Suppose that $A\in \mathbb R^{n\times n}$ is symmetric positive
semi-definite, i.e. $x^TAx \geq 0$ for all $x\in \mathbb R^n$. Then
$A$ is Hermitian positive semi definite.
Clearly $A$ is Hermitian but how can I show
$$\forall x\in \mathbb{R}^n: x^T Ax \geq 0 \Rightarrow \forall z\in \mathbb{C}^n :z^HAz\geq 0.$$
Best Answer
Notice that \begin{align}z^HAz&=(\Re z)^TA(\Re z)+i(\Re z)^TA(\Im z)-i(\Im z)^TA(\Re z)-i^2(\Im z)^TA(\Im z)=\\&=(\Re z)^TA(\Re z)+(\Im z)^TA(\Im z)\end{align} where $\Re z,\Im z\in\Bbb R^n$ are the obvious notions of real-and-imaginary part of a complex vector.