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Help finding a transition matrix between the Jordan form and a general form

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data

    First year-linear algebra (Proof based... and this is my first exposure to proofs so I'm like... lol). This question is pretty computational though.

    Find J, The Jordan Canonical form of a Given Matrix A, and an invertible Matrix Q such that J = Q(A)(Q^-1)



    2. Relevant equations
    The matrix is a 3x3 matrix with entries
    (0 1 -1)
    (-4 4 -2) = A
    (-2 1 1)


    3. The attempt at a solution

    I think I've got the first part, but I really want to understand this stuff thoroughly so I'd like to have my "justifications" checked.

    First we find the eigenvalues of A, by calculating det ([tex]\lambda[/tex]I-A) and factoring the characteristic polynomial. In this case the Characteristic polynomial is [tex]\lambda^{3}[/tex]-5[tex]\lambda^{2}[/tex]+8[tex]\lambda[/tex]-4. Factoring gives the eigenvalues 1, and 2 multiplicity 2.

    Since 1 has multiplicity one, the corresponding eigenspace cannot have dimension greater than one, therefore there is a single eigenvector of 1 which spans the entire space.

    However, 2 has multiplicity of 2, so \exists some v_{1}, v_{2} in Ker (A-2I)^{2} and some u_{1}, u_{2} in Ker (A-2I) such that (A-2I)v_{1}=u_{1}, and similarly for v_{2}. Which implies there are vectors in Ker (A-2I) such that Av_{1}= u_{1} + 2v_{1}, similarly for v_{2}

    (1 0 0)
    (0 2 1) = J
    (0 0 2)

    (In this particular case, I think there exist two linearly independent vectors in Ker (A-2I), so perhaps generalized eigenvectors are not necessary? I'd like confirmation, as in this case the matrix would be diagonal)

    Now after finding J-form, I need to find a matrix Q which satisfies the relation in the problem statement. I tried adjoining the eigenvectors I calculated {(1,2,1),(1,0,2),(0,1,1)} but that was ineffectual. Then I tinkered around with row operations and found one matrix which produced the desired effect, but that's not helpful. I think I might need to find a particular basis, but I'm not sure what properties my basis needs to satisfy (My class is using Axler's LA done right, and it doesn't really have much in the way of algorithms) / how to go about "choosing" in order to construct the transition matrix. I'd prefer if possible, a "tip" in the right direction rather than an outright solution.

    Thanks!
     
    Last edited: Nov 15, 2008
  2. jcsd
  3. Nov 15, 2008 #2
    Ok, I just re-worked my calculations.

    For Ker (T-I) I got the condition v_{1}=2v_{2}=v_{3}, implying Ker (T-I) is spanned by (1, 2, 1) = the eigenvector.

    For Ker (T-2I) the the set of spanning vectors to be (1,0,2), and (0,1,1). So why isn't the U_{1}[tex]\oplus[/tex]U_{2}[tex]\oplus[/tex]U_{3} the transition matrix in question? (The matrix Q which satisfies that relation...)

    Or am I missing something still?
     
  4. Nov 15, 2008 #3

    morphism

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    (1,0,2) is not an eigenvector of A.

    And the Jordan form of A is a diagonal matrix, because dimker(A-2) = 2 = algebraic multiplicity of 2.
     
  5. Nov 16, 2008 #4
    Blah, typo, meant (1,0,-2) for the second eigenvector.

    But their direct sum still doesn't satisfy the condition for the Matrix of Q.

    What am I missing?
     
    Last edited: Nov 16, 2008
  6. Nov 16, 2008 #5

    morphism

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    Homework Helper

    The direct sum of what?

    If you let

    [tex]Q = \begin{pmatrix} 1 & 1 & 0 \\ 2 & 0 & 1 \\ 1 & -2 & 1 \end{pmatrix},[/tex]

    then

    [tex]Q^{-1}AQ = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}.[/tex]
     
  7. Nov 16, 2008 #6
    Really? That's the matrix I calculated... I guess I must've just been messing up the matrice multiplication or something.

    Thanks.
     
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