Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-11
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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.

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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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End⁡F(V) is a ring and matrix representation is a ring isomorphism End⁡F(V)≅Mn(F)

Statement

Let V be an n-dimensional vector space over F. Then End⁡F(V):=L(V,V) is a ring under pointwise addition and composition, and for every ordered basis B the map

T⟼[T]BB

is a ring isomorphism End⁡F(V)≅Mn(F).

Facts & Assumptions

Given: A finite-dimensional F-vector space V and an ordered basis B of length n.

[L1]

L(V,V) is a vector space under pointwise operations (L(V,W) is a vector space over the common scalar field).

[L2]

Matrix representation is a vector-space isomorphism and sends composition to matrix multiplication (T↦[T]BC is a vector-space isomorphism L(V,W)≅Mm×n(F), [S∘T]BD=[S]CD[T]BC).

Proof

technique · direct
1.1

Composition of endomorphisms is associative, has id⁡V as identity, and distributes over pointwise addition; together with the additive group from [L1], this makes End⁡F(V) a ring.

givenL1
2.1

By [L2], matrix representation is a bijective linear map, so it preserves addition and zero.

step 1.1L1L2
3.1

It preserves products by the composition formula in [L2], and [id⁡V]BB=In by coordinate action. Thus it is a bijective unital ring homomorphism and hence a ring isomorphism.

step 2.1L2∎

Depends on

Used by

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Dependency tree · two levels

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Sources