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Powers of a Matrix and Eigenvalues proof

  1. May 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that if A is an nxn matrix with eigenvector v, then v is an eigenvector for Ak where kε(all positive integers)


    2. Relevant equations

    Av=λv

    3. The attempt at a solution

    Av=λv
    A(Av)=A(λv)
    Akv=λ(Av)

    i know i may not be doing it right but this is what i can think of so far
     
  2. jcsd
  3. May 16, 2013 #2

    Office_Shredder

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    Gold Member

    What if I rewrote that as
    A(Av) = λ(Av)
     
  4. May 16, 2013 #3

    statdad

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    If you want to show that v is an eigenvector of a power of A (given that v is an eigenvector of A itself), you need to know what happens when you multiply v by some power of A. Consider the second power to begin with - the pattern you see continues for higher powers. So,

    if Av = λv, what do you get when you examine

    A(Av) = A(λv)

    * and simplify the left side?
    * and simplify the right side (remember that A(λv) = λ(Av)) when you simplify the right?

    If you correctly simplify these two pieces you should be able to show that v IS an eigenvector of A-squared, AND you will know the associated eigenvalue. The latter is important, because it will give you a major hint about the link between A^k, v, and the corresponding eigenvalue. Once you see that link, you will know how to answer your question.
     
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