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Math Help - Eigenvalues of A^3-A^(-1)

  1. #1
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    Eigenvalues of A^3-A^(-1)

    I have to find the Eigenvalues of a Matrix A. This is pretty easy but the second part of the question is:
    What are the corresponding eigenvalues and eigenvectors for A3-A-1?
    I know that the eigenvalues of A3 are equal to the third power of the eigenvalues of A, so this is no problem.

    If the Eigenvalues of A are -2 and 3 like in my case, the Eigenvalues for A3 are -8 and 27.

    But I don't know how to deal with the subtraction of the inverse, is there any rule or do I have to take the inverse of the matrix manually and subtract it from A3?
    In this case I would have to do a lot of matrix multiplication which takes a lot of time...
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  2. #2
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    Re: Eigenvalues of A^3-A^(-1)

    If v is an eigenvector of A with eigenvalue \lambda, then Av= \lambda v. Taking A^{-1} of both sides, A^{-1}(Av)= A^{-1}\lambda v so that v= \lambda A^{-1}v and A^{-1}v= (1/\lambda)v. In other words, if v is an eigenvector of A with eigenvalue \lambda then it is also an eigenvector of A^{-1} with eigenvalue 1/\lambda. Enough?
    Last edited by HallsofIvy; June 7th 2012 at 03:04 PM.
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  3. #3
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    Re: Eigenvalues of A^3-A^(-1)

    So if I have eigenvalues following eigenvalues for A: 3 and 7.

    For A3 I have eigenvalues of 9 and 27, than for A3-A-1 I have eigenvalues:

    9-1/3 and 27-1/7

    right?

    Thank you!
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  4. #4
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    Re: Eigenvalues of A^3-A^(-1)

    (s there a rule for the Inverse too?

    For example if I have to find the eigenvalue of A3-I
    Can I simply take the eigenvalue of A3 and subtract 1?
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  5. #5
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    Re: Eigenvalues of A^3-A^(-1)

    well, let's see if that works:

    suppose an eigenvalue of A is λ, with eigenvector v.

    (A3 - I)(v) = A3(v) - I(v) = A2(Av) - v

    = A2(λv) - v = A(A(λv)) - v = A(λ(A(v))) - v = A(λ(λv)) - v = A(λ2v) - v = λ2(A(v)) - v

    = λ2(λv) - v = λ3v - v = (λ3 - 1)v.

    in other words, for all integers k, Ak has eigenvalue λk (even when k = 0) (if Ak is even defined for k < 0, that is, if A is invertible).

    for example, let A be the matrix:

    [2 4]
    [0 2], which has eigenvalue 2. then A3 - I =

    [7 48]
    [0 ..7], which clearly has eigenvalue 7 = 23 - 1.
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  6. #6
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    Re: Eigenvalues of A^3-A^(-1)

    Can't we just use the Cayley hamilton theorem and then the evalues of A are 1, -1, i, -i?
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  7. #7
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    Re: Eigenvalues of A^3-A^(-1)

    Thank you for your explaination, but:
    you got the following Matrix for A^3-I
    [7 48]
    [0 ..7]

    if I calculate it I get:
    [7 64]
    [0 ..7], why do you have the 48? Is there any rule I didn't took into consideration?
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  8. #8
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    Re: Eigenvalues of A^3-A^(-1)

    [2 4]3
    [0 2] =

    [2 4][2 4][2 4]
    [0 2][0 2][0 2] =

    [4 16][2 4]
    [0 ..2][0 2] = (in the upper-right we have 2*4 + 4*2 = 8 + 8 = 16)

    [8 48]
    [0 ..8] (in the upper-right we have 4*4 + 16*2 = 16 + 32 = 48)


    you don't just "cube each entry".
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    Re: Eigenvalues of A^3-A^(-1)

    can you please tell me how 48 comes in the matrix without using matrix multiplication.
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  10. #10
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    Re: Eigenvalues of A^3-A^(-1)

    A3 *is* matrix multiplication (it's A times A times A). there's no simpler way to describe the entries of powers of a matrix, except by doing the multiplication (unless the matrix is diagonal, or of some other "special form").
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