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Nilpotent similarity is classified by the ranks of all powers
Statement
Let and be nilpotent endomorphisms of finite-dimensional vector spaces and over the same field , where similar means that some -linear isomorphism satisfies . Then and are similar if and only if The 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 and over have rank at and rank for every , and there is no linear isomorphism between them at all.
Facts & Assumptions
Given: Finite-dimensional nilpotent endomorphisms and over the same field.
The ranks of all powers determine every nilpotent Jordan-block multiplicity (Power ranks determine every nilpotent Jordan-block multiplicity).
Similarity is change of basis for 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
If and are similar, conjugating gives ; an invertible change of coordinates preserves image dimension, so their ranks agree for every .
Conversely, equality of all power ranks makes the two Jordan-block multisets equal by [L1]. Choose Jordan-string bases of and of realising those multisets; since the multisets agree, both bases have the same length and and have the same block diagonal matrix in them. Because and are spaces over the same field , sending the th vector of to the th vector of defines an -linear isomorphism , and matching the block matrices entry by entry gives , hence ; [L2] identifies this with similarity.
Steps 1.1 and 1.2 establish both directions, with the empty block multiset covering zero-dimensional spaces.
Depends on
Used by
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Sources
- S. Treil, Linear Algebra Done Wrong, Chapter 9, Sections 4.3-4.4 (standard reference, not scraped)