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I was given the following statement:

If matrices $A$ and $B$ are positive definite, then it is impossible for $AB$ to be negative definite.

I'm trying to generate a counter-example. This question was similar (it showed that $AB$ doesn't have to be positive definite) but my question is a step further.

So, how do I construct an example where matrices $A$ and $B$ are both positive definite, so for every real non-zero vector $x$: \begin{gather*} x^T A x > 0\\ x^T B x > 0 \end{gather*} but $AB$ is negative definite, so: \begin{gather*} x^T (AB)x < 0 \end{gather*}

I know how to compute the eigenvalues of a matrix, but I'm not sure how to go about constructing this counterexample.

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Let $x$ be an eigenvector of $B$ with corresponding eigenvalue $\lambda$. Since $B$ is positive definite, $\lambda >0$. Then $$ x^T(AB)x = (x^TA)(Bx) = (x^TA)(\lambda x) = \lambda x^T Ax > 0 $$ since $A$ is positive definite. So negative definite will definitely not happen...

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Not that there is much to add to the solution of gt6989b, but note that if $A$ and $B$ are positive definite and $AB$ is self-adjoint then $AB$ is positive definite. Proof: $AB$ self-adjoint implies $A$ and $B$ commute, so there exists a basis of common eigenvectors of $A$ and $B$, which are also eigenvectors of $AB$. Moreover the eigenvalues of $AB$ are products of (certain) eigenvalues of $A$ and of $B$, hence strictly positive.

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