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Jordan Normal Form Problem

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data
    f, g endomorphisms on R3.
    f non-diagonalizable and g injective.
    h = g o f
    dim h-1 { (0,0,0) } = 2
    f(1,0,1) = f(1,1,2)
    f(1,1,-1) = (1,-1,0)

    I have to find:
    a.- J, Jordan normal form of f
    b.- A, canonical basis matrix for f and transition matrix P (J = P-1AP)
    c.- inverse of (A20 + I)

    2. Relevant equations

    3. The attempt at a solution
    a.- dim Ker h = 2 with g injective so dim Ker f = 2.
    Therefore 0 is an eigenvalue with multiplicity 2.
    I don't know how to find the other eigenvalue.
    (I mean without using A, which is found in the next question)

    b.- dim Ker f = 2 so dim Im f = 1
    f(1,1,-1) = (1,-1,0)
    Therefore: Im f = L { (1,-1,0) }
    f(1,0,1) - f(1,1,2) = f(0,-1,-1) = (0,0,0)
    Therefore Ker f = L { (1,0,-1) (0,1,1) }
    (I thought any vector that's linearly independent with (0,1,1) would be OK)
    f(1,1,-1) - f (1,0,-1) = f(0,1,0) = (1,-1,0)
    f(0,1,1) - f(0,1,0) = f(0,0,1) = (-1,1,0)
    f(1,0,-1) + f(0,0,1) = f(1,0,0) = (-1,1,0)
    Therefore:
    [tex]A=\[ \left( \begin{array}{ccc}
    -1 & 1 & -1 \\
    1 & -1 & 1 \\
    0 & 0 & 0 \end{array} \right)\] [/tex]
    But this can't be OK, because then dim Ker A2 = 2 and then 0's multiplicity would be 1.

    c.- I suppose it's done using the Caley-Hamilton theorem once I have the eigenvalues, which I don't...
     
  2. jcsd
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