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Minimum polynomial and canonical form

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi all.

    I have no clue on how to do this problem because I missed the class where he covered this so could someone please walk me through it.

    A = [2 2 -5; 2 7 2; -5 -15 -4] where ; means new column

    p(x) = (x-3)(x-1)^2

    1) what are the choices of m(x)
    2) find the minimum polynomial
    3) find the jordan canonical form
    d) find P s.t. P^(-1)AP




    2. Relevant equations



    3. The attempt at a solution

    1)m(x) = (x-3)(x-1)^2

    and m(x) = (x-3)(x-1)

    2) no clue.. m(x) = (x-3)(x-1)?

    do i have to plug in A for x or something?

    3) no clue

    4) no clue


    I'm really sorry for this half a**ed attempt but I really do not have any material on this stuff. I tried researching the minimum polynomial problem but none of the examples are similar to mine. I just need a push in the right direction please


    Thanks!
     
  2. jcsd
  3. May 21, 2009 #2

    dx

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    Homework Helper
    Gold Member

    Hi squaremeplz,

    You found that the possible minimal polynomials are m1(x) = (x-3)(x-1)² and m2(x) = (x-3)(x-1).

    m2 has the smaller degree, so if m2(A) = 0, then m2 is the minimal polynomial. Otherwise, m1 is the minimal polynomial. So all you have to do is check if (A - 3)(A - 1) = 0.

    (Do you know the definition of minimal polynomial?)
     
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