CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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$\operatorname{End}_F(V)$ is a ring and matrix representation is a ring isomorphism $\operatorname{End}_F(V)\cong M_n(F)$

Statement

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

$T\longmapsto[T]_{\mathcal B}^{\mathcal B}$

is a ring isomorphism $\operatorname{End}_F(V)\cong M_n(F)$ .

Facts & Assumptions

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

[L1]

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

[L2]

Matrix representation is a vector-space isomorphism and sends composition to matrix multiplication ( $T\mapsto[T]_{\mathcal B}^{\mathcal C}$ is a vector-space isomorphism $\mathcal L(V,W)\cong M_{m\times n}(F)$ , $[S\circ T]_{\mathcal B}^{\mathcal D}=[S]_{\mathcal C}^{\mathcal D}[T]_{\mathcal B}^{\mathcal C}$ ).

Proof

technique · direct

1.1

Composition of endomorphisms is associative, has $\operatorname{id}_V$ as identity, and distributes over pointwise addition; together with the additive group from [L1], this makes $\operatorname{End}_F(V)$ 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 $[\operatorname{id}_V]_{\mathcal B}^{\mathcal B}=I_n$ 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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Sources

S. Axler, Linear Algebra Done Right, 4th ed., §3C (standard reference, not scraped)

End⁡F(V)\operatorname{End}_F(V)EndF​(V) is a ring and matrix representation is a ring isomorphism End⁡F(V)≅Mn(F)\operatorname{End}_F(V)\cong M_n(F)EndF​(V)≅Mn​(F)