Initially it felt like a provable statement, because $A^*A$ is unitarily diagonalizable, because it's hermitian and therefore normal. When I tried proving it's non-negative, I ran into some problems.
The statement is as such:
Let $A\in M_n(\mathbb{C})$. therefore $A^*A$ is a non-negative unitarily diagonalizable matrix.
I've tried explaining that (according to previous statements I've seen) exists $H\in M_c(\mathbb{C})$ such that $H=\sqrt{A^*A}$ therefor $H^2=A^*A$.
$A^*A$ is hermitian therefore normal therefore unitarily diagonalizable so exists invertible $P$ and diagonal $D$ such that $A^*A=P^{-1}DP$ and all of $A^*A$ eigenvalues are real.
When I tried plugging in $H^2$ instead of $A^*A$ I realized that some of the values in $\sqrt{D}$ might be negative.
EDIT
I was able to prove it, I found out that: if $\lambda$ is eigenvalue of $A$ that fits some vector $v\in\mathbb{C}^n$, $\overline{\lambda}$ is eigenvalue of $A^*$ that fits the same vector $v$.
So: $$ v^tA^*Av=v^tA^*\lambda v=\lambda\overline{\lambda}v^tv=\lambda\overline{\lambda}\cdot<v,v> $$ because $\lambda\overline{\lambda}\geq 0$, and $<v,v> \geq0$ therefor $v^tA^*Av\geq 0$ and therefore $A^*A$ is non-negative (which is just the same as saying that it's positive($\geq 0$))
I hope my proof is somewhat correct, any reviews/ideas of improvements are always good for improving my understanding.
Thanks in advance!