Prove that $B$ is invertible:$B = A_{11} - A_{12}A_{22}^{-1}A_{21}$ if...

Question

Let $$A = \begin{bmatrix}A_{11}&A_{12} \\ A_{21}&A_{22} \\ \end{bmatrix}$$

Prove that $B$ is invertible:$$B = A_{11} - A_{12}A_{22}^{-1}A_{21}$$

$$$$

in case of $A$ and $A_{22}$ being invertible matrices.

I tried the following:

$\det(A)=\det(A_{11}A_{22}) - \det(A_{12}A_{21}) \not= 0 $

$$B = A_{11}I - A_{12}A_{22}^{-1}A_{21}I=A_{11}A_{22}A_{22}^{-1} - A_{12}A_{22}^{-1}A_{21}A_{22}A_{22}^{-1}$$

$$ = [A_{11}A_{22} - A_{12}A_{22}^{-1}A_{21}A_{22}]A_{22}^{-1} $$

I also tried to invert $A$ with Gauss-Jordan, but that got me nowhere.

Answer 1

According to Schur complement: $$ |A|=|A_{22}|\cdot|B| $$ Since $|A|\neq 0,|A_{22}|\neq 0$, $|B|\neq 0$ and $B$ is invertible.

Answer 2

This is not an answer, but note that the matrix below shows that you can't generally assume that $\det(A)=\det(A_{11}A_{22})-\det(A_{12}A_{21})$.

$\begin{pmatrix}0&1&0&0\\0&0&1&0\\1&0&0&0\\0&0&0&1\end{pmatrix}$