Given two symmetric matrices $A,B \in \mathbb{R}^{n\times n}$ which are positive definite, and positive semidefinite respectively.
Can I conclude that
$$x^T(AB)x \ge0, \mbox{for all } x\in \mathbb{R}^n$$
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Probably you need some more hypotesis. In fact $AB$ is not symmetric: $(AB)^t=B^tA^t=BA$. It could be helpful have that $AB = BA$ – TEuler27 Feb 01 '18 at 22:21
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Maybe this question could help: https://math.stackexchange.com/questions/113842/is-the-product-of-symmetric-positive-semidefinite-matrices-positive-definite – TEuler27 Feb 01 '18 at 22:38
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Doesn't the wanted inequality mean that $AB$ is positive semidefinite? If this is the case, $AB=BA$ actually even gives that $AB$ is positive definite, @edo1998 – Richard Feb 01 '18 at 22:57
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Can you link me some proof of this fact? Anyway if $AB$ isn't symmetric then it wouldn't have sense to have a symmetric bilinear form associated to it – TEuler27 Feb 01 '18 at 23:11
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@edo1998 No sorry, I actually mixed something up. But your link shows that if positive semidefinite $A $ and $B $ commute then $AB $ is positive semidefinite, i.e. the conclusion here. In general though, it doesn't hold: – Richard Feb 02 '18 at 00:07
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$\begin{bmatrix}2 &1 \ 1&2 \end{bmatrix}$ is positive definite, $\begin{bmatrix}0 &0 \ -1&0 \end{bmatrix}$ is positive semidefinite, but $\begin{bmatrix}2 &1 \ 1&2 \end{bmatrix} \begin{bmatrix} 0&0 \ -1&0 \end{bmatrix}= \begin{bmatrix} -1&0 \ -2&0 \end{bmatrix}$ is negative semidefinite – Richard Feb 02 '18 at 00:12