PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Singular values are well defined because the positive square root of T^*T is unique

Statement

For a linear map $T : V \to W$ between finite-dimensional real or complex inner product spaces, the multiset of singular values depends only on $T$ .

Facts & Assumptions

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

[L1]

The singular values of $T$ are defined as the eigenvalues of the unique non-negative square root $∣ T ∣ = \sqrt{T^{*} T}$ (The singular values of a linear map as the eigenvalues of the positive square root of T^*T).

[L2]

A non-negative operator has a unique non-negative square root (A non-negative operator has a unique non-negative square root).

Proof

technique · direct

1.1L1L2

The operator $T^{*} T$ is determined by $T$ , and [L2] gives it a unique non-negative square root. Therefore the operator $∣ T ∣ = \sqrt{T^{*} T}$ in [L1] is determined uniquely by $T$ .

2.1L1step 1.1∎

The singular values are the eigenvalues of that uniquely determined operator $∣ T ∣$ , counted with multiplicity. Hence they depend only on $T$ .

Depends on

Used by

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition Theorem

Dependency tree · two levels

5 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)