TITLE: Why is the largest element of symmetric, positive semidefinite matrix on the diagonal?
QUESTION [10 upvotes]: I know the very well know equivalence of the properties of a positive, semidefinite matrix:

$A$ is positive semidefinite,
$A = U^T U$ for some matrix $U$,
$\mathbf{x}^T A \mathbf{x}\geq 0$ for every $\mathbf{x} \in \mathbb{R}^n$,
All principal minors $A$ are nonnegative.

But how can you derive from this that the largest entry of the matrix $A$ appears on the diagonal and why - when a diagonal entry is equal to zero - are all the entries of the corresponding row and column also equal to zero?

REPLY [13 votes]: Suppose $A = [A_{ij}]$ is symmetric, and let $s$ be an arbitrary real number. If $x = se_{i} - e_{j}$, then
$$
x^tAx = s^{2} A_{ii} - 2sA_{ij} + A_{jj}.
\tag{1}
$$

Assume, contrapositively, that the strictly largest entry is $A_{ij} = A_{ji}$ with $i \neq j$. Taking $s = 1$, i.e., $x = e_{i} - e_{j}$, (1) gives
$$
x^tAx = A_{ii} - 2A_{ij} + A_{jj}
  = (A_{ii} - A_{ij}) + (A_{jj} - A_{ij})
  < 0.
$$
If $A_{ii} = 0$ but $A_{ij} \neq 0$ for some $j$, (1) gives
$$
x^tAx = -2sA_{ij} + A_{jj},
$$
which is negative some $s$ of sufficiently large absolute value.