Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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.

If W is T-invariant, then W⊥ is T∗-invariant

Statement

Let T be an endomorphism of a finite-dimensional inner product space and let W be T-invariant. Then W⊥ is T∗-invariant.

Facts & Assumptions

Given: An endomorphism T, a T-invariant subspace W, a vector v∈W⊥, and w∈W.

[L1]

The adjoint identity says ⟨Tx,y⟩=⟨x,T∗y⟩ for all x,y (The adjoint T∗:W→V is characterised by ⟨Tv,w⟩W=⟨v,T∗w⟩V).

[L2]

A vector belongs to W⊥ exactly when it pairs to zero with every vector of W (The orthogonal complement W⊥={v:⟨v,w⟩=0 for all w∈W}).

Proof

technique · direct
1.1

Since W is T-invariant, Tw∈W. As v∈W⊥, [L2] and conjugate symmetry give ⟨Tw,v⟩=⟨v,Tw⟩‾=0.

givenL2
2.1

By [L1] and conjugate symmetry, ⟨T∗v,w⟩=⟨w,T∗v⟩‾=⟨Tw,v⟩‾=0. This holds for every w∈W, so [L2] gives T∗v∈W⊥.

step 1.1L1L2∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 15 results over 7 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources