I was reading through a least squares text book and I am curious to why this is true. Let A ∈ R m×n have rank(A) = r, which has full rank. Show that $||A^{+}||_{2}$ = $1/σ_{r}$. $σ_{r}$ is the smallest singular value of A.
2026-02-22 21:43:46.1771796626
Solution to least squares problem
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