I know why in a symmetric positive definite matrix every entry on the trace is positive entry $a_{ii}>0$. However I don't how to show that the largest value of the matrix is also on it's trace, meaning a $z$ exist with $a_{zz} = max \vert a_{ij} \vert $ for $1\le i,j\le n$.
Thanks for help.
Best Answer
First, every diagonal element of $A$ is positive since $$ 0<e_i^\top A e_i=a_{ii} $$
Let $a_{kk}$ be the largest diagonal element of $A$ and let $i\neq j$. Since $A$ is positive-definite, every principal submatrix of $A$ is positive-definite. In particular, $$ 0<\det \begin{bmatrix} a_{ii} & a_{ij} \\ a_{ji} & a_{jj} \end{bmatrix} =a_{ii}a_{jj}-a_{ji}a_{ij}=a_{ii}a_{jj}-a_{ij}^2<a_{kk}^2-a_{ij}^2 $$ What can we conclude?