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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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For a positive-dimensional finite-dimensional operator, Tadj(T)=adj(T)T=det(T)I

Statement

Let T:VV be a linear operator on a finite-dimensional vector space over a field, with dimV1. Then

Tadj(T)=adj(T)T=det(T)IV.

Facts & Assumptions

Given: T,V as in the statement and an ordered basis B.

[L1]

For a positive-sized square matrix A, Aadj(A)=adj(A)A=det(A)I (For every positive-sized square matrix over a commutative ring, Aadj(A)=adj(A)A=det(A)I).

[L2]

The matrix of a composite is the product of the representing matrices ([ST]BD=[S]CD[T]BC).

Proof

technique · direct
1.1

Put A=[T]B. By [F1] and [L2], the matrices of Tadj(T) and adj(T)T are respectively Aadj(A) and adj(A)A.

F1L2
2.1

By [L1], both matrices in step 1.1 equal det(A)In, which equals det(T)In by [L3].

step 1.1L1L3
3.1

Equality of representing matrices gives both asserted operator identities.

step 2.1L4

Depends on

Used by

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Dependency tree · next 3 levels

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Sources