A question about the eigenvactors of positive semi-definite matrices

Question

Assume that $A\in\mathbb{C}^{n\times n}$ is a Hermitian positive semi-definite matrix and $z\in\mathbb{C}^{n}$ such that $z\neq 0$. If $z^*Az=0$, is it true to conclude that $z$ is an eigenvector of $A$ associated with eigenvalue $0$?

Answer 1

This is true. Using diagonalization we can construct a positive semi-definite matrix $B$ such that $B^{2}=A$. Then $z^{*}Az=\|Bz||^{2}$ so we get $Bz=0$ which implies $Az=0$.