# Hermitian Matrix question

1. Aug 5, 2006

### interested_learner

The question was:

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

$$B^2$$ 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 $$\lambda^2$$?

Last edited: Aug 5, 2006
2. Aug 6, 2006

### Parlyne

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

3. Aug 6, 2006

### 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?

Last edited: Aug 6, 2006
4. Aug 7, 2006

### 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.

5. Aug 7, 2006

### 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.

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

6. Aug 8, 2006

### interested_learner

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