4
$\begingroup$

If I have a large (e.g. 6000x6000), sparse, positive definite matrix $M$ (which may have individual entries everywhere, but most non-zero entries are on / around the diagional). Divide $M$ into blocks (with $A$, $D$, $F$ square):

$M = \begin{pmatrix}A&B&C\\B^T&D&E\\C^T&E^T&F\end{pmatrix}$

typically, $A$ is medium-sized (e.g. 200x200), $D$ is very small (e.g. 6x6), and $F$ is very large (e.g. 5794x5794).

Now if I already have the Schur complement of the large block $F$

$S_F = \begin{pmatrix}A'&B'\\B'^T&D'\end{pmatrix} = \begin{pmatrix}A&B\\B^T&D\end{pmatrix} - \begin{pmatrix}C\\E\end{pmatrix} F^{-1} \begin{pmatrix}C^T&E^T\end{pmatrix}$

I can, with little required computations, include more rows / columns in the complement, e.g.

$S_{DF} = A' - B' D'^{-1}B'^T$ ( = complement of $\begin{pmatrix}D&E\\E^T&F\end{pmatrix}$ in $M$)

Is it possible to (efficiently) do the same thing the other way round, that is, compute $S_F$ from $M$ and $S_{DF}$, without having to invert $F$ again?

($M$ comes from pose-graph optimization)

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.