Show that $\lambda_{2}(zz^\ast)=\lambda_{n-1}(zz^\ast)=0$ where $z\in \mathbf{C}^n$ and $n\geq 2$.
I know that $zz^\ast$ is a Hermitian matrix, and all the eigenvalues are nonnegative, but how to prove that $zz^\ast$ has only one positive eigenvalue (which is $z^\ast z$) ?