# How do I prove positive definiteness for a matrix difference?

Let $A$ be a symmetric positive semidefinite matrix. Let $W$ be a diagonal matrix with the entries $w_i \in (0,1)$.

I think $$A - WAW$$ should be positive semidefinite, but I don't know how to prove it or how to find a counterexample. I think it makes sense to think about $WAW$ as a rescaling of $A$, and since all the coefficients are less than $1$, it should be less than $A$ in the appropriate sense. How would I prove this?

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How about $A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ and $W=\begin{bmatrix}1&0\\0&0\end{bmatrix}$? – Rahul May 10 '12 at 17:35
Maybe you guys might fancy a criterion in the $2\times2$ case? Given positive definite $\mathbf A=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ and $\mathbf W=\mathrm{diag}(p,q)$, if $$\frac{b^2}{c}\leq a<\frac{b^2(pq-1)^2}{c(p^2-1)(q^2-1)}$$ then $\mathbf A-\mathbf W\mathbf A\mathbf W$ is not positive definite. – J. M. May 10 '12 at 18:05
Thanks guys, that clears my question up! If you submit an answer, I'll accept it. – John Salvatier May 10 '12 at 18:16
Would the criterion (to not be $>0$) be just $ab \leq bc \frac{(1-pq)^2}{(1-p^2)(1-q^2)}$? – copper.hat May 10 '12 at 18:21
More generally, for any symmetric, positive semidefinite but not positive definite $A$ whose diagonal elements are all positive, there is some diagonal $W$ with diagonal entries in $(0,1)$ such that $A - W A W$ is not positive definite. – Robert Israel May 10 '12 at 18:31

Choose $A = \begin{bmatrix} 200 & 100 \\ 100 & 100 \end{bmatrix}$, $W = \frac{1}{10}\begin{bmatrix} 1 & 0 \\ 0 & 9 \end{bmatrix}$, then $$A-WAW = \begin{bmatrix} 198 & 91 \\ 91 & 19 \end{bmatrix},$$ which is not $\geq 0$ (since $198\cdot19-91^2 < 0$).