LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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.

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For every polynomial $p$ , both $\ker p (T)$ and $im p (T)$ are $T$ -invariant

Statement

Let $T : V \to V$ be an endomorphism and $p \in F [x]$ . Both $\ker p (T)$ and $im p (T)$ are invariant under $T$ .

Facts & Assumptions

Given: An endomorphism $T$ and a polynomial $p \in F [x]$ .

[L1]

Polynomial evaluation is $p (T) = \sum a_{k} T^{k}$ , with $T^{0} = I$ and $T^{k + 1} = T \circ T^{k}$ (Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ ).

Proof

technique · direct

1.1L1algebra

By [L1] and associativity of composition, $T p (T) = p (T) T$ .

2.1step 1.1

If $v \in \ker p (T)$ , then $p (T) (T v) = T (p (T) v) = 0$ , so $T v \in \ker p (T)$ .

3.1step 1.1∎

If $v \in im p (T)$ , write $v = p (T) u$ ; then $T v = p (T) (T u)$ by step 1.1, so $T v \in im p (T)$ .

Depends on

Polynomial evaluation at an endomorphism: $p(T)=\sum_k a_kT^k$

Used by

Primary components ker q(T)ᵉ and generalised eigenspaces G_λ⁽ᵉ⁾(T)=ker(T-λ I)ᵉ Definition

Dependency tree · two levels

4 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

Anthony W. Knapp, Basic Algebra, 2nd ed., Ch. V, §5 (standard reference, not scraped)

For every polynomial pp, both ker⁡p(T)kerp(T) and im⁡p(T)imp(T) are TT-invariant