# All eigenvalues 0 implies nilpotent

1. Nov 28, 2009

### altcmdesc

1. The problem statement, all variables and given/known data
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?

3. 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.

2. Nov 29, 2009

### 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?

3. Nov 30, 2009

### altcmdesc

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

Last edited: Nov 30, 2009
4. Nov 30, 2009

### 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?

5. Dec 1, 2009

### 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.

Last edited: Dec 1, 2009
6. Dec 1, 2009

### 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.

7. Dec 1, 2009

### 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?

8. Dec 1, 2009

### Dick

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.