how to find: $A$ is invertible

Question

i have been given these matrices. $A\in K^{m \times m}$, $B\in K^{m \times n}$, $C\in K^{n \times m}$, $D\in K^{n \times n}$.

i need to show that if $A$ and $D-CA^{-1}B$ invertible, then $\begin{pmatrix} A & B \\ C & D \end{pmatrix} \in K^{(m+n) \times (m+n)}$ is invertible.

i know what invertible means: $A$ is invertible if $A A' = E$ where $E$ has dimension of $A$.

is there a light trick to know if it is invertible before calculation many things?

thanks a lot

Answer 1

Notice $$\begin{pmatrix}A & B\\C & D\end{pmatrix} \begin{pmatrix}I & -A^{-1}B\\0 & I\end{pmatrix} = \begin{pmatrix}A & 0\\C & D - CA^{-1}B\end{pmatrix}$$ we have $$\det\begin{pmatrix}A & B\\C & D\end{pmatrix} = \det(A)\det( D - CA^{-1}B) \ne 0$$ whenever $A$ and $D - CA^{-1}B$ invertible.

Answer 2

You are actually defining the Schur complement. There is a very standard way of dealing with this question (you can find it anywhere on the web).

Answer 3

Note that a matrix $P$ is invertible iff $Px=0$ has only trivial solution. Now consider $ \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. $ As $A$ is invertible, $x_1 = -A^{-1}Bx_2$. And substituting this in the next equation, we get $(D - CA^{-1}B)x_2=0$, which has only trivial solution if $D - CA^{-1}B$ is invertible. Hence the original system of equations has only trivial solution.

Answer 4

(Improved my comment into an answer)

It might be useful do observe that $$ \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} A' & B' \\ C' & D' \end{bmatrix} = \begin{bmatrix} AA'+BC' & AB'+BD' \\ CA'+DC' & DB'+DD' \end{bmatrix} $$ so that you have 4 equations to solve whose variables are the matrices $A',B',C',D'$: $$ \begin{cases} AA'+BC' = E_m \\ AB'+BD' = 0 \\ CA'+DC' = 0 \\ DB'+DD' = E_n \end{cases} $$ Since $A$ is invertible, from the first equation we deduce that $$ A' = A^{-1}(E_m-BC') $$ so that the system turns into $$ \begin{cases} A'=A^{-1}-A^{-1}BC' \\ AB'+BD' = 0 \\ CA^{-1} + \big(D - CA^{-1}B\big) C' = 0 \\ DB'+DD' = E_n \end{cases} $$ Now, let $M=\big(D - CA^{-1}B\big)^{-1}$ so that from the third equation it follows that $C'=-MCA^{-1}$, and so on...