False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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FALSE: Geometric multiplicity alone determines Jordan block sizes

Statement

False claim. The geometric multiplicity of an eigenvalue determines all Jordan block sizes at that eigenvalue.

Facts & Assumptions

Given: For any $λ \in F$ , $A = J_{3} (λ) \oplus J_{1} (λ), B = J_{2} (λ) \oplus J_{2} (λ) .$

[L1]

The geometric multiplicity is the number of Jordan blocks for the eigenvalue (For split operators, Jordan blocks read off eigenspace multiplicities and both canonical polynomials).

[L2]

Ranks of shifted powers determine every Jordan block size (Ranks of shifted powers determine Jordan form up to block order).

Refutation

technique · counterexample

1.1

Each matrix has exactly two $λ$ -blocks, so [L1] gives geometric multiplicity two for both.

L1algebra

1.2

Yet $(A - λ I)^{2}$ has rank one, contributed by $J_{3} (0)^{2}$ , while $(B - λ I)^{2} = 0$ ; [L2] therefore distinguishes the block multisets ${3, 1}$ and ${2, 2}$ .

L2algebra

2.1

Equal geometric multiplicity has not determined the sizes, so the claim is false.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

S. Treil, Linear Algebra Done Wrong, Chapter 9, Sections 4.3-4.4 (standard reference, not scraped)