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I am solving an overdetermined system of equations: $$Ax= b$$

Using QR factorization, we can solve this system easily by posing it as:

$$Rx= Q'b$$

I would like to regularize my estimate of $x$. I can use ridge regression with the normal equations being:

$$(A'A+\lambda\Gamma'\Gamma)x=A'b$$

My question is, how can I directly transform $A$ to matrix $\Delta$ such that solving the system $\Delta x=b$ via QR factorization yields the regularized estimate for $x$.

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  • $\begingroup$ What do you mean "directly transform $A$ to $\Delta$"? The relation is $\Delta=A'A+ \lambda \Gamma' \Gamma$. Do you mean how to obtain the QR factorization of $\Delta$ from that of $A$? $\endgroup$
    – Manos
    Commented Feb 10, 2013 at 17:12
  • $\begingroup$ I voted you up here as you gave a name to my own personal "discovery" and can now study some literature on the subject. That being said, I am not sure that I understand your question. But I write an answer I think helps... $\endgroup$
    – adam W
    Commented Feb 10, 2013 at 17:25
  • $\begingroup$ Yes I meant QR factorization of $\Delta$ from $A$. Poster Adam W has guided me in the correct direction. Thanks. $\endgroup$
    – Atlas
    Commented Feb 11, 2013 at 7:42

1 Answer 1

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Try writing it as (using $\lambda = 1$ for display simplicity) $$B = \pmatrix{ A \\ \Gamma}$$ then you have for the equation $Bx = \pmatrix{b \\ 0}$

\begin{align} B'Bx &= B' \pmatrix{b \\ 0} \\ \pmatrix{ A' & \Gamma'}\pmatrix{ A \\ \Gamma}x &= \pmatrix{ A' & \Gamma'} \pmatrix{b \\ 0} \\ (A'A + \Gamma'\Gamma)x &= A'b \end{align}

Therefore in terms of the QR factorization, take the QR factors of $$B = \pmatrix{ A \\ \Gamma}$$

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