Let $A\in \mathbb R ^{n\times n}$ be positive definite. Prove that this implicates that all elements on the main diagonal are positive.
So $A$ is positive definite means, that $v^T Av>0$, where $v^T=[v_1,...,v_n]$.
I got that: $$ v^T Av=(v_1^2A_{1,1}+v_1 v_2 A_{2,1}+...+v_1 v_n A_{n,1})+...+(v_1 v_n A_{1,n}+...+v_n^2 A_{n,n}) $$
The only thing I noticed is that coefficient by $A_{i,i}$(so the main diagonal) is always $v_i^2$, so if one of the elements on the main diagonal were non positive, there would be at least one non-positive factor in the equation. Still I have no idea why would this imply that they all must be positive. Any tips?
Choose $v = e_j$, the $j$th column of the identity matrix to see that $a_{jj} = e_j^T A e_j > 0$.