Linear Algebra - Characteristic Polynomials and Nilpotent Operators

[glacier302]

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 : )

 

[Dick]

If p(t) is the characteristic polynomial of a matrix T, then p(T)=0, yes? Now what does 'nilpotent' mean?

 

[glacier302]

Nilpotent means that T^p = 0 for some positive integer p.

 

[glacier302]

Oh wow. I just realized how easy this question is...I just didn't remember the Cayley-Hamilton theorem. Thank you so much!

 

[glacier302]

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?

 

[Dick]

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?

 

[glacier302]

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!