Is square of a matrix positive definite?

Question

Given a $n\times n$ matrix $A$, and its Hermitian conjugate $A^{\dagger}$, I am trying to see if $A^{\dagger} A$ is semi-positive definite. Is this true ?

Answer 1

Hint: $\langle u, A^\dagger v\rangle = \langle Au, v\rangle$.

Answer 2

Yes. $A^{\dagger}A$ =$A^2$.
Recall the result that if $\lambda $ is an eigenvalue of $A$ the $f (\lambda) $ is an eigenvalue of $f (A)$. Here take $f(x)=x^2$.
From this you can see that $A^{\dagger}A $ is positive semi-definite.