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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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    I meant that M also satisfies x^T M x > 0.2012-09-02

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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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    No. $\ \ \ \ \ $2017-06-27