# Equivalence in positive definiteness of $AB$ and $BA$

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.

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:
$$\begin{bmatrix}1 & 3\end{bmatrix}\begin{bmatrix} 2 & 1 \\ -2 & 0 \end{bmatrix}\begin{bmatrix}1 \\ 3 \end{bmatrix} = \begin{bmatrix}1 & 3\end{bmatrix}\begin{bmatrix}5 \\ -2 \end{bmatrix} = -1$$
• @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. – mechanodroid Jul 13 '18 at 20:30