show that a symmetric matrix is positive semidefinite

Question

Let $A$ be a symmetric positive definite matrix, and $D$ be a diagonal matrix with non-negative entries. Show that

$2A-D\succeq 0\implies 2D-DA^{-1}D\succeq 0$

EDIT: the implication is true even if $D\succeq0$ is not diagonal

Answer 1

Let's denote by $A^\frac{1}{2}$ (resp. $D^\frac{1}{2}$) the square root of $A$ (resp. $D$), which is symmetric and positive definite (resp. semidefinite).

$0\preceq2A-D\implies0\preceq2I-A^{-\frac{1}{2}}D A^{-\frac{1}{2}}=2I-(A^{-\frac{1}{2}}D^{\frac{1}{2}})(D^{\frac{1}{2}}A^{-\frac{1}{2}})\implies2I-D^{\frac{1}{2}}A^{-1}D^{\frac{1}{2}}\succeq0\implies2D-D A^{-1}D\succeq0$

using the fact that $2I-ST$ and $2I-TS$ have the same eigenvalues.

Note that we have used the fact that $D\succeq0$ but we have not used the diagonal structure of $D$.

Answer 2

1.) Suppose $D \succ \mathbf 0$
$ D^{\frac{1}{2}}\big(2\cdot D^{\frac{-1}{2}} AD^{\frac{-1}{2}}-I\big)D^{\frac{1}{2}} = 2\cdot A-D \succeq \mathbf 0$
$\implies \big(2\cdot A^{\frac{1}{2}} D^{-1}A^{\frac{1}{2}}-I\big)\succeq \mathbf 0$
$\implies \big(2\cdot D^{-1}-A^{-1}\big)\succeq \mathbf 0$
$\implies \big(2\cdot D-DA^{-1}D\big)\succeq \mathbf 0$

2.) Suppose $D$ is singular, and fix some $\delta \gt 0$
$D'_n := D + \frac{\delta}{n}I \text{ and } A'_n := A + \frac{\delta}{2n}I$
$ 2\cdot A_n'-D_n' =2\cdot A-D \succeq \mathbf 0$
$\implies \big(2\cdot D_n'-D_n'(A_n')^{-1}D_n\big)\succeq \mathbf 0$
by the above argument. Now the matrix on the LHS is a continuous function of $\delta$ which implies its eigenvalues vary continuously with $\delta$ and thus

$ \big(2\cdot D-DA^{-1}D\big)=\lim_{n\to\infty}\big(2\cdot D_n'-D_n'(A_n')^{-1}D_n\big)\succeq \mathbf 0$