# Linear Algebra) If A is similar to inverse of A, must all the eigenvalues equal 1?

## Homework Statement

If A is similar to A^(-1) (=inverse of A), must all the eigenvalues equal 1 or -1?

## The Attempt at a Solution

I don't know why the textbook gives me the specific value 1 or -1.
If A is similar to its inverse, are the eigenvalues really 1 or -1? Why? Help!

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jbunniii
Homework Helper
Gold Member

Yes, the eigenvalues will be 1 or -1.

First, if A and B are similar matrices, what can you say about their eigenvalues?

Second, if $\lambda$ is an eigenvalue of A, what number must be an eigenvalue of A^(-1)?

Dick
Homework Helper

Yes, the eigenvalues will be 1 or -1.

First, if A and B are similar matrices, what can you say about their eigenvalues?

Second, if $\lambda$ is an eigenvalue of A, what number must be an eigenvalue of A^(-1)?
Suppose the eigenvalues are say, 2 and 1/2?

jbunniii
Homework Helper
Gold Member

Suppose the eigenvalues are say, 2 and 1/2?
Oops, you're right.

Dick
Homework Helper

Sanglee, I think the 1 and -1 are just to lead you into giving a wrong answer by thinking too quickly. It's true if A is a 1x1 matrix. Just think of a 2x2 matrix that is similar to its inverse without the diagonal entries being 1 or -1. Diagonal matrices will do.

Last edited:

Oh~~~~~~~~I got it!!! it's really simple question, but I thought it was complicated. hehe

So, A and inverse of A are similar, so their eigenvalues are same.
if one of A's eigenvalues is n, a eigenvalues of its inverse will be 1/n.
But the two matrices are similar, so n=1/n
Then, n^2=1, so n=1or-1

Is it right?
Thanks guys!

Dick
Homework Helper

Oh~~~~~~~~I got it!!! it's really simple question, but I thought it was complicated. hehe

So, A and inverse of A are similar, so their eigenvalues are same.
if one of A's eigenvalues is n, a eigenvalues of its inverse will be 1/n.
But the two matrices are similar, so n=1/n
Then, n^2=1, so n=1or-1

Is it right?
Thanks guys!
The two eigenvalues don't have to be equal. That's the mistake jbunniii made and the one the poser of the problem assumed you might make. Look at post #3.

Oh, i understand it now :) Thanks!