# Block Triangular Inverses

## 2020/06/30

Consider the 2x2 block triangular matrices where $$A, B, C, D$$ are each matrices of appropriate dimensions:

$$\displaystyle U = \left( \begin{array}{cc} A & B\\ 0 & D \end{array} \right), \quad L = \left( \begin{array}{cc} A & 0\\ C & D \end{array} \right)$$

These matrices can be inverted using only the inverses of their diagonal blocks $$A^{- 1}$$ and $$D^{- 1}$$ (assuming those inverses exist). Let's try multiplying the upper triangular case by some matrix to try and obtain the identity:

$$\displaystyle UU^{- 1} = \left( \begin{array}{cc} A & B\\ 0 & D \end{array} \right) \left( \begin{array}{cc} A^{- 1} & X\\ 0 & D^{- 1} \end{array} \right) = \left( \begin{array}{cc} I_1 & AX + BD^{- 1}\\ 0 & I_2 \end{array} \right)$$

where $$X$$ is unknown and $$I_1, I_2$$ are identity matrices denoted separately since they may have different sizes. To get the identity, we must set the upper right block to $$0$$:

\begin{eqnarray*} AX + BD^{- 1} = 0 & \Rightarrow & X = - A^{- 1} BD^{- 1} \end{eqnarray*}

Plugging this value of $$X$$ back in gives us the result:

$$\displaystyle \left( \begin{array}{cc} A & B\\ 0 & D \end{array} \right)^{- 1} = \left( \begin{array}{cc} A^{- 1} & - A^{- 1} BD^{- 1}\\ 0 & D^{- 1} \end{array} \right)$$

A similar process gives the lower triangular result:

$$\displaystyle \left( \begin{array}{cc} A & 0\\ C & D \end{array} \right)^{- 1} = \left( \begin{array}{cc} A^{- 1} & 0\\ - D^{- 1} CA^{- 1} & D^{- 1} \end{array} \right)$$