Are the Eigenvalues of A Similar to Inverse of A Really 1 or -1?

In summary, the question asks whether, if a matrix A is similar to its inverse, the eigenvalues must all be 1 or -1. The answer is yes, because if one of A's eigenvalues is n, then the eigenvalue of its inverse will be 1/n. But since the two matrices are similar, n=1/n, leading to n^2=1 and n=1 or -1. This is true for diagonal matrices, and the mistake in the problem is assuming it to be true for all matrices.
  • #1
Sanglee
6
0

Homework Statement


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


Homework Equations





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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  • #2


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 [itex]\lambda[/itex] is an eigenvalue of A, what number must be an eigenvalue of A^(-1)?
 
  • #3


jbunniii said:
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 [itex]\lambda[/itex] is an eigenvalue of A, what number must be an eigenvalue of A^(-1)?

Suppose the eigenvalues are say, 2 and 1/2?
 
  • #4


Dick said:
Suppose the eigenvalues are say, 2 and 1/2?

Oops, you're right.
 
  • #5


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:
  • #6


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!
 
  • #7


Sanglee said:
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.
 
  • #8


Oh, i understand it now :) Thanks!
 

FAQ: Are the Eigenvalues of A Similar to Inverse of A Really 1 or -1?

1. What is Linear Algebra?

Linear Algebra is a branch of mathematics that deals with linear equations, linear transformations, and vector spaces. It involves the study of systems of linear equations and their solutions, as well as the properties and operations of vectors and matrices.

2. Why is Linear Algebra important in science?

Linear Algebra is important in science because it provides a powerful framework for representing and solving complex systems of equations. It is widely used in fields such as physics, engineering, computer science, and statistics to model and analyze real-world phenomena.

3. What is the difference between a vector and a matrix?

A vector is a quantity that has both magnitude and direction, represented by an ordered list of numbers. A matrix is a rectangular array of numbers that can be used to represent linear transformations between vectors.

4. How is Linear Algebra used in data analysis?

Linear Algebra is used in data analysis to perform tasks such as dimensionality reduction, clustering, and regression. It allows us to represent and manipulate large datasets efficiently, making it an essential tool for data scientists.

5. If A is similar to the inverse of A, must all the eigenvalues equal 1?

No, not necessarily. Similarity and eigenvalues are related, but not equivalent concepts. While similar matrices have the same eigenvalues, the converse is not always true. Therefore, the eigenvalues of A and its inverse do not have to be equal for A to be similar to its inverse.

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