Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Eigenvalue kind of nilpotent problem

  1. Jun 29, 2008 #1
    If a vector [tex]v\in V[/tex] and a linear mapping [tex]T:V\to V[/tex] are fixed, and there exists numbers [tex]\lambda_1\in\mathbb{C}[/tex], [tex]n_1\in\mathbb{N}[/tex] so that

    (T - \lambda_1)^{n_1}v = 0,

    is it possible that there exists some [tex]\lambda_2\neq\lambda_1[/tex], and [tex]n_2\in\mathbb{N}[/tex] so that

    (T - \lambda_2)^{n_2}v = 0?

    (Here complex numbers are interpreted as multiplication operators [tex]V\to V[/tex], [tex]v\mapsto \lambda v[/tex], as usual.)
  2. jcsd
  3. Jun 29, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    That happens iff v=0. This is the statement that the intersection of different generalized eigenspaces is {0}.
  4. Jun 30, 2008 #3


    User Avatar

    Furthermore, if f,g are any two polynomials with no common factor and f(T)v=g(T)v=0 then you can conclude that v=0.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Eigenvalue kind nilpotent Date
A Eigenvalue Problem and the Calculus of Variations Jan 8, 2018
I Eigenvalues of Circulant matrices Oct 1, 2017
I Eigenvalues of block matrices Sep 10, 2017
A Numerically Calculating Eigenvalues Aug 26, 2017
I What kind of space is the space of spinors? May 27, 2016