PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-16

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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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Polynomial evaluation commutes with restriction and invariant quotients

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional vector space and let $W \leq V$ be $T$ -invariant. For every $p \in F [x]$ , $p (T ∣_{W}) = p (T) ∣_{W}, p (\overset{ˉ}{T}) (v + W) = p (T) v + W .$ Consequently, the minimal polynomials of $T ∣_{W}$ and $\overset{ˉ}{T}$ both divide $μ_{T}$ .

Facts & Assumptions

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

[L1]

The induced quotient endomorphism is linear and satisfies $\overset{ˉ}{T} π = π T$ (Invariance makes the induced quotient operator well defined and linear, with $π T = \overset{ˉ}{T} π$ ).

[L2]

Polynomial evaluation is $p (T) = \sum_{k} a_{k} T^{k}$ , with $T^{0} = I$ and only finitely many nonzero coefficients (Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ ).

[L3]

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

[L4]

For an invariant $W$ , $T ∣_{W}$ maps $W$ to itself and $\overset{ˉ}{T} (v + W) = T (v) + W$ (Invariant subspaces, restrictions, and induced quotient operators).

Proof

technique · direct

1.1L1L2L4

Induction on $k$ gives $(T ∣_{W})^{k} = T^{k} ∣_{W}$ and ${\overset{ˉ}{T}}^{k} (v + W) = T^{k} v + W$ , beginning with the identity at $k = 0$ and using [L1] at the successor step; summing with the coefficients of $p$ yields both displayed identities.

2.1step 1.1L2L3L4∎

Taking $p = μ_{T}$ makes $p (T) = 0$ , so step 1.1 gives $p (T ∣_{W}) = 0$ and $p (\overset{ˉ}{T}) = 0$ ; [L3] then gives $μ_{T ∣_{W}} ∣ μ_{T}$ and $μ_{\overset{ˉ}{T}} ∣ μ_{T}$ , including $W = 0$ , $W = V$ , and $V = 0$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

K. Hoffman and R. Kunze, Linear Algebra, 2nd ed., Section 6.4 (standard reference, not scraped)