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Suppose $A$ is an arbitrary invertible matrix. Does there always exist a square matrix $H$ (does not have to be symmetric), such that $H^TA+A^TH$ is strictly positive definite?

I know as long as $A$ and $-A^T$ do not have common eigenvalue, by the existence of the solution to Sylvester equation, it can be concluded that we can always find such $H$. But does this hold for the general case?

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One can take $H=A$. In this case $H^TA+A^TH=A^TA+A^TA=2A^TA$.

Since $A$ is invertible, $Ax=0$ iff $x=0$, thus, for any $x\ne 0$ $$ x^T (2A^TA) x= 2(Ax)^T (Ax)= 2\|Ax\|_2^2>0. $$ It means that $H^TA+A^TH=2A^TA$ is strictly positive definite.

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