MATH 235 Solutions Assignment 4b

1. Let A be an n × n matrix. Also, let α, β, γ be three distinct eigenvalues of A havingcorresponding eigenvectors x,y, z, respectively. Consider the vector v = x + y + z.Can v be an eigenvector of A corresponding to an eigenvalue λ (possibly different fromα, β, γ? Explain.

Solution:

Given A⇀x= α

⇀x, A

⇀y= β

⇀y , A

⇀z= γ

⇀z and

⇀v=

⇀x +

⇀y +

⇀z , then

A⇀v = A

⇀x +A

⇀y +A

⇀z= α

⇀x +β

⇀y γ

⇀z . (1)

Assume⇀v is an eigenvector of A corresponding to the eigenvalue λ. Then A

⇀v= λ

⇀v .

From (1), it follows that

α⇀x +β

⇀y +γ

⇀z = λ

(

⇀x +

⇀y +

⇀z

)

⇒ (α − λ)⇀x +(β − λ)

⇀y +(γ − λ)

⇀z=

⇀

0 . (2)

Since the eigenvectors⇀x,

⇀y ,

⇀z belong to distinct eigenvalues, the eigenvectors

⇀x,

⇀y ,

⇀z

are linearly independent. Thus, (2) gives that

α − λ = β − λ = γ − λ = 0

⇒ α = β = γ = λ

which is a contradiction (since α, β and γ are distinct). Hence, it follows that⇀v cannot

be an eigenvector of A.

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2. Recall that the trace of an n× n matrix A = [aij], denoted by tr(A), is the sum of thediagonal elements, that is, tr(A) =

∑ni=1

aii.

(a) Let C and D be any two n × n matrices. Prove that tr(CD) = tr(DC).

(b) Prove that if A and B are similar n × n matrices, then tr(A) = tr(B).

Solution:

(a) Let A = CD and B = DC, then

tr(A) =n

∑

i=1

aii =n

∑

i=1

n∑

j=1

cijdji

=n

∑

i=1

n∑

j=1

djicij

=n

∑

j=1

{

n∑

i=1

djicij

}

=n

∑

j=1

bjj = tr(B)

Hence, tr(CD) = tr(DC) as required.

(b) If A is similar to B, there exists an invertible matrix P such that P−1AP = B.

Then,

tr(B) = tr(

P−1AP)

= tr(

P−1[AP ])

= tr(

[AP ]P−1)

from (a)

= tr(

APP−1)

= tr(A) as required.

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3. Let M, N be n × n matrices and suppose that M is invertible. Denote A = MN andB = NM . Prove that A is similar to B.

Solution:

We must show that P−1AP = B for some invertible matrix P .

Since M is invertible, N = M−1A and N = BM−1.

Equating these expressions gives

M−1A = BM−1

⇒ M−1AM = B

Thus, A is similar to B (with P = M).

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4. Let A and B be invertible n × n matrices. Prove that AB and BA have the sameeigenvalues.

Solution:

If AB and BA satisfy the same characteristic equation, then they will have the sameeigenvalues. Thus, we will show that AB and BA have the same characteristic poly-nomial.

Now,

PAB = det (AB − λI)

= det(

ABAA−1− λAA−1

)

= det[

A(BA − λI)A−1]

= det(A) det(BA − λI)1

det(A)

= det (BA − λI)

= PBA as required.

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5. Determine if the following pair of matrices are similar to each other:

A =

1 1 50 2 00 0 2

, B =

1 7 00 2 70 0 2

.

Solution:

It is clear that A and B have the same eigenvalues since A and B are both uppertriangular matrices. This does not imply that A and B are similar since theorem 4on page 315 of the text states: If A and B are similar, then A and B have the sameeigenvalues. But is the converse true? This question provides a counterexample toshow that the converse is not true.

Assume A and B are similar, that is, B = P−1AP for some invertible matrix P .

Let P =

a b c

d e f

g h j

then from PB = AP we can easily find that d = e = g = h = 0.

Thus, matrix P has the form P =

a b c

0 0 f

0 0 j

⇒ det(P ) = 0 ⇒ P−1 does not exist

and therefore A and B are not similar.

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