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I hope you can help me out with this. I have to find the solution x to an inverse system

$$ x=A^{-1}b $$

This inverse problem is basically a least square problem with a rank-1 update.

$$ x=[uv^{T}+H^{H}H]^{-1}H^{H}b $$

There are two objectives, one can be achieved with the least square problem and the other with the rank-1 update. I need to find the best tradeoff between the two.

The least square solution (without the rank-1 update) needs to be regularized, so the smallest singular values of the matrix $[H^{H}H]$ would need to be replaced by zeroes.

Normally, I should introduce a parameter (real positive scalar) in front of the rank-1 update $uv^{T}$ and hand-tune it until the solution leads to the best compromise between the two objectives. Nevertheless, I am thinking: could I just do a singular value decomposition of the matrix $[H^{H}H]$ and replace one of my useless elements (one that will be replaced by zero after regularization) by my rank-1 update?* Wouldn't that save me from having to include a parameter that has to be tuned? Once I have my new matrix A, I think the solution could be easily regularized by applying Selective Singular Value Decomposition...

  • The reason why I think this might be done is that there is only one non-zero singular value in $uv^{T}$ and, at the same time, some (the smallest) singular values of $[H^{H}H]$ will be rejected, so why not create a new matrix A such that one of its svd components is the (non-zero) $svd(uv^{T})$?

Thanks in advance

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  • $\begingroup$ ask at scicomp se $\endgroup$
    – guest
    Apr 21, 2014 at 16:16
  • $\begingroup$ I do not get what your tradeoffs are exactly... Also, what is a rank-one decomposition? $\endgroup$
    – Dirk
    Apr 21, 2014 at 18:34
  • $\begingroup$ I edited the question. I hope it is clearer. There are two objectives. The error associated to one of them is minimized with the least square solution. The error associated to the other objective is minimized by selecting a solution x equal to the eigenvector corresponding to the non-zero eigenvalue of the rank-1 update. $\endgroup$
    – user49843
    Apr 21, 2014 at 20:26

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