TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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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Split matrices are similar exactly when their Jordan block multisets agree

Statement

Let $A, B \in M_{n} (F)$ have characteristic polynomials that split over $F$ . Then $A$ and $B$ are similar if and only if their Jordan canonical forms have the same multiset of Jordan blocks. The blocks may occur in different orders.

Facts & Assumptions

Given: Matrices $A, B \in M_{n} (F)$ with split characteristic polynomials.

[L1]

A split characteristic polynomial is equivalent to existence of Jordan form over the base field (Jordan form over the base field exists exactly when the characteristic polynomial splits).

[L2]

The ranks of shifted powers determine Jordan form uniquely up to block order (Ranks of shifted powers determine Jordan form up to block order).

[L3]

Similarity models two ordered-basis matrices of one endomorphism and is an equivalence relation (Similarity is an equivalence relation, and two matrices represent the same endomorphism in two bases exactly when they are similar).

Proof

technique · direct

1.1L1L2L3

If $A$ and $B$ are similar, [L3] realises them as matrices of one endomorphism in two bases; [L1] supplies Jordan forms and [L2] makes their block multisets equal.

1.2L1L3construct

Conversely, enumerate the common block multiset once; after permuting blocks, both block diagonal matrices equal the matrix from that enumeration, and each block permutation is conjugation by a permutation matrix. Each of $A$ and $B$ is therefore similar to that common matrix, so the equivalence-relation clause in [L3] makes $A$ similar to $B$ .

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove both directions, including $n = 0$ .

Depends on

Used by

FALSE: Equal characteristic and minimal polynomials imply similarity 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. Treil, Linear Algebra Done Wrong, Chapter 9, Sections 4-5 (standard reference, not scraped)