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Homework Help: Linear Algebra - Eigenvalues/Eigenvectors?

  1. Jul 18, 2010 #1
    1. The problem statement, all variables and given/known data
    a.) If A2 = I, what are the possible eigenvalues of A?
    b.) If this A is 2 by 2, and not I or -I, find its trace and determinant.
    c.) If the first row is (3,-1), what is the second row?

    2. Relevant equations
    None was given, but I think:

    1. det(A) = 1 for A2 = I or A-1 = A

    We're studying about Matrix diagonalization and the topic is called "Diagonalization of a Matrix". The equation S[tex]\Lambda[/tex]S-1 = A is supposed to be relevant.

    3. The attempt at a solution
    a.) I got the possible eigenvalues to be: [tex]\lambda[/tex]1 x [tex]\lambda[/tex] 2 x ... x [tex]\lambda[/tex]n = 1
    b.) tr(A) = [tex]\lambda[/tex] + [tex]1/\lambda[/tex]
    det(A) = 1
    c.) This is where I'm stuck... I know I'm supposed to use the equation det(A) = 1 but there are two unknowns and I only know one equation. I was going to use the equation S[tex]\Lambda[/tex]S-1 = A but then I realized that I need the eigenvectors and I can't find the eigenvectors since I don't know the whole matrix.
    Last edited: Jul 18, 2010
  2. jcsd
  3. Jul 18, 2010 #2


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    Welcome to PF!

    Hi KIDRoach! Welcome to PF! :smile:

    You're making this very complicated. :redface:

    Hint: if det(A) = a, and det(B) = b, what is det(AB)? :wink:

    (alternatively, just write A relative to a basis in which A is diagonal)
  4. Jul 18, 2010 #3
    Hi Tim!

    Thanks for answering my question. I tried that too I think, but I don't think it's right, or at least either that's wrong or one of the earlier answer is wrong. :(

    I know that det(AB) = det(A) x det(B) = ab

    When I tried it to solve the equation:

    \[ \left( \begin{array}{ccc}
    3 & -1 \\
    0 & x \end{array} \right)\]

    I get x = 1/3, which, when I square the A matrix, I don't get the I matrix, so the equation is inconsistent/ something is wrong... But I can't figure out which part I'm wrong at... :(
    Last edited: Jul 18, 2010
  5. Jul 18, 2010 #4


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    ok, so if A2 = I, and if A ≠ I or -I, then det(A) = … ? :smile:
  6. Jul 18, 2010 #5


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    The eigenvalue equation is [tex]Ax=\lambda x[/tex]. Multiply by A to obtain
    A^{2}x=\lambda Ax=>x=\lambda^{2}x
    So for this to hold [tex]\lambda =\pm 1[/tex]
    In order to compute the third part, you have two equations, one coming from the determinant equation and the other one coming from the trace equation.
  7. Jul 18, 2010 #6
    Just wondering, so my answers in part a and b is wrong then? No one has mentioned it yet so I'm not sure... I know that part b is wrong now, that det(A) = -1 instead of 1 ?


    If A is 2x2 and not I or -I, then det(A) = -1 right? I still don't get how to get from that point to the third though. I still get 1/3 when I use det(A) = -1 to solve the diagonal matrix equation:
    Sorry for being so dumb... :frown:

    Wouldn't it be A2x = [tex]\lambda^{2}[/tex]x instead?
    I tried using the trace equation. The thing is, the trace equation brings in the two [tex]\lambda[/tex] into the equation and adds significant complexity when I'm computing the eigenvectors.
  8. Jul 18, 2010 #7


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    From the question, we know that A^2 =I. One thing which you may not know is that you can diagonalise a matrix via the eigenvectors. So the determinant is just the product of the eigenvalues, hence det(A)=(+1)(-1)=1. Off the top of my head, I think that the tr(A) is given by the sum of the eigenvalues, so tr(A)=0. Write you matrix as:
    -3 & 1 \\
    x & y
    Use the two equations which I mentioned to obtain your solution.
  9. Jul 18, 2010 #8


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    Yes, they're wrong. Hunt_mat showed you that the eigenvalues have to be +1 or -1. So think about how A looks in the basis where it's been diagonalized. There are only four possible matrices.

    Remember that A2=I, so no more A's on the lefthand side.
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