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PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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Polynomial evaluation commutes with restriction and invariant quotients

Statement

Let T:VV be an endomorphism of a finite-dimensional vector space and let WV be T-invariant. For every pF[x], p(TW)=p(T)W,p(Tˉ)(v+W)=p(T)v+W. Consequently, the minimal polynomials of TW and Tˉ both divide μT.

Facts & Assumptions

Given: A finite-dimensional F-vector space V, an endomorphism T, a T-invariant subspace W, and pF[x].

[L1]

The induced quotient endomorphism is linear and satisfies Tˉπ=πT (Invariance makes the induced quotient operator well defined and linear, with πT=Tˉπ).

[L2]

Polynomial evaluation is p(T)=kakTk, with T0=I and only finitely many nonzero coefficients (Polynomial evaluation at an endomorphism: p(T)=kakTk).

[L3]

For a finite-dimensional endomorphism S, q(S)=0 exactly when μS divides q, and on the zero space μS=1 (The annihilator ideal is nonzero and has a unique monic generator; p(T)=0 if and only if μTp).

[L4]

For an invariant W, TW maps W to itself and Tˉ(v+W)=T(v)+W (Invariant subspaces, restrictions, and induced quotient operators).

Proof

technique · direct
1.1

Induction on k gives (TW)k=TkW and Tˉk(v+W)=Tkv+W, beginning with the identity at k=0 and using [L1] at the successor step; summing with the coefficients of p yields both displayed identities.

L1L2L4
2.1

Taking p=μT makes p(T)=0, so step 1.1 gives p(TW)=0 and p(Tˉ)=0; [L3] then gives μTWμT and μTˉμT, including W=0, W=V, and V=0.

step 1.1L2L3L4

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