Block matrix pseudoinverse: symmetry of the inverse of a symmetric matrix In the wiki page for block matrix pseudoinverses, there is a formula
$$
\begin{pmatrix}A & B \\ C & D\end{pmatrix}^{-1}=\begin{pmatrix}
(A-BD^{-1}C)^{-1}         & -A^{-1}B(D-CA^{-1}B)^{-1}\\
-D^{-1}C(A-BD^{-1}C)^{-1} & (D-CA^{-1}B)^{-1}
\end{pmatrix}\cdot
$$
Let's call $M=\begin{pmatrix}A & B \\ C & D\end{pmatrix}$ and $N$ the matrix on the RHS above. I can verify that $MN=I$ but I am stuck when I try to show that $NM=I$. Similarly, when $M$ is symmetric ($A^T=A$, $C^T=B$, and $D^T=D$), $N$ should also be symmetric but I can't verify that the off-diagonal blocks of $N$ are transposes of one another. Can you please help?
 A: The verification of $NM=I$ is more involved, as multiplication is done in the less favorable order.
Let me show only 
$$
(A-BD^{-1}C)^{-1} A - A^{-1}B(D-CA^{-1}B)^{-1}C = I.
$$
Let's go:
$$
(A-BD^{-1}C)^{-1} A - I=(A-BD^{-1}C)^{-1} (A-(A-BD^{-1}C)) = (A-BD^{-1}C)^{-1}BD^{-1}C.
$$
Let me abbreviate $T:=(A-BD^{-1}C)^{-1} A-I - A^{-1}B(D-CA^{-1}B)^{-1}C$. We want to show $T=0$.
Then
$$\begin{split}
(A-BD^{-1}C)T &= BD^{-1}C - (A-BD^{-1}C)A^{-1}B(D-CA^{-1}B)^{-1}C\\
&=BD^{-1}C -B(D-CA^{-1}B)^{-1}C + BD^{-1}\underbrace{CA^{-1}B(D-CA^{-1}B)^{-1}}_{=D(D-CA^{-1}B)^{-1}-I}C\\
&=0.
\end{split}$$
The computation of the other block entries of $NM$ shold be analogous.

Now towards symmetry: Using $A=A^T$, $B=C^T$, $D=D^T$, it remains to show
$$
-(A^{-1}B(D-B^TA^{-1}B)^{-1} )^T = -\underbrace{D^{-1}B^T(A-BD^{-1}B^T)^{-1}}_{=:Y}.
$$
Since $A$ and $D-B^TA^{-1}B$ are symmetric, we have
$$
(A^{-1}B(D-B^TA^{-1}B)^{-1} )^T = (D-B^TA^{-1}B)^{-1} B^TA^{-1}=:X.
$$
Then 
$$\begin{split}
X(A-BD^{-1}B^T)&=(D-B^TA^{-1}B)^{-1} B^T - \underbrace{(D-B^TA^{-1}B)^{-1}
B^TA^{-1}B}_{=(D-B^TA^{-1}B)^{-1}D-I}D^{-1}B^T
\\
&=D^{-1}B^T = Y(A-BD^{-1}B^T)
\end{split}$$,
hence $X=Y$ and symmetry follows.
