CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck 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 $\dim \ker (T - λ I)$ is the number of those blocks, and the exponent of $x - λ$ in $μ_{T}$ is the size of the largest such block. Thus $χ_{T} = \prod_{λ} (x - λ)^{\sum_{j} m_{λ, j}}, μ_{T} = \prod_{λ} (x - λ)^{\max_{j} m_{λ, j}} .$ 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 $m_{λ, j}$ .

[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 $\ker (T - λ I)$ contains a nonzero vector, and that kernel is its eigenspace (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism).

[L3]

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

Proof

technique · direct

1.1L1L2algebra

A block $J_{m} (λ)$ contributes $m$ copies of $x - λ$ to the characteristic polynomial and one independent initial vector to $\ker (T - λ I)$ ; blocks at other eigenvalues contribute no kernel because their shifted blocks are invertible. This proves the algebraic- and geometric-multiplicity claims.

1.2L3algebra

On $J_{m} (λ)$ , $(T - λ I)^{a}$ vanishes exactly when $a \geq m$ . If $p = (x - λ)^{a} q$ with $q (λ) \neq 0$ , then $q (J_{m} (λ))$ is invertible by a finite geometric-series inverse for its nonzero scalar part; hence $p$ annihilates that block exactly when $a \geq m$ . Applying this to every block and using [L3] gives the displayed minimal polynomial.

2.1step 1.1step 1.2L3∎

Multiplying the block contributions gives the characteristic-polynomial formula, and the empty block list gives $χ_{T} = μ_{T} = 1$ on the zero space.

Depends on

Used by

Recovering a 5×5 Jordan form from shifted power ranks at two eigenvalues Example
FALSE: Equal characteristic and minimal polynomials imply similarity False statement
FALSE: Geometric multiplicity alone determines Jordan block sizes False statement

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

S. Axler, Linear Algebra Done Right, 4th ed., Section 8C (standard reference, not scraped)