ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Recovering a $5 \times 5$ Jordan form from shifted power ranks at two eigenvalues

Example

Let $T = J_{3} (0) \oplus J_{2} (1)$ . Its shifted-power ranks are $ρ_{k} (0) : 5, 4, 3, 2, 2, \dots, ρ_{k} (1) : 5, 4, 3, 3, 3, \dots .$ Their second differences recover one size-three block at $0$ and one size-two block at $1$ . Accordingly $χ_{T} = μ_{T} = x^{3} (x - 1)^{2} .$

Facts & Assumptions

Given: The displayed block diagonal matrix $T$ .

[L1]

The exact-size $k$ block count at $λ$ is $ρ_{k - 1} (λ) - 2 ρ_{k} (λ) + ρ_{k + 1} (λ)$ (Ranks of shifted powers determine Jordan form up to block order).

[L2]

Jordan block sizes give the characteristic and minimal polynomials by total and maximum exponents, respectively (For split operators, Jordan blocks read off eigenspace multiplicities and both canonical polynomials).

Verification

technique · computation

1.1algebra

At $λ = 0$ , the $J_{2} (1)$ summand stays invertible while the ranks of powers of $J_{3} (0)$ are $3, 2, 1, 0, 0, \dots$ , giving $5, 4, 3, 2, 2, \dots$ . At $λ = 1$ , the $J_{3} (0) - I$ summand stays invertible while the nilpotent part of $J_{2} (1)$ has ranks $2, 1, 0, 0, \dots$ , giving $5, 4, 3, 3, \dots$ .

2.1step 1.1L1L2algebra∎

Applying [L1] gives one exact size-three block at $0$ , one exact size-two block at $1$ , and zero other blocks; [L2] then gives both displayed polynomials.

Depends on

Used by

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Dependency tree · two levels

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Recovering a 5×55×5 Jordan form from shifted power ranks at two eigenvalues