An interesting property of positive definite and essentially non-negative matrix

linear algebramatricespositive definitepositive-semidefinite

Let $S \in \mathbb{R}^{k\times k}$ be a symmetric matrix that is

  1. positive definite;

  2. essentially non-negative, i.e., all the off-diagonal entries are non-negative.

I observe the following phenomenon: for any vector $v \ge 0$ such that $Sv > 0$ (element-wisely), the following matrix

$$ T:= S^{-1} – \text{diag}\left(\frac{(v)_i}{(Sv)_i}\right) $$

is positive semidefinite. Here, $(v)_i$ stands for the $i$-th coordinate of vector $v$.

I can prove for small $k$ (e.g., $k=2$ or $3$) via brute force. But I do not have a good idea how to prove for general $k$. I did run some simulations and the claim is very likely to be true.

Any comments or suggestions would be much appreciated!

Best Answer

Denote by $u\circ v$ and $u\oslash v$ the Hadamard product and entrywise division of two vectors respectively. Let $D=\operatorname{diag}(\sqrt{v\oslash Sv})$ and $u=\sqrt{v\circ Sv}$. Then $$ \begin{aligned} DSDu &=DS\operatorname{diag}(\sqrt{v\oslash Sv})(\sqrt{v\circ Sv})\\ &=DSv\\ &=\operatorname{diag}(\sqrt{v\oslash Sv})Sv\\ &=\sqrt{v\circ Sv}\\ &=u. \end{aligned} $$ Since $DSD$ is nonnegative, $u>0$ is its Perron vector and $\rho(DSD)=1$. As $DSD$ is also real symmetric, we have $DSD\preceq I$. Therefore $0\prec S\preceq D^{-2}$ and $T=S^{-1}-D^2\succeq0$.

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