Exercise 1

Exercise 1

Consider the matrices

A =





5 4 3

−1 0 −3

1 −2 1



, B =





3 1 −1 1 3 −1

0 0 2





(1) Compute the characteristic polynomial of A and B (2) Find the eigenvalues and eigenvectors

(3) Determine whether matrix A (respectively, B) is diagonalisable and, in this case, provide an invertible matrix H such that H⁻¹AH (respectively, H⁻¹BH) is diagonal

Exercise 1, answer

(1) ^I The characteristic polynomial of A is 

5 − t 4 3

−1 −t −3

1 −2 1 − t      

=

= (5 − t)(−t)(1 − t) − 12 + 6 + 3t + 4(1 − t) − 6(5 − t) =

= −t(t²− 6t + 5) − 6 + 3t + 4 − 4t − 30 + 6t =

= −t³+ 6t²− 32

I The characteristic polynomial of B is 

3 − t 1 −1

1 3 − t −1

0 0 2 − t

=(2 − t)[(3 − t)²− 1] =

=(2 − t)(t²− 6t + 8) =

= − t³+ 8t²− 20t + 16

Exercise 1, answer (cont.)

(2) ^I By inspection, −2 is a root of the characteristic polynomial of A.

This polynomial factorises as

(t + 2)(−t²+ 8t − 16) = −(t + 2)(t − 4)² so that the eigenvalues of A are −2, 4.

Exercise 1, answer (cont.)

The eigenvectors associated to −2 are the non-null solutions of the linear system

 7x + 4y + 3z = 0

−x + 2y − 3z = 0

The solutions to the system are the ordered triples (x , y , z) such that

x =    

−3z 4 3z 2    

18 = −z

y =    

7 −3z

−1 3z

18 = 4

3z The eigenspace associated with −2 is then

−1,⁴₃, 1
, while the set of eigenvectors is



−z,4 3z, z



| z 6= 0



Exercise 1, answer (cont.)

The eigenvectors associated with 4 are the non-null solutions to the system

 x + 4y + 3z = 0 x − 2y − 3z = 0

The solutions to the system are the ordered triples (x , y , z) such that

x =    

−3z 4

3z −2

−6 = z

y =    

1 −3z 1 3z

−6 = −z

The eigenspace associated to 4 is then h(1, −1, 1)i, while the set of eigenvectors is

{(z, −z, z) | z 6= 0}

Exercise 1, answer (cont.)

I The characteristic polynomial of B factorises as (2 − t)(t − 2)(t − 4) = (2 − t)²(4 − t) so the eigenvalues of B are 2, 4

Exercise 1, answer (cont.)

The eigenvectors associated to 2 are the non-null solutions of the equation

x + y − z = 0

The solutions are the ordered triples (x , y , z) such that z = x + y

Thus the eigenspace associated to 2 is h(1, 0, 1), (0, 1, 1)i, while the set of eigenvectors is {(x , y , x + y ) |x,y not both 0}.

Exercise 1, answer (cont.)

The eigenvectors associated to 4 are the non-null solutions of the system

 −x + y − z = 0

−2z = 0

The solutions are the ordered triples (x , y , z) such that x =y

z =0

Thus the eigenspace associated to 4 is h(1, 1, 0)i, while the set of eigenvectors is {(x , x , 0) | x 6= 0}.

Exercise 1, answer (cont.)

(3) ^I Matrix A is not diagonalisable, since the dimension of the eigenspace associated to the eigenvalue 4 is less than the multiplicity of 4 as a root of the characteristic polynomial.

I Matrix B is diagonalizable, as the dimension of all eigenspaces equal the multiplicity of the corresponding eigenvalues as roots of the characteristic polynomial.

To diagonalise B, notice that a basis of eigenvectors of R³is ((1, 0, 1), (0, 1, 1), (1, 1, 0))

Thus, letting H =





1 0 1

0 1 1

1 1 0



, one has

H⁻¹BH =





2 0 0

0 2 0

0 0 4





Exercise 2

Consider the following matrices Aλ





1 − 2λ −λ −λ

2λ 1 + λ λ

2λ λ 1 + λ



,









1 2 λ²− λ λ

2 1 0 0

0 0 0 0

0 0 0 3









where λ ∈ R.

(1) Find eigenvalues and eigenvectors of Aλ

(2) Find the values of λ for which A_λis diagonalisable (3) Find the eigenspaces of Aλ

(4) If Aλ is diagonalisable, find an invertible matrix Hλ such that H_λ⁻¹AλHλ is diagonal

Exercise 2, answer

(1) The characteristic polynomial of





1 − 2λ −λ −λ

2λ 1 + λ λ

2λ λ 1 + λ



is

1 − 2λ − t −λ −λ

2λ 1 + λ − t λ

2λ λ 1 + λ − t

=

= (1 − 2λ − t)(1 + λ − t)²− 2λ³− 2λ³+ 2λ²(1 + λ − t)+

+ 2λ²(1 + λ − t) − λ²(1 − 2λ − t) =

= (1 − 2λ − t)(1 + t²+ 2λ − 2t − 2λt) + 4λ²(1 − t) =

= (1 − 2λ − t)(t − 1)(t − 2λ − 1) + 4λ²(1 − t) =

= (1 − t)[t²− 2t + 1 − 4λ²+ 4λ²] = (1 − t)³ The unique eigenvalue is then 1.

Exercise 2, answer (cont.)

The eigenvectors are the non-null solutions of the equation 2λx + λy + λz = 0

Thus:

I If λ = 0, the eigenvectors are exactly the non-null vectors

I If λ 6= 0, the eigenvectors are the non-null solutions of the equation 2x + y + z = 0

that is the non-null ordered triples (x , y , z) such that z = −2x − y

In other words, the set of eigenvectors is {(x, y , −2x − y ) | x, y not both 0}

Exercise 2, answer (cont.)

The characteristic polynomial of the matrix









1 2 λ²− λ λ

2 1 0 0

0 0 0 0

0 0 0 3







 is

1 − w 2 λ²− λ λ

2 1 − w 0 0

0 0 −w 0

0 0 0 3 − w

= −w (3 − w )(1 − 2w + w²− 4) =

= −w (3 − w )(w²− 2w − 3) = w (w − 3)²(w + 1) so the eigenvalues are −1, 0, 3.

Exercise 2, answer (cont.)

The eigenvectors associated to the eigenvalue −1 are the non-null solutions to the system







2x + 2y = 0

z = 0

4t = 0

Thus the set of such eigenvectors is {(x , −x , 0, 0) | x 6= 0}.

The eigenvectors associated to the eigenvalue 0 are the non-null solutions of the system







x + 2y + (λ²− λ)z + λt = 0 2x + y = 0 3t = 0

Exercise 2, answer (cont.)

Solving with respect to z:

x =    

(λ − λ²)z 2

0 1

−3 =λ²− λ 3 z

y =    

1 (λ − λ²)z

2 0

−3 =2(λ − λ²)

3 z

t =0

Thus the set of such eigenvectors is

 λ²− λ

3 z,2(λ − λ²) 3 z, z, 0



| z 6= 0



= {((λ²− λ)z, 2(λ − λ²)z, 3z, 0) | z 6= 0}

Exercise 2, answer (cont.)

The eigenvectors associated to the eigenvalue 3 are the non-null solutions of the system











−2x + 2y + (λ²− λ)z + λt = 0 2x − 2y = 0

−3z = 0

0 = 0

Notice that such a system has the same solutions as







x − y = 0

z = 0

λt = 0

Exercise 2, answer (cont.)

Consequently:

I If λ = 0, the set of eigenvectors is {(x , x , 0, t) | x , t not both 0}

I If λ 6= 0, the set of eigenvectors is {(x , x , 0, 0) | x 6= 0}

Exercise 2, answer (cont.)

(2) ^I For the matrix





1 − 2λ −λ −λ

2λ 1 + λ λ

2λ λ 1 + λ



, since its unique

eigenvalue is 1, thus with algebraic multiplicity equal to 3, the matrix turns out to be diagonalisable if and only if the unique eigenspace has dimension 3, so it equals R³, which means that the matrix is the identity matrix.

So such a matrix is diagonalisable if and only if λ = 0.

I The matrix









1 2 λ²− λ λ

2 1 0 0

0 0 0 0

0 0 0 3









is diagonalisable if and only if the

eigenspace associated with 3 has dimension 2, so if and only if λ = 0.

Exercise 2, answer (cont.)

(3) ^I For the matrix





1 − 2λ −λ −λ

2λ 1 + λ λ

2λ λ 1 + λ



:

I If λ = 0, the eigenspace (associated with the unique eigenvalue 1) is R³

I If λ 6= 0, the eigenspace (associated with the unique eigenvalue 1) is h(1, 0, −2), (0, 1, −1)i

Exercise 2, answer (cont.)

I For the matrix









1 2 λ²− λ λ

2 1 0 0

0 0 0 0

0 0 0 3







 :

I The eigenspace associated to −1 is h(1, −1, 0, 0)i.

I The eigenspace associated to 0 is h(λ²− λ, 2(λ − λ²), 3, 0)i.

I The eigenspace associated to 3 is:

h(1, 1, 0, 0), (0, 0, 0, 1)i if λ = 0;

h(1, 1, 0, 0)i if λ 6= 0.

Exercise 2, answer (cont.)

(4) ^I For the matrix





1 0 0

0 1 0

0 0 1



: any non-singular matrix H is such that

H⁻¹





1 0 0

0 1 0

0 0 1



H =





1 0 0

0 1 0

0 0 1



.

For example, one can take

H =





1 0 0

0 1 0

0 0 1





Exercise 2, answer (cont.)

I For the matrix









1 2 0 0

2 1 0 0

0 0 0 0

0 0 0 3







: a basis of eigenvectors of R⁴is:

((1, −1, 0, 0), (0, 0, 3, 0), (1, 1, 0, 0), (0, 0, 0, 1))

thus a suitable matrix is H =









1 0 1 0

−1 0 1 0

0 3 0 0

0 0 0 1







 , so that

H⁻¹









1 2 0 0

2 1 0 0

0 0 0 0

0 0 0 3







 H =









−1 0 0 0

0 0 0 0

0 0 3 0

0 0 0 3









Exercise 3: the trace of an endomorphism

Define the trace tr (A) of a square matrix A as the sum of the entries on the diagonal of A. Show that for any invertible matrix P one has tr (A) = tr (P⁻¹AP) (hint: consider the characteristic polynomial of A and look at the coefficient of the monomial of degree n − 1). Derive that we can define the trace of an endomorphism as the trace of any matrix that represents it, after choosing a basis for the space (that is, the definition does not depend on the choice of the basis).

(1) Show that tr (A + B) = tr (A) + tr (B), and tr (αA) = αtr (A) for α ∈ R

(2) Show that tr (AB) = tr (BA) and that, in general tr (AB) 6= tr (A)tr (B)

Exercise 3, answer

Notice that A and P⁻¹AP have the same characteristic polynomial, since det(P⁻¹AP − λI ) = det(P⁻¹AP − λP⁻¹IP) =

= det(P⁻¹AP − P⁻¹λIP) =

= det(P⁻¹(A − λI )P) =

= det P⁻¹· det(A − λI ) · det P =

= det(A − λI )

The coefficient of degree n − 1 of the characteristic polynomial of A is (−1)ⁿ⁻¹tr (A). Consequently, since A and P⁻¹AP have the same characteristic polynomial, they have the same trace.

Since if A is the matrix of an epimorphism with respect to a basis, then with respect to any other basis the matrix of A is of the form P⁻¹AP for a suitable invertible matrix P, it follows that the definition of trace of an epimorphism does not depend on the choice of the matrix representing it.

Exercise 3, answer (cont.)

Let A = (aij), B = (bij). Then:

A + B =(aij+ bij) αA =(αaij) AB =

n

X

h=1

aihbhj

!

BA =

n

X

h=1

b_iha_hj

!

Exercise 3, answer (cont.)

Thus:

(1) ^I tr (A + B) =Pn

i =1(aii+ bii) =Pn

i =1aii+ bii= tr (A) + tr (B)

I tr (αA) =Pn

i =1αaii= αPn

i =1aii= αtr (A) (2) ^I

tr (AB) =

n

X

i =1 n

X

h=1

aihbhi=

=

n

X

h=1 n

X

i =1

bhiaih=

=tr (BA)

I Let A =1 0 0 2



, B =0 0 0 1

 . Then

tr (A) = 3, tr (B) = 1, tr (AB) = tr0 0 0 2



= 2 so that tr (AB) = 2 6= 3 = tr (A)tr (B).