Does $q I-Y-Y^T\succ 0\implies (Y^T-Y)(q I-Y-Y^T)^{-1}(Y-Y^T)\succ 0$?

Question

For $2\times 2$ real square matrices $Y$, the following is true:

If $q$ is a real number large enough that $q I-Y-Y^T$ is positive definite, then $(Y^T-Y)(q I-Y-Y^T)^{-1}(Y-Y^T)$ is also positive (semi-)definite.

To show this, the eigenvalues of $q I-Y-Y^T$ are $$\lambda_1=q-\sqrt{(y_{11}-y_{22})^2+(y_{12}+y_{21})^2}-y_{11}-y_{22}\\ \lambda_2=q+\sqrt{(y_{11}-y_{22})^2+(y_{12}+y_{21})^2}-y_{11}-y_{22}$$ By assumption, both are positive. The eigenvalues of $(Y^T-Y)(q I-Y-Y^T)^{-1}(Y-Y^T)$ are $$\frac{(y_{12}-y_{21})^2}{\lambda_2},\frac{(y_{12}-y_{21})^2}{\lambda_1}$$ and indeed, both are positive.

Question

I seek to generalize this result to $n\times n$ matrices. Any hints?

Answer 1

The result holds in general. We note that for any positive definite $A$ and any compatible matrix $M$, the matrix $M^TAM$ is positive semidefinite. It will be positive definite if and only if $M$ has linearly independent columns (even if $M$ is not square).

For your result, it suffices to take $A = qI - Y - Y^T$ and $M = Y - Y^T$.