CorollaryStatement: AI-adaptedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2

Statement

For compatible linear maps between finite-dimensional real or complex inner product spaces,

$∥ S T ∥ \leq ∥ S ∥ ∥ T ∥ .$

For every such linear map $T$ ,

$∥ T^{*} T ∥ = ∥ T ∥^{2} .$

Facts & Assumptions

Given: Compatible finite-dimensional linear maps $T$ and $S$ between real or complex inner product spaces.

[L1]

The operator norm is defined by the maximum over unit vectors (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).

[L2]

The operator norm equals the largest singular value (The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector).

[L3]

If $T e_{j} = s_{j} e_{j}^{'}$ is a singular value decomposition, then $T^{*} T e_{j} = s_{j}^{2} e_{j}$ (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

Proof

technique · direct

1.1L1algebra

If the domain of $T$ is zero, then $T$ and $S T$ are zero maps and both norms vanish by [L1]. Otherwise, for every unit vector $v$ in that domain, [L1] gives $∥ S T v ∥ \leq ∥ S ∥ ∥ T v ∥ \leq ∥ S ∥ ∥ T ∥$ . Taking the maximum over all unit $v$ yields $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ .

2.1L1L2L3algebra∎

If the domain of $T$ is zero, then $T^{*} T$ is the zero map on the zero space, so both sides are zero by [L1]. Otherwise, if $s_{1} \geq s_{2} \geq \dots$ are the singular values of $T$ , then [L3] shows that $T^{*} T$ has eigenvalues $s_{1}^{2}, s_{2}^{2}, \dots$ and is already non-negative. Hence its singular values are $s_{1}^{2}, s_{2}^{2}, \dots$ , so [L2] gives $∥ T^{*} T ∥ = s_{1}^{2} = ∥ T ∥^{2}$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)