Suppose that H = A + iB is Hermitian and positive definite with A and B being Real n by n matrices. This means that $x^HHx > 0$ whenever $x\neq0$. (a) Show that
C = $\left(\begin{array}{cc} A & -B\\ B & A \end{array}\right)$ is symmetric and positive definite.
I get the symmetric part as A has to be a symmetric itself to make a Hermitian for H and B needs to be skew symmetric for the same reason. But, how can we say something about the positive semi-definiteness of C. All I can write is $x^HAx + ix^HBx > 0$ since $x^HHx > 0$. I can't figure out how to move from there.
Let $z = x+iy$ where $x,y\in \mathbb{R}^n$. With slight abuse of notation, I shall also use $z$ for the concatenated coluimn vector $[x;y] \in \mathbb{R}^{2n}$. Notice that $z^T C z = z^* H z \ge 0$ where $z^* = x^T -iy^T$ is the complex conjugate of $z$.