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

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In summary, the conversation discusses a linear transformation N on a finite dimensional space V, where N^e=0 and the question of whether e\leqdim V is true. The participants agree that it is easy to show when e is the minimal positive integer satisfying N^e=0. They also mention that N can be represented as a nilpotent matrix in a basis representation of V, and that a strictly upper triangular matrix has the property that N^n is always zero.
  • #1
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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  • #2
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.
 
  • #3
I got it. It's part of a much larger proof. Thanks.
 
  • #4
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).
 
  • #5


I would respond by stating that the statement "e can be less than or equal to dim V" is not necessarily true for all cases. It is possible for e to be less than or equal to dim V, but it is not always true.

To prove this, we can consider an example where e is greater than dim V. Let V be a two-dimensional vector space and N be a linear transformation such that N^3=0. In this case, e=3, which is greater than dim V=2.

On the other hand, it is also possible for e to be less than dim V. For instance, let V be a three-dimensional vector space and N be a linear transformation such that N^2=0. In this case, e=2, which is less than dim V=3.

Therefore, it is not always true that e is less than or equal to dim V. It depends on the specific values of e and dim V, and it is possible for e to be either greater or less than dim V.
 

1. What is a nilpotent operator?

A nilpotent operator is a linear operator on a vector space whose powers become zero after a finite number of iterations. In other words, there exists a positive integer n such that the nth power of the operator is equal to the zero operator.

2. Can a nilpotent operator have a dimension less than or equal to the dimension of the vector space?

Yes, it is possible for a nilpotent operator to have a dimension less than or equal to the dimension of the vector space. This depends on the specific properties and structure of the vector space and the operator itself.

3. How is the dimension of a nilpotent operator related to its eigenvalues?

The dimension of a nilpotent operator is equal to the number of distinct eigenvalues with multiplicity. This means that the dimension of a nilpotent operator can never be greater than the number of distinct eigenvalues it has.

4. Can a nilpotent operator have non-zero eigenvalues?

No, a nilpotent operator can only have eigenvalues equal to zero. This is because the powers of a nilpotent operator eventually become zero, and the eigenvalues are the roots of the operator's characteristic polynomial which is raised to these powers.

5. How can a nilpotent operator be used in applications?

Nilpotent operators are used to study and understand the structure and behavior of linear transformations in various mathematical fields such as linear algebra, functional analysis, and differential equations. They also have applications in physics, engineering, and computer science.

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