Dr.

Ahlem Nemer 1

Course : Algebra 3 Year : 2023/2024

Chapter 3 : Endomorphisms Department of Computer Science

Solutions

Solution 0.1 .
1) The matrix D is similar to A, then
D = P −1 AP,
such that P is a square matrix and is nonsingular. This leads to

det D = det(P −1 AP )
= det(P −1 )(det A)(det P )
1
= (det A)(det P )
det P
= det A

2) We have that X is an eigenvector of A

AX = λX.
Then, we find
P AP −1 X = λX.
This yields
D(P −1 )X = λ(P −1 X).
If we take that
Z = P −1 X,
we get
DZ = λZ.
Finally, we deduce that Z = P −1 X is an eigenvector of D.

Solution 0.2 . We have that A and D are similar, then

D = P −1 AP,

which leads to

det(D − λI) = det(P −1 (A − λI)P )

= det(P −1 ) det(A − λI) det P
1
= det(A − λI) det P
det P
= det(A − λI)

Solution 0.3 .

det(A − λI) = det(A − λI)T

= det(AT − λI T )
= det(AT − λI)
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Solution 0.4 .
1) The characteristic polynomial of A is given by

p(λ) = det(A − λI)

4−λ 1
=
9 4−λ
= (4 − λ)2 − 9
= λ2 − 8λ + 7

2) We take
p(λ) = λ2 − 8λ + 7 = (λ − 1)(λ − 7) = 0,
then, the eigenvalues of A are λ1 = 1 and λ2 = 7.
The eigenvectors of A are given by
If λ1 = 1, then we get     
3 1 x1 0
(A − λ1 I)X1 = = .
9 3 x2 0
This yields

3x1 + x2 = 0
9x1 + 3x2 = 0

Here, we have x2 = −3x1 . Then, the eigenvector corresponding to λ1 is given by

 
1
X1 = .
−3

If λ2 = 7, then we get     
−3 1 x1 0
(A − λ2 I)X2 = = .
9 −3 x2 0
This yields

−3x1 + x2 = 0
9x1 − 3x2 = 0

Here, we have x2 = 3x1 . Then, the eigenvector corresponding to λ2 is given by

 
1
X2 = .
3

3) Yes, the matrix A is diagonalizable.

4) The matrix P is  
1 1
P = ,
−3 3
and  
−1 1 3 −1
P = ,
6 3 1
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such that  
1 0
D= .
0 7
Finally, we obtain
 
1 1
 2 −  4 1

1 1
 
1 0

−1
P AP =  1 6 = .
1  9 4 −3 3 0 7
2 6
Solution 0.5 .

1−λ 2 −3
det(A − λI) = 1 1−λ 2
1 0 3−λ
= (1 − λ)(λ − 2)2

Then, the eigenvalues of A are λ1 = 1 and λ2 = λ3 = 2. Now, we need to find the eigenvectors of A.
If λ1 = 1, we get     
0 2 −3 x1 0
(A − λ1 I)X1 =  1 0 2   x2  =  0  .
1 0 2 x3 0
3
Here, we have x1 = −2x3 , x2 = x3 . Then
2
−2
 
 3 
X1 =  .
2
1
If λ2 = λ3 = 2, we get
    
−1 2 −3 x1 0
(A − λ2 I)X2 =  1 −1 2   x2  =  0  .
1 0 1 x3 0

Here, we have x1 = −x3 and x2 = x3 . Then,

 
−1
X2 =  1  .
1

Finally, we deduce that A is not diagonalizable.

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Solution 0.6 .

−1 − λ 1 1
det(A − λI) = 0 3−λ 4
−9 4 −3 − λ
= (1 + λ)(4 − λ)(4 + λ)

Then, the eigenvalues of A are λ1 = −1, λ2 = 4 and λ3 = −4. Now, we need to find the eigenvectors of A.
If λ1 = −1, we get     
0 1 1 x1 0
(A − λ1 I)X1 =  0 4 4   x2  =  0  .
−9 4 −2 x3 0
−2
Here, we have x2 = −x3 and x1 = x3 . Then
3
−2
 
 3 
X1 =  −1 .
1

If λ2 = 4, we get     
−5 1 1 x1 0
(A − λ2 I)X2 =  0 −1 4   x2  =  0  .
−9 4 −7 x3 0
Here, we have x2 = 4x3 and x1 = x3 . Then,


1
X2 =  4  .
1

If λ3 = −4, we get     
3 1 1 x1 0
(A − λ3 I)X3 =  0 7 4   x2  =  0  .
−9 4 1 x3 0
4 1
Here, we have x2 = − x3 and x1 = − x3 . Then,
7 7
1 
−
 7
4

X3 = 
 −
.

7
1

Then, the matrix P is

2  1 
− 1 −
 3 7
4

P =
 −1 4 −
,

7
1 1 1
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and
32 8
 
− 0
 7 7 
−1 21  3 11 5 
P =−  − − .
40  7 21 21 
 5 5 
−5 −
3 3
Finally, we obtain
672 168
 
2 1
 − 280
 
0   − 1 −  
280  −1 1 1  3 7 −1 0 0
63 11 5
P −1 AP =  4
  
 0 3 4  = 0 4 0 .
 − 280 40 40   −1 4 − 
 105 105 105  −9 4 −3 7 0 0 −4
− 1 1 1
40 120 120
Solution 0.7 .

1−λ 3 1
det(A − λI) = 0 4−λ 2
26 24 6−λ
= (λ + 2)(λ + 1)(14 − λ)

Then, the eigenvalues of A are λ1 = −2, λ2 = −1 and λ3 = 14. Now, we need to find the eigenvectors of A.
If λ1 = −2, we get     
3 3 1 x1 0
(A − λ1 I)X1 =  0 6 2   x2  =  0  .
26 24 8 x3 0
Here, we have x3 = −3x2 and x1 = 0. Then
 
0
X1 =  1  .
−3

If λ2 = −1, we get     
2 3 1 x1 0
(A − λ2 I)X2 =  0 5 2   x2  =  0  .
26 24 7 x3 0
5 1
Here, we have x3 = − x2 and x1 = − x2 . Then,
2 4
 1 
−
 4 
 1
X2 =  .
5

−
2
If λ3 = 14, we get     
−13 3 1 x1 0
(A − λ3 I)X3 =  0 −10 2   x2  =  0  .
26 24 −8 x3 0
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8
Here, we have x3 = 5x2 and x1 = x2 . Then,
13
8
 

X3 =  13
1 .
 

Then, the matrix P is

 1 8 
0 −
 4 13 
 1
P = 1 1 ,
5

−3 − 5
2
and  
−2 0 0
D= 0 −1 0 .
0 0 14
Here, we have 0
Y = DY,
Then  
C1 exp(−2t)
Y =  C2 exp(−t)  ,
C3 exp(14t)
Finally, the solution is
 1 8 
0 − 
C1 exp(−2t)

 4 13
 1 1 1
X = PY =    C2 exp(−t)  .
5

−3 − 5 C3 exp(14t)
2
Then  1 8 
− C2 exp(−t) + C3 exp(14t)
 4 13 
X= C1 exp(−2t) + C2 exp(−t) + C3 exp(14t) .
5
 
−3C1 exp(−2t) − C2 exp(−t) + 5C3 exp(14t)
2