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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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