I have the following question:

Matrix $N$ is a diagonal matrix with all entries strictly positive (hence, $N$ is positive definite since it satisfies $x^T N x > 0$). Matrix $M$ is an asymmetric positive definite matrix with all entries non-negative.

Since $NM \neq MN$, it does not follow that the product $NM$ is positive definite. However, given the special structure of $N$, can we still show that $NM$ is positive definite? Or maybe, under certain additional conditions?

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    $\begingroup$ What do you mean by positive definite for an asymmetric matrix? There are different options. $\endgroup$ – Simon Markett Sep 2 '12 at 13:40
  • $\begingroup$ I meant that M also satisfies $x^T M x > 0$. $\endgroup$ – User2012 Sep 2 '12 at 14:43

It's not true in general: take $$ N=\begin{bmatrix}1&0\\0&1/5\end{bmatrix},\ \ \ M=\begin{bmatrix}1&1\\0&1\end{bmatrix}. $$ $M$ is positive-definite according to your definition, since $$ \begin{bmatrix}x\\ y\end{bmatrix}^TM\begin{bmatrix}x\\ y\end{bmatrix}=x^2+xy+y^2>0 $$ on nonzero vectors.

On the other hand $$ \begin{bmatrix}1\\-2\end{bmatrix}^TNM\begin{bmatrix}1\\-2\end{bmatrix}=-\frac15. $$

As for conditions, of course one can force $M$ to be trivial enough for the property to hold; but I'll be surprised if there is a meaningful condition on $M$ that guarantees that $NM$ is positive-definite.

  • $\begingroup$ Yes, it does look like that. Thanks for your response! $\endgroup$ – User2012 Sep 3 '12 at 19:37
  • $\begingroup$ Would the result hold if $M$ is also symmetric? $\endgroup$ – user_lambda Jun 27 '17 at 8:34
  • $\begingroup$ No. $\ \ \ \ \ $ $\endgroup$ – Martin Argerami Jun 27 '17 at 13:46
  • $\begingroup$ The following fact might be helpful. If both $M$ and $N$ are symmetric and p.d., and furthermore, $MN=NM$ (i.e., symmetric), then $MN$ must be p.d. as well. $\endgroup$ – XGS Sep 10 '20 at 15:43

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