PropositionStatement: 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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Invariance makes the induced quotient operator well defined and linear, with $π T = \overset{ˉ}{T} π$

Statement

Let $T : V \to V$ be linear and let $W \leq V$ be $T$ -invariant. Then $\overset{ˉ}{T} (v + W) : = T (v) + W$ is a well-defined linear endomorphism of $V / W$ , and the canonical projection satisfies $π \circ T = \overset{ˉ}{T} \circ π .$

Facts & Assumptions

Given: An endomorphism $T : V \to V$ and a $T$ -invariant subspace $W \leq V$ .

[L1]

$T$ -invariance means $T (W) \subseteq W$ , and the proposed induced map is $\overset{ˉ}{T} (v + W) = T (v) + W$ (Invariant subspaces, restrictions, and induced quotient operators).

[L2]

$v + W = v^{'} + W$ exactly when $v - v^{'} \in W$ ; the operations on $V / W$ are independent of the chosen representatives and make $V / W$ a vector space over $F$ ; and $π : V \to V / W$ is a surjective linear map with $\ker π = W$ (Coset equality, well-defined quotient operations, and the canonical projection with kernel $W$ ).

[L3]

The quotient operations are $(v + W) + (u + W) : = (v + u) + W$ and $a (v + W) : = (a v) + W$ , and the canonical projection is $π (v) : = v + W$ (The quotient vector space $V / W$ and its canonical projection).

Proof

technique · direct

1.1L1L2

If $v + W = v^{'} + W$ , then $v - v^{'} \in W$ , so $T (v) - T (v^{'}) = T (v - v^{'}) \in T (W) \subseteq W$ and therefore $T (v) + W = T (v^{'}) + W$ ; thus $\overset{ˉ}{T}$ is well defined.

2.1step 1.1L1L2L3∎

For scalars $a, b$ , the operations of [L3] give $a (v + W) + b (u + W) = (a v + b u) + W$ , so by [L1] and the linearity of $T$ , $\overset{ˉ}{T} (a (v + W) + b (u + W)) = T (a v + b u) + W = (a T (v) + b T (u)) + W = a (T (v) + W) + b (T (u) + W)$ , and $\overset{ˉ}{T}$ is linear; also $\overset{ˉ}{T} (π (v)) = \overset{ˉ}{T} (v + W) = T (v) + W = π (T (v))$ proves $\overset{ˉ}{T} π = π T$ ; the same calculation covers $W = 0$ , $W = V$ , and $V = 0$ .

Depends on

Used by

Cited to discharge well-definedness by Invariant subspaces, restrictions, and induced quotient operators.

Dependency tree · two levels

6 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

Cornell Math 4330, Quotient Spaces, Exercise QuoSpace 7 (standard reference, not scraped)

Invariance makes the induced quotient operator well defined and linear, with πT=TˉππT=Tˉπ