TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.
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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

1.1

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.

L1L2L3

1.2

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 $B$ .

L1L3construct

2.1

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

step 1.1step 1.2∎

Depends on

Used by

FALSE: Equal characteristic and minimal polynomials imply similarity False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 66 results over 16 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 4-5 (standard reference, not scraped)