Eigenvalues of the power of a matrix

[IniquiTrance]

Homework Statement

If \lambda_i are the eigenvalues of a matrix A^2, and A is symmetric, then what can you say about the eigenvalues of A?

Homework Equations

The Attempt at a Solution

I know how to prove that if \sqrt(\lambda_i) is an eigenvalue of A, then \lambda_i is an eigenvalue of A^2.

I don't know how to prove the converse though.

Thanks!

 

[micromass]

Let \lambda be an eigenvalue of A^2. What can you say about A^2-\lambda I?

Now, use that

A^2-\lambda I=(A-\sqrt{\lambda}I)(A+\sqrt{\lambda} I)

 

[IniquiTrance]

Let \lambda be an eigenvalue of A^2. What can you say about A^2-\lambda I?

Now, use that

A^2-\lambda I=(A-\sqrt{\lambda}I)(A+\sqrt{\lambda} I)

Then:

det(A-\sqrt{\lambda}I)*det(A+\sqrt{\lambda} I) = 0

Not sure how to proceed...

 

[micromass]

If xy=0, then what can you say about x or y?

 

[IniquiTrance]

If xy=0, then what can you say about x or y?

Oh so all we can say is that either \sqrt{\lambda} or -\sqrt{\lambda} is an eigenvalue of A, but not both?

 

[micromass]

Oh so all we can say is that either \sqrt{\lambda} or -\sqrt{\lambda} is an eigenvalue of A, but not both?

Right.

 

[IniquiTrance]

Thank you.

 

[IniquiTrance]

One more question, how do we know that all the eigenvalues of A^2 are positive? Since A is symmetric, if \lambda < 0 were an eigenevalue of A^2 we'd run into a problem...

 

[micromass]

We can do the same with \sqrt{\lambda} complex numbers, no?

 

[IniquiTrance]

We can do the same with \sqrt{\lambda} complex numbers, no?

But since A is symmetric, it must have real eigenvalues, no?

 

[micromass]

But since A is symmetric, it must have real eigenvalues, no?

Yes, sure. So what we do is we argue the same argument where \lambda might be negative and where \sqrt{\lambda} might be complex. Then we can deduce that either \sqrt{\lambda} or -\sqrt{\lambda} are eigenvalues of A. Since A is symmetric, we have that \sqrt{\lambda} must be real, so \lambda was positive to begin with.

 

[IniquiTrance]

Yes, sure. So what we do is we argue the same argument where \lambda might be negative and where \sqrt{\lambda} might be complex. Then we can deduce that either \sqrt{\lambda} or -\sqrt{\lambda} are eigenvalues of A. Since A is symmetric, we have that \sqrt{\lambda} must be real, so \lambda was positive to begin with.

Ah ok, thanks again!

 

[Ray Vickson]

Ah ok, thanks again!

No, you can have both. The matrix
A = \pmatrix{2 & 0 \\0 & -2} has
B = A^2 = \pmatrix{4 & 0 \\0 & 4},
and so both \sqrt{4} and -\sqrt{4} are eigenvalues of A.

RGV