## 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.