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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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Nilpotent similarity is classified by the ranks of all powers

Statement

Let N and M be nilpotent endomorphisms of finite-dimensional vector spaces V and W over the same field F, where similar means that some F-linear isomorphism φ:VW satisfies M=φNφ1. Then N and M are similar if and only if rankNk=rankMkfor every k0. The k=0 equality includes equality of the dimensions.

The common field is a hypothesis, not a convenience: rank sequences are integers and can agree across different fields, while similarity cannot. The zero endomorphisms of the one-dimensional spaces over F2 and over F3 have rank 1 at k=0 and rank 0 for every k1, and there is no linear isomorphism between them at all.

Facts & Assumptions

Given: Finite-dimensional nilpotent endomorphisms N and M over the same field.

[L1]

The ranks of all powers determine every nilpotent Jordan-block multiplicity (Power ranks determine every nilpotent Jordan-block multiplicity).

[L2]

Proof

technique · direct
1.1

If N and M are similar, conjugating Nk gives Mk; an invertible change of coordinates preserves image dimension, so their ranks agree for every k0.

L2algebra
1.2

Conversely, equality of all power ranks makes the two Jordan-block multisets equal by [L1]. Choose Jordan-string bases B of V and C of W realising those multisets; since the multisets agree, both bases have the same length and N and M have the same block diagonal matrix in them. Because V and W are spaces over the same field F, sending the ith vector of B to the ith vector of C defines an F-linear isomorphism φ:VW, and matching the block matrices entry by entry gives φN=Mφ, hence M=φNφ1; [L2] identifies this with similarity.

L1L2givenchoose
2.1

Steps 1.1 and 1.2 establish both directions, with the empty block multiset covering zero-dimensional spaces.

step 1.1step 1.2

Depends on

Used by

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Sources