All eigenvalues 0 implies nilpotent

[altcmdesc]

Homework Statement

How would I go about proving that if a linear operator T\colon V\to V has all eigenvalues equal to 0, then T must be nilpotent?

The Attempt at a Solution

I know that this follows trivially from the Cayley-Hamilton theorem (the characteristic polynomial is x^n and hence T^n=0), but, unfortunately, I'm not allowed to use that theorem. How else could I go about proving this? I've been trying to show that any vector in V is also in the kernel of some power of T and hence T is nilpotent, but I'm getting stuck.

 

[Dick]

T^(k+1)(V) is a subset of T^k(V), right? If T isn't nilpotent then you eventually have a k such that T^(k+1)(V)=T^k(V) with T^k(V) not {0}. Consider the restricted map T:T^k(V)->T^k(V). What can you say about it?

 

[altcmdesc]

Is it that the restricted map doesn't have any eigenvalues/eigenvectors and is bijective? Am I on the right track?

 

[Dick]

You are on the right track. Yes, the restricted T is bijective. It must have an eigenvector. What can you say about the corresponding eigenvalue?

 

[altcmdesc]

Why must the restricted map have an eigenvector? I know where to go from there, but I can't seem to understand why that must be true since there isn't any assumption made that the base field of the vector space V is algebraically closed.

 

[Dick]

The restricted T maps the vector space T^m(V) to itself. Look at it's characteristic polynomial. It must have a root r, mustn't it? That means r*I-T is singular.

 

[altcmdesc]

My point is that if the vector space is over the reals, for example, not every characteristic polynomial has roots (e.g. x^2+1), so if T^m(V) isn't over an algebraically closed field, we don't necessarily have that the restricted map has an eigenvalue...right?

 

[Dick]

My point is that if the vector space is over the reals, for example, not every characteristic polynomial has roots (e.g. x^2+1), so if T^m(V) isn't over an algebraically closed field, we don't necessarily have that the restricted map has an eigenvalue...right?

Right. But then what exactly do you mean by "T has all eigenvalues equal to zero"? Take the matrix T=[[0,0,0],[0,0,-1],[0,1,0]]. T has one REAL eigenvalue and that's 0. T is not nilpotent. I don't think I would describe T as having all eigenvalues equal to 0.