TITLE: Largest entry in symmetric positive definite matrix
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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?