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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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The simple-fold complement restriction is an isomorphism

Statement

For the data and notation of Complement splitting and range projection for a simple fold, J∣X:X→Y is a linear isomorphism. The map P=I−qwT satisfies P2=P,range⁡P=Y,ker⁡P=span⁡{q},PJ=J,Pb=0.

Facts & Assumptions

Given: J,v,w,b,a,X,Y,q,P and their hypotheses in Complement splitting and range projection for a simple fold.

[F1]

Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T: for a map between finite-dimensional spaces, dim⁡V=dim⁡ker⁡T+dim⁡im⁡T.

[F2]

In finite dimension, W⊥⊥=W and dim⁡W+dim⁡W⊥=dim⁡V: dim⁡W+dim⁡W⊥=dim⁡V in finite-dimensional inner-product spaces.

[F6]

Rank and nullity of a linear map with finite-dimensional domain: rank of a map with finite-dimensional domain is the dimension of its image.

Proof

1.1F2F5givenalgebra

Since v is unit it is nonzero. If αv=0, then α∥v∥22=0, so α=0; thus {v} is independent, and it spans span⁡{v} by definition. It is therefore a basis and that span has dimension one by [F5]. The orthogonal-complement formula [F2] gives dim⁡X=n−1.

1.2F1F4givenalgebra

Since wTq=1, the map wT is onto R. Rank-nullity [F1] and dim⁡Rn=n [F4] give dim⁡Y=n−1; [F4] also gives dim⁡Rn=n.

1.3givenalgebra

Since wTq=1, direct multiplication gives P2=(I−qwT)2=I−qwT=P. Also wTP=0, so range⁡P⊆Y, while Py=y for every y∈Y; hence range⁡P=Y. The equation Pz=0 is equivalent to z=q(wTz), so ker⁡P=span⁡{q}.

2.1F1F3F6step 1.1step 1.2givenalgebra

From wTJ=0 we have im⁡J⊆Y, and J has rank n−1 by [F6]. Thus dim⁡im⁡J=dim⁡Y, so [F3] gives im⁡J=Y. Moreover, if x∈X∩ker⁡J, then x=αv and 0=vTx=α, so J∣X is injective by [F3]. Its image has dimension n−1 by [F1] and therefore equals Y by [F3]. This proves that J∣X is an isomorphism.

3.1givenstep 1.3algebra∎

Finally PJ=J−q(wTJ)=J, and Pb=b−q(wTb)=b−qa=0 by the definitions.

Depends on

Used by

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Sources