Nilpotent Operator: Can e be Less Than or Equal to dim V?

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Discussion Overview

The discussion revolves around the properties of nilpotent operators in finite-dimensional vector spaces, specifically examining whether the exponent \( e \) in the condition \( N^e = 0 \) can be less than or equal to the dimension of the vector space \( V \). The focus includes theoretical implications and mathematical reasoning related to linear transformations and matrix representations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether it is possible to show that \( e \leq \text{dim } V \) for a nilpotent operator \( N \) such that \( N^e = 0 \).
  • Another participant asserts that it is easy to show this, particularly when \( e \) is the minimal positive integer for which \( N^e = 0 \).
  • A later reply confirms understanding and indicates that this is part of a larger proof.
  • Another participant claims that if \( V \) is represented as \( \mathbb{R}^n \) (where \( n = \text{dim } V \)), then \( N \) can be represented as a strictly upper triangular matrix, which leads to the conclusion that \( N^n \) is always zero.

Areas of Agreement / Disagreement

While some participants express confidence in the statement regarding \( e \) and its relationship to the dimension of \( V \), the discussion does not reach a consensus on the proof or implications of this relationship. Multiple viewpoints and interpretations remain present.

Contextual Notes

The discussion includes assumptions about the representation of nilpotent operators and the properties of strictly upper triangular matrices, which may not be universally applicable without further clarification or conditions.

Treadstone 71
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Suppose V is finite dimensional and N:V->V is a linear transformation such that N^e=0. Is it possible to show that e[tex]\leq[/tex]dim V? Is it even true?
 
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Yes, it is very easy to show this, or rather the correct statement when e is the minimal positive integer such that N^e=0, so try it.
 
I got it. It's part of a much larger proof. Thanks.
 
Yes it's true. If V were represented via its "basis" as R^n, (n=dim V) then, N can be represented as a nilpotent matrix. We knew that with appropriate representation, N is a strictly upper triangular (diagonal and lower triangular elements are all zeroes) matrix. And now we just have to show that a strictly upper triangular matrix the property that N^n is always zero. (Sorry, I use N (the operator, as the matrix).
 

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