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.