Let $A,B,C,D \in \mathbb{R}^{n×n}$. Show that if $A, C, B−AC^{−1}D,$ and $D−CA^{−1}B$ are nonsingular then that the following matrix has the matrix:

Question

Let $A,B,C,D \in \mathbb{R}^{n×n}$. Show that if $A, C, B−AC^{−1}D,$ and $D−CA^{−1}B$ are nonsingular then

$\left[ \begin{smallmatrix} A&B\\ C&D \end{smallmatrix} \right]^{-1} = \left[ \begin{smallmatrix} A^{-1} - A^{-1}B(B-AC^{-1}D)^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\ (B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1} \end{smallmatrix} \right]$

So i think that probably i have to use The Determinant and the adjacency matrix but then i get determinant = $AD - BC $

Adjacency matrix $= \left[ \begin{smallmatrix} D&-B\\ -C&A \end{smallmatrix} \right]$

But i cant divide the determinant by the adjacency matrix because AD - BC doesnt have an inverse so what do i do?emphasized text

Answer 1

To prove that $$\left[ \begin{smallmatrix} A&B\\ C&D \end{smallmatrix} \right]^{-1} = \left[ \begin{smallmatrix} A^{-1} - A^{-1}B(B-AC^{-1}D)^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\ (B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1} \end{smallmatrix} \right]$$

I would first try to calculate

$$\left[ \begin{smallmatrix} A&B\\ C&D \end{smallmatrix} \right]\cdot \left[ \begin{smallmatrix} A^{-1} - A^{-1}B(B-AC^{-1}D)^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\ (B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1} \end{smallmatrix} \right]$$

and show that this equals the identity matrix.

Answer 2

Let $I$ denote the $n$-by-$n$ identity matrix. For simplicity, let $$X:=\begin{bmatrix}A&B\\C&D\end{bmatrix}\,.$$ As $A$ is invertible, $$X=\begin{bmatrix}I&0\\CA^{-1}&I\end{bmatrix}\,\begin{bmatrix}A&B\\0&D-CA^{-1}B\end{bmatrix}\,.$$ Since $D-CA^{-1}B$ is invertible, $$X=\begin{bmatrix}I&0\\CA^{-1}&I\end{bmatrix}\,\begin{bmatrix}A&0\\0&D-CA^{-1}B\end{bmatrix}\,\begin{bmatrix}I&A^{-1}B\\0&I\end{bmatrix}\,.$$ Consequently, $$X^{-1}=\begin{bmatrix}I&-A^{-1}B\\0&I\end{bmatrix}\,\begin{bmatrix}A^{-1}&0\\0&(D-CA^{-1}B)^{-1}\end{bmatrix}\,\begin{bmatrix}I&0\\-CA^{-1}&I\end{bmatrix}\,.$$ That is, $$X^{-1}=\begin{bmatrix}A^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\0&(D-CA^{-1}B)^{-1}\end{bmatrix}\,\begin{bmatrix}I&0\\-CA^{-1}&I\end{bmatrix}\,.$$ Hence, $$X^{-1}=\begin{bmatrix}A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1}\end{bmatrix}\,.\tag{*}$$

Similarly, $$X=\begin{bmatrix}I&AC^{-1}\\0&I\end{bmatrix}\,\begin{bmatrix}0&B-AC^{-1}D\\C&D\end{bmatrix}\,.$$ Thus, $$X=\begin{bmatrix}I&AC^{-1}\\0&I\end{bmatrix}\,\begin{bmatrix}0&B-AC^{-1}D\\C&0\end{bmatrix}\,\begin{bmatrix}I&C^{-1}D\\0&I\end{bmatrix}\,.$$ Because $B-AC^{-1}D$ is invertible, $$X^{-1}=\begin{bmatrix}I&-C^{-1}D\\0&I\end{bmatrix}\,\begin{bmatrix}0&C^{-1}\\(B-AC^{-1}D)^{-1}&0\end{bmatrix}\,\begin{bmatrix}I&-AC^{-1}\\0&I\end{bmatrix}\,.$$ Ergo, $$X^{-1}=\begin{bmatrix}-C^{-1}D(B-AC^{-1}D)^{-1}&C^{-1}\\(B-AC^{-1}D)^{-1}&0\end{bmatrix}\,\begin{bmatrix}I&-AC^{-1}\\0&I\end{bmatrix}\,.$$ Thence, $$X^{-1}=\begin{bmatrix}-C^{-1}D(B-AC^{-1}D)^{-1} & C^{-1}+C^{-1}D(B-AC^{-1}D)^{-1}AC^{-1}\\ (B-AC^{-1}D)^{-1}& -(B-AC^{-1}D)^{-1}AC^{-1}\end{bmatrix}\,.\tag{#}$$

From (*) and (#), we conclude that $$ (D-CA^{-1}B)^{-1}=-(B-AC^{-1}D)^{-1}AC^{-1}\,.$$ That is, $$(D-CA^{-1}B)^{-1}CA^{-1}=-(B-AC^{-1}D)^{-1}\,.$$ Hence, $$A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}=A^{-1}-A^{-1}B(B-AC^{-1}D)^{-1}\,.$$ This shows that $$X=\begin{bmatrix}A^{-1}-A^{-1}B(B-AC^{-1}D)^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\(B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1}\end{bmatrix}\,.$$

---

If $D$ is invertible, then $$X=\begin{bmatrix}I&BD^{-1}\\0&I\end{bmatrix}\,\begin{bmatrix}A-BD^{-1}C&0\\C &D\end{bmatrix}=\begin{bmatrix}I&BD^{-1}\\0&I\end{bmatrix}\,\begin{bmatrix}A-BD^{-1}C&0\\0 &D\end{bmatrix}\,\begin{bmatrix}I&0\\D^{-1}C&I\end{bmatrix}\,.$$ Ergo, $$X^{-1}=\begin{bmatrix}I&0\\-D^{-1}C&I\end{bmatrix}\,\begin{bmatrix}(A-BD^{-1}C)^{-1}&0\\0 &D^{-1}\end{bmatrix}\,\begin{bmatrix}I&-BD^{-1}\\0&I\end{bmatrix}\,,$$ provided that $B-D^{-1}C$ is nonsingular. Thus, $$X^{-1}=\begin{bmatrix}(A-BD^{-1}C)^{-1} & 0\\-D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}\end{bmatrix}\,\begin{bmatrix}I&-BD^{-1}\\0&I\end{bmatrix}\,,$$ or $$X=\begin{bmatrix} (A-BD^{-1}C)^{-1} & -(A-BD^{-1}C)^{-1} BD^{-1}\\ -D^{-1}C(A-BD^{-1}C)^{-1}& D^{-1}+D^{-1}C(A-BD^{-1}C)^{-1}BD^{-1}\end{bmatrix}\,.\tag{@}$$

If $B$ is invertible, then $$X=\begin{bmatrix}I&0\\DB^{-1}&I\end{bmatrix}\,\begin{bmatrix}A&B\\C -DB^{-1}A&0\end{bmatrix}=\begin{bmatrix}I&0\\DB^{-1}&I\end{bmatrix}\,\begin{bmatrix}0&B\\C-DB^{-1}A &0\end{bmatrix}\,\begin{bmatrix}I&0\\B^{-1}A&I\end{bmatrix}\,.$$ Ergo, $$X^{-1}=\begin{bmatrix}I&0\\-B^{-1}A&I\end{bmatrix}\,\begin{bmatrix}0&(C-DB^{-1}A )^{-1}\\B^{-1} &0\end{bmatrix}\,\begin{bmatrix}I&0\\-DB^{-1}&I\end{bmatrix}\,,$$ provided that $C-DB^{-1}A$ is nonsingular. Thus, $$X^{-1}=\begin{bmatrix}0 & (C-DB^{-1}A )^{-1}\\B^{-1} & -B^{-1}A(C-DB^{-1}A )^{-1}\end{bmatrix}\,\begin{bmatrix}I&0\\-DB^{-1}&I\end{bmatrix}\,,$$ or $$X=\begin{bmatrix} -(C-DB^{-1}A )^{-1}DB^{-1} & (C-DB^{-1}A )^{-1}\\B^{-1}+B^{-1}A(C-DB^{-1}A )^{-1}DB^{-1}& -B^{-1}A(C-DB^{-1}A )^{-1}\end{bmatrix}\,.\tag{\$}$$

In the special case where $A,B,C,D$ are invertible matrices such that $A-BD^{-1}C$, $B-AC^{-1}D$, $C-DB^{-1}A$, and $D-CA^{-1}B$ are all nonsingular, we have the following compact form of $X^{-1}$: $$X^{-1}=\begin{bmatrix}(A-BD^{-1}C)^{-1}&(C-DB^{-1}A)^{-1}\\(B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1}\end{bmatrix}\,.$$ While this looks nice, I think it is computationally more expensive than using (*), (#), (@), or (\$) alone.

Answer 3

This is what you get from performing a blockwise Gaussian elimination on colums, assuming everything is invertible when you need it:

Starting with $\left[\begin{smallmatrix}A&B\\C&D\end{smallmatrix}\right]$ you want to multiply the first block of colums by $A^{-1}$, you obtain $$ \left[\begin{smallmatrix}A&B\\C&D\end{smallmatrix}\right] \left[\begin{smallmatrix}A^{-1}&0\\0&I\end{smallmatrix}\right] = \left[\begin{smallmatrix}I&B\\CA^{-1}&D\end{smallmatrix}\right]. $$ Now subtract the first block column $B$-times from the second: $$ \left[\begin{smallmatrix}I&B\\CA^{-1}&D\end{smallmatrix}\right] \left[\begin{smallmatrix}I&-B\\0&I\end{smallmatrix}\right] = \left[\begin{smallmatrix}I&0\\CA^{-1}&D-CA^{-1}B\end{smallmatrix}\right]. $$ Now you want to get an $I$ in the lower right block, so you multiply the second column block by $(D-CA^{-1}B)^{-1}$ to obtain $$ \left[\begin{smallmatrix}I&0\\CA^{-1}&D-CA^{-1}B\end{smallmatrix}\right] \left[\begin{smallmatrix}I&0\\0&(D-CA^{-1}B)^{-1}\end{smallmatrix}\right] = \left[\begin{smallmatrix}I&0\\CA^{-1}&I\end{smallmatrix}\right]. $$ Finally subtract the second block column $CA^{-1}$-times from the first block column $$ \left[\begin{smallmatrix}I&0\\CA^{-1}&I\end{smallmatrix}\right] \left[\begin{smallmatrix}I&0\\-CA^{-1}&I\end{smallmatrix}\right] = \left[\begin{smallmatrix}I&0\\0&I\end{smallmatrix}\right]. $$ In summary, you obtained \begin{align*} \left[\begin{smallmatrix}A&B\\C&D\end{smallmatrix}\right]^{-1} &= \left[\begin{smallmatrix}A^{-1}&0\\0&I\end{smallmatrix}\right] \left[\begin{smallmatrix}I&-B\\0&I\end{smallmatrix}\right] \left[\begin{smallmatrix}I&0\\0&(D-CA^{-1}B)^{-1}\end{smallmatrix}\right] \left[\begin{smallmatrix}I&0\\-CA^{-1}&I\end{smallmatrix}\right] \\ &= \left[\begin{smallmatrix}A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1}\end{smallmatrix}\right] \end{align*} using only the assumptions that $A$ and $D-CA^{-1}B$ are invertible.

Assuming that $C$ and $B-AC^{-1}D$ are also invertible, you can simplify this to the matrix in your question by using $Y^{-1} Z = (Z^{-1} Y)^{-1}$ a few times.