If $A$ is any matrix then $A^*A$ and $AA^*$ are Hermitian with non-negative eigenvalues How can I show that if $A$ is any matrix then $A^*A$ and $AA^*$ are Hermitian with non-negative eigenvalues? 
I am stuck and any help will be greatly appreciated.
 A: By definition, $A^*A=\overline{A^T}A$. Then
$$
(A^*A)^*=\overline{(\overline{A^T}A)^T}=\overline{A^T\overline{A}}=\overline{A^T}A=A^*A
$$
So $A^*A$ is Hermitian. Let $x$ be any vector, then
$$
\bar{x^T} \overline{A^T}Ax=\overline{(Ax)^T}Ax=\|Ax\|^2\geqslant 0
$$
So $\overline{A^T}A$ ($A^*A$) is semi-positive definite and has non-negative eigenvalues.
The same reasoning can be applied for $AA^*$.
A: There is a way to show it simply. Let there be an inner product on $\mathbb{C}^n$ denoted by $\big<x, y\big> \; \forall x, y \in \mathbb{C}$. The inner product is deffined as: $\big<x, y\big> = \sum_{i=1}^{n} \bar{x}_i y_i$. There are also matrices called Hermitian, that they doesn't change the inner product, not every matrix (operator) can be Hermitian, there is at least one condition $\big<Ax, y\big> = \big<x, Ay\big>$ . So now let the operator have it's eigenvalue $\lambda$ and eigenvector $v$. From the equation above is sure that:
$$
\bar{\lambda}\big<v, v\big> = \big<\lambda v, v\big> = \big<Av, v\big> = \big<v, Av\big> = \big<v, \lambda v\big> = \lambda \big<v, v\big>
$$
So than:
$$
\bar{\lambda} = \lambda
$$
$$
\lambda \in \mathbb{R}
$$
