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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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First isomorphism theorem for vector spaces: V/kerT is isomorphic to imT

Statement

For every linear map T:VU, the formula T~(v+kerT):=T(v) defines a linear isomorphism T~:V/kerTimT.

Facts & Assumptions

Given: A linear map T:VU.

[L1]

A linear map whose kernel contains a subspace W factors uniquely through V/W by v+WT(v) (Universal property of the quotient vector space).

[L2]

The image of a linear map is a linear subspace, and a linear map is injective exactly when its kernel is the zero subspace (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial).

Proof

technique · direct
1.1

Apply [L1] with W=kerT and codomain restricted to imT to obtain the linear map T~(v+kerT)=T(v); it is surjective by the definition of imT.

L1L2
2.1

Its kernel consists of cosets v+kerT with T(v)=0, hence only the zero coset kerT; [L2] makes T~ injective, so it is an isomorphism, including the zero map and the zero-space case.

step 1.1L2

Remarks

For finite-dimensional V, taking dimensions in this isomorphism gives dimV=dimkerT+dimimT, the equality recorded independently as Rank-nullity: dimFV=nullityT+rankT. This is an agreement record, not a premise in the proof above.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 15 results over 7 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