$A\ge0$ equal to $\left( \begin{matrix} \mathrm{Re}A& \left( \mathrm{Im}A \right) ^T\\ \mathrm{Im}A& \mathrm{Re}A\\ \end{matrix} \right) \ge 0$? If we have semi-definite positive matrix $A$, can we say it's equivalent to matrix $\left( \begin{matrix}  \mathrm{Re}A&  \left( \mathrm{Im}A \right) ^T\\  \mathrm{Im}A&  \mathrm{Re}A\\ \end{matrix} \right)$ is also semi-definite positive?
 A: Yes. Let $A=X+iY$ where $X$ and $Y$ are real matrices. Then
\begin{aligned}
X^T=\operatorname{Re}(A)^T=\operatorname{Re}(A^T)=\operatorname{Re}(\overline{A^\ast})=\operatorname{Re}(\overline{A})=X,\\
Y^T=\operatorname{Im}(A)^T=\operatorname{Im}(A^T)=\operatorname{Im}(\overline{A^\ast})=\operatorname{Im}(\overline{A})=-Y.\\
\end{aligned}
Therefore $X$ is real symmetric and $Y$ is real skew-symmetric. It follows that $\pmatrix{X&Y^T\\ Y&X}$ is real symmetric and we can speak of its definiteness (or the lack of it). Also, since $X$ is symmetric and $Y$ is skew-symmetric, we have
\begin{aligned}
-u^TYv&=u^T(-Y)v=u^TY^Tv,\\
u^TXv&=(u^TXv)^T=v^TX^Tu=v^TXu,\\
u^TYu&=v^TYv=0.\\
\end{aligned}
Therefore
\begin{aligned}
&(u+iv)^\ast A(u+iv)\\
&=(u-iv)^T(X+iY)(u+iv)\\
&=(u^TXu+v^TXv+v^TYu-u^TYv)+i(u^TXv-v^TXu+u^TYu+v^TYv)\\
&=u^TXu+v^TXv+v^TYu+u^TY^Tv\\
&=\pmatrix{u^T&v^T}\pmatrix{X&Y^T\\ Y&X}\pmatrix{u\\ v}
\end{aligned}
and $A$ is positive semidefinite if and only if $\pmatrix{X&Y^T\\ Y&X}$ is positive semidefinite.
