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Homework Help: Strange question regarding eigenvectors / eigenvalues

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Suppose that the 2x2 matrix A has eigenvalues lambda = 1,3 with corresponding eigenvectors [2,-1]^T and [3,2]^T. Find a formula for the entries of A^n for any integer n. And then, find A and A^-1 from your formula.

    2. Relevant equations

    Ax = lambda X
    (P^-1)AP = D
    A = PDP^-1

    3. The attempt at a solution

    I've been trying to figure this out for a long time. I'm really not sure where to start... If anyone could provide some sort of guidance on how to begin this problem... that would be really helpful. I've tried using the above formulas, but I'm just not sure how to get a generalized formula. I could find A and A^-1 individually without the generalized formula, but the problem asks for it; thanks in advance for even the slightest hint.
  2. jcsd
  3. Nov 12, 2008 #2


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    D=P^(-1)AP is [[1,0],[0,3]], right? And you have P and P^(-1), I hope. Then isn't A^n=P(D^n)P^(-1)? Since A^n=PDP^(-1)*PDP^(-1)*... n times. D^n is pretty easy to figure out. It's not a strange question at all. It's easy.
  4. Nov 12, 2008 #3
    Hi Dick.

    Thanks for your explanation! So, my formula would be:
    A^(n) = P(D^n)P^(-1)

    P is obviously just
    [2 3
    -1 2]

    and P^(-1) would then just be its inverse, or
    [(2/7) (-3/7)
    (-1/7) (2/7)]

    Then you just use this P, P^(-1) and the given D^n to get the answer of A^n.

    So, then with A^(-1), I would just get the inverse of D, and plug it into the formula...?

  5. Nov 12, 2008 #4


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    D^n is just [[1,0],[0,3^n]], right? Sure, then use P and P^(-1) to convert it back to A^n. Why do you need to invert D? Ohh. I see what you mean. Sure you would.
  6. Nov 12, 2008 #5

    Thanks. Just for clarification though, why would I not say

    A^(n) = (PDP^(-1))^n

    Why is it just D that I raise to the n, and not the whole right of the equation?
  7. Nov 12, 2008 #6


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    Because they are both the same thing and it's much easier to raise D^n than A^n. Take A^3. A^3=(PDP^(-1))*(PDP^(-1))*(PDP^(-1)). Don't you see how the P^(-1)*P parts cancel in the middle?
  8. Nov 13, 2008 #7
    Ahhh, ok. Thanks for clearing that up for me; understand it now.

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