1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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:
    [tex]

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

    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

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

    hunt_mat

    User Avatar
    Homework Helper

    The eigenvalue equation is [tex]Ax=\lambda x[/tex]. Multiply by A to obtain
    [tex]
    A^{2}x=\lambda Ax=>x=\lambda^{2}x
    [/tex]
    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 ?

    Tim,

    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:

    hunt_mat
    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

    hunt_mat

    User Avatar
    Homework Helper

    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:
    [tex]
    A=\left(\begin{array}{cc}
    -3 & 1 \\
    x & y
    \end{array}\right)
    [/tex]
    Use the two equations which I mentioned to obtain your solution.
     
  9. Jul 18, 2010 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear Algebra - Eigenvalues/Eigenvectors?
Loading...