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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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For split operators, Jordan blocks read off eigenspace multiplicities and both canonical polynomials

Statement

Let T be an endomorphism whose characteristic polynomial splits. For each eigenvalue λ, its algebraic multiplicity is the sum of the sizes of the λ-Jordan blocks, its geometric multiplicity dimker(TλI) is the number of those blocks, and the exponent of xλ in μT is the size of the largest such block. Thus χT=λ(xλ)jmλ,j,μT=λ(xλ)maxjmλ,j. For V=0, both products are empty and equal 1.

Facts & Assumptions

Given: A finite-dimensional endomorphism T with split characteristic polynomial and Jordan block sizes mλ,j.

[L1]

The Jordan block multiset is determined up to order (Ranks of shifted powers determine Jordan form up to block order).

[L2]

A scalar λ is an eigenvalue when ker(TλI) contains a nonzero vector, and that kernel is its eigenspace (Eigenvalues, eigenvectors, eigenspaces Eλ(T)=ker(TλI), and the spectrum σF(T) of an endomorphism).

[L3]

A polynomial annihilates T exactly when it is divisible by μT, with μT=1 on the zero space (The annihilator ideal is nonzero and has a unique monic generator; p(T)=0 if and only if μTp).

Proof

technique · direct
1.1

A block Jm(λ) contributes m copies of xλ to the characteristic polynomial and one independent initial vector to ker(TλI); blocks at other eigenvalues contribute no kernel because their shifted blocks are invertible. This proves the algebraic- and geometric-multiplicity claims.

L1L2algebra
1.2

On Jm(λ), (TλI)a vanishes exactly when am. If p=(xλ)aq with q(λ)0, then q(Jm(λ)) is invertible by a finite geometric-series inverse for its nonzero scalar part; hence p annihilates that block exactly when am. Applying this to every block and using [L3] gives the displayed minimal polynomial.

L3algebra
2.1

Multiplying the block contributions gives the characteristic-polynomial formula, and the empty block list gives χT=μT=1 on the zero space.

step 1.1step 1.2L3

Depends on

Used by

Dependency tree · next 3 levels

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