[All Steps] Weighted least squares. In least squares, the objective (to be minimized) is \|A x-b\|^2=∑_i=1^m(\tildea_i^T x-b_i)^2 where \tildea_i^T i


Question: Weighted least squares. In least squares, the objective (to be minimized) is

\[\|A x-b\|^{2}=\sum_{i=1}^{m}\left(\tilde{a}_{i}^{T} x-b_{i}\right)^{2}\]

where \(\tilde{a}_{i}^{T} i\) are the rows of \(A\), and the \(n\) -vector \(x\) is to chosen. In the weighted least squares problem, we minimize the objective

\[\sum_{i=1}^{m} w_{i}\left(\tilde{a}_{i}^{T} x-b_{i}\right)^{2}\]

where \(w_{i}\) are given positive weights. The weights allow us to assign different weights to the different components of the residual vector.

  1. Show that the weighted least squares objective can be expressed as \(\|D(A x-b)\|^{2}\) for an appropriate diagonal matrix \(D\). This allows us to solve the weighted least squares problem as a standard least squares problem, by minimizing \(\|B x-d\|^{2}\), where \(B=D A\) and \(d=D b\).
  2. Show that when \(A\) has linearly independent columns, so does the matrix \(B\).
  3. The least squares approximate solution is given by \(\hat{x}=\left(A^{T} A\right)^{-1} A^{T} b\). Give a similar formula for the solution of the weighted least squares problem. You might want to use the matrix \(W=\operatorname{diag}(w)\) in your formula.

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