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

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 $$n\ge 2$$?

• Is $A D$ even symmetric? If not, then discussing its PSD-ness is kind of silly Jun 9 at 8:00

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.

• Thank you for showing a counterexample and pointing out my proof is false. Jun 9 at 3:20