What is the Moore-Penrose pseudoinverse for a hermitian block-matrix with one zero block? Given a block matrix of the form
\begin{pmatrix}
  A & B^* \\ B & 0
\end{pmatrix}
where $A$ is singular (otherwise one could simply use the well-known block matrix inverse), is there a simpler formula for the Moore-Penrose inverse than the general one?

This question is merely a special case of another one with $C=B^*$ (actually, $B$ and $B^*$ are swapped, but I'll implicitly do that). So user1551's answer can be adapted:

If you are looking for a closed-form formula in terms of $A,B$ and
  $C$, I am very skeptical about its usefulness. Yet that doesn't mean
  there isn't one: since $X$ is invertible, $$ X^{-1} = (X^TX)^{-1}X^T
> =\begin{bmatrix}A^TA+C^TC & A^TB\\ B^TA & B^TB\end{bmatrix}^{-1} \begin{bmatrix}A^T & C^T\\ B^T & 0\end{bmatrix}. $$ As $B$ has full
  column rank, $B^TB$ is invertible. Therefore you can use the formula
  for Schur complement to calculate the inverse of the block matrix
  on the RHS above, and the result is $$ X^{-1}= \begin{bmatrix}
  S^{-1} & -S^{-1} A^TB (B^TB)^{-1} \\
 -(B^TB)^{-1} B^TA S^{-1} & (B^TB)^{-1} + (B^TB)^{-1} B^TA S^{-1} A^TB (B^TB)^{-1} \end{bmatrix} \begin{bmatrix}A^T & C^T\\ B^T &
 0\end{bmatrix}, $$ where $S=A^TA+C^TC-A^TB(B^TB)^{-1}B^TA$.

In this case that means:
\begin{align}
  S &= A^TA + B^TB - A^TB^*(\bar B B^*)^{-1}\bar BA\quad \text{where}\ \bar B=(B^*)^T\text{, i.e. the complex conjugate},
\\ X^{-1} &= \begin{bmatrix}
  S^{-1} & -S^{-1}A^TB^*(\bar BB^*)^{-1}
\\ -(\bar BB^T)^{-1}\bar BAS^{-1} & (\bar BB^*)^{-1}+(\bar BB^*)^{-1}\bar BAS^{-1}A^TB^*(\bar BB^*)^{-1}
\end{bmatrix}\begin{bmatrix} A^T & B^T \\ \bar B & 0 \end{bmatrix}.
\end{align}
That is unfortunately not really simpler...
