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Canonical form and change of coordinates for a matrix

  1. Mar 25, 2009 #1
    Hello! I'm trying to do some linear algebra. I have an insane Russian teach whose English is, uh, lacking.. so I'd appreciate any help with these I can get here!

    1. The problem statement, all variables and given/known data
    Find the canonical forms for the following linear operators and the matrices for the corresponsing change of coordinates.

    Here is the 6x6 matrix:
    0 1 0 0 0 0
    0 0 1 0 0 0
    0 0 0 1 0 0
    0 0 0 0 1 0
    0 0 0 0 0 1
    -1 0 0 -2 0 0


    2. Relevant equations



    3. The attempt at a solution
    I know I have to do subtract [tex]\lambda[/tex] on the diagonal, take the determinant, find the roots by solving for the [tex]\lambda[/tex] values, and then plug them in one at a time to find the different [tex]\zeta[/tex], turn that into a change of coordinates, and then depending on case, put it into canonical form...

    Unfortunately, my professor has only shown us the various [tex]\lambda[/tex] cases for 2 x 2 matrices and because we can "look everything up on google," we have no book!!!

    A couple questions: Can I simplify this or maybe turn it into the Jordan block? Can anyone point me to a similar problem, even? I've been searching for two hours, have searched through three free linear algebra e-books and am still lost =(

    Thanks so much!
     
  2. jcsd
  3. Mar 26, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi jejaques! Welcome to PF! :smile:

    (have a lambda: λ :wink:)
    Just put -λ down the diagonal, and calculate the determinant :smile:
     
  4. Mar 26, 2009 #3
    Re: Welcome to PF!

    Hello, and thanks for the welcome...

    Yeah, my reasoning was in my "attempt at a solution" section. I subtracted [tex]\lambda[/tex] from the diagonal and did the determinant; I just thought it was too much tedious stuff to post here, as I'm having problems further on.

    The determinant is [tex]\lambda[/tex]6 - 2[tex]\lambda[/tex]3 + 1

    To factor roots, I set the determinant equal to zero and factored, as follows:
    0 = ([tex]\lambda[/tex]3 - 1)2
    = ([tex]\lambda[/tex]-1)([tex]\lambda[/tex]5 + [tex]\lambda[/tex]4 + [tex]\lambda[/tex]3 - [tex]\lambda[/tex]2 - [tex]\lambda[/tex] - 1)
    = ([tex]\lambda[/tex] - 1)([tex]\lambda[/tex] - 1)([tex]\lambda[/tex]4 + 2[tex]\lambda[/tex]3 + 3[tex]\lambda[/tex]2 + 2[tex]\lambda[/tex] + 1)
    = ([tex]\lambda[/tex] - 1)2([tex]\lambda[/tex]2 + [tex]\lambda[/tex] + 1)2

    It has identical real roots at... [tex]\lambda1[/tex] = [tex]\lambda2[/tex] = 1, and identical complex roots at [tex]\lambda3[/tex] = [tex]\lambda4[/tex] = 1/2 + [tex]\sqrt{3}[/tex]i[tex]/[/tex]2 and [tex]\lambda5[/tex] = [tex]\lambda6[/tex] = 1/2 - [tex]\sqrt{3}[/tex]i[tex]/[/tex]2

    But the issue is, with a 6 x 6 matrix, which case should I evaluate and how should I go about finding the eigenvectors?

    I know complex roots evaluate to the canonical form A[tex]\bar{}[/tex]:
    [tex]\alpha[/tex] [tex]\beta[/tex] 0
    -[tex]\beta[/tex] [tex]\alpha[/tex] 0
    0 0 1

    But do I need to evaluate each of the positive and negative complex roots separately, and where do I throw in the [tex]\lambda1[/tex] = [tex]\lambda2[/tex] canonical form in that big 6 x 6?

    Thanks!
     
  5. Apr 29, 2009 #4
    Hey, buddy are you in sergey nikitin class at ASU?
     
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