Linear Algebra : Excercises, Exercises of Linear Algebra

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Linear Algebra: Exercises

Alessandro Berarducci, Oscar Papini

November 24, 2016

ii

iv CONTENTS

Chapter 1

Linear Systems

Exercise 1.1. Let A be the matrix

A =

(a) Determine if the system Ax = 0 has zero, one or infinitely many solutions,

and compute a basis of the space of solutions.

(b) Is it true that the system Ax = b has a solution for any b ∈ R

3 ? If so, prove

the statement, otherwise find a counterexample.

Solution. (a) We have to find ker A. At first we row-reduce A:

(second row) − (first row)

(third row) − 2(first row)

(third row) − (second row)

Solution. The matrix associated to the system is

1 2 a

2 − 19 a

We get the row echelon form of the matrix subtracting the first row from the

third, and then subtracting four times the first row from the second:

0 0 a

2 − 16 a − 4

Notice that the roots of a

2 − 16 = (a + 4)(a − 4) are ± 4.

• If a 6 = ± 4 , the term a

2 − 16 is not zero and we have a unique solution.

• If a = − 4 , the third equation of the system in echelon form becomes 0 z = − 8

and there are no solutions.

• If a = 4, the third equation of the system in echelon form becomes 0 z = 0,

which is satisfied by any value of z. Therefore the system has infinitely

many solutions. 

Exercise 1.3. Determine the number of solutions of the following system

x 1 + kx 2 + (1 + 4k)x 3 = 1 + 4k

2 x 1 + (k + 1)x 2 + (2 + 7k)x 3 = 1 + 7k

3 x 1 + (k + 2)x 2 + (3 + 9k)x 3 = 1 + 9k

depending on the parameter k ∈ R.

Solution. If we row-reduce the complete matrix of the system, we get

A =

1 k 4 k + 1 4 k + 1

0 1 − k −k −k − 1

0 0 −k −k

• If k 6 = 0 and k 6 = 1, the matrix A has three non-zero pivots in the first three

columns. Therefore the coefficients matrix is invertible, and the system has

a unique solution.

• If k = 0, the matrix A has two non-zero pivots in the first two columns,

and we have only zeros in the third row. This means that the rank of

the coefficients matrix is 2 and it is the same of the one of the complete

matrix, but it is not the maximum rank. Thus the system has infinitely

many solutions.

4 CHAPTER 1. LINEAR SYSTEMS

• If k = 1, the matrix A is no longer row reduced, so we can go on and get

This matrix has a pivot in the last column, which then cannot be a linear

combination of the others. In this case the system has no solutions. 

Exercise 1.4. Consider the following system, where the variables are x, y, z:

(k + 2)x + 2ky − z = 1

x − 2 y + kz = −k

y + z = k.

(a) For which values of k ∈ R does the system have a unique solution?

(b) For which values of k ∈ R does the system have infinitely many solutions?

If any, compute for those values of k all the solutions of the system.

Solution. An n×n system has a solution if and only if the matrix of the coefficients

and the complete matrix have the same rank; in particular, the solution is unique

if this rank is n, and there are infinitely many solutions if this rank is less than

n.

In this case, the complete matrix is

k + 2 2 k − 1 1

1 − 2 k −k

0 1 1 k

Elementary row operations change neither the set of the solutions, nor the ranks

of the matrices. After some computation we get to the row echelon form:

6 CHAPTER 1. LINEAR SYSTEMS

In conclusion, the set of solutions is

{(− 1 − λ, − 1 − λ, λ) | λ ∈ R}. 

Chapter 2

Vector Spaces

Exercise 2.1. (I) Find the coordinates of the vector c =

[

]

with respect to the

basis (u 1 , u 2 ) of R

2 , where

u 1 =

[

]

and u 2 =

[

]

(II) Find the coordinates of the polynomial p(x) = 2 + x in the space R[x]≤ 1 of

polynomials with real coefficients and degree less than or equal to 1 , with respect

to the basis (q 1 (x), q 2 (x)), where q 1 (x) = 3 + 2x and q 2 (x) = 2 + 3x (in this

order).

Solution. (I) We have to find two numbers x 1 , x 2 ∈ R such that x 1 u 1 + x 2 u 2 = c,

i.e.

x 1

[

]

• x 2

[

]

[

]

that is equivalent to solve the system

[ 3 2

2 3

] [

x 1

x 2

]

[

]

We row-reduce the complete matrix

[ 3 2 2

2 3 1

] [

]

From the second equation we obtain x 2 = − 1 / 5 , which we substitute in the first

equation to get, after a quick computation, x 1 = 4/ 5. Therefore the coordinates

of c are (4/ 5 , − 1 /5).

(II) We use the basis (e 1 (x), e 2 (x)) of R[x]≤ 1 , where e 1 (x) = 1 and e 2 (x) = x, to

transfer the problem in R

2 :

• p(x) = 2 + x = 2e 1 (x) + e 2 (x) which corresponds to the vector

[

]

Exercise 2.3. Let

v 1 =

, v 2 =

, v 3 =

(a) Is it true that v 1 , v 2 and v 3 are linearly independent, seen as vectors in R

4 ?

Write a basis of span(v 1 , v 2 , v 3 ) and complete it to a basis of R

4 .

(b) Is it true that v 1 , v 2 and v 3 are linearly independent, seen as vectors in

(Z 3 )

4 ? Write a basis of span(v 1 , v 2 , v 3 ) and complete it to a basis of (Z 3 )

4 .

Solution. (a) In order to test the linear dependence, we put the three vectors in

a matrix and row-reduce it:

(second row) − (first row)

(third row) − (first row)

(fourth row) − 2(first row)

(third row) + 2(second row)

(fourth row) + (second row)

(fourth row) − (third row)

10 CHAPTER 2. VECTOR SPACES

Since in the row reduced form there are three pivots, v 1 , v 2 and v 3 are linearly

independent over R and they are a basis of their span.

Now, to complete them to a basis of R

4 , we add a system of generators

A =

and row-reduce the matrix, obtaining



The pivots are in the first, second, third and fifth columns, therefore the corre-

sponding columns of A, i.e.



v 1 , v 2 , v 3 ,

form a basis of R

4 .

(b) Over Z 3 , the previous row reduction is still valid until we reach the matrix

(?). From there, since 3 = 0 in Z 3 , we can write (?) as



and that is the row reduced form. In this case, we have only two pivots, so the

vectors are linearly dependent and a basis of span(v 1 , v 2 , v 3 ) is (v 1 , v 2 ), i.e. the

vectors corresponding to the pivots.

As before, to complete it to a basis of (Z 3 )

4 we add a set of generators and

row-reduce the resulting matrix (remember that all operations are done in Z 3 ):

B =

The pivots are in the first, second, fourth and fifth columns, therefore the corre-

sponding columns of B, i.e.



v 1 , v 2 ,

12 CHAPTER 2. VECTOR SPACES

Chapter 3

Linear Subspaces

Exercise 3.1. Let K be a field, and let V = K[x]≤ 3. Consider the subspace

W ⊆ V given by

W = {p ∈ V | p(1) = 0}.

(a) Compute dim W , justifying the answer.

(b) Compute the cardinality of W for K = Z 13.

Solution. (a) We know that dim V = 4. It is clear that W is a subspace of V :

• the polynomial 0 belongs to W ;
• if p, q ∈ W , then (p + q)(1) = p(1) + q(1) = 0 + 0 = 0, so p + q ∈ W ;
• if p ∈ W and λ ∈ K, then (λp)(1) = λ · p(1) = λ · 0 = 0, so λp ∈ W.

Now, if p ∈ V , we can write p = ax

3

• bx

2

• cx + d with a, b, c, d ∈ K, and

W = {ax

3

• bx

2

• cx + d ∈ V | a + b + c + d = 0}.

If we identify V with K

4 through the isomorphism that sends a polynomial to

the vector of its coefficients, we can read W as a subspace of K

4 :

W =

a

b

c

d

∈ K

4

a + b + c + d = 0

It is easy to see that we can use b, c and d as free variables and set a = −b − c − d;

in other words

W =

−b − c − d

b

c

d

b, c, d ∈ K

thus dim W = 3.

(b) Every n-dimensional K-vector space is isomorphic to K

n , therefore if K = Z 13

we have W ' (Z 13 )

3 and its cardinality is 13

3

. 

• Let w 1 , w 2 ∈ W , that is w 1 , w 2 ∈ V and

w 1 =

a

b

, w 2 =

c

d

for some a, b, c, d ∈ R. Then w 1 + w 2 ∈ V because V is a vector space, and

w 1 + w 2 =

a + c

b + d

so it belongs to W.

• Let w ∈ W and k ∈ R. With the same argument as above, kv ∈ V (because

v ∈ V ) and its first two components are k · 0 = 0, so kv ∈ W.

(c) Notice that W = V ∩ Z, where Z is the set of the solutions of the system

{

x 1 = 0

x 2 = 0.

Then we look for a system of equations for V : a generic vector belongs to V if

and only if

rk

1 2 1 x 1

1 2 1 x 2

1 2 2 x 3

1 3 − 1 x 4

We row-reduce the matrix, obtaining



1 2 1 x 1

0 1 2 x 4 − x 1

0 0 1 x 3 − x 1

0 0 0 x 2 − x 1

thus its rank is not 4 if and only if x 2 − x 1 = 0. Therefore W = V ∩ Z is the set

of the solutions of (^) 



x 2 − x 1 = 0

x 1 = 0

x 2 = 0.

Now, by substitution we have that the first equation is redundant, because it

reduces to 0 = 0. So actually W = V ∩ Z = Z and a basis of it is



16 CHAPTER 3. LINEAR SUBSPACES

Exercise 3.4. In the vector space R

4 , consider the subspace V given by the

solutions of the system {

x + 2y + z = 0

−x − y + 3t = 0

and the subspace W generated by the vectors

w 1 =

and w 2 =

Compute dim(V ∩ W ) and dim(V + W ).

Solution. The two vectors spanning W are linearly independent, so dim W = 2.

On the other hand, it is easy to show that dim V = 2, because we can choose x and

y freely, and then z and t are uniquely determined. Moreover W * V because,

for instance, w 1 ∈/ V (x = 2, y = 0, z = 1 and t = 1 is not a solution of the

system (∗)). This means that dim(V ∩ W ) < dim W strictly, that is to say that

either dim(V ∩ W ) = 1 or 0. Notice that also w 2 ∈/ V , but this tells us nothing

more about dim(V ∩ W ). Then, we have to check when a linear combination of

w 1 and w 2 ,

a

• b

2 a + 3b

− 2 b

a − 2 b

a

belongs to V. By substituting the respective values in the system (∗), we obtain

{

(2a + 3b) + 2(− 2 b) + (a − 2 b) = 0

−(2a + 3b) − (− 2 b) + 3a = 0

which can be simplified to {

3 a − 3 b = 0

a − b = 0

that is equivalent to a = b. In other words, any vector of the form (†) with a = b

belongs to V ∩ W ; in particular, if a = b = 1 we get the vector



which is not the zero vector and belongs to V ∩ W , thus proving that dim(V ∩

W ) ≥ 1. Since we already know that dim(V ∩ W ) ≤ 1 , we can conclude that

dim(V ∩ W ) = 1.