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

M^5v given that M = [3 2 4 2 0 2 4 2 3]

  1. Jun 21, 2009 #1
    Hi,
    Just wondering how to calculate M^5v given that
    M = [3 2 4
    2 0 2
    4 2 3].
    and v = (1,0,-1). v being a column vector.
     
  2. jcsd
  3. Jun 21, 2009 #2
    Re: E-values/E-vectors

    v is an eigenvector of M. Find the eigenvalue that it corresponds to.
     
  4. Jun 21, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: E-values/E-vectors

    If [itex]Mv= \lambda v[/itex], then [itex]M^2v= M(Mv)= M(\lambda v)= \lambda Mv= \lambda^2 v[/itex]. Get the point?
     
  5. Jun 21, 2009 #4

    Pengwuino

    User Avatar
    Gold Member

    Re: E-values/E-vectors

    Also, remember that M^5 is just M acting on v 5 consecutive times. Eigenvectors make this quite simple :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: M^5v given that M = [3 2 4 2 0 2 4 2 3]
  1. Solve 2^n+3=m^m (Replies: 18)

Loading...