# Shae's Ramblings

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 :

• $L_{11}$ is invertible and thus $A = L_{11}^{-1}$
• $B = 0$
• $L_{22}$ is invertible and $D = L_{22}^{-1}$
• We have that $C = \, – L_{22}^{-1} L_{21} L_{11}^{-1}$

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.