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Math induction

  1. Apr 17, 2010 #1
    Let [tex]\lambda[/tex] be an eigenvalue of [tex]A[/tex] and let [tex]\mathbf{x}[/tex] be an eigenvector belonging to [tex]\lambda[/tex]. Use math induction to show that, for [tex]m\geq1[/tex], [tex](\lambda)^m[/tex] is an eigenvalue of [tex]A^m[/tex] and [tex]\mathbf{x}[/tex] is an eigenvector of [tex]A^m[/tex] belonging to [tex](\lambda)^m[/tex].

    [tex]A\mathbf{x}=\lambda\mathbf{x}[/tex]

    [tex]p(1): A^1\mathbf{x}=(\lambda)^1\mathbf{x}[/tex]\

    [tex]p(k): A^k\mathbf{x}=(\lambda)^k\mathbf{x}[/tex]

    [tex]p(k+1): A^{k+1}\mathbf{x}=(\lambda)^{k+1}\mathbf{x}[/tex]

    Assume p(k) is true.

    Since p(k) is true, [tex]p(k+1): A*(A^k\mathbf{x})[/tex]

    Not sure if this is correct path to take to the end result. Let alone, if it is if I am going about it right.
     
    Last edited: Apr 17, 2010
  2. jcsd
  3. Apr 17, 2010 #2

    Dick

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    You were doing ok, except you don't want A^(k+1)(lambda). You want p(k+1): A^(k+1)(x)=A*(A^k(x))=A*(lambda^k*x)=lambda^k*A(x)=???
     
  4. Apr 17, 2010 #3
    Ok I gotcha.
     
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