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.

  • $\begingroup$ Beautiful. Quick question: How is it immediately clear that $\mathcal{H}$ is Hermitian when we only know that $\mathcal{H}$ is positive semidefinite? $\endgroup$ – Benedict Voltaire Mar 8 at 3:37
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    $\begingroup$ 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. $\endgroup$ – David M. Mar 8 at 3:39
  • $\begingroup$ All good. I see what you're doing. Thank you! $\endgroup$ – Benedict Voltaire Mar 8 at 4:09

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