Stuff I find slightly meaningful and close to achievable

Inverse of a block triangular matrix

Suppose you have a block triangular matrix \(\mathrm{L}\) of size \((m+n)\times (m+n)\) so that

$$
\mathrm{L} =
\begin{bmatrix}
L_{11} & 0 \\
L_{21} & L_{22}
\end{bmatrix}
$$

where \(L_{11}\) is square \(n\times n\) and \(L_{22}\) is square \(m\times m\).
To find the inverse, we simply solve a system in the block-unknowns \(A,B,C,D\):

$$
\mathrm{L}\mathrm{L}^{-1} =
\begin{bmatrix}
L_{11} & 0 \\
L_{21} & L_{22}
\end{bmatrix}
\begin{bmatrix}
A & B \\
C & D
\end{bmatrix}
=
\begin{bmatrix}
\mathrm{I}_n & 0 \\
0 & \mathrm{I}_m
\end{bmatrix}
=
\mathrm{I}_{n+m}
$$

using block-wise matrix multiplication, we obtain the system

$$
\begin{cases}
&L_{11} A = \mathrm{I}_n\\
&L_{11} B = 0 \\
&L_{21}A = -L_{22}C \\
&L_{21}B + L_{22}D = \mathrm{I}_n
\end{cases}
$$

Therefore, necessary and sufficient conditions for the existence of \(\mathrm{L}^{-1}\) are :

Which can be summarized into the equality

$$
\mathrm{L}^{-1} =
\begin{bmatrix}
L_{11} & 0 \\
L_{21} & L_{22}
\end{bmatrix}^{-1}
=
\begin{bmatrix}
L_{11}^{-1} & 0 \\
– L_{22}^{-1} L_{21} L_{11}^{-1} & L_{22}^{-1}
\end{bmatrix}
$$

Note that upper-triangular matrices possess the same property, meaning that

$$
\mathrm{U}^{-1} =
\begin{bmatrix}
U_{11} & U_{12} \\
0 & U_{22}
\end{bmatrix}^{-1}
=
\begin{bmatrix}
U_{11}^{-1} & -U_{11}^{-1}U_{12}U_{22}^{-1}\\
0 & U_{22}^{-1}
\end{bmatrix}
$$

memorization of the two formula is not necessary, as one can transform \(U\) into a lower-block-triangular matrix \(\mathrm{L} = \mathrm{U}^\mathsf{T}\), apply the \(L\)-inverse formula to obtain the lower-left inverse block
$$\left[\mathrm{L}^{-1}\right]_{21} = – L_{22}^{-1} L_{21} L_{11}^{-1}$$
and then swap the blocks again: the \(1\)'s becomes \(2\)'s and vice-versa, giving that
$$\left[\mathrm{U}^{-1}\right]_{12} = – U_{11}^{-1} U_{12} U_{22}^{-1}$$
as expected.