Inverse of Partitioned Matrices

Theorem

Let

• $T$ : $m\times m$ nonsingular square matrix
• $U$ $m\times n$ matrix
• $V$ : $n\times m$ matrix
• $W$ : $n\times n$ square matrix

Let $Q=W-V{T}^{-1}U$

If $Q$ is invertible

Then $A$ and $B$ is nonsingular

$$\begin{array}{rl}{\left[\begin{array}{cc}T& U\\ V& W\end{array}\right]}^{-1}& =\left[\begin{array}{cc}{T}^{-1}+T^{-1}U{Q}^{-1}V{T}^{-1}& -T^{-1}U{Q}^{-1}\\ -Q^{-1}V{T}^{-1}& Q^{-1}\end{array}\right]\\ & =\left[\begin{array}{cc}{T}^{-1}& 0\\ 0& 0\end{array}\right]+\left[\begin{array}{c}-T^{-1}U\\ I_n\end{array}\right]Q^{-1}\left[\begin{array}{cc}-V{T}^{-1}& I_n\end{array}\right]\\ & \\ {\left[\begin{array}{cc}W& V\\ U& T\end{array}\right]}^{-1}& =\left[\begin{array}{cc}{Q}^{-1}& -Q^{-1}V{T}^{-1}\\ -T^{-1}U{Q}^{-1}& T^{-1}+T^{-1}U{Q}^{-1}V{T}^{-1}\end{array}\right]\\ & =\left[\begin{array}{cc}0& 0\\ 0& T^{-1}\end{array}\right]+\left[\begin{array}{c}{I}_n\\ -T^{-1}U\end{array}\right]Q^{-1}\left[\begin{array}{cc}{I}_n& -V{T}^{-1}\end{array}\right]\end{array}$$

Tip

To make it easier to memorize,

Let

• $t=T^{-1}$
• $u=T^{-1}U$
• $v=V{T}^{-1}$
• $q=Q^{-1}=(W-vU{)}^{-1}$

Then

$$\begin{array}{rl}{\left[\begin{array}{cc}T& U\\ V& W\end{array}\right]}^{-1}& =\left[\begin{array}{cc}t& 0\\ 0& 0\end{array}\right]+\left[\begin{array}{c}-u\\ I_n\end{array}\right]q\left[\begin{array}{cc}-v& I_n\end{array}\right]\\ & \\ {\left[\begin{array}{cc}W& V\\ U& T\end{array}\right]}^{-1}& =\left[\begin{array}{cc}0& 0\\ 0& t\end{array}\right]+\left[\begin{array}{c}{I}_n\\ -u\end{array}\right]q\left[\begin{array}{cc}{I}_n& -v\end{array}\right]\end{array}$$