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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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    The Schur Complement might be what you are looking for. http://en.wikipedia.org/wiki/Schur_complement2012-08-14
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    I don't know if there exists a solution that just involves $A^{-1}$, but I have seen [something similar](http://en.wikipedia.org/wiki/Woodbury_matrix_identity).2012-08-14
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    Cool, thanks for the pointers! I'll take a look at those links.2012-08-14

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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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    $\mu$ is the Schur Complement of $A$.2012-08-14
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    @copper, funny, we seem to have dealt with the determinantal version of this question before, haven't we?2012-08-14
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    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.)2012-08-14
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    "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.2012-08-14
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    True! the first times I encountered them explicitly was in a control theory class, next was as part of a Quasi-Newton update. It's probably a facet of my age, bit I think current education misses many basics.2012-08-14
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    Thanks a lot, guys. That helps a lot!2012-08-14
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    The lower left element of the RHS has to be transposed, otherwise dimension do not match for dim>1. I proposed an edit, but someone voted against it. Also compare with the first source posted by @J-M. There, $v_{k}$ is defined as row vector and left multiplied to $\mathbf{A}^{-1}$.2015-04-21
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    @Marco, fixed; thanks.2015-05-01
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    Holy chicken! HE'S ALIVE!!!! ALIVEEEEEE!!!!!2015-05-01
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    @Asaf, apparently so. :)2015-05-01
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    And just the other day I was wondering about you. Maybe in a couple of weeks Theo will come back as well. Who knows... :-) It's good to have you back, buddy.2015-05-01
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    @Asaf, anything is possible. I know you don't chat, so I don't have a way to drill you on what I've missed. :P2015-05-01
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    Not much, but also plenty. You should probably know by now that any catching up with people over several years is bound to be insanely broad strokes, and you stumble upon the details at later and later points... It's just good to have you back, anyway.2015-05-01