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