# Condition for non-vanishing trace

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?

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

• 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 ? – user83457 Nov 20 '17 at 13:51

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.