Simplifying matrix equation $A^{-1}B^T(BA^{-1}B^T)^{-1}BA^{-1}$ I want to simplify the inverse of a $2\times 2$ block matrix:
$$
\begin{bmatrix} 
A_{n\times n} & B^T_{n\times m}\\
B_{m\times n} & 0_{m\times m}
\end{bmatrix}
$$
$A_{n\times n}$ is square and invertible (and possibly positive definite), and $B_{m\times n} (m < n) $ is not square but full row rank. Using this eq. (2.2), the inverse can be written as:
$$
\begin{bmatrix} 
A & B^T\\
B & 0
\end{bmatrix} ^{-1}= 
\begin{bmatrix}
A^{-1}-A^{-1}B^T(BA^{-1}B^T)^{-1}BA^{-1}  &  A^{-1}B^T(BA^{-1}B^T)^{-1} \\
(BA^{-1}B^T)^{-1}BA^{-1} & -(BA^{-1}B^T)^{-1}
\end{bmatrix}
$$
I want to see is there a way to simplify this matrix? For example, can we simplify this term $A^{-1}B^T(BA^{-1}B^T)^{-1}BA^{-1}$, or other terms?
 A: I believe that there is no way to significantly simplify it. You may rewrite it in different ways, but (as far as I see) nothing will cancel out completely.
One way of rewriting. Denote $ C = BA^{-1/2},\; D = A^{-1/2}$. Then the inverse is equal to
$$
\begin{bmatrix}
D(I-C^+C)D & DC^+ \\
C^{\rm T+}D & -(CC^{\rm T})^{-1}
\end{bmatrix}.
$$
Here $^+$ denotes the Moore-Penrose inverse.
Another way of rewriting. You can use the formula
$$
(BA^{-1}B^{\rm T})^{-1} = B^{\rm T+} A^{1/2} XA^{1/2}B^+,
$$
where
$$
X = I-A^{1/2}(I-B^{+}B) \big( A^{1/2}(I-B^{+}B)\big)^+.
$$
This formula opens $(BAB^{\rm T})^{-1}$ and takes some stuff out. It may be useful sometimes, but hardly simplifies something in your case.
Using limits. You may use Woodbury matrix identity to write
$$
\begin{align}
A^{-1}-A^{-1}B^{\rm T}(BA^{-1}B^{\rm T})^{-1}BA^{-1} & = \lim_{\varepsilon \to +0} \Big( A^{-1}-A^{-1}B^{\rm T}(\varepsilon I + BA^{-1}B^{\rm T})^{-1}BA^{-1} \Big) \\
& = \lim_{\varepsilon \to +0} \Big(A + \frac{1}{\varepsilon}B^{\rm T}B\Big)^{-1}.
\end{align}
$$
This looks nice, but again is not a simplification.
A: This is really just an extended comment. One can derive the desired expression for the inverse by considering the matrix equation $$\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}=\begin{bmatrix} A & B^T \\ B & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$
which is equivalent to the system of equations $y_1=Ax_1+B^T x_2,\, y_2=Bx_1$. To solve this for $x_1,x_2$, observe that
$$BA^{-1}y_1 = BA^{-1}(Ax_1+B^T x_2)=Bx_1+(BA^{-1}B^T)x_2=y_2+(BA^{-1}B^T)x_2.$$
Since $A^{-1}$ exists and $B$ has full row rank, the matrix $BA^{-1}B^T$ is full rank. Hence we may solve for $x_2$ by inverting $BA^{-1}B^T$:
\begin{align}
x_2&=(BA^{-1}B^T)^{-1}(BA^{-1}y_1-y_2)\\
&=(BA^{-1}B^T)^{-1}BA^{-1}y_1-(BA^{-1}B^T)^{-1}y_2.
\end{align}
Furthermore, we have $A^{-1}y_1=x_1+A^{-1}B^T x_2$ and therefore
\begin{align}
x_1
&=A^{-1}y_1-A^{-1}B^T x_2\\
&=[A^{-1}-A^{-1}B^T (BA^{-1}B^T)^{-1}BA^{-1}]y_1+A^{-1}B^T (BA^{-1}B^T)^{-1}BA^{-1}y_2
\end{align}
Converting this system of equations back to a matrix equation yields precisely the stated matrix elements of $\begin{bmatrix} A & B^T \\ B & 0\end{bmatrix}^{-1}$.
