Block Matrix Inversion

Here are some formula for matrix inversion.

Lemma 1: For a block matrix M=\begin{pmatrix}A & B \\C &D\end{pmatrix}, 
M^{-1}=\begin{pmatrix}(A-B D^{-1} C)^{-1}& -A^{-1}B(D-CA^{-1}B)^{-1}\\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1}\end{pmatrix}

=\begin{pmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}& -A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{pmatrix},
where S=D-CA^{-1}B is basically the Schur's 
complement of block A.

Proof: Let M^{-1}=\begin{pmatrix}E&F\\G&H\end{pmatrix}, M M^{-1}=1 gives us

AE+BG=I
AF+BH=0
CE+DG=0
CF+DH=I

From the four equations, we have
E=(A-BD^{-1}C)^{-1}
F=-A^{-1}B(D-CA^{-1}B)
G=-D^{-1}C(A-BD^{-1}C)^{-1}
H=(D-CA^{-1}B)^{-1}

And similarly from M^{-1}M=I, we have

E=(A-BD^{-1}C)^{-1}
F=-(A-BD^{-1}C)^{-1}BD^{-1}
G=-(D-CA^{-1}B)^{-1}CA^{-1}
H=(D-CA^{-1}B)^{-1}

Together, they show the first inequality. Also note that AE+BG=I and thus E=A^{-1}-A^{-1}BG. Substituting G into above shows the second equality. \Box.

Lemma 2 (Matrix Inversion Formula/Woodbury Matrix Identity): 
(A+BDC)^{-1}=A^{-1}-A^{-1}B(D^{-1}+CA^{-1}B)^{-1}CA^{-1}

Proof: From the previous proof, we have E=(A-BD^{-1}C)^{-1}=A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}. We get the identity immediately as we flip the sign of B.  \Box

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