I'm having some trouble showing that the block matrix $$D = \pmatrix{A & B\\ C & I}$$ is nonsingular, given that $A-BC$ is nonsingular.

I have gotten close with the following by saying let $$T = \pmatrix{(A-BC)^{-1} & -B(A-BC)^{-1}\\-C(A-BC)^{-1} & A(A-BC)^{-1}}.$$ (Basically, I'm trying to extend the formula for the inverse of a $2\times2$ real matrix to block matrices.)

Then $$\begin{align}DT &= \pmatrix{A & B\\ C & I}\pmatrix{(A-BC)^{-1} & -B(A-BC)^{-1}\\-C(A-BC)^{-1} & A(A-BC)^{-1}} \\ &= \pmatrix{A(A-BC)^{-1}-BC(A-BC)^{-1} & -AB(A-BC)^{-1}+BA(A-BC)^{-1}\\ C(A-BC)^{-1}-C(A-BC)^{-1} & -CB(A-BC)^{-1} + A(A-BC)^{-1}}\\ &=\pmatrix{(A-BC)(A-BC)^{-1} & (BA - AB)(A-BC)^{-1}\\ O & (A - BC)(A-BC)^{-1}}\\ &=\pmatrix{I & (BA - AB)(A-BC)^{-1}\\ O & I}.\end{align}$$

I manage to get almost everything, except the top-right block. I want it to be $O$, but I'm not sure what information I have that makes it equal $O$. I don't necessarily know that $A$ and $B$ commute, i.e., $AB = BA$, which would imply the top-right block would zero out. Can anyone see if I made a mistake? If there is no mistake, can anyone give a hint on what to do to try and zero out the top-right block?


$$\pmatrix{I&-B\\O&I}\pmatrix{A&B\\C&I} =\pmatrix{A-BC&O\\C&I}.$$ The first matrix is clearly invertible. The last is iff $A-BC$ is.


We have $$ \pmatrix{A&B\\C&I}\pmatrix{I&0\\-C&I}=\pmatrix{A-BC&B\\0&I} $$ and $$\text{det}\left(\pmatrix{A-BC&B\\0&I}\right)=det(A-BC)\neq0$$


Block matrix determinant formula: if $D$ is invertible, $$ \det \pmatrix{A & B\cr C & D\cr} = \det(D) \det(A-BD^{-1} C)$$ Take $D=I$.


Curiously enough, this question can be answered quite simply from first principles, without reference to determinants; we merely need to show that any vector mapped to $0$ by the matrix

$ \mathscr M = \begin{bmatrix} A & B \\ C & I \end{bmatrix} \tag 1$

is itself $0$ ; so let $\mathscr X$ be a vector such that

$\mathscr M \mathscr X = \begin{bmatrix} A & B \\ C & I \end{bmatrix}\mathscr X = 0. \tag 2$

If the size of $A$ is $p$ and the size of $I$ is $q$, then the size of $\mathscr M$ is $p + q$ whilst $B$ is $p \times q$ and $C$ is $q \times p$.

We can then write $\mathscr X$ in terms of two vectors $\mathbf x$ and $\mathbf y$, where $\dim \mathbf x = p$ and $\dim \mathbf y = q$, thusly:

$\mathscr X = \begin{pmatrix} \mathbf x \\ \mathbf y \end{pmatrix}; \tag 3$

we have:

$\mathscr M \mathscr X = \begin{bmatrix} A & B \\ C & I \end{bmatrix}\begin{pmatrix} \mathbf x \\ \mathbf y \end{pmatrix} = \begin{pmatrix} A\mathbf x + B\mathbf y \\ C\mathbf x + I\mathbf y \end{pmatrix} = \begin{pmatrix} A\mathbf x + B\mathbf y \\ C\mathbf x + \mathbf y \end{pmatrix}; \tag 4$


$\mathscr M \mathscr X = 0, \tag 5$

we have from (4) that

$A\mathbf x + B\mathbf y = 0, \tag 6$


$C\mathbf x + \mathbf y = 0, \tag 7$


$A\mathbf x = -B\mathbf y \tag 8$


$C\mathbf x = -\mathbf y; \tag 9$


$A\mathbf x = -B\mathbf y = B(-\mathbf y) = B(C\mathbf x) = BC\mathbf x, \tag{10}$


$(A - BC)\mathbf x = 0; \tag{11}$

since we are given that $A - BC$ is nonsingular, we conclude that

$\mathbf x = 0, \tag{12}$

and from (9) that

$\mathbf y = -C\mathbf x = -C(0) = 0 \tag{13}$

as well; so

$\mathscr X = 0; \tag{14}$

since the only solution to (2) is (14), we conclude that $\mathscr M$ is itself a nonsingular matrix.


Let's try finding the inverse; so be it $$ E=\begin{pmatrix} P & Q \\ R & S\end{pmatrix} $$ Then $$ DE=\begin{pmatrix} A & B \\ C & I\end{pmatrix} \begin{pmatrix} P & Q \\ R & S\end{pmatrix}= \begin{pmatrix} AP+BR & AQ+BS \\ CP+R & CQ+S\end{pmatrix} $$ We need $R=-CP$; this leads to $AP-BCP=I$ that yields $P=(A-BC)^{-1}$.

Next we need $S=-CQ+I$, so we get $$ AQ+BS=AQ-BCQ+B=0 $$ that is, $(A-BC)Q+B=0$ or $$ Q=-(A-BC)^{-1}B $$ In conclusion, the inverse is $$ \begin{pmatrix} (A-BC)^{-1} & -(A-BC)^{-1}B \\ -C(A-BC)^{-1} & I+C(A-BC)^{-1}B \end{pmatrix} $$


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