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Eigenspace and dimension

  1. Nov 25, 2012 #1
    Linear transformation f:C^∞(R) -> C^∞(R)

    f(x(t)) = x'(t)


    a) I have to set up the eigenvalue-problem and solve it :

    My solution : ke^λt


    b) Now I have to find the dimension of the single eigen spaces when λ is

    -5 and 0.


    My solution :

    Eigenspaces :

    E-5 = ke^-5t

    E0=k (because ke^0t = k)

    But I don't know how to find the dimension of the single eigen spaces ?

    I'm used to working with vectors but now it's functions and I'm not sure about the dimension.
     
    Last edited: Nov 25, 2012
  2. jcsd
  3. Nov 27, 2012 #2
    Re: Eigenspace and dimension (solved)

    Solved !
     
  4. Nov 27, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Good! I presume that you realized that since every solution, [itex]Ce^{\lambda x}[/itex] is a constant, C, times the single function [itex]e^{\lambda x}[/itex], the space is one dimensional.
     
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