1, 1)}.Before we evaluate Im l, we recall two fundamental Theorems. Given two finite dimensional linear spacesL,M , any basis BL = {b1,b2, . . . ,bn} of L, and a linear transformation l : L→M :

i) Im l = span{l(b1), l(b2), . . . , l(bn)},ii) dimL = dim Ker l + dim Im l.

In our case, ii) implies that dim Im l = dim IR2 − dim Ker l = 1, and, from i) we get

Im l = span{l(1, 0), l(0, 1)} = span{(2,−1), (−2, 1)} = span{(2,−1)}

thus, a basis of Im l is BI = {(2,−1)}.We now need to find all vectors of IR2 that are mapped to (4,−2), this means to identify the set{

(x, y) ∈ IR2 : A

(xy

)=

(4−2

)}, i.e. the set of solutions of the non homogeneous system with augmented

matrix: (2 −2−1 1

∣∣∣∣∣∣∣∣ 4−2

)R1/2

R1/2 +R2∼

(1 −10 0

∣∣∣∣∣∣∣∣ 20).

After the simple reduction, we see that the system has solutions, and, knowing already the space of solutionsof the homogeneous system (Ker l), we just need to guess a particular solution xp of the nonhomogeneoussystem, for example xp = (3, 1), to write{

(x, y) ∈ IR2 : A

(xy

)=

(4−2

)}= xp + Ker l = (3, 1) + span{(1, 1)}.

47

A linear transformation l : L→M is invertible if and only if it is injective, if and only if Ker l = {0L}. Thisimplies that in our case, the given map l is not invertible.

5.5) Given l : IR3 → IR3, l(x1, x2, x3) = (x1 + 2x2 + 3x3, 4x1 + 5x2 + 6x3, x1 + x2 + x3), find Ker(l), Im(l), theirbases and dimensions.

Solution. We refer to the Solution of Exercise 3 for all relevant definitions and theorems.We consider the ordered canonical basis of IR3, C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, and we will find the matrixA = A(l, C, C) associated to the given linear transformation with respect to C. For this purpose, we need toevaluate:

l(1, 0, 0) =(1, 4, 1)

l(0, 1, 0) =(2, 5, 1)

l(0, 0, 1) =(3, 6, 1).

Now we can write A =

1 2 34 5 61 1 1

. To find Ker l =

{(x1, x2, x3) ∈ IR3 : A

x1x2x3

=

000

}, the space of

solutions of the homogeneous system associated to A, we first evaluate detA, because, in case detA 6= 0, weknow that the homogeneous system has a unique solution, the trivial one. Since detA = 5+12+12−15−6−8 =0, we will use the matrix form of Gaussian reduction and get: 1 2 3

4 5 61 1 1

∣∣∣∣∣∣∣∣∣∣∣∣000

R2 − 4R1

∼R3 −R1

1 2 30 −3 −60 −1 −2

∣∣∣∣∣∣∣∣∣∣∣∣000

−R2/3∼

R3 −R2/3

1 2 30 1 20 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

Setting x3 = α, α ∈ IR, we get from the second row (imagining the correspondent equation x2 + 2x3 = 0)x2 = −2α, and, subsequently, from the first row x1 = α. Thus we conclude that:Ker l = {(α,−2α, α), α ∈ IR} = span{(1,−2, 1)}, dim Ker l = 1, and a basis of Ker l is BK = {(1,−2, 1)}.From the Theorem on dimensions ii), we get dim Im l = dim IR3−dim Ker l = 2, and, from Theorem i) we get

Im l = span{l(1, 0, 0), l(0, 1, 0), l(0, 0, 1)} = span{(1, 4, 1), (2, 5, 1), (3, 6, 1)} = span{(1, 4, 1), (2, 5, 1)},

where the first two vectors were chosen randomly, just considering that, because they are linearly independent(they are not a multiple of each other), they must span the space Im l that has dimension 2. Thus, a basis ofIm l is BI = {(1, 4, 1), (2, 5, 1)}.


l(1, 1, 0) = (2,−1), l(0, 1, 2) = (−1, 1), l(2, 0, 0) = (2,−3).

Find Ker(l), its basis and dimension. Calculate l(1, 2,−2).

Solution. We recall that a linear transformation is uniquely determined by the image of the vectors in a basis ofthe starting linear space. Thus the given conditions define a linear transformation l : IR3 → IR2 if B ={(1, 1, 0), (0, 1, 2), (2, 0, 0)} is a basis of IR3. We can easily verify that the given vectors are linearly indepen-dent, by evaluating the determinant of the matrix B formed by the vectors of the set B written as columns:

det

1 0 21 1 00 2 0

= 4 6= 0;

thus, knowing that the dimension of IR3 is three, we can conclude that the given three vectors form a basisof IR3.In order to find the matrix A that represents the linear transformation l with respect to the canonical basisC of IR3, C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, we start observing that such matrix A has the property that:

A

110

=

(2−1

),

A

012

=

(−11

),

48

A

200

=

(2−3

),

that can be written, using the properties of matrix operations, in this compact way:

A

1 0 21 1 00 2 0

= AB =

(2 −1 2−1 1 −3

).

From the last matrix equation, knowing that B is certainly invertible because its determinant is non-zero, itfollows that

A =

(2 −1 2−1 1 −3

)B−1.

We now evaluate the inverse of B using the formula B−1 = 1detB adjB. We have already calculated the

determinant of B: detB = 4. We now proceed with the evaluation of adjB, calculating all cofactors of B:

B11 = (−1)1+1 · det

(1 02 0

)= 0

B12 = (−1)1+2 · det

(1 00 0

)= 0

B13 = (−1)1+3 · det

(1 10 2

)= 2

B21 = (−1)2+1 · det

(0 22 0

)= 4

B22 = (−1)2+2 · det

(1 20 0

)= 0

B23 = (−1)2+3 · det

(1 00 2

)= −2

B31 = (−1)3+1 · det

(0 21 0

)= −2

B32 = (−1)3+2 · det

(1 21 0

)= 2

B33 = (−1)3+3 · det

(1 01 1

)= 1

Now we can write B−1 = 14

0 0 24 0 −2−2 2 1

T

= 14

0 4 −20 0 22 −2 1

. We thus evaluate A, calculating the

indicated product:

A =

(2 −1 2−1 1 −3

)B−1 =

(2 −1 2−1 1 −3

)1

4

0 4 −20 0 22 −2 1

=1

4

(4 4 −4−6 2 1

)=

(1 1 −1− 3

212

14

).

The found matrix A represents the linear transformation l with respect to the canonical basis C of IR3, thuswe may evaluate:

l(x, y, z) = A

xyz

=1

4

(4 4 −4−6 2 1

)xyz

= (x+ y − z,−3

2x+

1

2y +

1

4z),

from which:

l(1, 2,−2) = (1, 1).

49

To find Ker l =

{(x, y, z) ∈ IR3 : A

xyz

=

(00

)}, the space of solutions of the homogeneous system

associated to A, we will use the matrix form of Gaussian reduction and get:(1 1 −1− 3

212

14

∣∣∣∣∣∣∣∣ 00)

R2 + 32R1

∼

(1 1 −10 2 − 5

4

∣∣∣∣∣∣∣∣ 00).

The rank of A is two, this implies that dim Ker l = dim IR3 − rankA = 1. We may easily guess a solution ofthe homogeneous system associated to the last matrix, setting z = 8, from the second row we get y = 5, andsubstituting these values into the first row, we get x = 3. We conclude that

Ker l = span{(3, 5, 8)}, and a basis of Ker l is BK = {(3, 5, 8)}.

5.9) Given the linear transformation l : IR3 → IR3 defined by l(1, 2, 3) = (−3,−8,−3), l(1, 1, 0) = (1, 5, 2) andsuch that Ker l = span{(1, 1, 1)}, find the matrix associated with l with respect to the standard basis, findIm l, its basis and dimension. Find all v ∈ IR3 such that l(v) = (2, 3, 1).

Solution. The linear transformation l is well defined if B = {(1, 2, 3), (1, 1, 0), (1, 1, 1)} is a basis of IR3. We can easilyverify if the three vectors are linearly independent by evaluating the determinant of the matrix B formed bythe vectors of the set B written as columns:

det

1 1 12 1 13 0 1

= 1 + 3 + 0− 3− 0− 2 = −1 6= 0;

thus, knowing that the dimension of IR3 is three, we can conclude that B is a basis of IR3. In order tofind the matrix A that represents the linear transformation l with respect to the canonical basis C of IR3,C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, we start observing that such matrix A has the property that:

A

123

=

−3−8−3

,

A

110

=

152

,

A

111

=

000

,

that, using the properties of matrix operations, can be written in this compact way:

A

1 1 12 1 13 0 1

= AB =

−3 1 0−8 5 0−3 2 0

.

From the last matrix equation, knowing that B is certainly invertible because its determinant is non-zero, itfollows that

A = ABB−1 =

−3 1 0−8 5 0−3 2 0

B−1.

We now evaluate the inverse of B using the formula B−1 = 1detB adjB. We have already calculated the

determinant of B: detB = −1. We now proceed with the evaluation of adjB, calculating all cofactors of B:

B11 = (−1)1+1 · det

(1 10 1

)= 1

B12 = (−1)1+2 · det

(2 13 1

)= 1

B13 = (−1)1+3 · det

(2 13 0

)= −3

50

B21 = (−1)2+1 · det

(1 10 1

)= −1

B22 = (−1)2+2 · det

(1 13 1

)= −2

B23 = (−1)2+3 · det

(1 13 0

)= 3

B31 = (−1)3+1 · det

(1 11 1

)= 0

B32 = (−1)3+2 · det

(1 12 1

)= 1

B33 = (−1)3+3 · det

(1 12 1

)= −1

Now we can write B−1 = −

1 1 −3−1 −2 30 1 −1

T

=

−1 1 0−1 2 −13 −3 1

. We thus evaluate A, calculating the

indicated product:

A =

−3 1 0−8 5 0−3 2 0

B−1 =

−3 1 0−8 5 0−3 2 0

−1 1 0−1 2 −13 −3 1

=

2 −1 −13 2 −51 1 −2

.

The found matrix A represents the linear transformation l with respect to the canonical basis C of IR3, thuswe may evaluate:

l(x, y, z) = A

xyz

=

2 −1 −13 2 −51 1 −2

xyz

= (2x− y − z, 3x+ 2y − 5z, x+ y − 2z).

As previously explained in the Solution of Exercise 3, the column of A, seen as vectors in IR3, are the imageof the vectors in the canonical basis B of IR3. Thus, recalling Theorem i) from the solution of Exercise 3, wecan write

Im l = span{(2, 3, 1), (−1, 2, 1), (−1,−5,−2)}.

Now, without using a Gaussian reduction to find a set of linearly independent vectors that have the same span,we may reason as follows. It was given at the beginning of the exercise the information that the dimension ofKer l is one, thus (Theorem ii) in Solution of Exercise 3) dim Im l = dim IR3 − dim Ker l = 2, we may chooseany two linearly independent vectors among the vectors that form the column of A and these will span theIm l. For example, Im l = span{(2, 3, 1), (−1, 2, 1)}.

We now need to find all vectors v of IR3 that are mapped to (2, 3, 1), this means to identify the set{(x, y, z) ∈ IR3 : A

xyz

=

231

}, i.e. the set of solutions of the non homogeneous system with augmented

matrix: 2 −1 −13 2 −51 1 −2

∣∣∣∣∣∣∣∣∣∣∣∣231

R3 → R1

R1 − 2R3

R2 − 3R3

∼

1 1 −20 −3 30 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣100

−R2/3∼

R2 − 3R3

1 1 −20 1 −10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣100

.

We now see that the system has solutions, and, knowing already the space of solutions of the homogeneoussystem (Ker l), we just need to guess a particular solution xp of the nonhomogeneous system, for examplexp = (1, 0, 0), to write:

{v ∈ IR3 : l(v) = (2, 3, 1)} =

{(x, y, z) ∈ IR3 : A

xyz

=

231

} = xp + Ker l = (1, 0, 0) + span{(1, 1, 1)}.

5.11) Given the linear transformation l : IR3 → P1, l(a, b, c) = b− c+ (2a− c)x, find the matrix associated with lwith respect to the standard bases, evaluate l(2, 1,−1), find Ker l its basis and dimension. Is l surjective?

51

Solution. We will consider C, the standard basis of IR3, C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, and T , the standard basis ofP1, T = {1, x}. Using the approach described at the beginning of Solution of Exercise 3, we create the matrixA associated to l : IR3 → P1, l(a, b, c) = b− c+ (2a− c)x, with respect to C and T as follows:

l(1, 0, 0) = 2x = 0(1) + 2(x),

l(0, 1, 0) = 1 = 1(1) + 0(x),

l(0, 0, 1) = −1− x = −1(1)− 1(x),

from which

A = A(l, C, T ) =

(0 1 −12 0 −1

).

To evaluate l(2, 1,−1) we just use the definition of l with (a, b, c) = (2, 1,−1): l(2, 1,−1) = 2 + 5x.

To find Ker l =

{(x, y, z) ∈ IR3 : A

xyz

=

(00

)}, the space of solutions of the homogeneous system

associated to A, we will use the matrix form of Gaussian reduction and get:(0 1 −12 0 −1

∣∣∣∣∣∣∣∣ 00)

R1 ↔ R2

∼

(2 0 −10 1 −1

∣∣∣∣∣∣∣∣ 00).

The rank of A is two, this implies that dim Ker l = dim IR3 − rankA = 1. We may easily guess a solution ofthe homogeneous system associated to the last matrix, setting z = 1, from the second row we get y = 1, andfrom the first row, we get x = 1

2 . We conclude that:

Ker l = span{(1

2, 1, 1)} = span{(1, 2, 2)}, and a basis of Ker l is BK = {(1, 2, 2)}.

From the Theorem on dimensions we know that dim IR3 = dim Ker l + dim Im l, from which it follows thatdim Im l = 2 = dimP1. This implies that l is surjective.

5.13) Given the matrix M =

(1 23 4

), we define the linear transformation l :M2,2 →M2,2 by l(X) = M ·X.

Is l an isomorphism? If possible, find its inverse.

Solution. We consider the standard basis ofM2,2: C = {E1,E2,E3,E4} =

{(1 00 0

),

(0 10 0

),

(0 01 0

),

(0 00 1

)}.

We are looking for A = A(l, C, C), the matrix associated to the linear transformation l with respect to thebasis C. We will follow the approach described at the beginning of Solution of Exercise 3, and evaluate:

l(E1) = M ·E1 =

(1 23 4

)(1 00 0

)=

(1 03 0

)= 1E1 + 0E2 + 3E3 + 0E4,

l(E2) = M ·E2 =

(1 23 4

)(0 10 0

)=

(0 10 3

)= 0E1 + 1E2 + 0E3 + 3E4,

l(E3) = M ·E3 =

(1 23 4

)(0 01 0

)=

(2 04 1

)= 2E1 + 0E2 + 4E3 + 0E4,

l(E4) = M ·E4 =

(1 23 4

)(0 00 1

)=

(0 20 4

)= 0E1 + 2E2 + 0E3 + 4E4.

From this, it follows that:

A =

1 0 2 00 1 0 23 0 4 00 3 0 4

.

To determine if l is an isomorphism we will check if detA 6= 0, indeed l is invertible if and only if A isinvertible, and in this case the matrix associated to l−1 would be B = B(l−1, C, C) = A−1. We evaluate thedeterminant of A on the first row:

detA = 1 det

1 0 20 4 03 0 4

+ 0 + 2 det

0 1 23 0 00 3 4

+ 0 = −8 + 12 = 4 6= 0.

52

Thus, l is an isomorphism and is invertible. We can directly evaluate the inverse of l observing that fromY = l(X) = M · X, and the fact that M is invertible because detM = 4 − 6 = −2 6= 0, it follows that

X = l−1(Y) = M−1Y. We immediately evaluate M−1 = 1detM adjM = − 1

2

(4 −2−3 1

), and conclude that

l−1 :M2,2 →M2,2 is defined as

l−1(Y) = M−1Y =1

2

(−4 23 −1

)·Y.

Observation. In general, after we establish that a linear transformation is invertible, we may evaluate thematrix associated to the linear transformation l−1 recalling that this is equal to A−1. For the sake of solvingthe exercise in a second way, we will show also this method. We use the formula A−1 = 1

detA adjA. We willthus evaluate all cofactors of the given matrix.

A11 = (−1)1+1 · det

1 0 20 4 03 0 4

= −8

A12 = (−1)1+2 · det

0 0 23 4 00 0 4

= 0

A13 = (−1)1+3 · det

0 1 23 0 00 3 4

= 6

A14 = (−1)1+4 · det

0 1 03 0 40 3 0

= 0

A21 = (−1)2+1 · det

0 2 00 4 03 0 4

= 0

A22 = (−1)2+2 · det

1 2 03 4 00 0 4

= −8

A23 = (−1)2+3 · det

1 0 03 0 00 3 4

= 0

A24 = (−1)2+4 · det

1 0 23 0 40 3 0

= 6

A31 = (−1)3+1 · det

0 2 01 0 23 0 4

= 4

A32 = (−1)3+2 · det

1 2 00 0 20 0 4

= 0

A33 = (−1)3+3 · det

1 0 00 1 20 3 4

= −2

A34 = (−1)3+4 · det

1 0 20 1 00 3 0

= 0

A41 = (−1)4+1 · det

0 2 01 0 20 4 0

= 0

53

A42 = (−1)4+2 · det

1 2 00 0 23 4 0

= 4

A43 = (−1)4+3 · det

1 0 00 1 23 0 0

= 0

A44 = (−1)4+4 · det

1 0 20 1 03 0 4

= −2.

Now we can write A−1 = 14

−8 0 6 00 −8 0 64 0 −2 00 4 0 −2

T

= 12

−4 0 2 00 −4 0 23 0 −1 00 3 0 −1

.

We conclude that, for every Y =

(a bc d

)∈M2,2:

[l(Y)]C = A−1[Y]C =1

2

−4 0 2 00 −4 0 23 0 −1 00 3 0 −1

abcd

=1

2

−4a+ 2c−4b+ 2d3a− 2c3b− 2d

.

From this it follows that:

l(Y) =1

2

(−4a+ 2c −4b+ 2d3a− 2c 3b− 2d

)=

1

2

(−4 23 −1

)(a bc d

)=

1

2

(−4 23 −1

)·Y = M−1Y.

5.15) Given the linear transformations

f : IR3 → IR2, f(x, y, z) = (2x− y, 2y + z), and g : IR2 → IR3, g(u, v) = (u, u+ v, u− v),

find the matrix associated to f ◦g and g◦f with respect to the standard basis. Find rank(f ◦g) and rank(g◦f),is one of the two compositions an isomorphism? If yes find its inverse, otherwise identify Kernel and Image.

Solution. We recall that, given two linear transformations l : L → M , h : M → N , their composition h ◦ l : L → Nis linear, moreover, if we fix bases BL, BM , BN of the linear spaces L,M,N , then the matrices associated tothe linear transformations are such that:

Ah◦l = A(h ◦ l, BL, BN ) = A(h,BM , BN ) ·A(l, BL, BM ) = Ah ·Al.

We will thus work with the matrices associated to the given linear transformations f , g with respect to thecanonical bases B2 and B3 of IR2 and IR3 respectively: B2 = {(1, 0), (0, 1)}, B3 = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}.Following the approach explained in Solution of Exercise 3, we evaluate:

f(1, 0, 0) = (2, 0), f(0, 1, 0) = (−1, 2) f(0, 0, 1) = (0, 1).

It follows that

Af = A(f,B3, B2) =

(2 −1 00 2 1

).

Similarly, for the linear function g we have:

g(1, 0) = (1, 1, 1), g(0, 1) = (0, 1,−1).

It follows that

Ag = A(g,B2, B3) =

1 01 11 −1

.

We are now ready to evaluate the matrix associated to f ◦ g : IR2 → IR2:

Af◦g = Af ·Ag =

(2 −1 00 2 1

) 1 01 11 −1

=

(1 −13 1

).

54

From this we have the explicit definition of f ◦ g:

(f ◦ g)(u, v) =

(1 −13 1

)(uv

)= (u− v, 3u+ v).

Similarly, we evaluate the matrix associated to g ◦ f : IR3 → IR3:

Ag◦f = Ag ·Af =

1 01 11 −1

( 2 −1 00 2 1

)=

2 −1 02 1 12 −3 −1

.

Thus, the explicit definition of g ◦ f is:

(g ◦ f)(x, y, z) =

2 −1 02 1 12 −3 −1

xyz

= (2x− y, 2x+ y + z, 2x− 3y − z).

By definition, the rank of a linear transformation is the dimension of its Image.In order to find dim Im(f ◦ g) and dim Im(g ◦ f), we will first evaluate the determinants of the matrices thatrepresents these linear transformations, because, in case of non-zero determinants, the linear transformationsare isomorphisms and the dimension of their Images is maximal.

detAf◦g = det

(1 −13 1

)= 4 6= 0,

detAg◦f = det

2 −1 02 1 12 −3 −1

= −2− 2 + 6− 2 = 0.

Thus, f ◦ g is an isomorphism, Ker(f ◦ g) = {(0, 0)}, Im(f ◦ g) = IR2, rank(f ◦ g) = 2. f ◦ g is invertible andthe matrix associated to (f ◦ g)−1 is A(f◦g)−1 = A−1f◦g, that can be immediately evaluated using the formula

A−1 = 1detA adjA:

A(f◦g)−1 = A−1f◦g =

(1 −13 1

)−1=

1

4

(1 1−3 1

).

From this, we can write the explicit definition of (f ◦ g)−1 : IR2 → IR2 :

(f ◦ g)−1(u, v) =1

4

(1 1−3 1

)(uv

)=

(1

4u+

1

4v,−3

4u+

1

4v

).

The composition g ◦ f is not an isomorphism. Let’s identify its Kernel,

Ker(g ◦ f) = {(x, y, z) : (g ◦ f)(x, y, z) = (0, 0, 0)} =

{(x, y, z) : A(g◦f)

xyz

=

000

},this is the space of solutions of the homogeneous system associated to A(g◦f). We will use the matrix formof Gaussian reduction and get: 2 −1 0

2 1 12 −3 −1

R2 −R1

∼R3 −R1

2 −1 00 2 10 −2 −1

R3 +R2

∼

2 −1 00 2 10 0 0

.

We have that rankA(g◦f) = rank(g ◦ f) = 2, this implies that dim Ker(g ◦ f) = dim IR3 − rankA = 1. Wemay easily guess a solution of the homogeneous system associated to the last matrix, setting y = 1, from thesecond row we get z = −2, and from the first row, we get x = 1

2 . We conclude that:

Ker(g ◦ f) = span{(1

2, 1,−2)} = span{(1, 2,−4)}, and a basis of Ker(g ◦ f) is BK = {(1, 2,−4)}.

From the Theorem on dimensions we get again that dim IR3 = dim Ker(g ◦ f) + dim Im(g ◦ f), from which itfollows that dim Im(g ◦ f) = 2, as we already observed.

Im(g ◦ f) = span{(g ◦ f)(1, 0, 0), (g ◦ f)(0, 1, 0), (g ◦ f)(0, 0, 1)} =

= span{(2, 2, 2), (−1, 1,−3), (0, 1,−1)} = span{(−1, 1,−3), (0, 1,−1)},

55

where the last two vectors were chosen randomly, just considering that, because they are linearly independent(they are not a multiple of each other), they must span the space Im(g ◦ f) that has dimension 2. Thus, abasis of Im(g ◦ f) is B(g ◦ f) = {(−1, 1,−3), (0, 1,−1)}.

5.17) For the space IR3 are given two ordered bases:

B = {(1, 0, 0), (1, 2, 0), (1, 2, 1)}, S = {(2,−2, 0), (2,−1, 0), (0, 2,−2)}.

Find PB→S the matrix of transition from basis B to basis S.(Hint: use C, the standard basis of IR3, and the fact that PB→S = P−1S→CPB→C).

Solution. The matrix of transition from a basis B to a basis S, of the same linear space L, is a matrix PB→S with theproperty that, given any vector v ∈ L:

PB→S [v]B = [v]S ,

where [v]X are the coordinates of the vector v with respect to the basesX (seen as a column). Thus, the matrixPB→S = A(Id,B, S) can be seen as the matrix associated to the linear transformation Id : (L,B)→ (L, S),where Id(v) = v is the identity map.Let’s start considering the map Id : (IR3, B) → (IR3, C), where C is the standard basis of IR3, C ={(1, 0, 0), (0, 1, 0), (0, 0, 1)}. To evaluate PB→C = A(Id,B,C), following the method introduced in Solutionof Exercise 3, we calculate

Id(1, 0, 0) = (1, 0, 0) Id(1, 2, 0) = (1, 2, 0) Id(1, 2, 1) = (1, 2, 1),

thus PB→C = A(Id,B,C) =

1 1 10 2 20 0 1

.

Similarly, considering the map Id : (IR3, S)→ (IR3, C), we get PS→C = A(Id, S, C) =

2 2 0−2 −1 −20 0 −2

.

The Identity map is obviously an isomorphism, thus invertible, in particular the map Id : (IR3, S)→ (IR3, C)has as inverse the map Id : (IR3, C) → (IR3, S). For the theorem on matrices associated to inverse maps wehave A(Id, C, S) = A(Id, S, C)−1, thus PC→S = P−1S→C .Consider the following diagram:

Id(IR3, B)−→(IR3, S)

Id↘ ↗ Id(IR3, C)

illustrating how the map Id : (IR3, B)→ (IR3, S) can be obtained as a composition of two Identity maps:

Id : (IR3, B)→ (IR3, S) =[Id : (IR3, C)→ (IR3, S)

]◦[Id : (IR3, B)→ (IR3, C)

].

From this, and our previous considerations, we get:

A(Id,B, S) = A(Id, C, S) ·A(Id,B,C),

or, equivalently:PB→S = PC→SPB→C = P−1S→CPB→C .

We need to evaluate P−1S→C =

2 2 0−2 −1 −20 0 −2

−1. Let us indicate PS→C = P. Since P−1 = 1detP adjP, we

calculate: detP = det

2 2 0−2 −1 −20 0 −2

= −4, and then we proceed evaluating all cofactors of P:

P11 = (−1)1+1 · det

(−1 −20 −2

)= 2

P12 = (−1)1+2 · det

(−2 −20 −2

)= −4

56

P13 = (−1)1+3 · det

(−2 −10 0

)= 0

P21 = (−1)2+1 · det

(2 00 −2

)= 4

P22 = (−1)2+2 · det

(2 00 −2

)= −4

P23 = (−1)2+3 · det

(2 20 0

)= 0

P31 = (−1)3+1 · det

(2 0−1 −2

)= −4

P32 = (−1)3+2 · det

(2 0−2 −2

)= 4

P33 = (−1)3+3 · det

(2 2−2 −1

)= 2.

Now we can write P−1 = − 14

2 −4 04 −4 0−4 4 2

T

= − 14

2 4 −4−4 −4 40 0 2

= 12

−1 −2 22 2 −20 0 −1

. We thus

evaluate A(Id,B, S) = PB→S , calculating the indicated product:

PB→S = A(Id,B, S) = A(Id, C, S) ·A(Id,B,C) =1

2

−1 −2 22 2 −20 0 −1

1 1 10 2 20 0 1

=1

2

−1 −5 −32 6 40 0 −1

.

5.19) Let A(l) =

(1 1 11 −1 2

)be the matrix associated to a linear transformation l : IR3 → IR2 with respect to

the standard bases of IR3 and IR2. Find the matrix associated to the given transformation with respect tothe bases S and B, where:

S = {(2,−2, 2), (2,−1,−2), (0, 2,−2)}, and B = {(1, 1), (1,−1)}.

Solution. The given A(l) = A(l, C3, C2) is the matrix associated to the linear transformation l : (IR3, C3) → (IR2, C2)with respect to the canonical bases C3, C2 of IR3 and IR2 respectively. We are asked to find A(l, S,B).Let’s consider the following diagram:

(IR3, S)l−−−−→ (IR2, B)

Id

y xId(IR3, C3) −−−−→

l(IR2, C2)

where Id indicates the Identity map. This illustrates how the map l : (IR3, S)→ (IR2, B) can be obtained asa composition of Id : (IR3, S)→ (IR3, C3), l : (IR3, C3)→ (IR2, C2), and Id : (IR2, C2)→ (IR2, B):

l : (IR3, S)→ (IR2, B) =[Id : (IR2, C2)→ (IR2, B)

]◦[l : (IR3, C3)→ (IR2, C2)

]◦[Id : (IR3, S)→ (IR3, C3)

],

or, in terms of the matrices associated to the linear transformations:

A(l, B, S) = A(Id, C2, B) ·A(l, C3, C2) ·A(Id, S, C3).

Using the notation and the notions introduced in Solution of Exercise 17:

A(l, B, S) = PC2→B ·A(l) ·PS→C3 = P−1B→C2·A(l) ·PS→C3 ,

where A(l) is the given matrix, and, using the considerations explained in Solution of Exercise 17,:

PS→C3=

2 2 0−2 −1 22 −2 −2

and PB→C2=

(1 11 −1

).

57

We immediately evaluate P−1B→C2= 1

2

(1 11 −1

). Now, we get A(l, B, S) calculating the indicated product:

A(l, B, S) =1

2

(1 11 −1

)·(

1 1 11 −1 2

)·

2 2 0−2 −1 22 −2 −2

=

=1

2

(1 11 −1

)·(

2 −1 08 −1 −6

)=

1

2

(10 −2 −6−6 2 6

)=

(5 −1 −3−3 1 3

).

5.21) Given the linear transformation lh : IR3 → IR3, lh(x, y, z) = (2x+y, y−z, 2y+hz), write the matrix associatedto lh with respect to the standard basis of IR3, find Ker l, Im l, its bases and dimensions for every possiblevalue of h ∈ IR. Does there exist a value of h for which lh is an isomorphism?.

Solution. We consider the standard basis C of IR3, C = {(1, 0, 0), (0, 1, 1), (0, 0, 1)}, and, following the method explainedin Solution of Exercise 3, we evaluate Ah = Ah(lh, C, C) as follows:

l(1, 0, 0) = (2, 0, 0), l(0, 1, 0) = (1, 1, 2), l(0, 0, 1) = (0,−1, h),

from which we deduce:

Ah =

2 1 00 1 −10 2 h

.

To find Ker lh =

{(x, y, z) ∈ IR3 : Ah

xyz

=

000

}, the space of solutions of the homogeneous system

associated to Ah, we first evaluate detAh, because, for all values of h ∈ IR such that detAh 6= 0, we knowthat the homogeneous system has a unique solution, the trivial one, thus Ker lh = {0}.Since detAh = 2h+4 = 2(h+2), we conclude that, for h 6= −2, Ker lh = {(0, 0, 0)}, consequently Im lh = IR3,that means that lh is an isomorphism.For h = −2, we need to find a basis of

Ker l2 =

{(x, y, z) ∈ IR3 : A2

xyz

=

000

} =

{(x, y, z) ∈ IR3 :

2 1 00 1 −10 2 −2

xyz

=

000

},that means, we need to solve the homogeneous system: 2 1 0

0 1 −10 2 −2

∣∣∣∣∣∣∣∣∣∣∣∣000

R3 − 2R2

∼

2 1 00 1 −10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

Setting z = α, α ∈ IR, we get from the second row (imagining the correspondent equation y − z = 0) y = α,and, subsequently, from the first row x = − 1

2α. Thus we conclude that:Ker l2 = {(− 1

2α, α, α), α ∈ IR} = span{(− 12 , 1, 1)} = span{(−1, 2, 2)}, dim Ker l2 = 1, and a basis of Ker l2 is

BK2 = {(−1, 2, 2)}.From the Theorem on dimensions (ii) Solution of Exercise 3), we get dim Im l2 = dim IR3 − dim Ker l2 = 2,and, from Theorem i) we get

Im l2 = span{l(1, 0, 0), l(0, 1, 0), l(0, 0, 1)} = span{(2, 0, 0), (1, 1, 2), (0,−1,−2)} = span{(2, 0, 0), (1, 1, 2)},

where the first two vectors were chosen because they are linearly independent (they are not a multiple of eachother), thus they must span the space Im l2 that has dimension 2. A basis of Im l2 is BI2 = {(2, 0, 0), (1, 1, 2)}.

58

Chapter 66 Eigenvalues, eigenvectors and their applications

6.1) Determine if A =

(3 −11 1

), B =

(1 −12 1

)and C =

2 1 10 3 10 0 2

are diagonalizable (in IR).


1 −3 33 −5 36 −6 4

and B =

−3 1 −1−7 5 −1−6 6 −2

are diagonalizable.

Notice that the given matrices have the same characteristic polynomial but they are not similar.

6.3) Diagonalize the following matrices (for each eigenvalue find a basis of the corresponding eigenspace)

A =

−2 1 01 −2 0−1 1 1

, C =

0 1 41 −1 11 0 1

.

6.4) Diagonalize the following matrices (for each eigenvalue find a basis of the corresponding eigenspace)

A =

(1 14 1

), B =

2 1 11 2 1−2 −2 −2

.

6.5) Are the following matrices diagonalizable in IR?

A =

2 1 01 3 −1−1 2 3

, B =

1 1 0 02 0 0 00 0 1 20 0 2 1

, C =

2 2 2 20 1 1 10 0 2 20 0 0 1

.


A =

1 1 0 00 1 0 00 0 2 20 1 2 2

, B =

5 2 −22 5 −2−2 −2 5

.

6.7) Given the transformation l : IR3 → IR3, l(x, y, z) = (2x+ y, y − z, 2y + 4z), find all eigenvalues and a basis ofeach eigenspace. Is l diagonalizable? Determine if l is invertible.

6.8) Given the linear transformations from IR2 toIR2 defined below, find characteristic equation, real eigenvaluesand corresponding eigenvectors.a) l(x, y) = (x+ 5y, 2x+ 4y),b) f(1, 0) = (2,−5), f(0, 1) = (4, 4).

6.9) Given the linear transformation l : IR2 → IR2, l(x, y) = (x− y, x+ 3y), write the matrix A(l, S, S) associatedto l with respect to the basis S = {(1, 2), (2,−2)}. Determine if there exists a basis B of IR2 such thatA(l, B,B) is diagonal.

6.10) Given the linear transformation l : IR3 → IR3, l(x, y, z) = (2x+y+3z, x+3y+z, x) find a basisB of eigenvectorsof l, such that the matrix D associated to l with respect to B is diagonal. Verify that P−1AP = D, where Pis the transition matrix from basis B to the canonical one, and A is the matrix associated to l with respectto the canonical basis.

6.11) Given the transformation fh : IR3 → IR3 defined by fh(x, y, z) = (x− hz, x+ y − hz,−hx+ z), where h ∈ IRis a parameter, determine if fh is diagonalizable for some values of h. For such values of h, find a basis of IR3

of eigenvectors of fh.

6.12) Given the linear transformation l : IR3 → IR3 defined by

59

l(x, y, z) = (5x+ 6y − 3z, 6x+ 9y,−3x+ 9z),

find a diagonal matrix associated to l (if it exists) and the corresponding basis of IR3.

6.13) Find the linear transformation l : IR3 → IR3 stretching every vector on the plane x + 2y + 3z = 0 twice andreversing the vector (0, 1, 1).

6.14) Find the linear transformation l : IR3 → IR3 stretching twice the vector (1, 0, 0), three times the vector (1, 1, 0)and mapping (1, 0, 1) to (0, 0, 0).

6.15) Find the linear transformation l : IR2 → IR2 knowing that (1, 2) and (1, 3) are two of its eigenvectors andl(1, 0) = (4, 6)

6.16) Find the linear transformation l : IR2 → IR2 knowing that (2, 2) and (−1, 3) are two of its eigenvectors andl(0, 1) = (2, 1)

6.17) Given A =

(2 −34 −5

), use the eigenvalue method

a) to derive an explicit formula for An,b) to solve the system of differential equations

dx

dt= 2x− 3y

dy

dt= 4x− 5y.

6.18) Given A =

(2 11 2


a) to derive an explicit formula for An,b) to solve the system of differential equations

dx

dt= 2x+ y

dy

dt= x+ 2y.

6.19) Solve the system of recurrence relations

xn+1 = 2xn − ynyn+1 = −xn + 2yn

given that x0 = 0 and y0 = −1.


xn+1 = 3xn − ynyn+1 = −xn + 3yn

given that x0 = 1 and y0 = 2.

60

Solutions of odd number problems Eigenvalues, eigenvectors and their applications

Basic facts: Given a matrix A ∈ Mn,n, a real number λ ∈ IR is called an eigenvalue of A if there existsa nontrivial vector v ∈ IRn (v 6= 0), such that Av = λv. The vector v is said to be an eigenvector of A,associated to the eigenvalue λ. The set of all eigenvectors associated to the same eigenvalue λ, plus the zerovector, form a linear subspace of IRn, called eigenspace associated to the eigenvalue λ. Eigenvectors corre-sponding to different eigenvalues are linearly independent. If there exists a basis of IRn made of eigenvectorsof A, then A is diagonalizable and, if P is the matrix formed with the n linearly independent eigenvectors ofA written as columns, then P−1AP = D, where D is a diagonal matrix, with entries in the diagonal equalto the corresponding eigenvalues of A.In order to identify the eigenvalues of a matrix A, we observe that, looking for a nontrivial solution of theequation Av = λv is equivalent to finding nontrivial solutions of the homogeneous system (A−λI)v = 0, thuswe must find the possible values of λ ∈ IR such that det(A−λI) = 0. The last equation is called characteristicequation of A. Given one solution λ of the characteristic equation, the eigenvectors corresponding to λ areall nontrivial solutions of the system (A− λI)v = 0.


(3 −11 1

), B =

(1 −12 1

)and C =

2 1 10 3 10 0 2

are diagonalizable (in IR).

Solution. Let’s consider A =

(3 −11 1

), we start by finding the real solutions of det(A − λI) = 0. In this case, we

have:

det

((3 −11 1

)− λ

(1 00 1

))= det

(3− λ −1

1 1− λ

)= (3− λ)(1− λ) + 1 = λ2 − 4λ+ 4 = (λ− 2)2.

The characteristic equation of A has a unique solution λ = 2 with algebraic multiplicity 2. We now look forthe eigenvectors corresponding to λ = 2. We must find the nontrivial solutions of the homogeneous system(A− λI)v = 0, where λ = 2. Thus, we look for v = (x, y) 6= (0, 0) so that(

3− 2 −11 1− 2

)(xy

)=

(00

).

We solve the homogeneous system with augmented matrix:(1 −11 −1

∣∣∣∣∣∣∣∣ 00)

R2 −R1 ∼(

1 −10 0

∣∣∣∣∣∣∣∣ 00).

All the eigenvectors corresponding to λ = 2 are {t(1, 1) : t ∈ IR \ {0}} and the eigenspace associated to λ = 2is U2 = span{(1, 1)}. We found only one linearly independent eigenvector corresponding to the eigenvaluewith algebraic multiplicity two, thus the given matrix A is not diagonalizable.

Let’s now consider B =

(1 −12 1

). We solve det(B− λI) = 0.

det

((1 −12 1

)− λ

(1 00 1

))= det

(1− λ −1

2 1− λ

)= (1− λ)2 + 2.

The characteristic equation of B has no real solutions, thus we did not find any real eigenvalue, the matrix isnot diagonalizable on the reals.

Let’s now consider C =

2 1 10 3 10 0 2

. We solve det(C− λI) = 0.

det

( 2 1 10 3 10 0 2

− λ 1 0 0

0 1 00 0 1

) = det

2− λ 1 10 3− λ 10 0 2− λ

= (2− λ)2(3− λ).

The characteristic equation has solutions λ = 2 with algebraic multiplicity two, and λ = 3 with algebraicmultiplicity one. We observe that C is an upper triangular matrix, thus the eigenvalues are exactly the

61

numbers on the diagonal and their multiplicity corresponds with the number of times they appear on thediagonal.We now look for the eigenvectors corresponding to λ = 2. We must find the nontrivial solutions of thehomogeneous system (C− λI)v = 0, where λ = 2. The augmented matrix of the system is 0 1 1

0 1 10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

R2 −R1 ∼

0 1 10 0 00 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 2 are {t(0, 1,−1) + u(1, 0, 0) : t, u ∈ IR non both zero}, and theeigenspace corresponding to λ = 2 is U2 = span{(1, 0, 0), (0, 1,−1)}.We now look for the eigenvectors corresponding to λ = 3. We must find the nontrivial solutions of thehomogeneous system (C− λI)v = 0, where λ = 3. The augmented matrix of the system is−1 1 1

0 0 10 0 −1

∣∣∣∣∣∣∣∣∣∣∣∣000

R1 −R2

∼R3 +R2

−1 1 00 0 10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 3 are {s(1, 1, 0) : s ∈ IR \ {0}}, and the eigenspace correspondingto λ = 3 is U3 = span{(1, 1, 0)}. Thus the set B = {(1, 0, 0), (0, 1,−1), (1, 1, 0)} is basis of IR3, formed by 3linearly independent eigenvectors of C, and C is diagonalizable. One matrix P that diagonalizes C is:

P =

1 0 10 1 10 −1 0

,

and in this case:

P−1CP = D =

2 0 00 2 00 0 3

.

By changing the order of the vectors of B as columns of P, we may find several different other matrices thatdiagonalize C, for example:

P′ =

1 0 11 1 00 −1 0

,

in this case:

P′−1

CP′ = D′ =

3 0 00 2 00 0 2

.

6.3) Diagonalize the following matrices (for each eigenvalue find a basis of the corresponding eigenspace):

A =

−2 1 01 −2 0−1 1 1

, C =

0 1 41 −1 11 0 1

.

Solution. Let’s consider A =

−2 1 01 −2 0−1 1 1

. We start by finding the real solutions of det(A− λI) = 0. In this case,

we have:

det

(−2 1 01 −2 0−1 1 1

− λ 1 0 0

0 1 00 0 1

) = det

−2− λ 1 01 −2− λ 0−1 1 1− λ

= (1− λ)(3 + λ)(1 + λ).

The characteristic equation of A has three distinct solutions λ = 1, λ = −3, λ = −1, with algebraicmultiplicity 1. In this case, we may already conclude that A is diagonalizable, indeed, corresponding to eacheigenvalue we will choose an eigenvector, and, since the eigenvalues are distinct, the chosen eigenvectors arelinearly independent and form a basis of IR3.We now look for the eigenvectors corresponding to λ = 1. We must find the nontrivial solutions of thehomogeneous system (A− λI)v = 0, where λ = 1. Thus, we solve the homogeneous system with augmentedmatrix: −3 1 0

1 −3 0−1 1 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

62

All the eigenvectors corresponding to λ = 1 are {t(0, 0, 1) : t ∈ IR \ {0}} and the eigenspace associated toλ = 1 is U1 = span{(0, 0, 1)}.We consider λ = −3, and find the corresponding eigenvectors solving the system: 1 1 0

1 1 0−1 1 4

∣∣∣∣∣∣∣∣∣∣∣∣000

R2 −R1

∼R3 +R1

1 1 00 0 00 2 4

∣∣∣∣∣∣∣∣∣∣∣∣000

.

It is easy to deduce that all the eigenvectors corresponding to λ = −3 are {t(2,−2, 1) : t ∈ IR \ {0}} and theeigenspace associated to λ = −3 is U−3 = span{(2,−2, 1)}.Finally, we consider the eigenvalue λ = −1, and find the corresponding eigenvectors solving the system:−1 1 0

1 −1 0−1 1 2

∣∣∣∣∣∣∣∣∣∣∣∣000

R2 +R1

∼R3 +R2

−1 1 00 0 00 0 2

∣∣∣∣∣∣∣∣∣∣∣∣000

.

The eigenvectors corresponding to λ = −1 are {t(1, 1, 0) : t ∈ IR \ {0}} and the eigenspace associated toλ = −1 is U−1 = span{(1, 1, 0)}.

We conclude that the given matrix A is similar to a diagonal matrix, and the matrix P =

0 2 10 −2 11 1 0

is

such that

P−1AP = D =

1 0 00 −3 00 0 −1

.

Let’s consider C =

0 1 41 −1 11 0 1

. Looking for real solutions of det(C− λI) = 0, we have:

det

( 0 1 41 −1 11 0 1

− λ 1 0 0

0 1 00 0 1

) = det

−λ 1 41 −1− λ 11 0 1− λ

= −λ3 + 6λ+ 4.

To find the solution of the characteristic equation, we must find the roots of the polynomial p(λ) = −λ3+6λ+4.Among the integer divisors of 4, we choose −2, and with the use of a Horner’s scheme we evaluate:

−1 0 6 4−2 2 − 4 − 4

−1 2 2 0

Thus −2 is a root of p(λ) and:

−λ3 + 6λ+ 4 = (λ+ 2)(−λ2 + 2λ+ 2).

The remaining two roots can be found using the formula for quadratic equations

−λ3 + 6λ+ 4 = −(λ+ 2)(λ− 1 +√

3)(λ− 1−√

3).

The characteristic equation of C has three distinct solutions λ = −2, λ = 1−√

3, λ = 1 +√

3, with algebraicmultiplicity 1. Again, we may already conclude that C is diagonalizable because the eigenvalues are threedistinct numbers. We now look for the corresponding eigenspaces.For λ = −2, we solve (C−λI)v = 0, where λ = −2. Thus, we solve the homogeneous system with augmentedmatrix: 2 1 4

1 1 11 0 3

∣∣∣∣∣∣∣∣∣∣∣∣000

R1 −R2 −R3

∼

2 1 40 0 01 0 3

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = −2 are {t(−3, 2, 1) : t ∈ IR \ {0}} and the eigenspace associated toλ = −2 is U−2 = span{(−3, 2, 1)}.For λ = 1 −

√3, we solve (C − λI)v = 0, where λ = 1 −

√3. Thus, we solve the homogeneous system with

augmented matrix:−1 +√

3 1 41 −2 +

√3 1

1 0√

3

∣∣∣∣∣∣∣∣∣∣∣∣000

(1−√

3)R2 +R1

∼R2 −R3

−1 +√

3 1 40 −4 + 3

√3 5−

√3

0 2−√

3√

3− 1

∣∣∣∣∣∣∣∣∣∣∣∣000

∼63

(1 + 2√

3)R3 −R2

∼R3 ↔ R2

−1 +√

3 1 40 2−

√3√

3− 10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

After simple calculation, we find that all the eigenvectors corresponding to λ = 1−√

3 are {t(√

3, 1+√

3,−1) :t ∈ IR \ {0}} and the eigenspace associated to λ = 1−

√3 is U1−

√3 = span{(

√3, 1 +

√3,−1)}.

Finally, for λ = 1 +√

3, we solve (C− λI)v = 0, where λ = 1 +√

3. Thus, we solve the homogeneous systemwith augmented matrix:−1−

√3 1 4

1 −2−√

3 11 0 −

√3

∣∣∣∣∣∣∣∣∣∣∣∣000

(1 +√

3)R2 +R1

∼R3 −R2

−1−√

3 1 40 −4− 3

√3 5 +

√3

0 2 +√

3 −√

3− 1

∣∣∣∣∣∣∣∣∣∣∣∣000

∼(1− 2

√3)R3 −R2

∼R3 ↔ R2

−1 +√

3 1 40 2 +

√3 −

√3− 1

0 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

After simple calculation, we find that all the eigenvectors corresponding to λ = 1+√

3 are {t(√

3,−1+√

3, 1) :t ∈ IR \ {0}} and the eigenspace associated to λ = 1−

√3 is U1+

√3 = span{(

√3,−1 +

√3, 1)}.

We conclude that the given matrix C is similar to a diagonal matrix D, and the matrix P formed by thefound eigenvectors:

P =

−3√

3√

32 1 +

√3 −1 +

√3

1 −1 1

,

is such that

P−1CP = D =

−2 0 00 1−

√3 0

0 0 1 +√

3

.


A =

2 1 01 3 −1−1 2 3

, B =

1 1 0 02 0 0 00 0 1 20 0 2 1

, C =

2 2 2 20 1 1 10 0 2 20 0 0 1

.

Solution. We first consider the matrix A =

2 1 01 3 −1−1 2 3

, and look for its eigenvalues solving the characteristic

equation det(A− λI) = 0. In this case, we have:

det

( 2 1 01 3 −1−1 2 3

− λ 1 0 0

0 1 00 0 1

) = det

2− λ 1 01 3− λ −1−1 2 3− λ

= −λ3 + 8λ2 − 22λ+ 20.

To find the solution of the characteristic equation, we must find the roots of the polynomial p(λ) = −λ3 +8λ2 − 22λ + 20. Among the integer divisors of 20, we choose 2, and with the use of a Horner’s scheme weevaluate:

−1 8 − 22 202 − 2 12 − 20

−1 6 − 10 0

Thus 2 is a root of p(λ) and:

−λ3 + 8λ2 − 22λ+ 20 = (λ− 2)(−λ2 + 6λ− 10).

The polynomial −λ2 + 6λ − 10 is irreducible, thus the characteristic equation has only one real root. Thisimplies that the given matrix A is not diagonalizable on the reals.

We now consider the matrix B =

1 1 0 02 0 0 00 0 1 20 0 2 1

, and look for its eigenvalues solving the characteristic

equation det(B− λI) = 0. We have:

64

det

(1 1 0 02 0 0 00 0 1 20 0 2 1

− λ

1 0 0 00 1 0 00 0 1 00 0 0 1

) = det

1− λ 1 0 0

2 −λ 0 00 0 1− λ 20 0 2 1− λ

=

= (1− λ) det

−λ 0 00 1− λ 20 2 1− λ

− (1) det

2 0 00 1− λ 20 2 1− λ

=

= (1− λ)(−λ) det

(1− λ 2

2 1− λ

)− (1)(2) det

(1− λ 2

2 1− λ

)=

det

(1− λ 2

2 1− λ

)(− λ(1− λ)− 2

)=((1− λ)2 − 4

)(λ2 − λ− 2

)= (λ+ 1)2(λ− 3)(λ− 2).

Notice that, incidently, we have just proven that:

det

1− λ 1 0 0

2 −λ 0 00 0 1− λ 20 0 2 1− λ

= det

(1− λ 1

2 −λ

)det

(1− λ 2

2 1− λ

).

The characteristic equation of B has three real solutions: λ = 2 and λ = 3 with algebraic multiplicity one,and λ = −1 with algebraic multiplicity two. We now find the corresponding eigenvectors.For λ = 2, we solve the system (B − λI)v = 0, where λ = 2. Thus, we solve the homogeneous system withaugmented matrix:

−1 1 0 02 −2 0 00 0 −1 20 0 2 −1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

R2 + 2R1

∼R4 + 2R3

−1 1 0 00 0 0 00 0 −1 20 0 0 3

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

All the eigenvectors corresponding to λ = 2 are {t(1, 1, 0, 0) : t ∈ IR \ {0}} and the eigenspace associated toλ = 2 is U2 = span{(1, 1, 0, 0)}.For λ = 3, we solve the system (B − λI)v = 0, where λ = 3. Thus, we solve the homogeneous system withaugmented matrix:

−2 1 0 02 −3 0 00 0 −2 20 0 2 −2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

R2 +R1

∼R4 +R3

−2 1 0 00 −2 0 00 0 −2 20 0 0 0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

All the eigenvectors corresponding to λ = 3 are {t(0, 0, 1, 1) : t ∈ IR \ {0}} and the eigenspace associated toλ = 3 is U3 = span{(0, 0, 1, 1)}.For λ = −1, we solve the system (B − λI)v = 0, where λ = −1. Thus, we solve the homogeneous systemwith augmented matrix:

2 1 0 02 1 0 00 0 2 20 0 2 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

R2 −R1

∼R4 −R3

2 1 0 00 0 0 00 0 2 20 0 0 0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

The rank of the matrix is two, thus the eigenspace U−1 has dimension two. All the eigenvectors correspondingto λ = −1 are {t(1,−2, 0, 0)+u(0, 0, 1,−1) : t, u ∈ IR non both zero } and the eigenspace associated to λ = −1is U−1 = span{(1,−2, 0, 0), (0, 0, 1,−1)}.We conclude that the set B = {(1, 1, 0, 0), (0, 0, 1, 1), (1,−2, 0, 0), (0, 0, 1,−1)} is basis of IR4, formed by 4linearly independent eigenvectors of B, and B is diagonalizable. One matrix P that diagonalizes B is:

P =

1 0 1 01 0 −2 00 1 0 10 1 0 −1

,

65

and in this case:

P−1BP = D =

2 0 0 00 3 0 00 0 −1 00 0 0 −1

.

We now consider the matrix C =

2 2 2 20 1 1 10 0 2 20 0 0 1

, and evaluate its eigenvalues by solving the equation

det(C− λI) = 0. We have:

det

(2 2 2 20 1 1 10 0 2 20 0 0 1

− λ

1 0 0 00 1 0 00 0 1 00 0 0 1

) = det

2− λ 2 2 2

0 1− λ 1 10 0 2− λ 20 0 0 1− λ

= (2− λ)2(1− λ)2.

The characteristic equation of C has two real solutions: λ = 2 and λ = 1, both with algebraic multiplicitytwo. We now find the corresponding eigenspaces.For λ = 2, we solve the system (C − λI)v = 0, where λ = 2. Thus, we solve the homogeneous system withaugmented matrix:

0 2 2 20 −1 1 10 0 0 20 0 0 −1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

R1 + 2R2

∼R3 + 2R4

0 2 2 20 0 4 40 0 0 20 0 0 0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

All the eigenvectors corresponding to λ = 2 are {t(1, 0, 0, 0) : t ∈ IR \ {0}} and the eigenspace associatedto λ = 2 is U2 = span{(1, 0, 0, 0)}. Since the eigenvalue λ = 2 has algebraic multiplicity two, but thedimension of the corresponding eigenspace is one, we may already conclude that the given matrix C is notdiagonalizable. We anyway determine also the eigenspace corresponding to the eigenvalue λ = 1. We solvethe system (C− λI)v = 0, where λ = 1. This homogeneous system has augmented matrix:

1 2 2 20 0 1 10 0 1 20 0 0 0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣0000

.

All the eigenvectors corresponding to λ = 1 are {t(2,−1, 0, 0) : t ∈ IR \ {0}}, the eigenspace associated toλ = 1 is U1 = span{(2,−1, 0, 0)}, thus also U1 has dimension only one. C is not diagonalizable.

6.7) Given the transformation l : IR3 → IR3, l(x, y, z) = (2x+ y, y − z, 2y + 4z), find all eigenvalues and a basis ofeach eigenspace. Is l diagonalizable? Determine if l is invertible.

Solution. First of all, we will find the matrix A = A(l, C, C) associated to the linear transformation l with respect tothe canonical basis C of IR3, C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. For this purpose, we need to evaluate:

l(1, 0, 0) =(2, 0, 0)

l(0, 1, 0) =(1, 1, 2)

l(0, 0, 1) =(0,−1, 4).


2 1 00 1 −10 2 4

. The eigenvalues and eigenvectors of l correspond with the eigenvalues

and eigenvectors of A, and l is diagonalizable if and only if A is diagonalizable. We look for eigenvalues of Asolving the characteristic equation det(A− λI) = 0. In this case, we have:

det

( 2 1 00 1 −10 2 4

− λ 1 0 0

0 1 00 0 1

) = det

2− λ 1 00 1− λ −10 2 4− λ

=

= (2− λ)(1− λ)(4− λ) + 2(2− λ) = (2− λ)(λ2 − 5λ+ 6) = (2− λ)2(3− λ).

The characteristic equation has solutions λ = 2 with algebraic multiplicity two, and λ = 3 with algebraicmultiplicity one.

66

We now look for the eigenvectors corresponding to λ = 2. We must find the nontrivial solutions of thehomogeneous system (A− λI)v = 0, where λ = 2. The augmented matrix of the system is 0 1 0

0 −1 −10 2 2

∣∣∣∣∣∣∣∣∣∣∣∣000

−R2

∼R3 + 2R2

0 1 00 1 10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 2 are {t(1, 0, 0) : t ∈ IR \ {0}}, and the eigenspace correspondingto λ = 2 is U2 = span{(1, 0, 0)}. Corresponding to the eigenvalue λ = 2 with algebraic multiplicity two, wefound and eigenspace U2 with dimension one. This implies that A, and thus l, is not diagonalizable.

We anyway look for the eigenvectors corresponding to λ = 3. We must find the nontrivial solutions of thehomogeneous system (A− λI)v = 0, where λ = 3. The augmented matrix of the system is−1 1 0

0 −2 −10 2 1

∣∣∣∣∣∣∣∣∣∣∣∣000

−R2

∼R3 +R2

−1 1 00 2 10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 3 are {t(1, 1,−2) : t ∈ IR \ {0}}, and the eigenspace correspondingto λ = 3 is U3 = span{(1, 1,−2)}.It is left to answer the question on the existence of l−1. Using the fact that the eigenvalues of l are 2 and 3,and the value 0 is not an eigenvalue of l, we may conclude that Ker l = {(0, 0, 0)} and thus l is invertible.

6.9) Given the linear transformation l : IR2 → IR2, l(x, y) = (x− y, x+ 3y), write the matrix A(l, S, S) associatedto l with respect to the basis S = {(1, 2), (2,−2)}. Determine if there exists a basis B of IR2 such thatA(l, B,B) is diagonal.

Solution. We start by finding the matrix A = A(l, C, C) associated to l with respect to the canonical basis C ={(1, 0), (0, 1)} of IR2. For this purpose, we need to evaluate:

l(1, 0) =(1, 1)

l(0, 1) =(−1, 3).

Thus A = A(l, C, C) =

(1 −11 3

)is the matrix associated to the linear transformation l : (IR2, C)→ (IR2, C).

We are asked to find A(l, S, S). Using the notations introduced in Exercise 17 and 19 of the chapter on LinearTransformations, we consider the following diagram:

(IR2, S)l−−−−→ (IR2, S)

Id

y xId(IR2, C) −−−−→

l(IR2, C)

where Id indicates the Identity map. This illustrates how the map l : (IR2, S)→ (IR2, S) can be obtained asa composition of Id : (IR2, S)→ (IR2, C), l : (IR2, C)→ (IR2, C), and Id : (IR2, C)→ (IR2, S):

l : (IR2, S)→ (IR2, S) =[Id : (IR2, C)→ (IR2, S)

]◦[l : (IR2, C)→ (IR2, C)

]◦[Id : (IR2, S)→ (IR2, C)

],

or, in terms of the matrices associated to the linear transformations:

A(l, S, S) = A(Id, C, S) ·A(l, C, C) ·A(Id, S, C).

Using the notation and the notions introduced in Solution of Exercise 17 in the chapter of Linear Transfor-mations:

A(l, S, S) = PC→S ·A ·PS→C = P−1S→C ·A(l) ·PS→C ,

where A is the matrix we have already evaluated, and PS→C is the transition matrix from basis S to basisC, formed by the vectors of basis S written as columns:

PS→C =

(1 22 −2

)67

We immediately evaluate P−1S→C = 16

(2 22 −1

). Now, we get A(l, S, S) calculating the indicated product:

A(l, S, S) =1

6

(2 22 −1

)·(

1 −11 3

)·(

1 22 −2

)=

=1

6

(2 22 −1

)·(−1 47 −4

)=

1

2

(12 0−9 12

)=

(2 0−3/2 2

).

The basis S does not diagonalize the linear transformation. To verify if there exists a basis B that diagonalizesl, we must evaluate the eigenvalues of l (eigenvalues of A) and find a possible basis B of IR2 made ofeigenvectors of l. We look for eigenvalues of A solving the characteristic equation det(A − λI) = 0. In thiscase, we have:

det

((1 −11 3

)− λ

(1 00 1

))= det

(1− λ −1

1 3− λ

)=

= (1− λ)(3− λ) + 1 = λ2 − 4λ+ 4 = (λ− 2)2.

The characteristic equation has solution λ = 2 with algebraic multiplicity two.We now look for the eigenvectors corresponding to λ = 2. We must find the nontrivial solutions of thehomogeneous system (A− λI)v = 0, where λ = 2. The augmented matrix of the system is(

−1 −11 1

∣∣∣∣∣∣∣∣ 00) −R1

∼R2 + 2R1

(1 10 0

∣∣∣∣∣∣∣∣ 00).

All the eigenvectors corresponding to λ = 2 are {t(1,−1) : t ∈ IR \ {0}}, and the eigenspace correspondingto λ = 2 is U2 = span{(1,−1)}. Corresponding to the eigenvalue λ = 2 with algebraic multiplicity two, wefound and eigenspace U2 with dimension one. This implies that A, and thus l, is not diagonalizable: thereexists no basis B of IR2 such that A(l, B,B) is diagonal.

6.11) Given the transformation fh : IR3 → IR3 defined by fh(x, y, z) = (x− hz, x+ y − hz,−hx+ z), where h ∈ IRis a parameter, determine if fh is diagonalizable for some values of h. For such values of h, find a basis of IR3

of eigenvectors of fh.

Solution. We consider the canonical basis C of IR3, C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, and evaluate the matrix Ah

associated to the given linear transformation fh with respect to C. For this purpose, we need to calculate:

fh(1, 0, 0) =(1, 1,−h)

fh(0, 1, 0) =(0, 1, 0)

fh(0, 0, 1) =(−h,−h, 1).

Now we can write Ah =

1 0 −h1 1 −h−h 0 1

. The eigenvalues and eigenvectors of fh correspond with the

eigenvalues and eigenvectors of Ah, and fh is diagonalizable if and only if Ah is diagonalizable. We look foreigenvalues of Ah solving the characteristic equation det(Ah − λI) = 0:

det

( 1 0 −h1 1 −h−h 0 1

− λ 1 0 0

0 1 00 0 1

) =

1− λ 0 −h1 1− λ −h−h 0 1− λ

=

= (1− λ)3 − (1− λ)h2 = (1− λ[(1− λ)2 − h2]) = (1− λ)(1− λ− h)(1− λ+ h).

The solutions of the characteristic equation are λ = 1, λ = 1−h, and λ = 1+h, thus for any h 6= 0, the foundeigenvalues are distinct, each with algebraic multiplicity one, and thus the corresponding fh is diagonalizable.Let’s find the eigenvectors of fh, for any h ∈ IR \ {0}.For λ = 1, we must find the nontrivial solutions of the homogeneous system (Ah − λI)v = 0, where λ = 1.The augmented matrix of the system is 0 0 −h

1 0 −h−h 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 1 are {t(0, 1, 0) : t ∈ IR \ {0}}, and the eigenspace corresponding toλ = 1 is U1 = span{(0, 1, 0)}.

68

For λ = 1 − h, we must find the nontrivial solutions of the homogeneous system (Ah − λI)v = 0, whereλ = 1− h. The augmented matrix of the system is h 0 −h

1 h −h−h 0 h

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 1 − h are {t(1, 1 − 1h , 1) : t ∈ IR \ {0}}, and the eigenspace

corresponding to λ = 1− h is U1−h = span{(1, 1− 1h , 1)}.

For λ = 1 + h, we must find the nontrivial solutions of the homogeneous system (Ah − λI)v = 0, whereλ = 1 + h. The augmented matrix of the system is−h 0 −h

1 −h −h−h 0 −h

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 1 + h are {t(1, 1 + 1h ,−1) : t ∈ IR \ {0}}, and the eigenspace

corresponding to λ = 1 + h is U1+h = span{(1, 1 + 1h ,−1)}.

For any h 6= 0, we found a basis B of IR3 made of eigenvectors of fh:

B = {(0, 1, 0), (1, 1− 1

h, 1), (1, 1 +

1

h,−1)}.

Let’s now analyse the case when h = 0, and determine if A0 =

1 0 01 1 00 0 1

is diagonalizable. We have already

calculated the eigenvalues of A0: we found that λ = 1 is the only eigenvalue with algebraic multiplicity three.Let’s now evaluate the eigenvectors corresponding to λ = 1. We must find the nontrivial solutions of thehomogeneous system (A0 − λI)v = 0, where λ = 1. The augmented matrix of the system is 0 0 0

1 0 00 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All eigenvalues of A0 are {t(0, 1, 0) + u(0, 0, 1) : t, u ∈ IR non both zero}, and the eigenspace correspondingto λ = 1 is U1 = span{(0, 1, 0), (0, 0, 1)}. Since the eigenvalue λ = 1 has algebraic multiplicity three, but thecorresponding eigenspace has dimension two, we conclude that A0, and thus f0, is not diagonalizable. Thereis no basis of IR3 formed by eigenvectors of f0.

6.13) Find the linear transformation l : IR3 → IR3 stretching every vector on the plane x + 2y + 3z = 0 twice andreversing the vector (0,−1, 1).

Solution. We choose two linearly independent vectors lying on the plane x + 2y + 3z = 0, for example (2,−1, 0) and(3, 0,−1) (both vectors have coordinates that satisfy the equation of the plane, and since they are not multipleof each other, they are linearly independent). The vector (0, 1, 1) does not satisfy the equation of the plane,thus does not lie on the plane, and it is linearly independent with the chosen vectors. This implies thatB = {(2,−1, 0), (3, 0,−1), (0,−1, 1)} is a basis of IR3, moreover B is formed by eigenvectors and in particular:

l(2,−1, 0) =2(2,−1, 0)

l(3, 0,−1) =2(3, 0,−1)

l(0,−1, 1) =− (0,−1, 1),

thus the matrix D = A(l, B,B) associated to the linear transformation with respect to the basis B is diagonal.

We can write D = A(l, B,B) =

2 0 00 2 00 0 −1

. In order to find the matrix A = A(l, C, C) associated to l

with respect to the canonical basis C of IR3, C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, we just recall that

D = P−1AP

where P =

2 3 0−1 0 −10 −1 1

is the transition matrix from basis B to basis C, and thus

A = PDP−1.

69

In order to evaluate P−1 we use a Gauss reduction:

P =

2 3 0−1 0 −10 −1 1

1 0 00 1 00 0 1

R1 + 2R2

2 3 00 3 −20 −1 1

1 0 01 2 00 0 1

3R3 +R2

R1R2 2 0 20 3 −20 0 1

0 −2 01 2 01 2 3

R2 + 2R3

R1 − 2R3 2 0 00 3 00 0 1

−2 −6 −63 6 61 2 3

R1/22R2/3 1 0 0

0 1 00 0 1

−1 −3 −31 2 21 2 3

= P−1.

We are now ready to find the matrix A = A(l, C, C) associated to l with respect to the canonical basis C ofIR3:

A = PDP−1 =

2 3 0−1 0 −10 −1 1

2 0 00 2 00 0 −1

−1 −3 −31 2 21 2 3

=

2 0 03 8 8−3 −6 −7

.

l : IR3 → R3 is defined as follows:

l(x, y, z) =

2 0 03 8 8−3 −6 −7

xyz

= (2x, 3x+ 8y + 9z,−3x− 6y − 7z).

6.15) Find the linear transformation l : IR2 → IR2 knowing that (1, 2) and (1, 3) are two of its eigenvectors andl(1, 0) = (4, 6)

Solution. We must determine the two eigenvalues λ1, λ2 so that l(1, 2) = λ1(1, 2) and l(1, 3) = λ2(1, 3). Then weproceed in finding the matrix A associated to l with respect to the canonical basis C = {(1, 0), (0, 1)} of IR2,following the same method used in Solution of Exercise 13. In order to find λ1 and λ2, we can use the onlyinformation given about l, l(1, 0) = (4, 6), and the linearity of l. Let’s first write the vector (1, 0) as a linearcombination of (1, 2), and (1, 3), i.e., we look for numbers a, b ∈ IR so that

(1, 0) = a(1, 2) + b(1, 3).

This is equivalent to solving the system:a+ b = 1

2a+ 3b = 0.

After simple calculation, we find a = 3, and b = −2 as the only possible solution of the system, thus(1, 0) = 3(1, 2)−2(1, 3). We now use the linearity of l, the fact that (1, 2) and (1, 3) are eigenvectors associatedto λ1 and λ2, and write:

(4, 6) = l(1, 0) = l(3(1, 2)− 2(1, 3)

)= 3l(1, 2)− 2l(1, 3) = 3λ1(1, 2)− 2λ2(1, 3) = (3λ1 − 2λ2, 6λ1 − 6λ2).

This allows us to find λ1 and λ2, solving the system:

3λ1 − 2λ2 = 4

6λ1 − 6λ2 = 6.

After simple calculation, we find λ1 = 2 and λ2 = 1.We now know that l(1, 2) = 2(1, 2), l(1, 3) = 1(1, 3). Considering the basis B = {(1, 2), (1, 3)} of IR2, the

matrix associate to l with respect to B is diagonal, D = A(l, B,B) =

(2 00 1

). In order to find the matrix

A = A(l, C, C) associated to l with respect to the canonical basis C of IR2, C = {(1, 0), (0, 1)}, we recall that

D = P−1AP

70

where P =

(1 12 3

)is the transition matrix from basis B to basis C, and thus

A = PDP−1.

We immediately evaluate P−1 = 1detP adjP =

(3 −1−2 1

). We are now ready to find the matrix A:

A = A(l, C, C) = PDP−1 =

(1 12 3

)(2 00 1

)(3 −1−2 1

)=

(4 −16 −1

).

l : IR2 → R2 is defined as follows:

l(x, y) =

(4 −16 −1

)(xy

)= (4x− y, 6x− y).

6.17) Given A =

(2 −34 −5


a) to derive an explicit formula for An,b) to find a general solution of the system of differential equations

dx

dt= 2x− 3y

dx

dt= 4x− 5y.

Solution. a) We start by finding the eigenvalues of A, solving the characteristic equation det(A− λI) = 0, we get

det

(2− λ −3

4 −5− λ

)= (2− λ)(−5− λ) + 12 = λ2 + 3λ+ 2 = (λ+ 1)(λ+ 2).

A has distinct eigenvalues λ1 = −1, and λ2 = −2. We find corresponding eigenvectors. For λ1 = −1, wesolve the homogeneous system with associated matrix A− λ1I:(

3 −34 −4

∣∣∣∣∣∣∣∣ 00) R1/3

∼R2/4−R1/3

(1 −10 0

∣∣∣∣∣∣∣∣ 00).

The eigenspace associated to λ1 = −1 is U−1 = span{(1, 1)}. For λ2 = −2, we solve the homogeneous systemwith associated matrix A− λ2I: (

4 −34 −3

∣∣∣∣∣∣∣∣ 00)

R2 −R1

∼

(4 −30 0

∣∣∣∣∣∣∣∣ 00).

The eigenspace associated to λ2 = −2 is U−2 = span{(3, 4)}. Hence, if P =

(1 31 4

), we have P−1 =(

4 −3−1 1

), and

P−1AP = D =

(−1 00 −2

).

Thus

An =(PDP−1

)n= PDP−1PDP−1 . . .PDP−1 = PDnP−1 =

=

(1 31 4

)(−1 00 −2

)n(4 −3−1 1

)=

=

(1 31 4

)((−1)n 0

0 (−2)n

)(4 −3−1 1

)=

=(−1)n(

1 3 · 2n1 4 · 2n

)(4 −3−1 1

)=

=(−1)n(

4− 3 · 2n −3 + 3 · 2n4− 4 · 2n −3 + 4 · 2n

)= (−1)n

(4 −34 −3

)+ (−2)n

(−3 3−4 4

).

71

b) Consider a linear system of differential equations −→x ′(t) = A−→x (t), where −→x (t) = (x1(t), x2(t), . . . , xn(t))T ,x1(t), x2(t), . . . , xn(t) are unknown functions, A is a real n×n matrix, and −→x ′(t) = (x′1(t), x′2(t), . . . , x′n(t))T .We recall that, if A has n distinct eigenvalues λ1, λ2, . . . , λn, and −→v 1,

−→v 2, . . . ,−→v n are eigenvectors corre-

sponding to λ1, λ2, . . . , λn, then a general solution −→x (t) of the system is

−→x (t) = c1−→v 1e

λ1t + c2−→v 2e

λ2t + · · ·+ cn−→v neλnt, c1, . . . cn ∈ IR.

We now find a general solution of the system of differential equations

dx

dt= 2x− 3y

dy

dt= 4x− 5y.

In a) we found λ1 = −1, −→v 1 =

(11

), and λ2 = −2, −→v 2 =

(34

). Hence, the general solution of the given

system is: (x(t)y(t)

)= c1

(11

)e−t + c2

(34

)e−2t =

(c1e−t + 3c2e

−2t

c1e−t + 4c2e

−2t

).


xn+1 = 2xn − ynyn+1 = −xn + 2yn

given that x0 = 0 and y0 = −1.

Solution. Let’s consider X =

(xnyn

). Then, the given system can be written in matrix form as

Xn+1 = AXn

where

A =

(2 −1−1 2

), and X0 =

(0−1

).

We can prove, with an easy induction, that:

X1 =AX0

X2 =AX1 = AAX0 = A2X0

. . .

Xn =AnX0

We thus need to evaluate An, and we will do so using the eigenvalues and eigenvectors of A as in Solution ofExercise 17.We start by finding the eigenvalues of A, solving the characteristic equation det(A− λI) = 0, we get

det

(2− λ −1−1 2− λ

)= (2− λ)2 − 1 = (2− λ− 1)(2− λ+ 1) = (λ− 1)(λ− 3).

A has distinct eigenvalues λ1 = 1, and λ2 = 3. We find corresponding eigenvectors. For λ1 = 1, we solve thehomogeneous system with associated matrix A− λ1I:(

1 −1−1 1

∣∣∣∣∣∣∣∣ 00)

R1 +R2

∼

(1 −10 0

∣∣∣∣∣∣∣∣ 00).

The eigenspace associated to λ1 = 1 is U1 = span{(1, 1)}, thus −→v 1 =

(11

). For λ2 = 3, we solve the

homogeneous system with associated matrix A− λ2I:(−1 −1−1 −1

∣∣∣∣∣∣∣∣ 00)

R2 −R1

∼

(1 10 0

∣∣∣∣∣∣∣∣ 00).

72

The eigenspace associated to λ2 = 3 is U3 = span{(1,−1)}, thus −→v 2 =

(1−1

). Hence, if P =

(1 11 −1

), we

have P−1 = 12

(1 11 −1

), and

P−1AP = D =

(1 00 3

).

Thus

An =PDnP−1 =

=1

2

(1 11 −1

)(1 00 3

)n(1 11 −1

)=

=1

2

(1 11 −1

)(1 00 3n

)(1 11 −1

)=

=1

2

(1 3n

1 −3n

)(1 11 −1

)=

=1

2

(1 + 3n 1− 3n

1− 3n 1 + 3n

).

We now conclude evaluating the solution of the given system of recurrence relations:(xnyn

)= Xn = AnX0 =

1

2

(1 + 3n 1− 3n

1− 3n 1 + 3n

)(0−1

)=

( −1+3n

2−1−3n

2

).

73

Chapter 7Euclidean spaces

7.1) Given the vectors v = (1, 5), u = (3, 4) in IR2, with the standard inner product, find 〈u,v〉, ||u||, ||v||.

7.2) Given the vectors v = (1, 5, 2), u = (3, 4, 0) in IR3, with the standard inner product, find 〈u,v〉, ||u||, ||v||.

7.3) Verify that, on IR2, 〈x,y〉 = 2x1y1 + x1y2 + x2y1 + x2y2, where x = (x1, x2), y = (y1, y2), is a well definedinner product.

7.4) In P2 with the standard inner product, i.e. 〈f, g〉 =1∫0

f(t)g(t) dt, consider

f(t) = t− 3, g(t) = −15

7t+ 1, h(t) = 3t− 2.

Verify that f is orthogonal to g and h has norm 1.

7.5) In C[0, 1] with the standard inner product, i.e. 〈f, g〉 =1∫0

f(t)g(t) dt, consider the functions

f(t) = t+ 2, g(t) = 3t− 2, h(t) = et.

Find 〈f, g〉, 〈f, h〉, ||f ||, ||h||.

7.6) In IR3 verify that the vectors x = (2, 3, 5) and y = (1,−4, 3) are not orthogonal but form an acute angle.

7.7) In IR3 are given u = (1, 1, 1), v = (1, 2,−3), w = (1,−4, 3), verify that u is orthogonal to v, u is orthogonalto w, but v is not orthogonal to w.

7.8) Find the angle θ between the vectors u,v ∈ IR4, u = (1, 3,−5, 4), v = (2,−3, 4, 1),

7.9) Find the angle θ between the vectors A,B ∈M2,3, if A =

(9 8 76 5 4

), B =

(1 2 34 5 6

).

7.10) Find k ∈ IR such that u = (1, 2, k, 3) and v = (3, k, 7,−5) are orthogonal in IR4.

7.11) Find a nonzero vector that is orthogonal to u = (1, 2, 1) and v = (2, 5, 4) in IR3.

7.12) Find k ∈ IR such that A =

(1 3 k1 1 1

)and B =

(k 3 21 2 1

)are orthogonal in M2,3.

7.13) Verify that cos t, sin t are orthogonal vectors in C[0, 2π], with respect to the standard inner product.

7.14) In C[0, 2π], prove that the set S = {1, sin t, cos t, sin 2t, cos 2t, . . . , sin kt, cos kt, . . . } is an orthogonal set,normalize each vector of S to create an orthonormal set.

7.15) Given, in IR5, W = span{(1, 2, 3,−1, 2), (2, 4, 7, 2,−1)}, find a basis of the orthogonal complement W⊥.

7.16) In IR4 is given the space W = span{(1, 2, 3, 1)}, find an orthogonal basis of the orthogonal complement W⊥.

7.17) For the subspace u⊥ of IR3, where u = (1, 3,−4), find a basis, an orthogonal basis, an orthonormal basis.

7.18) Find the Fourier coefficient c = 〈v,w〉||w||2 and the projection of v = (1,−2, 3,−4) along w = (1, 2, 1, 2) with

respect to the standard inner product in IR4.

7.19) In IR3 verify that B = {u1 = (1, 2, 1),u2 = (2, 1,−4),u3 = (3,−2, 1)} is an orthogonal basis, find [v]B , thecoordinates of v = (7, 1, 9) with respect to B.

7.20) In IR4 is given the set S = {(1, 1, 0,−1), (1, 2, 1, 3), (1, 1,−9, 2), (16,−13, 1, 3)}. Show that S is orthogonal,and it forms a basis of IR4. Find the coordinates of v = (a, b, c, d) with respect to S.

7.21) In IR4 consider U = span{(1, 1, 1, 1), (1, 1, 2, 4), (1, 2,−4,−3)}. Use the Gram-Schmidt algorithm to find anorthogonal basis of U , then find an orthonormal basis of U .

74

7.22) Consider the space P2(t) with the inner product 〈f, g〉 =∫ 1

0f(t)g(t) dt.

a) Find 〈f, g〉 where f(t) = t+ 2, g(t) = t2 − 3t+ 4.b) Apply Gram-Schmidt algorithm to the set {1, t, t2} to obtain an orthogonal set {f0, f1, f2} of polynomialswith integer coefficients.

7.23) Prove that every symmetric 2× 2 matrix is diagonalizable. Given l : IR2 → IR2, l(x, y) = (2x+ 2y, 2x+ 5y),show that l is diagonalizable and find a basis of IR2 made of orthonormal eigenvectors of l.

7.24) Given the symmetric matrix A =

2 1 11 2 11 1 2

find an orthonormal real matrix P such that PtAP is diagonal.

(Remember that P is constructed with orthonormal eigenvectors of A.)

7.25) Verify that the linear transformation l : IR3 → IR3, defined by l(x, y, z) = (x + 3y + 4z, 3x + y, 4x + z) issymmetric, prove that it is diagonalizable and there exists a basis of IR3 made of three orthogonal eigenvectorsof l.

7.26) Verify that the linear transformation l : IR2 → IR2, defined by l(x, y) = (x+ 2y, 2x+ 4y) is symmetric, provethat it is diagonalizable and there exists a basis of IR2 made of two orthogonormal eigenvectors of l.

7.27) Identify the curve with equation 12x2 − 12xy + 7y2 + 60x− 38y + 31 = 0.

7.28) Identify the curve with equation x2 − 4xy + 4y2 + 5y − 9 = 0.

75

Solutions of odd number problems - Euclidean spaces

7.1) Given the vectors v = (1, 5), u = (3, 4) in IR2, with the standard inner product, find 〈u,v〉, ||u||, ||v||.

Solution. On IRn, given x = (x1, . . . , xn), y = (y1, . . . , yn), we define the standard inner product 〈x,y〉 = u·v =n∑i=1

xiyi.

Moreover, we define the norm of x as the nonnegative number ||x|| = 〈x,x〉 12 =√〈x,x〉. Thus:

〈u,v〉 =< (1, 5), (3, 4) >= 1 · 3 + 5 · 4 = 3 + 20 = 23,

||u|| =√

12 + 52 =√

26,

||v|| =√

32 + 42 = 5.

7.3) Verify that, on IR2, 〈x,y〉 = 2x1y1 + x1y2 + x2y1 + x2y2, where x = (x1, x2), y = (y1, y2), is a well definedinner product.

Solution. By definition, given a linear space L on IR, we call inner product on L a function 〈, 〉 : L × L → IR,〈, 〉 : (x,y)→ 〈x,y〉 satisfying:i) 〈x,y〉 = 〈y,x〉 for every x,y ∈ L,ii) 〈x,y + z〉 = 〈x,y〉+ 〈x, z〉 for every x,y, z ∈ L,iii) c〈x,y〉 = 〈cx,y〉 for every x,y ∈ L, and every c ∈ IR,iv) 〈x,x〉 ≥ 0 for every x ∈ L, and 〈x,x〉 = 0 if and only if x = 0L.

The given inner product on IR2, can be defined with the use of the matrix product as follows:

〈x,y〉 = 2x1y1 + x1y2 + x2y1 + x2y2 = (x1 x2 )

(2 11 1

)(y1y2

)= (x1 x2 )A

(y1y2

),

where A =

(2 11 1

)∈ M2,2. Due to the properties of matrix product and the fact that A is symmetric

(A = AT ), we can easily verify that all properties i)–iv) are satisfied. Indeed:i) 〈x,y〉 = 2x1y1 + x1y2 + x2y1 + x2y2 = 2y1x1 + y1x2 + y2x1 + y2x2 = 〈y,x〉,ii) 〈x,y+z〉 = 2x1(y1 +z1)+x1(y2 +z2)+x2(y1 +z1)+x2(y2 +z2) = (2x1y1 +x1y2 +x2y1 +x2y2)+(2x1z1 +x1z2 + x2z1 + x2z2) = 〈x,y〉+ 〈x, z〉,iii) c〈x,y〉 = c(2x1y1 + x1y2 + x2y1 + x2y2) = 2cx1y1 + cx1y2 + cx2y1 + cx2y2 = 〈cx,y〉,iv) 〈x,x〉 = 2x1x1 + x1x2 + x2x1 + x2x2 = 2x21 + 2x1x2 + x22 = x21 + (x1 + x2)2 ≥ 0and 〈x,x〉 = x21 + (x1 + x2)2 = 0 if and only if x1 = x2 = 0, i.e. x = (0, 0).

7.5) In C[0, 1] with the standard inner product, i.e. 〈f, g〉 =1∫0

f(t)g(t) dt, consider the functions

f(t) = t+ 2, g(t) = 3t− 2, h(t) = et.

Find 〈f, g〉, 〈f, h〉, ||f ||, ||h||.

Solution. We evaluate:

〈f, g〉 =

1∫0

f(t)g(t) dt =

1∫0

(t+ 2)(3t− 2) dt =[t3 + 2t2 − 4t

]10

= −1.

〈f, h〉 =

1∫0

f(t)h(t) dt =

1∫0

(t+ 2)et dt =[et(t+ 1)

]10

= 2e− 1.

||f || =1∫

0

f(t)2 dt =

1∫0

(t+ 2)2 dt =[t3/3 + 2t2 + 4t

]10

=19

3.

||h|| =1∫

0

h(t)2 dt =

1∫0

(et)2 dt =

1∫0

e2t dt =[e2t

2

]10

=1

2(e2 − 1).

76

7.7) In IR3 are given the vectors u = (1, 1, 1), v = (1, 2,−3), w = (1,−4, 3). Verify that u is orthogonal to v, u isorthogonal to w, but v is not orthogonal to w.

Solution. Since no particular inner product is given, it is understood that we consider IR3 with the standard innerproduct. By definition, two vectors x,y in an Euclidean space V are orthogonal if 〈x,y〉 = 0, thus weevaluate:

〈u,v〉 = u · v = (1, 1, 1) · (1, 2,−3) = 1 + 2− 3 = 0, i.e. u ⊥ v,

〈u,w〉 = u ·w = (1, 1, 1) · (1,−4, 3) = 1− 4 + 3 = 0, i.e. u ⊥ w,

〈v,w〉 = v ·w = (1, 2,−3) · (1,−4, 3) = 1− 8 + 9 = 2, i.e. v is not orthogonal to w.

7.9) Find the angle θ between the vectors A,B ∈M2,3, if A =

(9 8 76 5 4

), B =

(1 2 34 5 6

).

Solution. We recall that in Mm,n, the standard inner product between two matrices A,B is defined as 〈A,B〉 =tr(BTA), where ”tr” stands for ”trace”, i.e. the sum of the diagonal entries of a matrix. Moreover, in areal Euclidean space we define the angle between two nonzero vectors u,v as the angle 0 ≤ θ ≤ π such that

cos θ = 〈u,v〉||u||·||v|| . Thus we need to evaluate

〈A,B〉 = tr(BTA) = tr

1 42 53 6

( 9 8 76 5 4

)= tr

9 + 24 . . . . . .. . . 16 + 25 . . .. . . . . . 21 + 24

= 119,

||A||2 = 〈A,A〉 = tr(ATA) = tr

9 68 57 4

( 9 8 76 5 4

)= tr

81 + 36 . . . . . .. . . 64 + 25 . . .. . . . . . 49 + 16

= 271,

||B||2 = 〈B,B〉 = tr(BTB) = tr

1 42 53 6

( 1 2 34 5 6

)= tr

1 + 16 . . . . . .. . . 4 + 25 . . .. . . . . . 9 + 36

= 91,

and finally

cos θ =〈A,B〉||A|| · ||B||

=119√

271 ·√

91' 0, 75777,

from which we get: θ ' arccos(0, 75777).

7.11) Find a nonzero vector that is orthogonal to u = (1, 2, 1) and v = (2, 5, 4) in IR3.

Solution. We are looking for a vector x = (x, y, z) ∈ IR3 such that x ·u = 0 and x ·v = 0, thus the coordinates of x area nontrivial solutions of the homogeneous system:

x+ 2y + z = 02x+ 5y + 4z = 0

R2 − 2R1

∼x+ 2y + z = 0

y + 2z = 0.

We may easily guess a solution: by setting z = 1, we get y = −2 and subsequently x = 3. The vectorx = (3,−2, 1), or any multiple of x is orthogonal to both u and v.

7.13) Verify that cos t, sin t are orthogonal vectors in C[0, 2π], with respect to the standard inner product.

Solution. In C[0, 2π] the standard inner product is 〈f, g〉 =2π∫0

f(t)g(t) dt , thus we just verify that:

2π∫0

cos t sin t dt =[1

2(sin t)2

]2π0

=1

2

[(sin(2π))2 − (sin 0)2

]= 0− 0 = 0,

and this implies that cos t, sin t are orthogonal vectors in C[0, 2π].

7.15) Given, in IR5, W = span{(1, 2, 3,−1, 2), (2, 4, 7, 2,−1)}, find a basis of the orthogonal complement W⊥.

Solution. Let W be a subset of an Euclidean space E, then W⊥ is defined as follows:

W⊥ = {v ∈ E : v ⊥ u, for every u ∈W}.

77

Moreover, we recall that W⊥ is a subspace of E. In case W is a subspace of E, then E = W ⊕W⊥, wesay that E is the direct sum of W and W⊥, i.e. every v ∈ E can be written in one and only one way asv = w + w⊥, where w ∈W and w⊥ ∈W⊥.

We need to find a basis of W⊥ =[

span{(1, 2, 3,−1, 2), (2, 4, 7, 2,−1)}]⊥

. We are looking for all possible

vectors x = (x, y, z, t, w) ∈ IR5, so that:

(1, 2, 3,−1, 2) · (x, y, z, t, w) = 0(2, 4, 7, 2,−1) · (x, y, z, t, w) = 0.

We are looking for a basis of the solution set of the homogeneous system:

x+ 2y + 3z − t+ 2w = 02x+ 4y + 7z + 2t− w = 0

R2 − 2R1

∼x+ 2y + 3z − t+ 2w = 0

z + 4t− 5w = 0.

Choosing y = α, t = β, w = γ, as parameters, α, β, γ ∈ IR, from the last equation we get z = −4β + 5γ, thatsubstituted into the first equation gives x = −2α+13β−17γ. Thus, all solutions of the previous homogeneoussystem can be written as

x = (x, y, z, t, w) =(−2α+ 13β − 17γ, α,−4β + 5γ, β, γ) =

=α(−2, 1, 0, 0, 0) + β(13, 0,−4, 1, 0) + γ(−17, 0, 5, 0, 1).

ThusW⊥ = span{(−2, 1, 0, 0, 0), (13, 0,−4, 1, 0), (−17, 0, 5, 0, 1)}.

7.17) For the subspace u⊥ of IR3, where u = (1, 3,−4), find a basis, an orthogonal basis, an orthonormal basis.

Solution. We recall that, given an Euclidean space V , and a non-zero vector u ∈ V , the set u⊥ = {v ∈ V : u ⊥ v} isa subspace of V . In our case, we may denote U = span{u}, then, since dimU = 1, and IR3 = U ⊕ u⊥, wededuce that u⊥ has dimension two. To find a basis of u⊥, we need two linearly independent vectors of IR3,v1 and v2, that are both perpendicular to the given u. We may choose the first v1 = (x1, y1, z1), just takingcare that the condition

u · v1 = 0, that is (1, 3,−4) · (x1, y1, z1) = x1 + 3y1 − 4z1 = 0,

must be satisfied. Thus, for example, we may choose v1 = (1, 1, 1). We now look for v2, in a way that{v1,v2} will form an orthogonal basis of u⊥, so we look for v2 = (x2, y2, z2), satisfying both conditions

u · v2 = (1, 3,−4) · (x2, y2, z2) = 0,

andv1 · v2 = (1, 1, 1) · (x2, y2, z2) = 0,

that is, satisfying the system:

x2 + 3y2 − 4z2 = 0x2 + y2 + z2 = 0

∼R1 −R2

x2 + 3y2 − 4z2 = 02y2 − 5z2 = 0.

We may easily guess a possible solution v2 = (−7, 5, 2). The set B = {v1,v2} = {(1, 1, 1), (−7, 5, 2)} is abasis of u⊥ made of orthogonal vectors, thus, it is an orthogonal basis of u⊥. To find an orthonormal basisof u⊥, it is sufficient to normalise v1 and v2.

B ={ v1

‖v1‖,

v2

‖v2‖

}={ (1, 1, 1)√

3,

(−7, 5, 2)√78

}is an orthonormal basis of u⊥.

7.19) In IR3, verify that B = {u1 = (1, 2, 1),u2 = (2, 1,−4),u3 = (3,−2, 1)} is an orthogonal basis, find [v]B , thecoordinates of v = (7, 1, 9) with respect to B.

Solution. Let’s recall an important Theorem: Given an Euclidean space V and B = {v1, . . . ,vn} an orthogonal basisof V , then for every vector x ∈ V ,

x =n∑i=1

civi where ci =〈x,vi〉〈vi,vi〉

.

78

The coefficients ci = 〈x,vi〉〈vi,vi〉 are called Fourier coefficients of x with respect to B, and the coordinates of x

with respect to B are

[x]B = (c1, . . . , cn).

Let’s verify that the given set B = {u1 = (1, 2, 1),u2 = (2, 1,−4),u3 = (3,−2, 1)} is an orthogonal basisof IR3. Since orthogonal vectors are linearly independent, it is enough to verify the given three vectors areorthogonal. Indeed:

u1 · u2 = (1, 2, 1) · (2, 1,−4) = 2 + 2− 4 = 0,

u1 · u3 = (1, 2, 1) · (3,−2, 1) = 3− 4 + 1 = 0,

u3 · u2 = (3,−2, 1) · (2, 1,−4) = 6− 2− 4 = 0.

Now we may evaluate the Fourier coefficients of v = (7, 1, 9) with respect to B.

c1 =〈v,u1〉〈u1,u1〉

=(7, 1, 9) · (1, 2, 1)

(1, 2, 1) · (1, 2, 1)=

7 + 2 + 9

1 + 4 + 1= 3,

c2 =〈v,u2〉〈u2,u2〉

=(7, 1, 9) · (2, 1,−4)

(2, 1,−4) · (2, 1,−4)=

14 + 1− 36

4 + 1 + 16= −1,

c3 =〈v,u3〉〈u3,u3〉

=(7, 1, 9) · (3,−2, 1)

(3,−2, 1) · (3,−2, 1)=

21− 2 + 9

9 + 4 + 1= 2.

We conclude that the coordinates of v = (7, 1, 9) with respect to B are:

[v]B = (c1, c2, c3) = (3,−1, 2).

7.21) In IR4 consider U = span{(1, 1, 1, 1), (1, 1, 2, 4), (1, 2,−4,−3)}. Use the Gram-Schmidt algorithm to find anorthogonal basis of U , then find an orthonormal basis of U .

Solution. Given any basis B = {v1, . . . ,vn} of a finite dimensional Euclidean space V , we may create a new orthogonalbasis S = {y1, . . . ,yn} of V using the following process, the Gram-Schmidt orthogonalization process:

y1 = v1

y2 = v2 −〈v2,y1〉〈y1,y1〉

y1

y3 = v3 −〈v3,y2〉〈y2,y2〉

y2 −〈v3,y1〉〈y1,y1〉

y1

. . .

yn = vn −〈vn,yn−1〉〈yn−1,yn−1〉

yn−1 − · · · −〈vn,y2〉〈y2,y2〉

y2 −〈vn,y1〉〈y1,y1〉

y1.

In our case, V is the given subspace of IR4, V = U = span{(1, 1, 1, 1), (1, 1, 2, 4), (1, 2,−4,−3)}, and B ={(1, 1, 1, 1), (1, 1, 2, 4), (1, 2,−4,−3)}. The process can be used starting from any set of vectors, nevertheless,for the sake of exercise, we may easily verify that the given vectors v1 = (1, 1, 1, 1), v2 = (1, 1, 2, 4), v3 =(1, 2,−4,−3) are linearly independent, and thus form a basis of V . Indeed, using Gauss reduction and thesymbol ∼ to denote the coincidence of the corresponding spans, we have:

(1, 1, 1, 1)(1, 1, 2, 4)

(1, 2,−4,−3)

R2 −R1

∼R3 −R1

(1, 1, 1, 1)(0, 0, 1, 34)

(0, 1,−5,−4).

Since the last three vectors are obviously linearly independent, it follows that the given ones are also linearlyindependent. We now proceed with the orthogonalization process:

79

y1 =v1 = (1, 1, 1, 1),

y2 =v2 −〈v2,y1〉〈y1,y1〉

y1 = (1, 1, 2, 4)− (1, 1, 2, 4) · (1, 1, 1, 1)

(1, 1, 1, 1) · (1, 1, 1, 1)(1, 1, 1, 1) =

=(1, 1, 2, 4)− 1 + 1 + 2 + 4

1 + 1 + 1 + 1(1, 1, 1, 1) = (1, 1, 2, 4)− 2(1, 1, 1, 1) = (−1,−1, 0, 2),

y3 =v3 −〈v3,y2〉〈y2,y2〉

y2 −〈v3,y1〉〈y1,y1〉

y1 =

=(1, 2,−4,−3)− (1, 2,−4,−3) · (−1,−1, 0, 2)

(−1,−1, 0, 2) · (−1,−1, 0, 2)(−1,−1, 0, 2)− (1, 2,−4,−3) · (1, 1, 1, 1)

(1, 1, 1, 1) · (1, 1, 1, 1)(1, 1, 1, 1) =

=(1, 2,−4,−3)− −1− 2− 6

1 + 1 + 4(−1,−1, 0, 2)− 1 + 2− 4− 3

1 + 1 + 1 + 1(1, 1, 1, 1) =

=(1, 2,−4,−3) +3

2(−1,−1, 0, 2) + (1, 1, 1, 1) =

1

2(1, 3,−6, 2).

An orthogonal basis of U is B = {y1,y2,y3} = {(1, 1, 1, 1), (−1,−1, 0, 2), 12 (1, 3,−6, 2)}, to create an or-

thonormal basis B of U , we need only to normalise the found vectors:

B =

{y1

‖y1‖,

y2

‖y2‖,

y3

‖y3‖

}=

{(1

4,

1

4,

1

4,

1

4

),

(−1

6,−1

6, 0,

1

3

),

(1

100,

3

100,−6

100,

2

100

)}.

7.23) Prove that every symmetric 2× 2 matrix is diagonalizable. Given l : IR2 → IR2, l(x, y) = (2x+ 2y, 2x+ 5y),show that l is diagonalizable and find a basis of IR2 made of orthonormal eigenvectors of l.

Solution. Every symmetric matrix is diagonalizable, the proof of this general theorem is not so easy, but the Exercise

asks to prove that a 2 × 2 symmetric matrix is always diagonalizable. Let’s suppose that A =

(a bb c

),

a, b, c ∈ IR, is the given matrix. If b = 0, then A is already diagonal, there is nothing to prove. Let’s supposethat b 6= 0, we will prove that A has two distinct eigenvalues, thus two linearly independent eigenvectors, andthis proves that A is diagonalizable. To find the eigenvalues of A, we must solve the characteristic equationdet(A− λI) = 0, in our case, we get:

det(A− λI) = det

((a bb c

)− λ

(1 00 1

))= det

(a− λ bb c− λ

)= (λ− a)(λ− c)− b2 = 0.

The equationλ2 − (a+ c)λ+ (ac− b2) = 0

has two distinct roots if and only if the discriminant is strictly greater than zero. We have:

∆ = (a+ c)2 − 4(ac− b2) = a2 + 2ac+ c2 − 4ac+ 4b2 = a2 − 2ac+ c2 + 4b2 = (a− c)2 + (2b)2.

Since this is a sum of squares, it is nonnegative. Since b 6= 0, it is actually positive. This proves that thecharacteristic polynomial has two distinct real roots, so the symmetric matrix A has two distinct eigenvalues,thus it is diagonalizable.

Let’s now consider l : IR2 → IR2, l(x, y) = (2x+ 2y, 2x+ 5y).

Observation. By definition, a linear transformation l : V → V , V Euclidean space, is symmetric if〈l(v),w〉 = 〈v, l(w)〉, for every v,w ∈ V . We prove that a linear transformation l : IRn → IRn is sym-metric if and only if the matrix A(l) associated to l with respect to the standard basis of IRn is symmetric.

We consider the ordered canonical basis of IR2, C = {(1, 0), (0, 1)}, and we will find the matrix A = A(l, C, C)associated to the given linear transformation with respect to C. For this purpose, we need to evaluate:

l(1, 0) =(2, 2)

l(0, 1) =(2, 5).

The matrix associated to l with respect to the canonical basis of IR2 is symmetric:

A = A(l, C, C) =

(2 22 5

).

80

This implies that l is symmetric. To diagonalize A, we start by solving the characteristic equation:

det(A− λI) = det

((2 22 5

)− λ

(1 00 1

))= det

(2− λ 2

2 5− λ

)= (λ− 2)(λ− 5)− 22 = 0.

Incidentally, we verify that, for a symmetric 2× 2 matrix A, the characteristic equation has the form

λ2 − trAλ+ detA = 0,

indeed, we get:λ2 − 7λ+ 6 = 0.

The eigenvalues are λ1 = 6 and λ2 = 1. We now look for the corresponding eigenvectors.For λ1 = 6, we must find the nontrivial solutions of the system:

(A− λ1I)v = 0.

The augmented matrix of this homogeneous system is(−4 22 −1

∣∣∣∣∣∣∣∣ 00)

R1 + 2R2

∼

(0 02 −1

∣∣∣∣∣∣∣∣ 00).

All the eigenvectors corresponding to λ1 = 6 are {t(1, 2) : t ∈ IR \ {0}}, and the eigenspace corresponding toλ1 = 6 is U6 = span{(1, 2)}.To find the eigenvectors corresponding to the second eigenvalue λ2 = 1, it is enough to recall that, forsymmetric matrices, eigenvectors corresponding to distinct eigenvalues are orthogonal, thus the eigenspacecorresponding to λ2 = 1 is U1 = span{(2,−1)}.A basis of IR2 made of orthonormal eigenvectors of A is

B =

{( 1√5,

2√5

),( 2√

5,−1√

5

)}.

The matrix P formed by the vectors in B, written as columns, is unitary, and has the property that P−1 = PT :

P =

(1√5

2√5

2√5

−1√5

).

Moreover,

PTAP = D =

(6 00 1

).

7.25) Verify that the linear transformation l : IR3 → IR3, defined by l(x, y, z) = (x + 3y + 4z, 3x + y, 4x + z) issymmetric, prove that it is diagonalizable and there exists a basis of IR3 made of three orthogonal eigenvectorsof l.

Solution. We consider the ordered canonical basis of IR3, C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, and we will find the matrixA = A(l, C, C) associated to the given linear transformation with respect to C. For this purpose, we need toevaluate:

l(1, 0, 0) =(1, 3, 4)

l(0, 1, 0) =(3, 1, 0)

l(0, 0, 1) =(4, 0, 1).


1 3 43 1 04 0 1

. The matrix associated to l with respect to the canonical basis of IR3 is

symmetric, this implies that l is symmetric.To diagonalize A, we start by solving the characteristic equation:

det(A− λI) = det

( 1 3 43 1 04 0 1

− λ 1 0 0

0 1 00 0 1

) = det

1− λ 3 43 1− λ 04 0 1− λ

=

=(1− λ)3 − 16(1− λ)− 9(1− λ) = (1− λ)[(1− λ)2 − 25] = (1− λ)(−4− λ)(6− λ) = 0.

81

The characteristic equation of A has three distinct solutions λ = 1, λ = −4, λ = 6, with algebraic multiplicity1. In this case, we may already conclude that A is diagonalizable, indeed, corresponding to each eigenvaluewe will choose an eigenvector, and, since the eigenvalues are distinct, the chosen eigenvectors are linearlyindependent and form a basis of IR3. We will verify that, being A symmetric, the found basis is indeedformed by orthogonal vectors.Thus, we look for the eigenvectors corresponding to λ = 1. We must find the nontrivial solutions of thehomogeneous system (A − λI)v = 0, where λ = 1. We solve the homogeneous system with augmentedmatrix: 0 3 4

3 0 04 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 1 are {t(0, 4,−3) : t ∈ IR \ {0}} and the eigenspace associated toλ = 1 is U1 = span{(0, 4,−3)}.We consider λ = −4, and find the corresponding eigenvectors solving the system: 5 3 4

3 5 04 0 5

∣∣∣∣∣∣∣∣∣∣∣∣000

3R1 − 5R2

∼4R1 − 5R3

5 3 40 −16 120 12 −9

∣∣∣∣∣∣∣∣∣∣∣∣000

R2/4∼

R2/4 +R3/3

5 3 40 −4 30 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

It is easy to deduce that all the eigenvectors corresponding to λ = −4 are {t(−5, 3, 4) : t ∈ IR \ {0}} and theeigenspace associated to λ = −4 is U−4 = span{(−5, 3, 4)}.Finally, we consider the eigenvalue λ = 6, and find the corresponding eigenvectors solving the system:−5 3 4

3 −5 04 0 −5

∣∣∣∣∣∣∣∣∣∣∣∣000

3R1 + 5R2

∼4R1 + 5R3

−5 3 40 −16 120 12 −9

∣∣∣∣∣∣∣∣∣∣∣∣000

R2/4∼

R2/4 +R3/3

−5 3 40 −4 30 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

The eigenvectors corresponding to λ = 6 are {t(5, 3, 4) : t ∈ IR \ {0}} and the eigenspace associated to λ = 6is U6 = span{(5, 3, 4)}.A basis of IR3 made of eigenvectors of l is B = {(0, 4,−3), (−5, 3, 4), (5, 3, 4)}. We easily verify that the foundeigenvectors are orthogonal:

(0, 4,−3) · (−5, 3, 4) = 0 + 12− 12 = 0,

(0, 4,−3) · (5, 3, 4) = 0 + 12− 12 = 0,

(−5, 3, 4) · (5, 3, 4) = −25 + 9 + 16 = 0.

We conclude that the given matrix A is similar to a diagonal matrix, and the matrix P =

0 −5 54 3 3−3 4 4

is

such that

P−1AP = D =

1 0 00 −4 00 0 6

.

If we normalise the vectors in B, and create a new basis B of IR3 made of orthonormal eigenvectors of A,

B =

{1

5(0, 4,−3),

1

5√

2(−5, 3, 4),

1

5√

2(5, 3, 4)

},

the matrix P, made of vectors in B written as columns,

P =

0 −1√2

1√2

45

35√2

35√2

−35

45√2

45√2

is unitary, has the property that P

−1= P

T, thus:

P−1

AP = PTAP = D =

1 0 00 −4 00 0 6

.

7.27) Identify the curve with equation 12x2 − 12xy + 7y2 + 60x− 38y + 31 = 0.

82

Solution. We will solve this exercise following a general approach, see for example the lecture notes Elementary LinearAlgebra, by K.R.Matthews, Department of Mathematics, University of Queensland.Given the curve with equation ax2 + 2hxy + by2 + cx+ dy + e = 0, the expression ax2 + 2hxy + by2 is calleda quadratic form in x and y. Using matrices, we have the identity

ax2 + 2hxy + by2 = (x y )

(a hh b

)(xy

)= XTAX,

where X =

(xy

), A =

(a hh b

). A is called the matrix of the quadratic form, and it is a symmetric 2 × 2

real matrix. It follows that A is diagonalizable and there exists a matrix P made of orthonormal eigenvectorsof A such that

PTAP = D

is a diagonal matrix with nonzero diagonal entries λ1 and λ2, the real eigenvalues of A. The quadratic formsatisfies the following:

ax2 + 2hxy + by2 = XTAX = XT (PDPT )X = (PTX)TD(PTX),

and if Y =

(x1y1

)= PTX, then X = PY and

ax2 + 2hxy + by2 = YTDY = (x1 y1 )

(λ1 00 λ2

)(x1y1

)= λ1x

21 + λ2x

22,

that expresses the fact that the transformation (change of basis)(xy

)= P

(x1y2

)that can be shown to be a rotation of axes, diagonalize the quadratic form corresponding to A.Then, under the rotation X = PY, the original quadratic equation becomes

λ1x21 + λ2y

21 + cx1 + dy1 + e = 0.

Completing the square in x1 and y1, after a suitable transformation (a translation of axes)

x2 = x1 + α, y2 = y1 + β

the original equation will reduce to

ux22 + vy22 = w, u, v, w ∈ IR, not all zero,

from which we will be able to classify the original curve.Let’s work out the detailed computation for the given curve 12x2 − 12xy + 7y2 + 60x− 38y + 31 = 0. In thiscase,

A =

(12 −6−6 7

).

The characteristic equation of A is λ2 − (trA)λ+ detA = 0, or

λ2 − 19λ+ 48 = (λ− 16)(λ− 3) = 0.

A has distinct eigenvalues λ1 = 16 and λ2 = 3.For λ1 = 16, we look for corresponding eigenvector solving (A − λ1I)v = 0, thus we solve the homogeneoussystem: (

−4 −6−6 −9

∣∣∣∣∣∣∣∣ 00)

−R1/2R1/2−R3/3

∼(

2 30 0

∣∣∣∣∣∣∣∣ 00)

All the eigenvectors corresponding to λ1 = 16 are {t(3,−2) : t ∈ IR \ {0}}, and the eigenspace correspondingto λ1 = 16 is U16 = span{(3,−2)}.To find the eigenvectors corresponding to the second eigenvalue λ2 = 3, it is enough to recall that, forsymmetric matrices, eigenvectors corresponding to distinct eigenvalues are orthogonal, thus the eigenspacecorresponding to λ2 = 3 is U3 = span{(2, 3)}.

83

We normalize the found eigenvectors and create a basis B of IR2 made of orthonormal eigenvectors of A,

B ={(

3√13, −2√

13

),(

2√13, 3√

13

)}. Using B, we create the unitary matrix

P =

(3√13

2√13

−2√13

3√13

)

that has the property that P−1 = PT , and PTAP = D =

(16 00 3

). Moreover, the fact that detP = 1

guarantees that the rotation X = PY preserves the orientation of the axes. More explicitly the transformation

X =

(xy

)= PY = P

(x1y1

)can be written as:

x =3x1 + 2y1√

13, y =

−2x1 + 3y1√13

.

Under the rotation X = PY, the original quadratic equation becomes

16x21 + 3y21 +60√13

(3x1 + 2y1)− 38√13

(−2x1 + 3y1) + 31 = 0

or

16x21 + 3y21 +256√

13x1 +

6√13y1 + 31 = 0.

We now complete the square in x1 and y1:

16(x21 +

16√13x1

)+ 3(y21 +

2√13y1

)+ 31 = 0

16(x1 +

8√13

)2+ 3(y1 +

1√13

)2= 48.

Now, if we perform a translation of axes to the new origin (x1, y1) =(− 8√

13,− 1√

13

):

x2 = x1 +8√13, y2 = y1 +

1√13,

the quadratic equation reduces to16x22 + 3y22 = 48

orx223

+y2216

= 1.

This equation is now in standard form and represent an ellipse centred at (x2, y2) = (0, 0) and with axes ofsymmetry the x2 and y2 axes. In terms of the original x, y coordinates, knowing that (x1, y1) =

(− 8√

13,− 1√

13

),

we find that the center is

x =3x1 + 2y1√

13= −2, y =

−2x1 + 3y1√13

= 1.

Also, from (x, y) =(

3x1+2y1√13

, −2x1+3y1√13

), we deduce (x1, y1) =

(3x−2y√

13, 2x+3y√

13

). Thus the y2-axis is given by

0 = x2 = x1 +8√13

=3x− 2y√

13+

8√13

or 3x− 2y + 8 = 0. Similarly, the x2-axis is given by

0 = y2 = y1 +1√13

=2x+ 3y√

13+

1√13

or 2x+ 3y + 1 = 0.

84

Chapter 8Analytic geometry in three dimensions

8.1) Find the parametric and canonical equation of the line p passing through the points A = [1, 0, 2] and B =[3, 1,−2]; check whether the point M = [7, 3, 1] lies on p.

8.2) Verify that the line p lies on the plane σ, where:

p : X = [1, 0, 1] + t(2, 1, 3), t ∈ IR, σ : 2x− 2y − z − 1 = 0.

8.3) Find the equation of the planes ρ and σ, verify that they are not parallel and find the vector equation of theline p, intersection of ρ and σ, where:

ρ is the plane containing the point M = [1,−2, 3] and orthogonal to the vector n = (4, 5,−6);σ is the plane passing through the points A = [2, 5,−1], B = [2,−3, 3], and C = [4, 5, 0].

8.4) Given the plane σ and the point A, verify that A does not belong to σ and find the equation of the plane ρthrough A parallel to σ, where:

σ : 2x+ 7y − 3z = 1 A = [1, 1, 1].

8.5) Find the angle between the planes ρ and σ, if ρ passes through the points M1 = [−2, 2, 2], M2 = [0, 5, 3] andM3 = [−2, 3, 4], and σ has equation 3x− 4y + z + 5 = 0.

8.6) Find the angle between the lines p and q, if:

p : X = [1, 2, 1] + t(1,−2,−3), t ∈ IR, q : contains the points A = [2,−3, 1], B = [−3, 0, 2].

8.7) Find the distance between the point A = [8,−7, 1] and the plane with equation 2x+ 3y − 4z + 5 = 0.

8.8) Find the distance between the point A = [−1,−1, 1] and the line p, intersection of the planes 2x−y+3z+2 = 0,and x+ 2y − z + 1 = 0.

8.9) Given A = [2, 9, 8], B = [6, 4,−2] and C = [7, 15, 7], show that−→AB and

−→AC are perpendicular, then find D

so that ABCD forms a rectangle.

8.10) Given the line p, X = [1, 2, 3] + t(1, 2,−2) and the point A = [3, 3, 5], find the points on p that have distance3√

5 from A.

8.11) Given the line p passing through the points A = [1, 2, 1] and B = [2,−1, 3], find the point P on p closest tothe origin and the shortest distance from the origin to p.

8.12) Find the plane σ going through the points A = [2, 1,−3], B = [−3, 2, 0], C = [0, 3,−5] and determine the

angle between σ and the line p :x− 2

−1=y + 1

1=z − 3

2.

8.13) Show that the planes x+ y − 2z = 1 and x+ 3y − z = 4 intersect in a line and find the distance between thepoint C = [1, 0, 1] and this line.

8.14) Find the line p going through the points A = [2, 1, 0] and B = [1,−3, 2] and the plane σ going throughP = [4, 1, 1] that has the properties that d(P, p) = d(p, σ), where d indicates the distance.

8.15) Find an equation for the plane through P = [1, 0, 1] and containing the line of intersection of the planesx+ y + 2z = 1 and x+ 3y + z = 4.

8.16) Find an equation for the plane passing through the point A = [1, 0,−2] and containing the line p with vectorequation X = [1, 1,−1] + t(3, 2, 0), t ∈ IR.

8.17) Find an equation for the plane passing through P = [6, 0, 2] and perpendicular to the line of intersection ofthe planes x+ y − 2z = 4 and 3x− 2y + z = 1.

85

8.18) Find the intersection of the line p : X = [1,−2, 0] + t(3, 4, 2) t ∈ IR with the plane σ containing the pointS = [19, 13, 9] and the line s : X = [−3,−1, 2] + u(2, 0, 3), u ∈ IR.

8.19) Prove that the lines p and q are skewed, find their distance d(p, q), and find also the points P ∈ p, Q ∈ q suchthat d(p, q) = d(P,Q), where

p : X = [1, 2, 1] + t(5,−2,−3), t ∈ IR, q : X = [−2, 6, 3] + s(4,−3,−1), s ∈ IR.

8.20) Given the line p through A = [1, 2, 1] and B = [3,−1, 2] and the line q through C = [1, 0, 2] and D = [2, 1, 3],prove that the distance between p and q is 16√

62.

8.21) Given the line p and two points A and B, find a point M on p that is equally far away from A and B, where:

p = ρ ∩ σ, ρ : x+ y − z − 3 = 0, σ : 3y − 2z + 1 = 0, A = [1, 0,−1], and B = [3, 4, 5].

8.22) Prove that the lines p and q are skewed, and find an equation of the line r that intersects p and q and isorthogonal to both, where:

p : X = [−4, 4, 1] + t(2,−1,−2), t ∈ IR, q : X = [−5, 5, 5] + s(4,−3,−5), s ∈ IR.

8.23) Find the pointR symmetric to P = [−4, 5, 8] with respect to the line p throughA = [9, 4, 10] andB = [−6, 1, 1].

8.24) Find the point R symmetric to P = [−1, 5, 0] with respect to the plane σ through A = [4, 2, 1], B =[−4,−3,−1] and C = [0, 1, 3].

8.25) A line with direction vector v = (0, 9,−1) intersects lines p and q, find the coordinates of the points ofintersection, where

p :x− 8

5=y − 5

1=

z

−1, and q :

x

1=y − 1

−2=z + 1

1.

8.26) Find the line p parallel to s :x− 3

1=y + 1

3=z − 3

−2that intersects the lines

q :x− 3

3=y − 1

−7=z − 2

1, and r : X = [1, 4, 4] + t(3, 2,−1), t ∈ IR.

8.27) Find the distance of the point A = [3, 2, 1] from the plane containing the lines p and q, where

p :x+ 1

1=y − 3

2=z − 2

2, and q : X = [0, 4, 2] + t(1, 1, 0), t ∈ IR.

8.28) Find the plane going through the point A = [2,−1, 1] and orthogonal to the line p of intersection of the planes

σ : X = [0, 4, 2]+t(1, 1, 0)+u(1, 1, 2), t, u ∈ IR, ρ : X = [1,−1, 1]+s(−1, 2, 2)+v(3,−4, 1), s, v ∈ IR.

86

Solutions of odd number problems - Analytic geometry in three dimensions

8.1) Find the parametric and the vector equation of the line p passing through the points A = [1, 0, 2] andB = [3, 1,−2]; check whether the point M = [7, 3, 1] lies on p.

Solution. We recall that a line p, passing through a point A = [a, b, c], with direction vector v = (v1, v2, v3), has vectorequation:

X = A+ tv, t ∈ IR,

and consequently (fixing X = [x, y, z]) parametric equations:

x = a+ tv1

y = b+ tv2

z = c+ tv3, t ∈ IR.

In our case, A = [1, 0, 2] is given, and we may evaluate v = B − A = [3, 1,−2] − [1, 0, 2] = (2, 1,−4). ThusX = [1, 0, 2] + t(2, 1,−4), t ∈ IR is the required vector equation of p, and its parametric equation is:

x = 1 + 2t

y = t

z = 2− 4t, t ∈ IR.

To verify if the point M = [7, 3, 1] lies on p, it is enough to substitute the coordinates of M in one of theequations of p and verify if there exists a value of t ∈ IR for which the equation(s) is(are) satisfied. Wesubstitute [7, 3, 1] = [x, y, z] into the parametric equation of p, and get:

7 = 1 + 2t

3 = t

1 = 2− 4t, t ∈ IR.

This is equivalent to:

6 = 2t

3 = t

1 = 4t, t ∈ IR,

the value t = 3 satisfies the first two equations but not the third, thus there exists no value of t that solvesthe given system, M = [7, 3, 1] does not lie on p.

Observation. An equivalent result is obtained with the following reasoning: the point M belongs to line p

if and only if the vector−−→AM is parallel to the vector

−→AB, i.e. if

−−→AM = M −A = [7, 3, 1]− [1, 0, 2] = (6, 3,−1)

is a multiple of the vector−→AB = B − A = [3, 1,−2]− [1, 0, 2] = (2, 1,−4). This is not the case, thus M does

not belong to p.

8.3) Find the equation of the planes ρ and σ, verify that they are not parallel and find the vector equation of theline p, intersection of ρ and σ, where:ρ is the plane containing the point M = [1,−2, 3] and orthogonal to the vector n = (4, 5,−6);σ is the plane passing through the points A = [2, 5,−1], B = [2,−3, 3], and C = [4, 5, 0].

Solution. The point-normal equation of a plane containing a point A = [x0, y0, z0] and orthogonal to a vector n = (a, b, c)

is obtained posing the condition that, if X = [x, y, z] is any other point of the plane, then the vector−→AX

must be orthogonal to n. Thus, using the scalar product:

−→AX · n = 0, ⇔ (X −A) · n = 0 ⇔ a(x− x0) + b(y − y0) + c(z − z0) = 0.

For the plane ρ, we get

ρ :−−→MX · n = 0, ⇔

([x, y, z]− [1,−2, 3]

)· (4, 5,−6) = 0 ⇔ 4x+ 5y − 6z + 24 = 0.

87

To find the equation of the plane σ, we first need to identify a vector m orthogonal to σ. Knowing that

the points A, B, and C lie on σ, we consider the vectors−→AB and

−→AC, and create m =

−→AB × −→AC. We now

evaluate: −→AB = B −A = [2,−3, 3]− [2, 5,−1] = (0,−8, 4),

−→AC = C −A = [4, 5, 0]− [2, 5,−1] = (2, 0, 1),

m =−→AB ×−→AC = (0,−8, 4)× (2, 0, 1) = det

i j k0 −8 42 0 1

= (−8, 8, 16) = −8(1,−1,−2).

We choose to use the shorter vector m1 = (1,−1,−2), still orthogonal to the plane σ, and the equation of σis obtained:

σ :−→AX ·m1 = 0, ⇔

([x, y, z]− [2, 5,−1]

)· (1,−1,−2) = 0 ⇔ x− y − 2z + 1 = 0.

To verify that ρ and σ are not parallel, it is enough to observe that the vectors n = (4, 5,−6) and m1 =(1,−1,−2) are not parallel, since they are not proportional. We now need to find the equation of the linep = ρ∩ σ. The direction vector v of p can be obtained as the cross product n×m1, because v lies on both ρand σ, thus must be orthogonal to both n and m1:

v = n×m1 = det

i j k4 5 −61 −1 −2

= (−16, 2,−9).

We now need a point P belonging to ρ ∩ σ, thus we look for a solution of the system:

ρ : 4x+ 5y − 6z + 24 = 0σ : x− y − 2z + 1 = 0

∼ x− y − 2z + 1 = 09y + 2z + 20 = 0,

and the point P = [xp, yp, zp] can be easily guessed starting from the last equation, posing yp = 0, we getzp = −10, and, substituting these values into the first equation, xp = −21, thus P = [−21, 0,−10]. Therefore:

p : X = P + tv, t ∈ IR ⇐⇒ X = [−21, 0,−10] + t(−16, 2,−9), t ∈ IR.

Observation. The equation of p is given by all solutions of the system formed with the equations of theintersecting planes: [−21, 0,−10] is a particular solution, and {t(−16, 2,−9), t ∈ IR} are all solutions of thehomogeneous system.

8.5) Find the angle between the planes ρ and σ, if ρ passes through the points M1 = [−2, 2, 2], M2 = [0, 5, 3] andM3 = [−2, 3, 4], and σ has equation 3x− 4y + z + 5 = 0.

Solution. The angle between two planes is the same as the angle between the orthogonal vectors to the planes. For σ

we can deduce from its equation that nσ = (3,−4, 1). For ρ, we obtain nρ =−−−−→M1M2 ×

−−−−→M1M3

nρ =(M2 −M1

)×(M3 −M1

)=([0, 5, 3]− [−2, 2, 2]

)×([−2, 3, 4]− [−2, 2, 2]

)=

= (2, 3, 1)× (0, 1, 2) = det

i j k2 3 10 1 2

= (5,−4, 2).

Thus, the cosine of the required angle θ is

cos θ =nρ · nσ

‖nρ‖ · ‖nσ‖=

(3,−4, 1) · (5,−4, 2)

‖(3,−4, 1)‖ · ‖(5,−4, 2)‖=

33√26√

45=

11√130

.

8.7) Find the distance between the point A = [8,−7, 1] and the plane σ with equation 2x+ 3y − 4z + 5 = 0.

Solution. We may solve the problem in several different ways, we will present a first short and elegant solution thatrelies on the properties of inner product, and a second more geometrical and standard way.First method. Let’s consider the vector n = (2, 3,−4) orthogonal to the plane and any point Q on theplane, for example Q = (−1,−1, 0), that we can guess from the equation of σ. From the properties of innerproduct (given any two vectors v,u, we have u · v = ‖v‖ · ‖u‖ cos θ, where θ is the angle in between u and

88

v), the vector−→AQ is such that its projection on the unit vector n

‖n‖ has length equal to the distance d(A, σ)

of A from the plane σ. Thus

d(A, σ) =

∣∣∣∣(Q−A) · n

‖n‖

∣∣∣∣ =|(9,−6, 1) · (2, 3,−4)|

‖n‖=| − 4|√

29=

4√29.

Second method. Let’s find the projection A′ of the point A on the plane σ, then d(A, σ) = ‖−−→AA′‖. We first

look for the equation of the line p going through A and orthogonal to the plane, thus p has direction vectorn = (2, 3,−4).

p : X = A+ tn, t ∈ IR or X = [8,−7, 1] + t(2, 3,−4), t ∈ IR.

Now we find A′ = p∩σ. We take the parametric equation of p, [x, y, z] = [8+2t,−7+3t, 1−4t], and substituteit into the equation of σ, to find the particular value (t0) of t that corresponds to the point A′.

[x, y, z] = [8 + 2t,−7 + 3t, 1− 4t]2x+ 3y − 4z + 5 = 0

=⇒ 2(8 + 2t) + 3(−7 + 3t)− 4(1− 4t) + 5 = 0 =⇒ t0 = − 4

29.

Therefore A′ = [8,−7, 1]− 429 (2, 3,−4), and

d(A, σ) = ‖−−→AA′‖ = ‖A−A′‖ =

4

29‖(2, 3,−4)‖ =

4√29.

8.9) Given A = [2, 9, 8], B = [6, 4,−2] and C = [7, 15, 7], show that−→AB and

−→AC are perpendicular, then find D

so that ABCD forms a rectangle.

Solution. Let’s evaluate:−→AB = B−A = [6, 4,−2]−[2, 9, 8] = (4,−5,−10),

−→AC = C−A = [7, 15, 7]−[2, 9, 8] = (5, 6,−1),

thus−→AB ·−→AC = (4,−5,−10) · (5, 6,−1) = (0) implies that

−→AB and

−→AC are perpendicular. Now we will find D

just posing the condition that the middle point M of the segment CB, that is one diagonal of the rectangleABCD, must also be the middle point of the other diagonal of the rectangle AD. Thus:

M =1

2(B + C) =

1

2([6, 4,−2] + [7, 15, 7]) =

1

2[13, 19, 5],

and, since 12 (D +A) = M implies D = 2M −A, we get

D = 21

2[13, 19, 5]− [2, 9, 8] = [11, 10,−3].

8.11) Given the line p passing through the points A = [1, 2, 1] and B = [2,−1, 3], find the point P on p closest tothe origin and the shortest distance from the origin to p.

Solution. To find the vector equation of p, we first calculate the direction vector v of p: v = B−A = [2,−1, 3]−[1, 2, 1] =(1,−3, 2). Now we can write

p : X = A+ tv, t ∈ IR =⇒ X = [1, 2, 1] + t(1,−3, 2), t ∈ IR.

First method. If X is a general point of the line p, its distance from the origin is :

d(X, 0) = ‖X − 0‖ =‖(1 + t, 2− 3t, 1 + 2t)‖ =√

(1 + t)2 + (2− 3t)2 + (1 + 2t)2 =

=√

1 + 2t+ t2 + 4− 12t+ 9t2 + 1 + 4t+ 4t2 =√

14t2 − 6t+ 6.

The value t0 that correspond to the minimum value of the distance is the critical point of the functionf(t) =

√14t2 − 6t+ 6, or also the critical point of the function that represents the distance squared: g(t) =

f2(t) = 14t2 − 6t+ 6, that is a parabola with vertex (and minimum) at t0 = 314 . We conclude that the point

P on p closest to the origin is:

P = [1, 2, 1] + t0(1,−3, 2) = [1, 2, 1] +3

14(1,−3, 2) =

[17

14,

19

14,

20

14

],

and its distance from the origin is: ∥∥∥∥(17

14,

19

14,

20

14

)∥∥∥∥ = f( 3

14

)=

√75

14.

89

Second method. We find the equation of the plane σ orthogonal to p passing through the origin. Since thenormal vector to σ is n = v = (1,−3, 2), we get

σ : x− 3y + 2z = 0.

The point P is the intersection of σ with p, thus we find it solving the system:

[x, y, z] = [1 + t, 2− 3t, 1 + 2t]x− 3y + 2z = 0

=⇒ t+ 1− 3(2− 3t) + 2(1 + 2t) = 0 =⇒ t =3

14.

Therefore P =[1714 ,

1914 ,

2014

], and

d(P, p) = ‖−→OP‖ =

∥∥∥∥(17

14,

19

14,

20

14

)∥∥∥∥ =

√75

14.

8.13) Show that the planes x+ y − 2z = 1 and x+ 3y − z = 4 intersect in a line and find the distance between thepoint C = [1, 0, 1] and this line.

Solution. The plene ρ with equation x + y − 2z = 1 has normal vector n = (1, 1,−2), and the plane σ with equationx + 3y − z = 4 has normal vector m = (1, 3,−1). Since n and m are not proportional, the planes are notparallel, thus they must intersect in a line. The equation of the line can be obtained as the set of all solutionsof the following system:

ρ : x+ y − 2z = 1σ : x+ 3y − z = 4

∼ x+ y − 2z = 12y + z = 3

.

Setting y = t, from the second equation we get z = −2t + 3, and, substituting these values into the firstequation, we get x = −5t+ 7. The equation of the line p is:

p : [x, y, z] = [−5t+ 7, t,−2t+ 3], t ∈ IR, or X = [7, 0, 3] + t(−5, 1,−2), t ∈ IR.

The line p has direction vector v = (−5, 1,−2), and is passing through the point A = [7, 0, 3]. To find the

distance between C and p, we will look for the projection C ′ of C on p and evaluate d(C, p) = ‖−−→CC ′‖. Looking

for C ′, we search for the unique particular point X of p, so that the vector−−→CX is orthogonal to v. Thus, we

pose the condition −−→CX · v = 0

and solve it for t.

−−→CX · v = 0 ⇐⇒ (X − C) · v = 0 ⇐⇒

([−5t+ 7, t,−2t+ 3]− [1, 0, 1]

)· (−5, 1,−2) = 0 ⇐⇒

⇐⇒ (−5t+ 6, t,−2t+ 2) · (−5, 1,−2) = 0 ⇐⇒ 30t− 34 = 0 ⇐⇒ t =17

15.

Therefore t0 = 1715 is the value of t that substituted into the equation of p will give the point C ′:

C ′ =(− 5

17

15+ 7,

17

15,−2

17

15+ 3)

=(20

15,

17

15,

11

15

).

Thus,

d(C, p) = ‖−−→CC ′‖ = ‖C ′ − C‖ =

∥∥∥∥(20

15,

17

15,

11

15

)− (1, 0, 1)

∥∥∥∥ =

∥∥∥∥( 5

15,

2

15,−4

15

)∥∥∥∥ =1

15

√45 =

1√5.

8.15) Find an equation for the plane through P = [1, 0, 1] and containing the line of intersection of the planesx+ y + 2z = 1 and x+ 3y + z = 4.

Solution. First method. Let’s call p the line of intersection of the planes x+ y + 2z = 1 and x+ 3y + z = 4. To getan equation for p we solve the system:

x+ y + 2z = 1x+ 3y + z = 4

∼ x+ y + 2z = 12y − z = 3

.

90

Setting y = t, from the second equation we get z = 2t − 3, and, substituting these values into the firstequation, we get x = −5t+ 7. The equation of the line p is:

p : [x, y, z] = [−5t+ 7, t, 2t− 3], t ∈ IR, or X = [7, 0,−3] + t(−5, 1, 2), t ∈ IR.

The line p has direction vector v = (−5, 1, 2), and is passing through the point A = [7, 0,−3]. The plane σpassing through P and containing p, contains the vector v and also all possible vectors joining any point of

p (for example A) with P , thus σ has normal vector n = v×−→AP :

n =v×−→AP = v× (P −A) = v× ([1, 0, 1]− [7, 0,−3]) = (−5, 1, 2)× (−6, 0, 4) =

= det

i j k−5 1 2−6 0 4

= (4, 8, 6) = 2(2, 4, 3).

The plane σ passing through P = [1, 0, 1] and with normal vector the ”shorter” n1 = (2, 4, 3), has equation:

σ : (X − P ) · n1 = 0, =⇒ σ : ([x, y, z]− [1, 0, 1]) · (2, 4, 3) = 0, =⇒ σ : 2x+ 4y + 3z − 5 = 0.

Second method. Every plane, containing the line of intersection of the two given planes, must have anequation that is a linear combination of the equations of the given plane (the system formed by the equationof this new plane and one of the given planes has the same set of solutions equal to all points in the line ofintersection). Thus we are looking for a plane with equation

α(x+ y + 2z − 1) + β(x+ 3y + z − 4) = 0.

In order to find α and β, it is enough to pose the condition that the plane passes through the given point P .

α(1 + 0 + 2− 1) + β(1 + 0 + 1− 4) = 0, ⇔ α = β.

Thus σ : 2x+ 4y + 3z − 5 = 0.

8.17) Find an equation for the plane passing through P = [6, 0, 2] and perpendicular to the line of intersection ofthe planes x+ y − 2z = 4 and 3x− 2y + z = 1.

Solution. A plane perpendicular to a line with direction vector v has normal vector n = v. Thus we need to find thedirection vector of the line of intersection of the planes σ : x + y − 2z = 4 and ρ : 3x − 2y + z = 1.Knowing than nσ = (1, 1,−2), and nρ = (3,−2, 1), we evaluate:

n = v = nσ × nρ = (1, 1,−2)× (3,−2, 1) = det

i j k1 1 −23 −2 1

= (−3,−7,−5).

The plane τ passing through P = [6, 0, 2] and with normal vector n1 = (3, 7, 5) (the opposite of n) hasequation:

τ : (X − P ) · n1 = 0, =⇒ τ : ([x, y, z]− [6, 0, 2]) · (3, 7, 5) = 0, =⇒ τ : 3x+ 7y + 5z − 28 = 0.

8.19) Prove that the lines p and q are skewed, find their distance d(p, q), and find also the points P ∈ p, Q ∈ q suchthat d(p, q) = d(P,Q), where

p : X = [1, 2, 1] + t(5,−2,−3), t ∈ IR, q : X = [−2, 6, 3] + s(4,−3,−1), s ∈ IR.

Solution. Two lines with vector equation:

p : X = A+ tv, t ∈ IR and q : X = B + su, s ∈ IR

are skewed if and only if the vectors u,v, and−→AB, are not on the same plane, i.e. if and only if the three

vectors are linearly independent, thus if and only if the determinant of the matrix formed with the threevectors as rows (or columns) has a non-zero determinant. In our case: v = (5,−2,−3), u = (4,−3,−1), and−→AB = [−2, 6, 3]− [1, 2, 1] = (−3, 4, 2), and the above determinant is:

det

5 −2 −34 −3 −1−3 4 2

= −21 6= 0,

91

thus, the given lines are skewed. In order to find the distance d(p, q), we may remember the formula expressing

this distance as the height of the parallelepiped V constructed with sides u,v and−→AB, and with base B, the

parallelogram formed by v and u:

d(p, q) =Volume(V )

Area(B)=

∣∣∣∣∣∣det

vu−→AB

∣∣∣∣∣∣‖v× u‖

=| − 21|

‖(−7,−7,−7)‖=

21

7√

3=√

3.

In case we don’t remember this geometrical fact, we may find directly the points P ∈ p, Q ∈ q such that

d(p, q) = d(P,Q). These points have the property that the vector−→PQ is orthogonal to v and also to u. Thus

we look for the particular values t0 and s0 such that P = [1, 2, 1]+t0(5,−2,−3), Q = [−2, 6, 3]+s0(4,−3,−1),−→PQ = Q− P = [−2, 6, 3] + s0(4,−3,−1)− [1, 2, 1] + t0(5,−2,−3) = (−3, 4, 2) + t0(5,−2,−3) + s0(4,−3,−1)and the following conditions of orthogonality are satisfied:(

(−3, 4, 2) + t0(5,−2,−3) + s0(4,−3,−1))· (5,−2,−3) = 0(

(−3, 4, 2) + t0(5,−2,−3) + s0(4,−3,−1))· (4,−3,−1) = 0.

This gives the following system:

−15− 8− 6 + t0(25 + 4 + 9) + s0(20 + 6 + 3) = 0

−12− 12− 2 + t0(20 + 6 + 3) + s0(16 + 9 + 1) = 0.

After simplification, we get:

38t0 + 29s0 = 29

29t0 + 26s0 = 26

that obviously has a unique solution t0 = 0, s0 = 1. Thus P = [1, 2, 1] + t0(5,−2,−3) = [1, 2, 1] = A, andQ = [−2, 6, 3] + s0(4,−3,−1) = [2, 3, 2]. From this, we again obtain:

d(p, q) = d(P,Q) = ‖−→PQ‖ = ‖[2, 3, 2]− [1, 2, 1]‖ = ‖(1, 1, 1)‖ =√

3.

8.21) Given the line p and two points A and B, find a point M on p that is equally far away from A and B, where:

p = ρ ∩ σ, ρ : x+ y − z − 3 = 0, σ : 3y − 2z + 1 = 0, A = [1, 0,−1], and B = [3, 4, 5].

Solution. A general point X = [x, y, z] of p has distance from A:

d(A,X) = ‖−→AX‖ = ‖[x, y, z]− [1, 0,−1]‖ = ‖(x− 1, y, z + 1)‖ =√

(x− 1)2 + y2 + (z + 1)2,

and distance from B:

d(B,X) = ‖−−→BX‖ = ‖[x, y, z]− [3, 4, 5]‖ = ‖(x− 3, y − 4, z − 5)‖ =√

(x− 3)2 + (y − 4)2 + (z − 5)2.

Because M lies on p = ρ∩ σ and has same distance from A and B, the point M that we are looking for musthave coordinates that satisfy the following three conditions:

x+ y − z − 3 =0

3y − 2z + 1 =0√(x− 1)2 + y2 + (z + 1)2 =

√(x− 3)2 + (y − 4)2 + (z − 5)2.

Thus we are looking for the solution of the system:

x+ y − z − 3 =0

3y − 2z + 1 =0

x2 − 2x+ 1 + y2 + z2 + 2z + 1 =x2 − 6x+ 9 + y2 − 8y + 16 + z2 − 10z + 25,

92

that, after simplification, gives a linear system for which we use a gaussian reduction:

x+ y − z = 33y − 2z = −1

x+ 2y + 3z = 12,

R3 −R1

∼

x+ y − z = 33y − 2z = −1y + 4z = 9,

−3R3 +R2

∼

x+ y − z = 33y − 2z = −1−14z = −28

.

Starting from the last equation, we get that the unique solution of the system (the coordinates of the pointM) is z = 2, y = 1, x = 4. The point M = [4, 1, 2] lies on p and:

d(A,M) = ‖(3, 1, 3)‖ = d(B,M) = ‖(1,−3,−3)‖ =√

19.

8.23) Find the pointR symmetric to P = [−4, 5, 8] with respect to the line p throughA = [9, 4, 10] andB = [−6, 1, 1].

Solution. We first find the projection P ′ of P on p, and then we look for the point R such that P ′ is the middle pointof the segment PR, i.e. P+R

2 = P ′ or equivalently R = 2P ′ − P . Looking for P ′, we observe that P ′ is thepoint of intersection of the line p with the plane ρ going through P and orthogonal to p. The plane ρ, thus,

has orthogonal vector n = v, where v =−→AB is the direction vector of p.

v =−→AB = B −A = [−6, 1, 1]− [9, 4, 10] = (−15,−3,−9) = −3(5, 1, 3).

We will use for p direction vector v1 = (5, 1, 3), parallel to v but with more ”convenient” coordinates.

p : X = A+ tv, t ∈ IR =⇒ X = [9, 4, 10] + t(5, 1, 3).

ρ : (X − P ) · v1 = 0 =⇒ (x+ 4, y − 5, z − 8) · (5, 1, 3) = 0 =⇒ 5x+ y + 3z − 9 = 0.

To find P ′, we look for the particular value t0 of t, such that the point of p with coordinates [9 + 5t0, 4 +t0, 10 + 3t0] satisfies also the equation of ρ:

5(9 + 5t0) + (4 + t0) + 3(10 + 3t0)− 9 = 0.

After simple computation we get:35t0 + 70 = 0,

which means that t0 = −2. Thus P ′ = [9 + 5(−2), 4− 2, 10 + 3(−2)] = [−1, 2, 4]. We can now find R:

R = 2P ′ − P = 2[−1, 2, 4]− [−4, 5, 8] = [2,−1, 0].

8.25) A line with direction vector v = (0, 9,−1) intersects lines p and q, find the coordinates of the points ofintersection, where

p :x− 8

5=y − 5

1=

z

−1, and q :

x

1=y − 1

−2=z + 1

1.

Solution. The given equations for p and q are equivalent to:

p : X = [8, 5, 0] + t(5, 1,−1), t ∈ IR and q : X = [0, 1,−1] + s(1,−2, 1), s ∈ IR.

We are looking for a point P ∈ p with coordinates P = [8 + 5t, 5 + t,−t] for a particular value of t, and a

point Q ∈ q, Q = [s, 1 − 2s,−1 + s], for a particular value of s, such that the vector−→PQ is parallel to the

given v, i.e.−→PQ = λv, for a particular value of λ. We get the following condition:

[s, 1− 2s,−1 + s]− [8 + 5t, 5 + t,−t] = λ(0, 9,−1),

that is equivalent to a system of linear equations:

s− 8− 5t = 01− 2s− 5− t = 9λ−1 + s+ t = −λ.

93

After reordering, we start a gaussian reduction:

s− 5t = 89λ+ 2s+ t = −4λ+ s+ t = 1

R3 ↔ R1

∼R2 − 8R3

λ+ s+ t = 1−7s− 8t = −13s− 5t = 8

R3 ↔ R2

∼R2 + 7R3

λ+ s+ t = 1s− 5t = 8−43t = 43.

Starting from the last equation, we get that the unique solution of the system is t = −1, s = 3, λ = −1.Thus, the sought points are P = [8 + 5t, 5 + t,−t] = [3, 4, 1], and Q = [s, 1− 2s,−1 + s] = [3,−5, 2].

8.27) Find the distance of the point A = [3, 2, 1] from the plane containing the lines p and q, where

p :x+ 1

1=y − 3

2=z − 2

2, and q : X = [0, 4, 2] + t(1, 1, 0), t ∈ IR.

Solution. First of all, we want to check that the given lines p and q really lie on the same plane and they are notskewed. Writing the equation of p in vector form: p : X = [−1, 3, 2] + s(1, 2, 2), s ∈ IR, we need to verifythat the vectors vp = (1, 2, 2), vq = (1, 1, 0) and w = ([−1, 3, 2] − [0, 4, 2]), are linearly dependent. Sincew = (−1,−1, 0) = −vq, the vectors are obviously linearly dependent, thus p and q lie on the same plane σ.In Solution of Example 7, we discussed two ways how to solve the general problem: find the distance of apoint A from a plane σ. Using the first method: we need to identify a point Q on σ and a vector n, orthogonalto σ. We can choose as Q the ”starting” point of line q: Q = [0, 4, 2]. To find n we will consider the directionvectors vp, vq of p and q, and set n = vp × vq. Since vp = (1, 2, 2) and vq = (1, 1, 0), we have:

n = vp × vq = det

i j k1 2 21 1 0

= (−2, 2,−1).

Now, we may conclude:

d(A, σ) =

∣∣∣∣−→AQ · n

‖n‖

∣∣∣∣ =

∣∣∣∣∣([0, 4, 2]− [3, 2, 1]

)· (−2, 2,−1)

‖(−2, 2,−1)‖

∣∣∣∣∣ =

∣∣∣∣ (−3, 2, 1) · (−2, 2,−1)

‖(−2, 2,−1)‖

∣∣∣∣ =

∣∣∣∣ 6 + 4− 1√4 + 4 + 1

∣∣∣∣ = 3.

94

Chapter 99. Exam-like problems

9.1) Consider the following system of linear equations

(S)

kx+ y − z = −k(1− k)y + z = h+ ky + (1− k)z = 2h+ 1

where h, k ∈ IR are real parameters.

9.1.a) Determine for what values of h, k ∈ IR the system has one unique solution, no solutions or infinitelymany solutions.

9.1.b) If Ak is the matrix associated to the system (S), so that the system can be written in the form

Ak

xyz

=

−kh+ k2h+ 1

write the linear transformation lk : IR3 → IR3 associated to Ak with respect to the canonical basis of IR3.

9.1.c) Determine for what values of k dim Im lk = 3.

9.1.d) Determine for what values of k dim Ker lk = 2.

9.1.e) For k = 0 is the transformation l0 diagonalizable?

9.2) Given the linear transformations f : IR4 → IR3, and g : IR3 → IR4,

f(x, y, z, t) = (x− t, x+ y, z + y), g(x, y, z) = (z − x, y, y, x+ t).

9.2.a) Find Ker f , Im f , Ker g, Im g,

9.2.b) Solve g(x, y, z) = (h,−1, h, 1) for any possible h ∈ IR,

9.2.c) Write down the general form of f ◦ g, g ◦ f and determine if they are isomorphisms,

9.2.d) Are f ◦ g, g ◦ f diagonalizable? Find all respective eigenvalues and a basis for each correspondingeigenspace.

9.3) Consider the linear transformation f : IR3 → IR3 defined by:

f(1, 0, 0) = (−1,−2,−3)

f(0, 1, 0) = (2, 2, 2)

f(0, 0, 1) = (0, 1, 2)

9.3.a) Is f invertible?

9.3.b) Find a basis of Ker(f) and a basis of Im(f).

9.3.c) Find all eigenvalues and corresponding eigenvectors of f . Is f diagonalizable?

9.3.d) Solve the system f2(v) = 0.

9.4) Given fh : P3(x)→M2,2 defined by

fh(a+ bx+ cx2 + dx3) =

(a+ b+ c+ hd b+ c−b− c− hd hb

),

where h ∈ IR is a parameter.

9.4.a) Give the definition of Kernel and Image of a linear transformation. Find, for all possible values of h,Ker(fh), Im(fh), their bases and dimensions.

9.4.b) Does there exist a value of h ∈ IR such that fh is not an isomorphism?

95

9.4.c) Determine f−1h (A) (i.e. the set of all polynomials mapped to A), where A =

(0 3−3 0

).

9.4.d) For h = 0, find all eigenvalues and corresponding eigenvectors of f0 determine if f0 is diagonalizable.

9.5) Given the transformation fh : IR3 → IR3 defined by

fh(x, y, z) = (x− hz, x+ y − hz,−hx+ z),


9.5.a) Find, for all possible values of h, Ker(fh), Im(fh), their bases and dimensions.

9.5.b) Is fh an isomorphism for some value of h?

9.5.c) Determine f−1h (1, 0, 1) = {(x, y, z) ∈ IR3 : fh(x, y, z) = (1, 0, 1)}.

9.5.d) Give the definition of eigenvalue and eigenvector of a linear transformation. Determine if fh is diago-nalizable for some values of h. For h = 1, find a basis of IR3 made of eigenvectors of f1.

9.6) Consider the matrix A =

1 0 −21 −3 h2 −h 1

.

9.6.a) Give the definition of regular matrix. Find all values of h for which the matrix A is regular.

9.6.b) For h = 1, solve the matrix equation A ·X = B,

where B =

1 0 −21 −1 12 −1 1

.

9.6.c) Find the solutions of the system A ·

xyz

=

8412

for all possible h ∈ IR.

9.6.d) Is the matrix A diagonalizable for h = 0?

9.6.e) For h = 3, find the Image and the Kernel of the linear transformation l that has A as associatedmatrix (with respect to the standard basis of IR3).

96

Linear Algebra - Solutions of odd number problems - Exam-like problems

9.1) Consider the following system of linear equations

(S)

kx+ y − z = −k(1− k)y + z = h+ ky + (1− k)z = 2h+ 1

where h, k ∈ IR are real parameters.

9.1.a) Determine for what values of h, k ∈ IR the system has one unique solution, no solutions or infinitelymany solutions.

9.1.b) If Ak is the matrix associated to the system (S), so that the system can be written in the form

Ak

xyz

=

−kh+ k2h+ 1

write the linear transformation lk : IR3 → IR3 associated to Ak with respect to the canonical basis of IR3.

9.1.c) Determine for what values of k dim Im lk = 3.

9.1.d) Determine for what values of k dim Ker lk = 2.

9.1.e) For k = 0 is the transformation l0 diagonalizable?

Solution. 9.1.a) Let’s consider the matrix Ak associated to the homogeneous system:

Ak =

k 1 −10 1− k 10 1 1− k

.

Evaluating detAk = k2(k − 2), we see that, for k 6= 0 and k 6= 2 (and any possible value of h), the systemhas a unique solution that can be evaluated using Cramer’s rule.For k = 0, the augmented matrix of the system is

A0 =

0 1 −10 1 10 1 1

∣∣∣∣∣∣∣∣∣∣∣∣

0h2h

.

Let’s use a matrix form of Gaussian reduction: with operations R2 −R1, and R3 −R2 we get: 0 1 −10 0 20 0 0

∣∣∣∣∣∣∣∣∣∣∣∣0hh

.

Now, using Frobenious theorem, we see that, for k = 0 and h 6= 0, the system has no solution since the rankof the augmented matrix is not equal to the rank of the matrix associated to the homogeneous system; whilefor k = 0 and h = 0 the system has infinitely many solution since the rank of the augmented matrix (equal tothe rank of the matrix associated to the homogeneous system) is two and the number of unknowns is three.In a similar, for k = 2, the augmented matrix of the system is

A2 =

2 1 −10 −1 10 1 −1

∣∣∣∣∣∣∣∣∣∣∣∣−2h+ 22h+ 1

.

We start a matrix form of Gaussian reduction with operations R3 +R2 and we get: 2 1 −10 −1 10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣−2h+ 23h+ 3

.

Again, using Frobenious theorem, we see that, for k = 2 and h 6= −1, the system has no solution since therank of the augmented matrix is not equal to the rank of the matrix associated to the homogeneous system;

97

while for k = −2 and h = −1 the system has infinitely many solution since the rank of the augmented matrix(equal to the rank of the matrix associated to the homogeneous system) is two and the number of unknownsis three.

9.1.b) We have seen that

Ak =

k 1 −10 1− k 10 1 1− k

.

To find the linear transformation lk : IR3 → IR3 associated to Ak with respect to the canonical basis of IR3,we just recall that, for every (x, y, z) ∈ IR3

lk(x, y, z) = Ak

xyz

=(kx+ y − z, (1− k)y + z, y + (1− k)z

).

9.1.c) We now need to find the values of k for which dim Im lk = 3. The theorem on dimensions of Kerneland Image of a linear transformation lk from IR3 to IR3 states that

3 = dim IR3 = dim Ker lk + dim Im lk,

thus the Image of lk has dimension three if and only if lk is an isomorphism. This happens for all values of kfor which detAk 6= 0, thus dim Im lk = 3 for all k 6= 0,−2.

9.1.d) We now determine for what values of k dim Ker lk = 2. We have seen that for k 6= 0,−2, lk is anisomorphism, moreover in 9.1.a) we have seen that for k = 0 and k = −2 the rank of Ak is two, this impliesthat the dimension of the kernel is one (since the dimension of the Kernel is equal to the dimension of the spaceof solutions of the homogeneous system). We conclude that there is no value of k for which dim Ker lk = 2.

9.1.e) For k = 0, the transformation l0(x, y, z) = (y − z, y + z, y + z) has associated matrix

A0 =

0 1 −10 1 10 1 1

.

To see if A0 is diagonalizable, we start by finding the real solutions of det(A0 − λI) = 0. In this case, wehave:

det

( 0 1 −10 1 10 1 1

− λ 1 0 0

0 1 00 0 1

) = det

−λ 1 −10 1− λ 10 1 1− λ

= λ2(2− λ).

The characteristic equation of A0 has two distinct solutions λ = 2 with algebraic multiplicity one and λ = 0with algebraic multiplicity two. We now look for the eigenvectors corresponding to λ = 0. We must find thenontrivial solutions of the homogeneous system (A0 − λI)v = 0, where λ = 0. The augmented matrix of thesystem is 0 1 −1

0 1 10 1 1

∣∣∣∣∣∣∣∣∣∣∣∣000

R2 −R1

∼R3 −R2

0 1 −10 0 20 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 0 are {t(1, 0, 0) : t ∈ IR \ {0}}, and the eigenspace corresponding toλ = 0 (that corresponds to the kernel of the linear transformation lk for k = 0, already discussed in 9.1.d))is U0 = span{(1, 0, 0)}. Corresponding to the eigenvalue λ = 0 with algebraic multiplicity two, we found andeigenspace U0 with dimension one. This implies that A0, and thus l0, is not diagonalizable.

9.3) Consider the linear transformation f : IR3 → IR3 defined by:

f(1, 0, 0) = (−1,−2,−3)

f(0, 1, 0) = (2, 2, 2)

f(0, 0, 1) = (0, 1, 2)

9.3.a) Is f invertible?

9.3.b) Find a basis of Ker(f) and a basis of Im(f).

9.3.c) Find all eigenvalues and corresponding eigenvectors of f . Is f diagonalizable?

9.3.d) Solve the system f2(v) = 0.

98

Solution. We consider the ordered canonical basis of IR3, C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, and we will find the matrixA = A(l, C, C) associated to the given linear transformation with respect to C. We are given that:

f(1, 0, 0) = (−1,−2,−3)

f(0, 1, 0) = (2, 2, 2)

f(0, 0, 1) = (0, 1, 2)

Thus we can write A =

−1 2 0−2 2 1−3 2 2

.

9.3.a) We evaluate detA, because, only in case detA 6= 0, we know that the linear transformation is anisomorphism thus invertible. Since detA = −4− 6 + 2 + 8 = 0, we conclude that f is not invertible.

9.3.b) We now find Ker l =

{(x1, x2, x3) ∈ IR3 : A

x1x2x3

=

000

}, the space of solutions of the homo-

geneous system associated to A. Since detA = 0, we know that the homogeneous system does not have aunique solution (the trivial one). We will use the matrix form of Gaussian reduction and get:−1 2 0

−2 2 1−3 2 2

∣∣∣∣∣∣∣∣∣∣∣∣000

R2 − 2R1

∼R3 − 3R1

−1 2 00 −2 10 −4 2

∣∣∣∣∣∣∣∣∣∣∣∣000

−R2

∼R3 − 2R2

−1 2 00 2 −10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

Setting x2 = α, α ∈ IR, we get from the second row (imagining the correspondent equation 2x2 − x3 = 0)x3 = 2α, and, subsequently, from the first row x1 = 2α. Thus we conclude that:Ker l = {(2α, α, 2α), α ∈ IR} = span{(2, 1, 2)}, dim Ker l = 1, and a basis of Ker l is BK = {(2, 1, 2)}.From the Theorem on dimensions, we get dim Im l = dim IR3 − dim Ker l = 2, and, knowing that

Im l = span{l(1, 0, 0), l(0, 1, 0), l(0, 0, 1)}

we getIm l = span{(−1,−2,−3), (2, 2, 2), (0, 1, 2)} = span{(−1,−2,−3), (2, 2, 2)},

where the first two vectors were chosen randomly, just considering that, because they are linearly independent(they are not a multiple of each other), they must span the space Im l that has dimension 2. Thus, a basis ofIm l is BI = {(−1,−2,−3), (2, 2, 2)}.

9.3.c) To find all eigenvalues and corresponding eigenvectors of f (or equivalently of A), we start by findingthe real solutions of det(A− λI) = 0. In this case, we have:

det

(−1 2 0−2 2 1−3 2 2

− λ 1 0 0

0 1 00 0 1

) = det

−1− λ 2 0−2 2− λ 1−3 2 2− λ

= −λ(λ− 1)(λ− 2).

The characteristic equation of A has three distinct solutions with algebraic multiplicity one: λ = 0, λ = 1and λ = 2 are the eigenvalues of A. We may already conclude that A (thus f) is diagonalizable.We now look for the eigenvectors corresponding to λ = 0. We must find the nontrivial solutions of thehomogeneous system (A− λI)v = 0, where λ = 0. The augmented matrix of the system is−1 2 0

−2 2 1−3 2 2

∣∣∣∣∣∣∣∣∣∣∣∣000

.

We don’t need to evaluate anything, since we have already done the math looking for the Kernel of f . Allthe eigenvectors corresponding to λ = 0 are {t(2, 1, 2) : t ∈ IR \ {0}}, and the eigenspace corresponding toλ = 0 is (the kernel of f) U0 = span{(2, 1, 2)}.Corresponding to the eigenvalue λ = 1, we look for eigenvectors, i.e. the nontrivial solutions of the homoge-neous system (A− λI)v = 0, where λ = 1. The augmented matrix of the system is−2 2 0

−2 1 1−3 2 1

∣∣∣∣∣∣∣∣∣∣∣∣000

R1/2R2 −R1

2R3 − 3R1

∼

−1 1 00 −1 10 −2 2

∣∣∣∣∣∣∣∣∣∣∣∣000

R3 − 2R2

∼

−1 1 00 −1 10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.


99

Finally, we look for the eigenvectors corresponding to the eigenvalue λ = 2, i.e. the look for the nontrivialsolutions of the homogeneous system (A− λI)v = 0, where λ = 2. The augmented matrix of the system is−3 2 0

−2 0 1−3 2 0

∣∣∣∣∣∣∣∣∣∣∣∣000

R3 −R1

∼

−3 2 0−2 0 10 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.


9.3.d) To solve the system f2(v) = 0, is equivalent to finding the Kernel of f2. Indeed, we recall that, givena linear transformation f , with associated matrix A, the transformation f2 = f ◦ f is again linear and hasassociated matrix A2 = A ·A. Let’s evaluate A2:

A2 = A ·A =

−1 2 0−2 2 1−3 2 2

·−1 2 0−2 2 1−3 2 2

=

−3 2 2−5 2 4−7 2 6

.

We know that rankA2 ≤ rankA (and also Ker f ⊆ Ker f2). Using a Gaussian reduction, we evaluate:−3 2 2−5 2 4−7 2 6

3R2 − 5R1

∼3R3 − 7R1

−3 2 20 −4 20 −8 4

R2/2∼

R3 − 2R2

−3 2 20 −2 10 0 0

.

From this, it follows that rankA2 = rankA, and thus Ker f2 = Ker f = span{(2, 1, 2)}.

9.5) Given the transformation fh : IR3 → IR3 defined by

fh(x, y, z) = (x− hz, x+ y − hz,−hx+ z),


9.5.a) Find, for all possible values of h, Ker(fh), Im(fh), their bases and dimensions.

9.5.b) Is fh an isomorphism for some value of h?

9.5.c) Determine f−1h (1, 0, 1) = {(x, y, z) ∈ IR3 : fh(x, y, z) = (1, 0, 1)}.

9.5.d) Give the definition of eigenvalue and eigenvector of a linear transformation. Determine if fh is diago-nalizable for some values of h. For h = 1, find a basis of IR3 made of eigenvectors of f1.

Solution. We consider the ordered canonical basis of IR3, C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, and we will find the matrixAh = Ah(fh, C, C) associated to the given linear transformation with respect to C. We evaluate:

fh(1, 0, 0) = (1, 1,−h)

fh(0, 1, 0) = (0, 1, 0)

fh(0, 0, 1) = (−h,−h, 1)

Thus, we can write

Ah =

1 0 −h1 1 −h−h 0 1

.

9.5.a) To find Ker fh =

{(x, y, z) ∈ IR3 : Ah

xyz

=

000

}, the space of solutions of the homogeneous

system associated to Ah, we first evaluate detAh, because, for all values of h ∈ IR such that detAh 6= 0, weknow that the homogeneous system has a unique solution, the trivial one, thus Ker fh = {0}.Since detAh = 1−h2 = (1−h)(1+h), we conclude that, for h 6= 1,h 6= −1, Ker fh = {(0, 0, 0)}, consequentlyIm fh = IR3, that means that fh is an isomorphism (9.5.b)).For h = 1, we need to find a basis of

Ker f1 =

{(x, y, z) ∈ IR3 : A1

xyz

=

000

} =

{(x, y, z) ∈ IR3 :

1 0 −11 1 −1−1 0 1

xyz

=

000

},100

that means, we need to solve the homogeneous system: 1 0 −11 1 −1−1 0 1

∣∣∣∣∣∣∣∣∣∣∣∣000

R2 −R1

∼R3 +R1

1 0 −10 1 00 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

Setting z = α, α ∈ IR, we get from the first row x = α, while y = 0 follows from the second row. Thus weconclude that:Ker f1 = {(α, 0, α), α ∈ IR} = span{(1, 0, 1)}, dim Ker f1 = 1, and a basis of Ker f1 is BK1

= {(1, 0, 1)}.From the Theorem on dimensions, we get dim Im f1 = dim IR3 − dim Ker f1 = 2, moreover:

Im f1 =span{f1(1, 0, 0), f1(0, 1, 0), f1(0, 0, 1)}=span{(1, 1,−1), (0, 1, 0), (−1,−1, 1)}=span{(1, 1,−1), (0, 1, 0)}.

Im f1 has dimension 2, and a basis of Im f1 is BI1 = {(1, 1,−1), (0, 1, 0)}.

For h = −1, we need to find a basis of

Ker f−1 =

{(x, y, z) ∈ IR3 : A−1

xyz

=

000

} =

{(x, y, z) ∈ IR3 :

1 0 11 1 11 0 1

xyz

=

000

},that means, we need to solve the homogeneous system: 1 0 1

1 1 11 0 1

∣∣∣∣∣∣∣∣∣∣∣∣000

R2 −R1

∼R3 −R1

1 0 10 1 00 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

Setting z = α, α ∈ IR, we get from the first row x = −α, while y = 0 follows from the second row. Thus weconclude that:Ker f−1 = {(α, 0,−α), α ∈ IR} = span{(1, 0,−1)}, dim Ker f−1 = 1, and a basis of Ker f−1 is BK−1

={(1, 0,−1)}.From the Theorem on dimensions, we get dim Im f−1 = dim IR3 − dim Ker f−1 = 2, moreover:

Im f−1 =span{f−1(1, 0, 0), f−1(0, 1, 0), f−1(0, 0, 1)}=span{(1, 1, 1), (0, 1, 0), (1, 1, 1)}=span{(1, 1, 1), (0, 1, 0)}.

Im f−1 has dimension 2, and a basis of Im f−1 is BI−1 = {(1, 1, 1), (0, 1, 0)}.

9.5.c) Now we determine f−1h (1, 0, 1) = {(x, y, z) ∈ IR3 : fh(x, y, z) = (1, 0, 1)}, for any possible value ofh ∈ IR.For h 6= 1, h 6= −1, we have seen that fh is an isomorphism, thus the system fh(x, y, z) = (1, 0, 1) has aunique solution. The augmented matrix of the system is 1 0 −h

1 1 −h−h 0 1

∣∣∣∣∣∣∣∣∣∣∣∣101

.

We may evaluate its unique solution (for h 6= 1, h 6= −1) using Cramer’s rule:

x =detC1

detAh=

det

1 0 −h0 1 −h1 0 1

(1− h)(1 + h)

=1 + h

(1− h)(1 + h)=

1

1− h

y =detC2

detAh=

det

1 1 −h1 0 −h−h 1 1

(1− h)(1 + h)

=h2 − 1

(1− h)(1 + h)= −1

z =detC3

detAh=

det

1 0 11 1 0−h 0 1

(1− h)(1 + h)

=1 + h

(1− h)(1 + h)=

1

1− h.

101

For h = 1, we will perform a Gaussian reduction starting from the augmented matrix of the given system,where we substitute the value h = 1: 1 0 −1

1 1 −1−1 0 1

∣∣∣∣∣∣∣∣∣∣∣∣101

R2 −R1

∼R3 +R1

1 0 −10 1 00 0 0

∣∣∣∣∣∣∣∣∣∣∣∣012

.

From the last row, using Frobenius theorem, we see that the system has no solution.For h = −1, we will perform a Gaussian reduction starting from the augmented matrix of the given system,where we substitute the value h = −1: 1 0 1

1 1 11 0 1

∣∣∣∣∣∣∣∣∣∣∣∣101

R2 −R1

∼R3 −R1

1 0 10 1 00 0 0

∣∣∣∣∣∣∣∣∣∣∣∣

0−10

.

The system has infinitely many solutions. We have already evaluated that Ker f−1 = {(α, 0,−α), α ∈ IR} =span{(1, 0,−1)}, guessing a particular solution of the system xp = (0,−1, 0), we conclude that all solutionsof the system are:

x = (0,−1, 0) + span{(1, 0,−1)}.

9.5.d) By definition, given a linear transformation l : M → L (M,L linear spaces), a real number λ ∈ IR iscalled an eigenvalue of l if there exists a nontrivial vector v ∈ M (v 6= 0), such that lv = λv. The vectorv is said to be an eigenvector of l, associated to the eigenvalue λ. The set of all eigenvectors associated tothe same eigenvalue λ, plus the zero vector, form a linear subspace of M , called eigenspace associated to theeigenvalue λ.The transformation fh(x, y, z) = (x− hz, x+ y − hz,−hx+ z), has associated matrix

Ah =

1 0 −h1 1 −h−h 0 1

.

To see if fh is diagonalizable, we start by finding the real solutions of det(Ah−λI) = 0. In this case, we have:

det

( 1 0 −h1 1 −h−h 0 1

− λ 1 0 0

0 1 00 0 1

) =

1− λ 0 −h1 1− λ −h−h 0 1− λ

=

= (1− λ)3 − (1− λ)h2 = (1− λ[(1− λ)2 − h2]) = (1− λ)(1− λ− h)(1− λ+ h).

The solutions of the characteristic equation are λ = 1, λ = 1 − h, and λ = 1 + h, thus for any h 6= 0, thefound eigenvalues are distinct, each with algebraic multiplicity one. It follows that the corresponding fh isdiagonalizable.

For h = 0, we must determine if A0 =

1 0 01 1 00 0 1

is diagonalizable. We have calculated the eigenvalues

of A0: we found that λ = 1 is the only eigenvalue with algebraic multiplicity three. Let’s now evaluatethe eigenvectors corresponding to λ = 1. We must find the nontrivial solutions of the homogeneous system(A0 − λI)v = 0, where λ = 1. The augmented matrix of the system is 0 0 0

1 0 00 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All eigenvalues of A0 are {t(0, 1, 0) + u(0, 0, 1) : t, u ∈ IR non both zero}, and the eigenspace correspondingto λ = 1 is U1 = span{(0, 1, 0), (0, 0, 1)}. Since the eigenvalue λ = 1 has algebraic multiplicity three, but thecorresponding eigenspace has dimension two, A0, and thus f0, is not diagonalizable.We conclude that fh is diagonalizable for any possible real value of h 6= 0.

Let’s now consider the value h = 1 and find the eigenvectors of f1. In this case the eigenvalues are λ = 1,λ = 0, and λ = 2.For λ = 1, we must find the nontrivial solutions of the homogeneous system (A1 − λI)v = 0, where λ = 1.The augmented matrix of the system is 0 0 −1

1 0 −1−1 0 0

∣∣∣∣∣∣∣∣∣∣∣∣000

.

102

All the eigenvectors corresponding to λ = 1 are {t(0, 1, 0) : t ∈ IR \ {0}}, and the eigenspace corresponding toλ = 1 is U1 = span{(0, 1, 0)}.For λ = 0, we must find the nontrivial solutions of the homogeneous system (A1 − λI)v = 0, where λ = 0.The augmented matrix of the system is 1 0 −1

1 1 −1−1 0 1

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 0 are {t(1, 0, 1) : t ∈ IR \ {0}}, and the eigenspace corresponding toλ = 0 is U0 = span{(1, 0, 1)}.For λ = 2, we must find the nontrivial solutions of the homogeneous system (A1 − λI)v = 0, where λ = 2.The augmented matrix of the system is −1 0 −1

1 −1 −1−1 0 −1

∣∣∣∣∣∣∣∣∣∣∣∣000

.

All the eigenvectors corresponding to λ = 2 are {t(1, 2,−1) : t ∈ IR \ {0}}, and the eigenspace correspondingto λ = 2 is U2 = span{(1, 2,−1)}.For h = 1, we found a basis B of IR3 made of eigenvectors of f1:

B = {(0, 1, 0), (1, 0, 1), (1, 2,−1)}.

103

Table of content

Chapter 1 - Polynomials and Gaussian elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 1

Solutions of odd number problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Chapter 2 - Linear speces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Solutions of odd number problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Chapter 3 - Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Solutions of odd number problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Chapter 4 - Systems of linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32


Chapter 5 - Linear transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43


Chapter 6 - Eigenvalues, eigenvectors and their applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58


Chapter 7 - Euclidean spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73


Chapter 8 - Analytic geometry in three dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .84


Chapter 9 - Exam-like problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .94


Bibliography

[1] M.Demlova, B. Pondelicek, Uvod do Algebry, Vydavatelstvi CVUT, 2003.

[2] P.Ptak Introduction to Linear Algebra, Nakladatelstvi CVUT, 2006.

[3] K.R.Matthews, Elementary Linear Algebra, Lecture Notes, Department of Mathematics, University ofQueensland, 2013, http://www.numbertheory.org/book/mp103.pdf.

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