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In the wiki page for block matrix pseudoinverses, there is a formula $$ \begin{pmatrix}A & B \\ C & D\end{pmatrix}^{-1}=\begin{pmatrix} (A-BD^{-1}C)^{-1} & -A^{-1}B(D-CA^{-1}B)^{-1}\\ -D^{-1}C(A-BD^{-1}C)^{-1} & (D-CA^{-1}B)^{-1} \end{pmatrix}\cdot $$ Let's call $M=\begin{pmatrix}A & B \\ C & D\end{pmatrix}$ and $N$ the matrix on the RHS above. I can verify that $MN=I$ but I am stuck when I try to show that $NM=I$. Similarly, when $M$ is symmetric ($A^T=A$, $C^T=B$, and $D^T=D$), $N$ should also be symmetric but I can't verify that the off-diagonal blocks of $N$ are transposes of one another. Can you please help?

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  • $\begingroup$ Is this really for the pseudoinverse and not the ordinary matrix inverse? The page you linked does not show this formula. $\endgroup$
    – kanso37
    Commented Oct 9, 2020 at 4:56

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The verification of $NM=I$ is more involved, as multiplication is done in the less favorable order.

Let me show only $$ (A-BD^{-1}C)^{-1} A - A^{-1}B(D-CA^{-1}B)^{-1}C = I. $$ Let's go: $$ (A-BD^{-1}C)^{-1} A - I=(A-BD^{-1}C)^{-1} (A-(A-BD^{-1}C)) = (A-BD^{-1}C)^{-1}BD^{-1}C. $$ Let me abbreviate $T:=(A-BD^{-1}C)^{-1} A-I - A^{-1}B(D-CA^{-1}B)^{-1}C$. We want to show $T=0$. Then $$\begin{split} (A-BD^{-1}C)T &= BD^{-1}C - (A-BD^{-1}C)A^{-1}B(D-CA^{-1}B)^{-1}C\\ &=BD^{-1}C -B(D-CA^{-1}B)^{-1}C + BD^{-1}\underbrace{CA^{-1}B(D-CA^{-1}B)^{-1}}_{=D(D-CA^{-1}B)^{-1}-I}C\\ &=0. \end{split}$$ The computation of the other block entries of $NM$ shold be analogous.


Now towards symmetry: Using $A=A^T$, $B=C^T$, $D=D^T$, it remains to show $$ -(A^{-1}B(D-B^TA^{-1}B)^{-1} )^T = -\underbrace{D^{-1}B^T(A-BD^{-1}B^T)^{-1}}_{=:Y}. $$ Since $A$ and $D-B^TA^{-1}B$ are symmetric, we have $$ (A^{-1}B(D-B^TA^{-1}B)^{-1} )^T = (D-B^TA^{-1}B)^{-1} B^TA^{-1}=:X. $$ Then $$\begin{split} X(A-BD^{-1}B^T)&=(D-B^TA^{-1}B)^{-1} B^T - \underbrace{(D-B^TA^{-1}B)^{-1} B^TA^{-1}B}_{=(D-B^TA^{-1}B)^{-1}D-I}D^{-1}B^T \\ &=D^{-1}B^T = Y(A-BD^{-1}B^T) \end{split}$$, hence $X=Y$ and symmetry follows.

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  • $\begingroup$ +1. Really nice. The observations that $\ldots = \ldots - I$ are brilliant! $\endgroup$
    – yurnero
    Commented Aug 11, 2015 at 10:29

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