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If the product of two matrices, N (a diagonal positive definite matrix) and M (a non-symmetric matrix), is positive definite i.e. $x^TNMx>0$, then is the matrix M positive definite i.e. is $x^TMx>0$ ?

This is a converse question for the one posted at: Is product NM positive definite when N is a diagonal positive definite matrix and M is an asymmetric positive definite matrix

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    $\begingroup$ Take the $N^{-1}$ and $NM$ from Martin Argerami's counterexample, and let them be your $N$ and $M$, respectively, and you get a counterexample. Or did I get your question wrong? $\endgroup$ – darij grinberg Mar 28 '13 at 1:03
  • $\begingroup$ You are right. Thanks a lot for the answer! $\endgroup$ – Nick Mar 28 '13 at 1:16
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Counterexample obtained by concatenating an answer by Martin Argerami with a comment by darij grinberg: $$N=\begin{pmatrix} 1 & 0 \\ 0 & 5 \end{pmatrix},\quad M =\begin{pmatrix} 1 & 1 \\ 0 & 1/5 \end{pmatrix}, \quad \text{hence } \ NM=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$ The positive-definiteness of (not necessarily symmetric) matrix $A$ in the sense of the OP amounts to positive definiteness of $A+A^T$ in the more usual sense. We have $$ M+M^T =\begin{pmatrix} 2 & 1 \\ 1 & 2/5 \end{pmatrix}, \quad NM+(NM)^T=\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} $$ where the first matrix has negative determinant, while the second matrix is positive definite.

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