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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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Every idempotent endomorphism is diagonalisable and is projection onto its image along its kernel

Statement

If an endomorphism T:VV of a finite-dimensional vector space satisfies T2=T, then it is diagonalisable and

V=imTkerT.

Under this decomposition, T is projection onto imT along kerT.

Facts & Assumptions

Given: An idempotent endomorphism T, so T2=T.

[L1]

For an endomorphism of a finite-dimensional space, a polynomial annihilates T exactly when it is divisible by μT (The annihilator ideal is nonzero and has a unique monic generator; p(T)=0 if and only if μTp).

[L2]

For an endomorphism of a finite-dimensional space, having a product of distinct linear factors as minimal polynomial is equivalent to diagonalisability (An endomorphism is diagonalisable if and only if its minimal polynomial is a product of distinct linear factors).

[L4]

The polynomial ring over a field is a unique factorisation domain (For every field F, F[x] is a unique factorisation domain).

Proof

technique · direct
1.1

The identity T2T=0 says x(x1) annihilates T, so [L1] gives μTx(x1). By unique factorisation [L4], this monic divisor is a product of a subset of the two distinct irreducibles x and x1; [L2] therefore makes T diagonalisable.

L1L2L4algebra
1.2

Every vV has v=Tv+(vTv), where TvimT and T(vTv)=TvT2v=0. Hence the image and kernel span V.

givenalgebra
2.1

If wimTkerT, write w=Tu and compute w=Tu=T2u=Tw=0. Thus [L3] and step 1.2 give the direct sum, and T is identity on its image and zero on its kernel. The cases T=0 and T=I are included.

step 1.2L3algebra

Depends on

Used by

Dependency tree · next 3 levels

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