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Let $A$ and $B$ be two full column rank real matrices of dimension $n \times m$, where $n \ge m$. Let $P$ be an $m\times m$ positive definite matrix.

Question: Does there always exist a symmetric $n \times n$ matrix $X$ such that the following holds?

$$\mathrm{tr}(P(A^\top XB+B^\top X A)) \ne 0$$

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  • $\begingroup$ since replacing X with -X multiplies trace by -1 it is the same as asking if there is a non-zero. It is a linear function of X and this will be the case unless all coefficients of X are 0. Have you tried writing them down ? $\endgroup$
    – user83457
    Commented Nov 20, 2017 at 13:51

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Your trace equals ${\rm tr} ((BPA^\top+APB^\top)X)$. This equals 0 for all symmetric matrices $X$ if and only if $C=BPA^\top+APB^\top=0$ (else take $X=C$, note that $C$ is symmetric). Of course, this is possible. For example, if $m=n$ and $P=\rm I$ is identity matrix, the product $BA^{\top}$ may be antisymmetric without problems.

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