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Given that I have two matrix $c$ and $C$, which are $m \times n$ and $m \times m$ respectively with $m << n$, is there an efficient method for finding $(c'Cc)^{-1}$ that does not involve evaluating the inverse of the full $n \times n$ matrix.

If instead I was instead evaluating $(A+c'Cc)^{-1}$ it is possible to use the Woodbury identity which if $A$ is diagonal only requires finding the inverse for the $m \times m$ matrix.

I feel like I am missing something obvious since the more complex case appears to have lower computational complexity.

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up vote 1 down vote accepted

In fact, there is a very easy answer: $c^T C c$ is not invertible. One way to see this is the fact it's rank is at most $m$, and therefore it doesn't have full rank.

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That makes a lot of sense, thank you. –  rundel Dec 4 '12 at 21:14
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