Suppose we have a complex block matrix, with $2 \times 2$ blocks:
$$M = \begin{bmatrix} A && B \\ B^T && C \end{bmatrix},$$
where
$$B = \begin{bmatrix} x && 0 \\ 0 && -x \end{bmatrix},$$
With $x$ being a real number. If $A \ge 0$ and $C \ge 0$ (they are both Hermitian), is it possible to find a necessary and sufficient condition on $x$ to make sure that the matrix $M$ is positive semidefinite?
When at least one of $A$ or $C$ is singular, clearly $M$ is PSD if and only if $x=0$.
When $A$ and $C$ are positive definite, $M$ is congruent to $(A-BC^{-1}B)\oplus C$. Let $D=\operatorname{diag}(1,-1)$. Then $$ \begin{aligned} M\succeq0 &\Leftrightarrow A\succeq BC^{-1}B\\ &\Leftrightarrow I\succeq A^{-1/2}BC^{-1}BA^{-1/2}\\ &\Leftrightarrow \rho(A^{-1/2}BC^{-1}BA^{-1/2})\le1\\ &\Leftrightarrow \rho(A^{-1}BC^{-1}B)\le1\\ &\Leftrightarrow x^2\le\frac{1}{\rho(A^{-1}DC^{-1}D)}.\\ \end{aligned} $$