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Let us assume we have a symmetric $n \times n$ matrix $A$. We know the inverse of $A$. Let us say that we now add one column and one row to $A$, in a way that the resulting matrix ($B$) is an $(n+1) \times (n+1)$ matrix that is still symmetric.

For instance,

$A = \begin{pmatrix}a & b \\b & d \\\end{pmatrix}$

and

$B = \begin{pmatrix}a & b & X \\b & d & Y \\X & Y & Z\end{pmatrix}$

Given that I know $A^{-1}$, is there any way of using this information to find $B^{-1}$ without having to compute this latter inverse from scratch? If an exact solution is not possible, approximations would also help.

Thanks,
Bruno

P.S. in case it makes any difference, both $A$ and $B$ are covariance matrices.

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    $\begingroup$ The Schur Complement might be what you are looking for. en.wikipedia.org/wiki/Schur_complement $\endgroup$
    – copper.hat
    Commented Aug 14, 2012 at 3:34
  • $\begingroup$ I don't know if there exists a solution that just involves $A^{-1}$, but I have seen something similar. $\endgroup$
    – Tunococ
    Commented Aug 14, 2012 at 3:36
  • $\begingroup$ Cool, thanks for the pointers! I'll take a look at those links. $\endgroup$
    – Bruno
    Commented Aug 14, 2012 at 14:09

1 Answer 1

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Certainly, one can use the bordering method for this (a special case of the usual formula for block inversion):

$$\begin{pmatrix}\mathbf A&\mathbf \delta\\\mathbf \delta^\top&Z\end{pmatrix}^{-1}=\begin{pmatrix}\mathbf A^{-1}+\frac{\mathbf A^{-1}\mathbf \delta\mathbf \delta^\top\mathbf A^{-1}}{\mu}&-\frac{\mathbf A^{-1}\mathbf \delta}{\mu}\\-\frac{\mathbf \delta^\top\mathbf A^{-1}}{\mu}&\frac1{\mu}\end{pmatrix}$$

where $\mathbf \delta^\top=(X\quad Y)$ and $\mu=Z-\mathbf \delta^\top\mathbf A^{-1}\mathbf \delta$.

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    $\begingroup$ $\mu$ is the Schur Complement of $A$. $\endgroup$
    – copper.hat
    Commented Aug 14, 2012 at 3:48
  • $\begingroup$ @copper, funny, we seem to have dealt with the determinantal version of this question before, haven't we? $\endgroup$ Commented Aug 14, 2012 at 3:59
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    $\begingroup$ Schur have!!! For some reason I always find rank updates and block inverse formulae particularly fascinating. (I try to search for previous stuff like this, but rarely have much success finding it, and wading through my comments is very tedious.) $\endgroup$
    – copper.hat
    Commented Aug 14, 2012 at 4:13
  • $\begingroup$ "I always find rank updates and block inverse formulae particularly fascinating." - you and me both. It's a damn shame things like Schur complements and the Sherman-Morrison-Woodbury formula are not more well-known. $\endgroup$ Commented Aug 14, 2012 at 4:17
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    $\begingroup$ Holy chicken! HE'S ALIVE!!!! ALIVEEEEEE!!!!! $\endgroup$
    – Asaf Karagila
    Commented May 1, 2015 at 17:49

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