Results 1 to 4 of 4

Math Help - Characteristic Polynomial from a matrix with an unknown...

  1. #1
    Junior Member
    Joined
    Nov 2006
    From
    Copenhagen
    Posts
    39

    Characteristic Polynomial from a matrix with an unknown...

    Hi, for the second time today, still linear algebra...

    Okay, if I have the matrix,

    A = [2a a 0]
    --- [ a a 0]
    --- [ 0 0 a]

    and I want to find the eigenvalues of the matrix, I know I need to set up the characteristic polynomial, which I find to be...

    (2a-t)(a-t)^2 - a^2(a-t)

    I have made various adjustments, but I still haven't found the roots of the polynomial.

    What I have come up with is either...

    -t^3 + 4at^2 - 4a^2t - a^3

    or

    (a-t)(t^2 - 3at + a^2)

    Okay, so... No matter which equation I look at, I still can't find the roots.
    I can see from the last one -- that one root should be t = a? But division comes up with an illegal answer.

    So, if anyone has any input, it is very welcome.

    Thanks.
    Simon DK.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by sh01by View Post
    Hi, for the second time today, still linear algebra...

    Okay, if I have the matrix,

    A = [2a a 0]
    --- [ a a 0]
    --- [ 0 0 a]

    and I want to find the eigenvalues of the matrix, I know I need to set up the characteristic polynomial, which I find to be...

    (2a-t)(a-t)^2 - a^2(a-t)

    I have made various adjustments, but I still haven't found the roots of the polynomial.

    What I have come up with is either...

    -t^3 + 4at^2 - 4a^2t - a^3

    or

    (a-t)(t^2 - 3at + a^2)

    Okay, so... No matter which equation I look at, I still can't find the roots.
    I can see from the last one -- that one root should be t = a? But division comes up with an illegal answer.

    So, if anyone has any input, it is very welcome.

    Thanks.
    Simon DK.
    what do you mean by ".. But division comes up with an illegal answer."

    yes, one eigenvalue must be t=a. for the other two, solve using quadratic formula.. i just wonder if a is unknown or just an arbitrary variable from you field..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2006
    From
    Copenhagen
    Posts
    39
    I tried to divide (t-a) with (t^2 - 3at + a^2), but in the end I had a^2 left, so I came up with no solution. I have tried once again to use the quadratic formula -- and this time the answer seems more plausible... I have tried to set a to 5 and plot it on my ti-83 and now the roots are okay. Strange how I found it to be wrong the first time.

    Thanks. You're quite active here.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    hehe.. if you divide your quadratic equation by a and got a remainder, then a must be a unique eigenvalue of your matrix..

    hmm, for some reasons, i am active because it saturday here and i have a lot of time checking the site.. but during weekdays, i am in school almost the whole day that is why when i come back home, it is already late and i have to sleep for tomorrow..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Characteristic polynomial of a matrix
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: August 12th 2010, 02:56 AM
  2. Replies: 2
    Last Post: August 11th 2010, 05:26 AM
  3. Characteristic Polynomial
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 21st 2010, 03:40 PM
  4. Replies: 1
    Last Post: December 15th 2009, 08:26 AM
  5. Replies: 1
    Last Post: November 6th 2009, 01:48 AM

Search Tags


/mathhelpforum @mathhelpforum