1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

All eigenvalues 0 implies nilpotent

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data
    How would I go about proving that if a linear operator [tex]T\colon V\to V[/tex] has all eigenvalues equal to 0, then [tex]T[/tex] must be nilpotent?

    3. The attempt at a solution

    I know that this follows trivially from the Cayley-Hamilton theorem (the characteristic polynomial is [tex]x^n[/tex] and hence [tex]T^n=0[/tex]), 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 [tex]V[/tex] is also in the kernel of some power of [tex]T[/tex] and hence [tex]T[/tex] is nilpotent, but I'm getting stuck.
     
  2. jcsd
  3. Nov 29, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  4. Nov 30, 2009 #3
    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
  5. Nov 30, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  6. Dec 1, 2009 #5
    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
  7. Dec 1, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  8. Dec 1, 2009 #7
    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?
     
  9. Dec 1, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: All eigenvalues 0 implies nilpotent
  1. Eigenvalue of 0 (Replies: 8)

Loading...