Suppose $A$ is symmetric real matrix that has non-negative eigenvalues, i.e. positive semi-definite (PSD), and let $D,D'$ be a diagonal matrices with positive entries on the diagonal such that $D'-D$ is PSD. Is it true that $D'AD'-DAD$ is PSD?
We can assume that $D'=D+\delta$ where $\delta$ is diagonal with non-negative entries. Then after rewriting
$$D'AD'-DAD = DA\delta + \delta AD + \delta A\delta.$$
Of course $\delta A\delta$ is PSD. So to show that $D'AD'-DAD$ is PSD, we need to prove that $DA\delta + \delta AD$ is PSD. To that end, let $x\in \mathbb{R}^n$ and consider the expression
$$x^T(DA\delta + \delta AD)x = (Dx)^TA(\delta x)+(\delta x)^TA(D x).$$
After some rewriting, this is equal to
$$ \sum_{i,j=1}^n (\delta_{ii}D_{jj}+\delta_{jj}D_{ii})x_iA_{ij}x_j.$$
We need to show that this is non-negative if $A$ is PSD. I wasn't able to show this, which made me question whether it is true or not. Thanks in advance for all the input.
Best Answer
No. E.g. when $D'=\operatorname{diag}(2,1),\,D=I_2$ and $A=\pmatrix{1&2\\ 2&5}$, the difference $$ D'AD'-DAD=\pmatrix{4&4\\ 4&5}-\pmatrix{1&2\\ 2&5}=\pmatrix{3&2\\ 2&0} $$ is indefinite.