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Eigenvalue/vector Proof

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Let A be n x n, λ ∈ ℂ, let v be n x 1, and suppose that A ⋅ v = λ ⋅ v. Show that A^jv = mathImg.gif ^j ⋅ v for each positive integer j.

    2. Relevant equations


    3. The attempt at a solution
    I haven't been able to get very far but,

    mathImg.gif ^j ⋅ v - A^jv = 0n x 1
    v( mathImg.gif ^j - A^j) = 0n x 1

    Not sure how to prove that for every positive j that this is true. Any thought would be appreciated. Thanks.
     
  2. jcsd
  3. Nov 11, 2014 #2

    RUber

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    What if you tried induction?
     
  4. Nov 11, 2014 #3
    Where j = 0

    v( mathImg.gif ^j - A^j) = 0n x 1

    v( mathImg.gif ^0 -
    A^0) = 0n x 1

    v(In-1) = 0n x 1

    v ⋅ In
    - v= 0n x 1

    v - v = 0n x 1


    Since v is n x 1

    0n x 1
    = 0n x 1


    Because this is true I can assume this for all j?
     
  5. Nov 11, 2014 #4

    RUber

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    I don't think so, since it would be true for just about anything raised to the zero.
    Start with the original form ##Av=\lambda v ##. This is your base case.
    ##A^2v=A \lambda v ##.
    And for any integer...
     
  6. Nov 11, 2014 #5
    I see what you're saying about
    and vice-versa for mathImg.gif but I don't see how that helps?
    My skills with induction are sub par to say the least.
     
  7. Nov 12, 2014 #6

    RUber

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    Assume ##A^n v = \lambda^n v## show that ##A^{n+1}v=A A^n v = \lambda^{n+1}v##.
    The argument for induction is: It works for j=1, assume it works for some j and show it is true for the next j.
     
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