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Let $S\in\mathbb{R}^{N\times N }$ be an orthogonal matrix and denote $S_{N,N-1}\in\mathbb{R}^{N \times N-1}$ as the matrix with the same elements of $S$ but without the last column of $S$.

Let $A\in\mathbb{R}^{N\times N}$ be of full rank.

I am searching for a representation of the Inverse of $K:=S_{N,N-1}'AS_{N,N-1}$

Tommi
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Stefan Voigt
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1 Answers1

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There is a lemma posted at https://math.stackexchange.com/a/17780/240529:

Lemma. If $A$ and $A+B$ are invertible, and $B$ has rank $1$, then let $g=\mathrm{trace}(BA^{-1})$. Then $g\neq -1$ and $$(A+B)^{-1} = A^{-1} - \frac{1}{1+g}A^{-1}BA^{-1}.$$

Maybe you can rewrite $S'_{N,N-1}AS_{N,N-1}$ as a sum and use the lemma?

Community
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dsforecast
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  • Thank you for your advice. However, even if I'd be able to decompose $A$ into a sum $A=A_1+A_2$ I'd stuck with the same problem as I would be required to compute $(S'A_1S)^{-1}$. On the other hand side I am not aware of a matrix decomposition $S'AS=A+B$. – Stefan Voigt May 20 '15 at 17:32