Linear Algebra Done Right · Problem 7F.22

Operators on Complex Vector Spaces — Sheldon Axler

Mathematics — Linear Algebra · singular-value-decomposition

Problem

Suppose T∈L(V,W)T \in \mathcal{L}(V, W). Let n=dim⁡Vn = \dim V and let s1≥⋯≥sns_1 \geq \cdots \geq s_n denote the singular values of TT. Prove that if 1≤k≤n1 \leq k \leq n, then min⁡{∥T∣U∥:U is a subspace of V with dim⁡U=k}=sn−k+1.\min\{\|T|_U\| : U \text{ is a subspace of } V \text{ with } \dim U = k\} = s_{n-k+1}.
Intuitive explanation

The key insight is that the operator norm of TT restricted to a kk-dimensional subspace is minimized when that subspace is "aligned" with the smallest singular directions. The singular values order the stretching factors along orthonormal directions: s1s_1 stretches the most, sns_n the least. When you restrict TT to a kk-dimensional subspace, you're choosing kk directions to keep; the worst-case (largest) stretching factor on that subspace is minimized by choosing the kk directions corresponding to the kk smallest singular values. This minimum stretching factor turns out to be exactly sn−k+1s_{n-k+1}, the largest of those kk smallest singular values.

Intuitive explanation

The key insight is that the operator norm of TT restricted to a kk-dimensional subspace is minimized when that subspace is "aligned" with the smallest singular directions. The singular values order the stretching factors along orthonormal directions: s1s_1 stretches the most, sns_n the least. When you restrict TT to a kk-dimensional subspace, you're choosing kk directions to keep; the worst-case (largest) stretching factor on that subspace is minimized by choosing the kk directions corresponding to the kk smallest singular values. This minimum stretching factor turns out to be exactly sn−k+1s_{n-k+1}, the largest of those kk smallest singular values.

Related theory

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