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

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.

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Yes, it does look like that. Thanks for your response! –  User2012 Sep 3 '12 at 19:37
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