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

  • Thread starter squenshl
  • Start date
  • #1
479
4
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.
 

Answers and Replies

  • #2
1,015
70


v is an eigenvector of M. Find the eigenvalue that it corresponds to.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966


If [itex]Mv= \lambda v[/itex], then [itex]M^2v= M(Mv)= M(\lambda v)= \lambda Mv= \lambda^2 v[/itex]. Get the point?
 
  • #4
Pengwuino
Gold Member
5,009
16


Also, remember that M^5 is just M acting on v 5 consecutive times. Eigenvectors make this quite simple :)
 

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