A matrix of the form
$$A=\begin{bmatrix} A_{11} & A_{12}\\ 0 & A_{22} \end{bmatrix}$$
is said to be block upper triangular. Assume that $A_{11}$ is $p \times p$, $A_{22}$ is $q \times q$ and $A$ is invertible. Find a formula for $A^{-1}$.
Could anyone help on this?
An easy one to remember is the following surprising result: $$\begin{bmatrix}I & X \\ & I\end{bmatrix}^{-1} = \begin{bmatrix}I & -X \\ & I\end{bmatrix}$$ which is straightforward to verify by multiplying it out, or understand by just thinking about it.
You can get your desired inverse by factoring out the diagonal blocks, $$\begin{bmatrix}A_{11} & A_{12} \\ & A_{22}\end{bmatrix} = \begin{bmatrix}A_{11} \\ & I\end{bmatrix}\begin{bmatrix}I & A_{11}^{-1}A_{12}A_{22}^{-1} \\ & I\end{bmatrix}\begin{bmatrix}I \\ & A_{22}\end{bmatrix},$$ then applying the identity above. Doing so yields the following explicit formula for the inverse: \begin{align} \begin{bmatrix}A_{11} & A_{12} \\ & A_{22}\end{bmatrix}^{-1} &= \begin{bmatrix}I \\ & A_{22}^{-1}\end{bmatrix}\begin{bmatrix}I & -A_{11}^{-1}A_{12}A_{22}^{-1} \\ & I\end{bmatrix}\begin{bmatrix}A_{11}^{-1} \\ & I\end{bmatrix} \\ &= \begin{bmatrix}A_{11}^{-1} & -A_{11}^{-1}A_{12}A_{22}^{-1} \\ & A_{22}^{-1}\end{bmatrix}. \end{align} In some numerical methods, it is prefferable to keep it in the factored form rather than multiplying the matrices together at the end.
Since $A$ is nonsingular, $A_{11}$ and $A_{22}$ have to be nonsingular. (You can convince yourself of this fact by say, reducing $A_{11}$ and $A_{22}$ to RREF. If either is singular, then its RREF has to have zero rows and zero columns. This means, $A$ has to have zero rows or zero columns, contradicting nonsingularity of $A$. The easier argument is by determinants: $|A|=| A_{11}|\cdot| A_{22}|$. Left hand side is nonzero if and only if both factor on the right hand side are nonzero.)
Thus, consider the matrix $B=\begin{bmatrix}A_{11}^{-1}& -A_{11}^{-1}A_{12}A_{22}^{-1}\\0& A_{22}^{-1}\end{bmatrix}$.
Multiplying:
$AB=\begin{bmatrix}A_{11}& A_{12}\\0&A_{22}\end{bmatrix}\cdot\begin{bmatrix}A_{11}^{-1}& -A_{11}^{-1}A_{12}A_{22}^{-1}\\0& A_{22}^{-1}\end{bmatrix}\\ =\begin{bmatrix}I_{p}& -A_{11}\cdot A_{11}^{-1}A_{12}A_{22}^{-1}+A_{12}A_{22}^{-1}\\0&I_q\end{bmatrix}=I.$
Hence, $B$ is the inverse of $A$.
Hint: You may consider the matrix $$B = \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{bmatrix},$$ such that: $$\begin{bmatrix} A_{11} & A_{12}\\ 0 & A_{22} \end{bmatrix} \cdot \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{bmatrix} = \begin{bmatrix} I_{p\times p} & 0 \\ 0 & I_{q\times q} \end{bmatrix},$$ where $B_{11}$ is a $p\times p$ matrix, $B_{12}$ is a $p\times q$ matrix, $B_{21}$ is a $q \times p$ matrix and $B_{22}$ is a $q \times q$ matrix.
