How do we determine the eigenvalues of B^2 if B is Hermitian?

[interested_learner]

The question was:

If B is Hermitian show that A=B^2 is positive semidefinite.

The answer was:

B^2 has eigenvalues \lambda_1 ^2, ... \lambda_n^2
(the square of B's eigenvalues) all non negative.

My question is:

Why do we know that B^2 has eigenvalues \lambda^2?

 

[Parlyne]

What happens if you act B^2 on the eigenvectors of B?

 

[interested_learner]

I am not quite sure what you mean by act, but I will assume you mean multiply together. If we take:

A= \left[ \begin{array} {cc} 1 & 1 \\ 1 & 1 \end {array} \right]

Then the eigenvalues are 2 and 0.

Then eigenvectors are:

B = \left[ \begin{array} {cc} 1 & 1 \\ 1 & -1 \end {array} \right]Then AB = \left[ \begin{array} {cc} 2 & 2 \\ 0 & 0 \end {array} \right]

Wow! The values in the matrix are the eigenvalues. Is this result general? Can we prove it?
How does that help solve the original question?

 

[BoTemp]

Let V be any eigenvector of B, and l be any eigenvalue.

B*V= l*V by definition. Starting from that equation you should be able to determine the eigenvalues of B^2. This is what Parlyne meant.

edit: l is the corresponding eigenvalue of V.

 

[nocturnal]

heres an alternative proof to show that the eigenvalues of A are all nonnegative (and thus A is positive semidefinite). let x be an eigenvector of A, and let \lambda be its corresponding eigenvalue. Then we have the following.

\lambda \langle x,x \rangle = \langle \lambda x, x \rangle = \langle Ax, x \rangle = \langle B^2 x, x \rangle = \langle B^{*}Bx, x \rangle = \langle Bx, Bx \rangle \geq 0

 

[interested_learner]

Oh gee! of course. it is obvious or should have been. Thank you.