Linear Algebra - Characteristic Polynomials and Nilpotent Operators

In summary, given a linear operator T with a characteristic polynomial of (-1)^n*t^n, it is likely that T is nilpotent. This can be proven using the Cayley-Hamilton theorem, which states that for any linear operator T with characteristic polynomial p(t), p(T)=0. However, this theorem only applies to finite-dimensional vector spaces, so it is important to clarify that the vector space in question is finite-dimensional.
  • #1
glacier302
35
0

Homework Statement



If the characteristic polynomial of an operator T is (-1)^n*t^n, is T nilpotent?


Homework Equations





The Attempt at a Solution



My first instinct for this question is that the answer is yes, because the matrix form of T must have 0's on the diagonal and must be either upper triangular or lower triangular. This is what I found when I tried to find 2x2 and 3x3 matrices with characteristic polynomial (-1)^n*t^n. However, I'm not sure how to actually prove this fact (especially for the nxn case), or how to show that T is nilpotent using this fact.

Any help would be much appreciated : )
 
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  • #2
If p(t) is the characteristic polynomial of a matrix T, then p(T)=0, yes? Now what does 'nilpotent' mean?
 
  • #3
Nilpotent means that T^p = 0 for some positive integer p.
 
  • #4
Oh wow. I just realized how easy this question is...I just didn't remember the Cayley-Hamilton theorem. Thank you so much!
 
  • #5
One question: Doesn't the Cayley-Hamilton Theorem only apply to linear operators on finite-dimensional vector spaces? What if the vector space is infinite-dimensional?
 
  • #6
glacier302 said:
One question: Doesn't the Cayley-Hamilton Theorem only apply to linear operators on finite-dimensional vector spaces? What if the vector space is infinite-dimensional?

Likely doesn't apply to infinite dimensional vector spaces. I'm not sure 'characteristic polynomial' applies to infinite dimensional vector spaces either. How would you define it?
 
  • #7
That's a good point. So I'm pretty sure that although it's not stated that the vector space is finite-dimensional, we're supposed to assume it. Thanks!
 

FAQ: Linear Algebra - Characteristic Polynomials and Nilpotent Operators

1. What is a characteristic polynomial?

A characteristic polynomial is a polynomial function that is associated with a square matrix. It is used to find the eigenvalues of a matrix, which are the values for which the matrix multiplied by a vector is equal to a scalar multiple of that vector.

2. How do you find the characteristic polynomial of a matrix?

To find the characteristic polynomial of a matrix, you first need to find the determinant of the matrix. Then, you create a polynomial by subtracting the determinant from the diagonal entries of the matrix. This polynomial is the characteristic polynomial.

3. What is a nilpotent operator?

A nilpotent operator is a linear transformation on a vector space where there exists a positive integer "n" such that the nth power of the operator is equal to zero. In other words, repeatedly applying the operator will eventually result in the zero vector.

4. How do nilpotent operators relate to characteristic polynomials?

Nilpotent operators are closely related to characteristic polynomials because the eigenvalues of a nilpotent operator are all equal to zero. This means that the characteristic polynomial of a nilpotent operator will have a factor of (x-0), which simplifies to just x. Therefore, the characteristic polynomial of a nilpotent operator is always x to the power of the dimension of the vector space.

5. What is the significance of characteristic polynomials and nilpotent operators in linear algebra?

Characteristic polynomials and nilpotent operators are important tools in linear algebra because they help us understand the behavior and properties of matrices and linear transformations. They provide information about the eigenvalues of a matrix, which can help with solving systems of linear equations and understanding the stability and dynamics of a system. Additionally, nilpotent operators play a key role in the Jordan canonical form, which is a useful way to represent matrices and study their properties.

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