$\mathbf{H}\mathbf{\Sigma}+\mathbf{\Sigma}\mathbf{H}=\mathbf{0}$ implying $\mathbf{H} = \mathbf{0}$ for real, positive, diagonal $\mathbf{\Sigma}$?

Question

I have a matrix equation

$$\mathbf{H}\mathbf{\Sigma}+\mathbf{\Sigma}\mathbf{H}=\mathbf{0}$$ where $\mathbf{\Sigma}$ is a square real-valued diagonal matrix with strictly positive entries, $\mathbf{H}$ is (complex-valued) Hermitian, while $\mathbf{H}\mathbf{\Sigma}$ is anti-Hermitian. Is it possible for me to conclude anything about $\mathbf{H}$?

I have seen this thread but I am not too sure if it applies to my case as well. If it does, then would it mean that \begin{align} \mathbf{H}\mathbf{\Sigma} &= \mathbf{\Sigma}\mathbf{H} = \mathbf{0}\\ \end{align} and if I know that $\mathbf{\Sigma} \ne \mathbf{0}$, then $$\mathbf{H}=\mathbf{0}$$ Am I approaching this correctly?

Answer 1

The mentioned result does apply, but you may prove your hypothesis directly: from $H\Sigma+\Sigma H=0$, we get $h_{ij}\sigma_j+\sigma_ih_{ij}=0$. Consequently, $h_{ij}=0$ for every $(i,j)$ because $\sigma_i+\sigma_j\ne0$. Note that this is true not only for a Hermitian $H$, but for any square matrix $H$.