$A\ge0$ equal to $\left( \begin{matrix} \mathrm{Re}A& \left( \mathrm{Im}A \right) ^T\\ \mathrm{Im}A& \mathrm{Re}A\\ \end{matrix} \right) \ge 0$?

Question

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?

Answer 1

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.