Positive semidefiniteness of the product of a symmetric positive semidefinite matrix and a nonnegative diagonal matrix

Question

Is it true that the product of a symmetric PSD matrix and nonnegative diagonal matrix?

For $A\in\Bbb R^{n\times n}$ and $D\in\Bbb R^{n\times n}$, let $A\succeq O$ and $D:=\mathrm{diag}(d_1,\dots,d_n)$, where $d_i\ge0$ for all $i$. Then, is $AD$ positive semidefinite?

It is known that when two psd matrices $A$ and $B$ are non-commuting, $AB$ is not psd. Although $AD\neq DA$, how about the case where $B$ is a nonnegative diagonal matrix?

I found that the hypothesis is true when $n=2$. Given

$$ A:=\left(\begin{array}{cc}a & b\\ b & c\end{array}\right)\succeq O,\quad D:=\left(\begin{array}{cc}d_1 & 0 \\ 0 & d_2\end{array}\right)\ge O. $$

Since $A\succeq O$, $a\ge 0$, $c\ge 0$, and $ac-b^2\ge 0$. The product $AD$ is

$$ AD=\left(\begin{array}{cc}d_1 a & d_2 b \\ d_1 b & d_2 c\end{array}\right). $$

All the determinants of principal minors for $AD$ are nonnegative because

$$ d_1a\ge0,\ d_2c\ge 0,\ d_1d_2(ac-b^2)\ge 0. $$

Is this result possible to generalize $n\ge 2$?

Answer 1

In this setting it is not true that AD is positive semi-definite. Take the following example:

$ A=\left(\begin{array}{cc} 2 & 1\\ 1 & 2 \end{array}\right) $

and

$ D=\left(\begin{array}{cc} 0.01 & 0\\ 0 & 1 \end{array}\right) $

Then

$ AD=\left(\begin{array}{cc} 0.02 & 1\\ 0.01 & 2 \end{array}\right) $

For $AD$ to be positive semi-definite it must be the case that for any vector $x$, $x^{T}ADx\geq0$. So consider

$ x=\left(\begin{array}{c} -4\\ 1 \end{array}\right) $

This gives

$ x^{T}ADx=-1.72 $

The issue is that AD is not symmetric, so showing that all the determinants of principal minors for AD are nonnegative is not sufficient.