Inverse and determinant of partitioned symmetric matrix

Theorem 1

$$(A+CBD)^{-1}=A^{-1}-A^{-1}C(B^{-1}+DA^{-1}C)^{-1}DA^{-1}$$

Proof:

$$(A+CBD)[A^{-1}-A^{-1}C(B^{-1}+DA^{-1}C)^{-1}DA^{-1}]$$

$$= (A+CBD)A^{-1}-(A+CBD)A^{-1}C(B^{-1}+DA^{-1}C)^{-1}DA^{-1}$$

$$= I+CBDA^{-1}-(C+CBDA^{-1}C)(B^{-1}+DA^{-1}C)^{-1}DA^{-1}$$

$$= I+CBDA^{-1}-CB(B^{-1}+DA^{-1}C)(B^{-1}+DA^{-1}C)^{-1}DA^{-1}$$

$$= I+CBDA^{-1}-CBDA^{-1}=I$$

Theorem 2 (inverse of a partitioned symmetric matrix)

Divide an $n\times n$ symmetric matrix $A$ into four blocks

$$A=\left[ \begin{array}{cc} A_{11} & A_{12} A_{21} & A_{22}... ...cc} A_{11} & A_{12} A_{12}^T & A_{22} \end{array} \right]$$

The inverse matrix $B=A^{-1}$ can also be divided into four blocks:

$$B=A^{-1}=\left[ \begin{array}{cc} B_{11} & B_{12} B_{21} &... ...cc} B_{11} & B_{12} B_{12}^T & B_{22} \end{array} \right]$$

Here we assume the dimensionalities of these blocks are:

• $A_{11}$ and $B_{11}$ are $p\times p$,
• $A_{22}$ and $B_{22}$ are $q\times q$,
• $A_{12}=A_{21}^T$ and $B_{12}=B_{21}^T$ are $p\times q$

with $p+q=n$. Then we have

$$\begin{array}{l} B_{11}=(A_{11}-A_{12}A_{22}^{-1}A_{12}^T)^{-... ...^{-1}A_{12}(A_{22}-A_{12}^TA_{11}^{-1}A_{12})^{-1} \end{array}$$

Proof:

$$I_n = AA^{-1}=AB=\left[\begin{array}{cc}A_{11}&A_{12} A_{12}^T&A_{22... ...ight] \left[\begin{array}{cc}B_{11}&B_{12} B_{12}^T&B_{22}\end{array}\right]$$

$$= \left[\begin{array}{cc}A_{11}B_{11}+A_{12}B_{12}^T&A_{11}B_{12}+A... ...22} \end{array}\right] =\left[\begin{array}{cc}I_p&0 0&I_q\end{array}\right]$$

i.e.,

$$A_{11}B_{11}+A_{12}B_{12}^T=I_p\;\;\;\;\mbox{or}\;\;\;\; B_{11}=A_{11}^{-1}-A_{11}^{-1}A_{12}B_{12}^T$$

$$A_{11}B_{12}+A_{12}B_{22}=0\;\;\;\;\mbox{or}\;\;\;\; B_{12}=-A_{11}^{-1}A_{12}B_{22}$$

$$A_{12}^TB_{11}+A_{22}B_{12}^T=0\;\;\;\;\mbox{or}\;\;\;\; B_{12}^T=-A_{22}^{-1}A_{12}^TB_{11}$$

$$A_{12}^TB_{12}+A_{22}B_{22}=I_q\;\;\;\;\mbox{or}\;\;\;\; B_{22}=A_{22}^{-1}-A_{22}^{-1}A_{12}^TB_{12}$$

Plug $B_{12}^T$ into $B_{11}$ to get

$$B_{11}=A_{11}^{-1}+A_{11}^{-1}A_{12}A_{22}^{-1}A_{12}^TB_{11}$$

or

$$(I-A_{11}^{-1}A_{12}A_{22}^{-1}A_{12}^T)B_{11}=A_{11}^{-1} \;\;\;\mbox{or}\;\;\; (A_{11}-A_{12}A_{22}^{-1}A_{12}^T)B_{11}=I_p$$

or

$$B_{11}=(A_{11}-A_{12}A_{22}^{-1}A_{12}^T)^{-1}$$

Applying theorem 1 to this expression, we also get the other expression in the theorem. Similarly we can get

$$B_{22}=(A_{22}-A_{12}^TA_{11}^{-1}A_{12})^{-1}$$

and

$$B_{12}^T=-A_{22}^{-1}A_{12}^T(A_{11}-A_{12}A_{22}^{-1}A_{12}^T)^{-1}$$

$$B_{12}=-A_{11}^{-1}A_{12}(A_{22}-A_{12}^TA_{11}^{-1}A_{12}^{-1}$$

Theorem 3 (Determinant of a partitioned symmetric matrix)

$$\vert A\vert=\left\vert \left[\begin{array}{cc}A_{11}&A_{12}\... ...=\vert A_{11}\vert \vert A_{22}-A_{12}^TA_{11}^{-1}A_{12}\vert$$

Proof: Note that

$$A=\left[\begin{array}{cc}A_{11}&A_{12} A_{21}&A_{22}\end{array} \right] = \left[\begin{array}{cc}A_{11}&0 A_{12}^T&I \end{array}\right] ... ...c}I&A_{11}^{-1}A_{12} 0&A_{22}-A_{12}^TA_{11}^{-1}A_{12} \end{array} \right]$$

$$= \left[\begin{array}{cc}I&A_{12} 0&A_{22} \end{array}\right] \l... ...cc}A_{11}-A_{12}A_{22}^{-1}A_{12}^T&0\\ A_{22}^{-1}A_{21}&I\end{array} \right]$$

The theorem is proved as we also know that

$$\vert AB\vert=\vert A\vert\;\vert B\vert$$

and

$$\left\vert \begin{array}{cc}B&0 C&D\end{array} \right\vert=... ...cc}B&C 0&D\end{array} \right\vert=\vert B\vert\;\vert D\vert$$