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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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  • 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.

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Tensor unfoldings preserve the Frobenius inner product

Statement

For tensors A,B of the same finite shape, every mode unfolding and every cut unfolding preserves the Frobenius inner product and norm.

Facts & Assumptions

Given: A field F∈{R,C}, an order d≥2, and tensors A,B∈Fn1×⋯×nd.

[F1]

Finite tensor arrays and the Frobenius inner product: each unfolding uses a fixed lexicographic ordering of the same tensor indices, and the Frobenius inner product sums conjugate-entry times entry over all indices.

Proof

1.1givenF1construct

For a mode-k unfolding, the map from a multi-index (i1,…,id) to the matrix position consisting of row ik and the ordered tuple of all other indices is a bijection: the row and column tuple recover every original index exactly once. Thus the entrywise sum defining ⟨A(k),B(k)⟩F is precisely the same finite sum as ⟨A,B⟩F.

2.1givenF1algebra∎

For a cut-k unfolding, splitting the tuple into (i1,…,ik) and (ik+1,…,id) is also a bijection with the original multi-index set. Its entrywise inner-product sum therefore equals ⟨A,B⟩F as well. Taking B=A and nonnegative square roots proves each norm identity.

Depends on

Used by

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Sources