Homework # 10 (Written) Solutions
Math 152, Fall 2014
Instructor: Dr. Doreen De Leon
p. 272: 26 (Section 5.1)
26. Show that if A2 is the zero matrix, then the only eigenvalue of A is 0.
Solution: Let λ be an eigenvalue of A. Then Ax = λx for some x 6= 0. So,
A(Ax) = A(λx)
A2 x = λ(Ax)
0 = λ(λx).
We therefore obtain
λ2 x = 0.
Since x 6= 0, λ2 = 0, which is true only if λ = 0.
p. 260-1: 18, 20, 24 (Section 5.2)
18. Find h in the matrix A below such that the eigenspace for λ = 4 is two-dimensional.


4 2 3 3
0 2 h 3 

A=
0 0 4 14 .
0 0 0 2
Solution: The dimension of the eigenspace

0
0
A − 4I = 
0
0
for λ = 4 is dim Nul (A − 4I). Since

2 3 3
−2 h 3 
,
0 0 14 
0 0 −2
we will do row reduction to determine for what value(s) of h the system (A − 4I)x = 0
has two free variables.






0 2 3 3
0 2
3
3
0 2
3
3
0 −2 h 3  r2 →r2 +r1 0 0 h + 3 6  r4 →r4 +2r3 0 0 h + 3 6

 −−−−→ 
 −−−−−−→ 
.
0 0 0 14  −−
0 0
0 0
1
0
1
0
1
r3 → 14
r3
0 0 0 −2
0 0
0
−2
0 0
0
0
1
We see that if h + 3 6= 0, then
dim Nul (A − 4I) = the number of free variables
= 1.
Since we need dim Nul (A − 4I) = 2, we need h + 3 = 0,


 
0 2 3 3
0 2

0 −2 h 3  0 0
Then 
 

0 0 0 14  ∼ 0 0
0 0 0 −2
0 0
or h = −3 .
 
3 3

0 1
 .

0 0 
0 0
20. Use a property of determinants to show that A and AT have the same characteristic
polynomial.
Solution:
det(AT − λI) = det(AT − λI T )
= det(A − λI)T
= det(A − λI)
(since I T = I)
(since (A + B)T = AT + B T )
(since det AT = det A).
24. Show that if A and B are similar, then det A = det B.
Solution: If A and B are similar, then there exists an invertible matrix P such that
A = P BP −1 . Then,
det A = det(P BP −1 )
= (det P )(det B)(det P −1 )
1
= (det P )(det B)
det P
= det B.
p. 287: 26, 32 (Section 5.3)
26. A is a 7 × 7 matrix with three eigenvalues. One eigenspace is two-dimensional, and one
of the other eigenspaces is three-dimensional. Is it possible that A is not diagonalizable?
Justify your answer.
Solution: It is possible that A is not diagonalizable. From the information given, A
has three eigenvalues. The eigenspace associated with one eigenvalue has dimension
two, so the eigenvalue has multiplicity of at least two. The eigenspace asscoiated with
another eigenvalue has dimension three, so the eigevalue has multiplicity of at least three.
Therefore, the third eigenvalue has multiplicity of either one or two. If the third eigenvalue
has multiplicity one, then we know that there are only six eigenvectors, which is not
enough. If the third eigenvalue has multiplicity two, then we may or may not have seven
eigenvectors, depending on the dimension of its eigenspace.
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32. Construct a nondiagonal 2 × 2 matrix that is diagonalizable but not invertible.
Solution: To do this, we need only construct a matrix with two distinct eigenvalues, one
of which is 0. For example,
1 2
A=
0 0
will work. Since det A = 0, we know that A is not invertible. We see that
1 − λ 2 = (1 − λ)(−λ),
0 = det(A − λI) = 0
−λ
so A has two eigenvalues λ = 0 and λ = 1. Since A has two distinct eigenvalues, A is
diagonalizable.
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