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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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Trace Norm Spectral Formula

Statement

For every linear map A:H→K (Linear map between vector spaces over the same field) between finite-dimensional complex inner-product spaces,

∥A∥1=tr⁡H ⁣(A∗A).

Here A∗A and its non-negative square root are endomorphisms of the domain H; the trace is taken on H.

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces H,K and a linear map A:H→K.

[F1]

The singular values of A are the eigenvalues of ∣A∣=A∗A, counted with multiplicity (The singular values of a linear map as the eigenvalues of the positive square root of T^*T).

[F3]

The trace of an endomorphism is the sum of its diagonal entries in any ordered basis, and the trace on the zero space is 0 (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

[F4]

The trace norm is the sum of the singular values on the domain, including zeros (Trace Norm).

Proof

1.1givenF1F2F3choose

The operator ∣A∣=A∗A in [F1] is non-negative. By [F2], choose an orthonormal eigenbasis of H; its eigenvalues, with multiplicity, are exactly the singular values of A by [F1]. In that basis [F3] gives tr⁡H(∣A∣)=∑j<dim⁡Hsj(A).

2.1F3F4step 1.1∎

The sum in step 1.1 is ∥A∥1 by [F4]. If H=0, both are the empty sum 0 by [F3]–[F4]. This proves the formula for rectangular maps and all finite dimensions.

Depends on

Used by

Cited to discharge well-definedness by Trace Norm.

Dependency tree · two levels

19 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