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Let $A,B$ any two real matrices, with $AB$ and $BA$ not necessarily symmetric. I know that $AB$ and $BA$ have the same eigenvalues. Is it true that $AB$ is positive definite iff $BA$ is positive definite? The definition I use for positive definiteness of a general real matrix is that $$ u^T A u > 0 $$ for all non-zero real vectors $u$. This is equivalent to the symmetric part of $A$ being positive definite.

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The answer is no.

Consider $A = \begin{bmatrix}2 & 1 \\ 0 & 1\end{bmatrix}$ and $B = \begin{bmatrix}1 & 0 \\ -1 & 1\end{bmatrix}$.

Then $AB = \begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}$ is positive definite:

$$\begin{bmatrix}x & y\end{bmatrix} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}=\begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}x+y \\ -x+y \end{bmatrix}=(x^2+xy)+(-xy+y^2)=x^2+y^2$$

But $BA = \begin{bmatrix} 2 & 1 \\ -2 & 0 \end{bmatrix}$ is not:

$$\begin{bmatrix}1 & 3\end{bmatrix}\begin{bmatrix} 2 & 1 \\ -2 & 0 \end{bmatrix}\begin{bmatrix}1 \\ 3 \end{bmatrix} = \begin{bmatrix}1 & 3\end{bmatrix}\begin{bmatrix}5 \\ -2 \end{bmatrix} = -1$$

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  • $\begingroup$ Thank you. May I ask how you went about finding a counter-example? $\endgroup$ – Nao Jul 13 '18 at 20:25
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    $\begingroup$ @Nao I found the non-symmetric positive definite matrix $\begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}$ here and I just calculated its LU decomposition. $\endgroup$ – mechanodroid Jul 13 '18 at 20:30

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