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Homework Help: Eigenvalues: Proof that A^3=A

  1. Feb 25, 2010 #1
    Hey, I'm wondering if I have a known set of eigenvalues (-1, +1, 0) for A, if I can prove that the matrix A = A3?

    I can prove that if A3 = A, that the eigenvalues would be −1, +1, and 0. The following is the proof:

    Since A=A3, A*k=A3*k
    lambda*k - lambda3*k = 0
    lambda - lambda3 = 0
    lambda*(1-lambda2) = 0
    lambda = 0, -1, +1

    Is there any way to prove it the other way around? If I know that the eigenvalues are 0, -1, and +1, can I prove that A3 = A?

  2. jcsd
  3. Feb 25, 2010 #2


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    Welcome to PF!

    Hi noon0788! Welcome to PF! :smile:

    Hint: A is linear, so if three eignevectors are p q and r, what is A3(ap + bq + cr) ? :wink:

    (alternatively, write A = QPQ-1)
  4. Feb 25, 2010 #3
    Thanks tiny-tim, but I'm not quite sure what you mean by A3(ap + bq + cr).

    When you say A = QPQ-1, do you mean I should diagonalize the matrix? Here's what I'm thinking:

    C-1AC = D (where d is the diagonalized matrix)

    D = [1, 0, 0; 0, −1, 0; 0, 0, 0]

    det(C-1)*det(A)*det(C) = det(D)
    det(C-1)*det(C)*det(A) = det(D)
    det(A) = det(D)
    det(D) = 0
    det(A) = 0

    Maybe I'm way off...
  5. Feb 25, 2010 #4


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    I mean where a b and c are any numbers.

    (And why are you using determinants?? :confused:)
  6. Feb 25, 2010 #5
    :/ I don't know why I'm using determinants. I guess what I'm confused by is your notation of R3(ap+bq+cr). Is that I just don't recognize that notation. I'm familiar with matrices, but maybe not as familiar as I thought I was.
  7. Feb 25, 2010 #6


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    A3(ap+bq+cr) is matrix A followed by matrix A followed by matrix A followed by the sum of a times vector p plus b times vector q plus c times vector r, with Ap = p, Aq = -q, Ar = 0.
  8. Feb 25, 2010 #7
    Thanks tiny-tim, but I think my question might just be a bit over my head. I'm confused as to why p q are r are vectors now. I thought they were eigenvalues. Also, the Ap=p stuff is confusing me too. Here's what I've got so far...

    A3(a(1) + b(-1) + c(0))
    A3a - A3b

    Thanks for your help.
  9. Feb 26, 2010 #8
    are a &b scalars or vectors?

    I'll try to explain the idea better: the three eigenvectors are independent, therefore they create a basis to a three-tuple column vectors. Let's call them [tex]e_{1},e_{0},e_{-1}[/tex] (the index corresponds the eigenvalue).
    Now any other vector will be:


    Now, what is Av and what is [tex]A^{3}v[/tex]?
  10. Feb 26, 2010 #9


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    Hi noon0788! :smile:

    (just got up :zzz: …)
    No, they're eigenvectors.

    (That's why Ap = p, with p's eigenvalue = 1 :wink:)

    ok, now, for any numbers a b and c, what is A(ap + bq + cr) ? :smile:
  11. Feb 27, 2010 #10
    A(ap + bq + cr)
    = A(ap) + A(bq) + A(cr)
    = a(Ap) + b(Aq) + c(Ar)
    = a + b
  12. Feb 27, 2010 #11


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    Well, no! For one thing, on the left side you have a vector, A(ap+ bq+cr), and on the right you have a number, a+ b. p, b, and q are eigenvectors of A with eigenvalues of 1, -1, and 0, respectively.

    A(ap+ bq+ cr)= (1)ap+ (-1)bq+ (0)cr= ap- bq. A2(ap+bq+cr)= A(ap- bq)= ap-(-bq)= ap+ bq. And, finally, A3(ap+ bq+ cr)= A(ap+bq)= ap- bq.

    My first thought was to use the fact that every matrix satisfies its own characteristic equation. Since A has eigenvalues 1, -1, and 0, its characteristic equation is x(x+1)(x- 1)= x^3- x.
  13. Feb 27, 2010 #12
    I think I understand it. Just one more question. Am I allowed to distribute A in A(ap + bq + cr)? So it would become aAp + bAq + cAr?
  14. Feb 27, 2010 #13


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    Yes, because A is linear (and a b and c are scalars). :smile:
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