I got the following question to solve:
Given the lower triangular matrix
\begin{bmatrix}
A_{11} & 0 \\
A_{21} & A_{22}
\end{bmatrix}
of size $n \times n$ (n is a power of 2) where $A_{11}$, $A_{21}$ and $A_{22}$ are matrices of size $(n/2) \times (n/2)$, show that the inverse is,
\begin{bmatrix}
A_{11}^{-1} & 0 \\
-A_{22}^{-1}A_{21}A_{11} & A_{22}^{-1}
\end{bmatrix}
how do I go about to solve this problem?
Edit: the matrix is invertible.
Edit: the second matrix should be changed to:
\begin{bmatrix}
A_{11}^{-1} & 0 \\
-A_{22}^{-1}A_{21}A_{11}^{\color{red}{-1}} & A_{22}^{-1}
\end{bmatrix}
The inverse was missing.
Best Answer
Clearly, you are having some trouble evaluating, and I think this is because of a typo! Evaluating with normal matrix multiplication I got $$ \begin{bmatrix} A_{11} & 0 \\ A_{21} & A_{22} \end{bmatrix}\begin{bmatrix} A_{11}^{-1} & 0 \\ -A_{22}^{-1}A_{21}A_{11} & A_{22}^{-1} \end{bmatrix} = \begin{bmatrix} A_{11}A_{11}^{-1} & 0 \\ A_{21} A_{11}^{-1} -A_{22}A_{22}^{-1}A_{21}A_{11} & A_{22}A_{22}^{-1} \end{bmatrix} $$ Everything evaluates trivially except for this term $$A_{21} A_{11}^{-1} - A_{21}A_{11}$$ Which clearly does not equal $0$ all the time. Thus I believe there was a typo made here and that the $A_{11}$ at the end should be an $A_{11}^{-1}$, as then the above expression reduces to the identity matrix, $$ \begin{bmatrix} I & 0 \\ 0 & I \end{bmatrix} $$