# Explanation on how a matrix $A$ expressed as a product involving a positive semidefinite matrix $\mathcal{H}$ is also positive semidefinite

Suppose we know that that a Hermitian $$n \times n$$ matrix $$A$$ can be expressed as the following matrix product $$A = \begin{bmatrix} z_1 & 0 & ... & 0 \\ 0 & z_2 & ... & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & ... & z_n \end{bmatrix} \, \mathcal{H} \, \begin{bmatrix} \overline{z_1} & 0 & ... & 0 \\ 0 & \overline{z_2} & ... & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & ... & \overline{z_n} \end{bmatrix}$$ where $$\mathcal{H}$$ is positive semidefinite and all the complex entries $$z_i\neq 0$$ across the main diagonal

Can we also deduce that $$A$$ is also positive semidefinite?

Apparently, we can infer that $$A$$ is positive semidefinite, but from the definition of positive semidefinite matrices and the fact that $$A$$ can be expressed as a product involving a positive semidefinite matrix $$\mathcal{H}$$, it is not clear to me how this is true.

For simplicity, let $$Z\equiv\operatorname{diag}(z_1,\dots,z_n)$$. It is clear that $$ZHZ^*$$ is Hermitian, because $$H$$ is. It suffices to show that $$x^*ZHZ^*x\geq0$$ for all $$x\in\mathbb{C}^n$$. Fix $$x\in\mathbb{C}^n$$ and let $$w\equiv{Z^*x}$$. Then $$w^*Hw\geq0$$ because $$H$$ is positive semidefinite. QED.
• Beautiful. Quick question: How is it immediately clear that $\mathcal{H}$ is Hermitian when we only know that $\mathcal{H}$ is positive semidefinite? – Benedict Voltaire Mar 8 at 3:37
• Oh shoot--I thought you were assuming $H$ was Hermitian. As I understand it, for complex matrices, a matrix must be Hermitian to be positive definite (or at least, that's what Wikipedia says--I don't deal in complex matrices much). Read here. – David M. Mar 8 at 3:39