Inverse of a block matrix of a special form Let $A,B\in\mathbb{R}^{m\times m}$ be nonsingular. Define $R\triangleq\left[ \begin{array}{cc}A&B\\-B&A\end{array}\right]\in\mathbb{R}^{2m\times 2m}$. Can we show that $R$ is nonsingular and $R^{-1}=\left[ \begin{array}{cc}C&D\\-D&C\end{array}\right]$, where $D,C\in\mathbb{R}^{m\times m}$ are nonsingular matrices?
Is there a closed-form relation for $C$ and $D$ in terms of $A$ and $B$?
It seems that $C={\rm Re~}(A+\jmath B)^{-1}$, and $D={\rm Im~}(A+\jmath B)^{-1}$.
Any proof or hint is appreciated. Thanks

Edit based on Reinhard Meier answer:
Let $G\in\mathbb{C}^{m\times m}$ be nonsingular, and define $R\triangleq\left[ \begin{array}{cc}{\rm Re~}G&{\rm Im~}G\\-{\rm Im~}G&{\rm Re~}G\end{array}\right]\in\mathbb{R}^{2m\times 2m}$. Then, $R$ is invertible, and 
\begin{equation}
R^{-1}=\left[ \begin{array}{cc}{\rm Re~}G^{-1}&{\rm Im~}G^{-1}\\-{\rm Im~}G^{-1}&{\rm Re~}G^{-1}\end{array}\right].
\end{equation}
 A: We want to accomplish that
$$
\begin{bmatrix} A & B \\ -B & A \end{bmatrix}\cdot
\begin{bmatrix} C & D \\ -D & C \end{bmatrix}=
\begin{bmatrix} I & 0 \\ 0 & I \end{bmatrix}
$$
with the $m\times m$ identity matrix $I.$ Therefore, we have to find $C$ and $D$ such that
$$
AC-BD=I \\
AD+BC=0
$$
This can easily be solved, and we find
$$
C=\left(A+BA^{-1}B\right)^{-1}
=B^{-1}\left(AB^{-1}+BA^{-1}\right)^{-1}
=\left(B^{-1}A+A^{-1}B\right)^{-1}B^{-1}  \\
-D=(B+AB^{-1}A)^{-1}
=A^{-1}\left(BA^{-1}+AB^{-1}\right)^{-1}
=\left(A^{-1}B+B^{-1}A\right)^{-1}A^{-1}
$$
As we can see, the invertibility of $AB^{-1}+BA^{-1}$ (and the invertibility of $A^{-1}B+B^{-1}A,$ which is equivalent, which we can see when we transpose the expression) is a sufficient condition for the invertibility of $R,$ if $A$ and $B$ are also invertible.
It is also a necessary condition. Assume $AB^{-1}+BA^{-1}$ was singular. Then there is $v\in\mathbb{R}^m,$ $v\neq 0$ such that $\left(AB^{-1}+BA^{-1}\right)v=0.$ With this $v,$ we can show that
$$
R\cdot \begin{bmatrix} B^{-1}v \\ A^{-1}v \end{bmatrix}
=\begin{bmatrix} A & B \\ -B & A \end{bmatrix}\cdot
 \begin{bmatrix} B^{-1}v \\ A^{-1}v \end{bmatrix}
=\begin{bmatrix}
\left(AB^{-1}+BA^{-1}\right)v \\
\left(-BB^{-1}+AA^{-1}\right)v
\end{bmatrix}=0
$$
It can be shown that $AB^{-1}+BA^{-1}$ is singular if $AB^{-1}$ has $i$ as an eigenvalue. This can be used to construct a simple example with invertible $A$ and $B$ but singular $R.$ Just take 
$$
A=\begin{bmatrix} 1 & 0  \\ 0 & 1 \end{bmatrix}
\;\;\;\mbox{and}\;\;\;
B=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}
$$
Edit
The solution above does not require complex numbers, but it is simpler with them:
Starting again from 
$$
AC-BD=I \\
AD+BC=0
$$
We now assume 
$C={\rm Re~}(A+\jmath B)^{-1}$, and $D={\rm Im~}(A+\jmath B)^{-1}$ and show that the equations above are fulfilled. Note that for any two complex matrices $U$ and $V,$ we have
$$
{\rm Re~}(UV)={\rm Re~}(U)\,{\rm Re~}(V)-{\rm Im~}(U)\,{\rm Im~}(V) \\
{\rm Im~}(UV)={\rm Re~}(U)\,{\rm Im~}(V)+{\rm Im~}(U)\,{\rm Re~}(V) \\
$$
With $A={\rm Re~}(A+\jmath B)$ and $B={\rm Im~}(A+\jmath B),$ we find
\begin{eqnarray*}
& & AC-BD \\
&=&{\rm Re~}(A+\jmath B)\,{\rm Re~}((A+\jmath B)^{-1})-{\rm Im~}(A+\jmath B)\,{\rm Im~}((A+\jmath B)^{-1})\\
&=&{\rm Re~}((A+\jmath B)\,(A+\jmath B)^{-1}) \\
&=&{\rm Re~}(I) \\
&=& I
\end{eqnarray*}
and
\begin{eqnarray*}
& & AD+BC \\
&=&{\rm Re~}(A+\jmath B)\,{\rm Im~}((A+\jmath B)^{-1})+{\rm Im~}(A+\jmath B)\,{\rm Re~}((A+\jmath B)^{-1})\\
&=&{\rm Im~}((A+\jmath B)\,(A+\jmath B)^{-1}) \\
&=&{\rm Im~}(I) \\
&=& 0
\end{eqnarray*}
