Proving if $A$ is an $n\times n$ positive semi-definite matrix, A is Hermitian with non-negative eigenvalues.

Question

I have a test on Monday and the professor hinted that this question might be relevant to the exam, unfortunately, I'm at a loss.

As the title states, I would like to prove that if $A$ is an $n\times n$ positive semi-definite matrix then $A$ is Hermitian with non-negative eigenvalues.

I would love to be able to work through this and I hope someone can lend a hand.

Thank you very much.

Answer 1

First, prove that if $E$ is a Hermitian matrix then $w^*Ew\in\mathbb{R}$, for every vector $w$.

Second, prove that every complex matrix $A$ can be writen as $A=B+iC$, where $B$ and $C$ are Hermitians.

Let $c$ be an eigenvalue of $C$ and $v$ an associated normalized eigenvector. Notice that $c\in\mathbb{R}$.

Thus, $v^*Av=v^*Bv+iv^*Cv=v^*Bv+ic$. Since a $v^*Av$ must be a positive real number, we must conclude that $c=0$(why?).

This implies $C=0$ (why?). Therefore A is Hermitian.

Answer 2

Let $(\alpha, v)$ be an eigenpair of $A$. Then

$$v^{*}Av = v^{*}(Av) = \alpha v^{*}v = \alpha||v||^{2}.$$ As $A$ is positive semi definite $\Rightarrow \alpha||v||^{2} \ge 0$. But $||v||^{2} > 0 \Rightarrow \alpha \ge 0$.