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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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    Cool, thanks for the pointers! I'll take a look at those links.2012-08-14

1 Answers 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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    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