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:

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