LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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

1.1

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

L1algebra

2.1

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

step 1.1

3.1

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

step 1.1∎

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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 8 results over 4 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

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