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I was studying the weighted least squares algorithm and came across this formula for calculating the weighted result in terms of the original.

Here $x_{LS}$ is the solution for $D=I$ Goloub, G. & Van Loan, C. Matrix computations pp. 265

I can't figure out the proof for this formula. I see that $x_D=(A^TD^2A)^{-1}A^TD^2b$ but haven't been able to derive the formula from the book. Does anyone know a demonstration?

Arthur
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1 Answers1

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Let $$u=\arg\min_x\|Ax-b\|^2,\qquad v=\arg\min_x\|DAx-Db\|^2.$$

It follows that $$A^*(Au-b)=0,\qquad (DA)^*(DAv-Db)=0.$$

Therefore, $$\begin{eqnarray*}(A^*D^2A)^{-1}A^*(D^2-I)(b-Au)&=&(A^*D^2A)^{-1}(A^*D^2(b-Au)-A^*(b-Au))\\&=&(A^*D^2A)^{-1}(A^*D^2(b-Au))\\&=& (A^*D^2A)^{-1}(A^*D^2b-A^*D^2Au)\\&=&(A^*D^2A)^{-1}(DA)^*(Db)-u\\&=&(A^*D^2A)^{-1}(DA)^*(DAv+(Db-DAv))-u\\&=&v-u.\end{eqnarray*}$$

You may find interesting discussions about related results such as Why are the four fundamental subspaces fundamental?, searching for "\(A^*(Au-b)=0\)" on SearchOnMath.

José C Ferreira
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