13

I know the very well know equivalence of the properties of a positive, semidefinite matrix:

  1. $A$ is positive semidefinite,
  2. $A = U^T U$ for some matrix $U$,
  3. $\mathbf{x}^T A \mathbf{x}\geq 0$ for every $\mathbf{x} \in \mathbb{R}^n$,
  4. 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?

levap
  • 65,634
  • 5
  • 79
  • 122
Filip M
  • 329
  • 6
    if $A = (a_{ij})$ is positive semi-definite, so does all two sub-matrices of the form $(\begin{smallmatrix} a_{ii} & a_{ij} \ a_{ji} & a_{jj}\end{smallmatrix})$. This implies $a_{ii} a_{jj} \ge |a_{ij}|^2$ and hence at least one of $a_{ii}, a_{jj}$ is larger than $|a_{ij}|$ – achille hui Aug 03 '15 at 12:03
  • @achillehui So using this reasoning, can we say that for hermitian positive definite matrices, the greatest element on the diagonal is $\textit{strictly}$ greater than the magnitude of any element not on the diagonal? – Helix Apr 13 '21 at 16:36

1 Answers1

16

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

  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 $$ x^tAx = A_{ii} - 2A_{ij} + A_{jj} = (A_{ii} - A_{ij}) + (A_{jj} - A_{ij}) < 0. $$

  2. 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.

Andrew D. Hwang
  • 78,195