Exercise · ZW-LINEAR-LEAST-SQUARES-009 · ★★☆ Intermediate
Why is it not advisable to compute the least squares solution as \(x=(A^\top A)^{-1}A^\top b\) when \(A^\top A\) is ill-conditioned?
Choose one option.
A Because \(A^\top b\) stops existing B Because the problem stops being convex C Because it forms \(A^\top A\) and squares the condition number D Because the inverse is always zero Show answer C. Because it forms \(A^\top A\) and squares the condition number← Back to the question
Explanation Forming \(A^\top A\) squares the condition number, which amplifies numerical errors. The convexity option is false, since the problem remains convex.
Optimization problem that minimizes the sum of squared residuals \(\|Ax-b\|_2^2\) to approximate a vector \(b\) by linear combinations of the columns of \(A\).
\[\min_{x}\ \|Ax-b\|_2^2 \quad \Rightarrow \quad A^\top A\,x = A^\top b\]
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