If $\mathbf{A} \succeq 0 \in M_n (\mathbb{C})$, why $a_{ii} a_{jj} \geq |a_{ij}|^2$?

Question

If $\mathbf{A} = [a_{ij}] \succeq 0 \in M_n (\mathbb{C})$ is positive semi-definite, why $a_{ii} a_{jj} \geq |a_{ij}|^2$ $ \forall i,j \in \left\{ 1,\cdots,n\right\}$?

Thank you in advance

Answer 1

I'll assume it's actually $M_n(\mathbb{R})$ so I don't have to check about sesquilinearity conventions. The argument is the same, I just always forget which argument is linear and which one is conjugate linear in the complex case.

Notice that $a_{ij}=e_i^T A e_j$, where $e_i$ are the columns of the identity matrix. Then the statement is essentially a case of Cauchy-Schwarz for the "inner product" $\langle x,y \rangle = x^T A y$. I say "essentially" because this is not really an inner product because it is only positive semidefinite instead of positive definite. Nevertheless the proof of (the nonstrict form of) Cauchy-Schwarz goes through the same with this weaker assumption.

Answer 2

This is Cauchy - Schwartz inequality for the inner product $(x,y) \to \langle Ax,y \rangle$ If you denote $\langle Ax,y \rangle$ by $\langle x,y \rangle '$ then the inequality simply states that $\langle e_i,e_j \rangle ' \leq \langle e_i,e_i \rangle '\langle e_j,e_j \rangle '$.