Solution method
I am repetitively solving sparse linear systems (for the need of ARNOLDI iterations) of the type:
$$\underbrace{\begin{bmatrix}
J_1 & J_2 \\
J_3 & J_4
\end{bmatrix}}_J
\underbrace{\begin{bmatrix}
x \\
y
\end{bmatrix}}_z=
\underbrace{\begin{bmatrix}
b_x \\
b_y
\end{bmatrix}}_b$$
I build the Schur-complement ($J_4$ is non-singular):
$$J_D=J_1-J_2J_4^{-1}J_3$$
Then, I solve for $x$:
$$J_Dx=b_x-J_2J_4^{-1}b_y$$
and then for $y$:
$$J_4y=b_y-J_3x$$
At this point I have a full solution of $z$.
Intermediate calculations
To calculate $J_2J_4^{-1}J_3$, I first compute $\tilde{J_2}=J_2J_4^{-1}$ as follows:
$$\tilde{J_2}^T=(J_2J_4^{-1})^T=(J_4^{-1})^TJ_2^T$$
Thus, I get $\tilde{J_2}$ by solving the system:
$$J_4^T\tilde{J_2}^T=J_2^T$$
and tacking the transpose.
Edit:
The above is implemented in Matlab and I use the conjugate transpose ' (mathworks.com/help/matlab/ref/ctranspose.html) for both the real and the complex. Also I use dot (mathworks.com/help/matlab/ref/dot.html) for the dot products.
Problem
This way of solving works perfectly when $J$ is real (even if $b$ is complex). That is, solving $Jz=b$ directly or by building the Schur-complement gives the same $z$ as expected. However, when $J_4$ becomes complex, then I get different results…
I suspect that I did something really wrong when taking the transpose for $\tilde{J_2}$ or doing the dot product for $J_2J_4^{-1}b_y$. Yet, I fail to detect my error.
Note: The reason I solve using the Schur-complement is that matrices $J_2,J_3,J_4$ have some particular structures that allow for good parallelization of the operations involved. E.g. $J_4$ is block diagonal and non-singular.
Best Answer
Your approach looks good, so you may have a bug somewhere in your implementation. As a validation, I ran the following simple test case involving a randomly generated dense complex matrix $J$ and a randomly generated real right-hand side $b$.
Running this code multiple times the approximate averaged values for $\texttt{err}$, $\texttt{res}$, and $\texttt{restrue}$ were $1\times 10^{-13}$, $1\times 10^{-13}$, and $4\times 10^{-14}$, respectively.