Inversion of a 2$\times$2 block matrix

Let us define the $2\times2$ block matrix M as follows:

$${\bf M}= \left( \begin{array} {cc} {\bf A} & {\bf B} \\ {\bf C} & {\bf D}\end{array}\right),$$ (87)

where A, B, C, and D are matrices. First, we consider the matrix equation

$$\left( \begin{array} {cc} {\bf A} & {\bf B} \\ {\bf C} & ... ...eft( \begin{array} {c} {\bf G} \\ {\bf H}\end{array}\right).$$ (88)

If we multiply the top row by $-{\bf CA^{-1}}$ and add it to the bottom, we have

$$({\bf D} - {\bf CA^{-1}B}){\bf F} = {\bf H} - {\bf CA^{-1}G}.$$ (89)

Then we can easily find F and E. The quantity $({\bf D} - {\bf CA^{-1}B})$ is called the Schur complement of A and, denoted as ${\bf S_A}$, appears often in linear algebra Demmel (1997). The derivation of F and E can be written in a matrix form

$$\left( \begin{array} {cc} {\bf A} & {\bf B} \\ {\bf C} & ... ...\bf I} & {\bf A^{-1}B} \\ {\bf 0} & {\bf I}\end{array}\right),$$ (90)

which resembles an LDU decomposition of M. Alternatively, we have the UDL decomposition

$$\left( \begin{array} {cc} {\bf A} & {\bf B} \\ {\bf C} & ... ...\bf I} & {\bf 0} \\ {\bf D^{-1}C} & {\bf I}\end{array}\right),$$ (91)

where ${\bf S_D}={\bf A}-{\bf BD^{-1}C}$ is the Schur complement of D. The inversion formulas are then easy to derive as follows:

$$\left( \begin{array} {cc} {\bf A} & {\bf B} \\ {\bf C} & ... ...\bf I} & {\bf 0} \\ {\bf -CA^{-1}} & {\bf I}\end{array}\right)$$ (92)

and

$$\left( \begin{array} {cc} {\bf A} & {\bf B} \\ {\bf C} & ... ...bf I} & {\bf -BD^{-1}} \\ {\bf 0} & {\bf I}\end{array}\right).$$ (93)

The decomposition of the matrix M offers opportunities for fast inversion algorithms. The final expressions for M are  

$$\left( \begin{array} {cc} {\bf A} & {\bf B} \\ {\bf C} & ... ...} \\ -{\bf S_A^{-1}CA^{-1}} & {\bf S_A^{-1}}\end{array}\right)$$ (94)

and  

$$\left( \begin{array} {cc} {\bf A} & {\bf B} \\ {\bf C} & ... ...[{\bf D^{-1}}+{\bf D^{-1}CS_D^{-1}BD^{-1}}]\end{array}\right).$$ (95)

Equations () and () yield the matrix inversion lemma

$$({\bf A} - {\bf BD^{-1}C})^{-1} = {\bf A^{-1}}+{\bf A^{-1}B({\bf D} - {\bf CA^{-1}B})^{-1}CA^{-1}}.$$ (96)