Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28
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The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector

Statement

Let T:V→W be a linear map between finite-dimensional real or complex inner product spaces.

If V=0, then

∥T∥=0.

If V≠0 and s1 is the largest singular value of T, then

∥T∥=s1.

In the nonzero case, the maximum in The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| is attained at a right-singular vector for s1.

Facts & Assumptions

Given: A linear map T:V→W between finite-dimensional real or complex inner product spaces.

[L1]

There is a singular value decomposition Tv=∑j=1rsj⟨v,ej⟩fj with s1≥⋯≥sr>0 (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

Proof

technique · direct
1.1L1algebra

If V=0, then The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| gives ∥T∥=0. Assume now that V≠0, and choose an SVD from [L1]: Tv=∑j=1rσj⟨v,ej⟩fj,σ1≥⋯≥σr>0. Extend this list by zeros, setting σr+1=⋯=σn=0; then the displayed formula gives Tej=σjfj for every j, so σ1 is the largest singular value of T. Writing a unit vector as v=∑jcjej, one gets ∥Tv∥2=∑j=1nσj2∣cj∣2≤σ12∑j∣cj∣2=σ12, so every unit vector satisfies ∥Tv∥≤σ1.

2.1L1step 1.1∎

For the right-singular vector e1, step 1.1 gives ∥Te1∥=∥σ1f1∥=σ1. Therefore the maximum defining ∥T∥ equals the largest singular value σ1, and it is attained at e1 when V≠0.

Depends on

Used by

Dependency tree · two levels

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Sources