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Finding Jordan canonical form of these matrices

  1. Mar 30, 2016 #1
    1. The problem statement, all variables and given/known data
    For each matrix A, I need to find a basis for each generalized eigenspace of ## L_A ## consisting of a union of disjoint cycles of generalized eigenvectors. Then I need to find the Jordan canonical form of A.
    The matrices are:
    ## a)
    \begin{pmatrix}
    1 & 1\\
    -1 & 3
    \end{pmatrix}

    b)
    \begin{pmatrix}
    1 & 2\\
    3 & 2
    \end{pmatrix}

    c)
    \begin{pmatrix}
    11 & -4 & -5\\
    21 & -8 & -11\\
    3 & -1 & 0
    \end{pmatrix}

    d)
    \begin{pmatrix}
    2 & 1 & & \\
    & 2 & 1 & \\
    & & 3 & \\
    & 1 & -1 & 3
    \end{pmatrix} ##

    2. Relevant equations
    ## P_A(t)=det(A-tI) ##
    ## K_\lambda = \left \{ (A-\lambda I)^j \right \}, 1\leq j \leq p ## where p is the minimum value for which ## (A-\lambda I)^j(x)=0## for a generalized eigenvector x

    3. The attempt at a solution
    I hope you don't mind if I link to a picture of what I have so far, as I don't want to go through the trouble of typing it all up.
    http://prntscr.com/am7si6
    The squiggles next to some of the matrices are just messily written words indicating that there is only one generalized eigenvector for a given basis.

    For (a), I determined that the only basis element was (1,1)
    For (b), I found 2 single-element basis which contained (1,-1) and (1,1.5)
    For (c), I once again found that each generalized eigenspace only had one generalized eigenvector. However, what I'm not sure of is how to determine what the Jordan blocks are in this case- i.e, I don't know how to determine if
    ## J=\begin{pmatrix}
    -1 & 1 & \\
    & -1 & \\
    & & 2
    \end{pmatrix} ## or
    ##J=\begin{pmatrix}
    -1 & & \\
    & 2 & 1\\
    & & 2
    \end{pmatrix}##
    For (d), I have not yet started. I just need to confirm that my current approach of raising the power of the generalized matrices to find generalized eigenvectors is correct, and I need to know how to figure out what the Jordan blocks are.
     
  2. jcsd
  3. Mar 31, 2016 #2

    Ray Vickson

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    The eigenvalues for (c) are -1,2,2 so the eigenvalue 2 has multiplicity two but eigenspace dimensionality one. Therefore, the Jordan block is for 2.
     
  4. Apr 1, 2016 #3

    vela

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    For (a), you have a 2x2 matrix, so the basis should comprise two vectors.
     
  5. Apr 5, 2016 #4
    How do I find the other one? For the matrix U, I don't know how I would find another eigenvector for what it is, and U2 is the zero matrix. 2 was also the only eigenvalue I was able to find.

    Not on any of my assignments, but how would I do it for a matrix with multiple blocks but only one eigenvalue?
     
  6. Apr 5, 2016 #5

    vela

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    If ##\vec{p}_1## is the eigenvector, the generalized eigenvector satisfies ##(A-\lambda I)\vec{p}_2 = \vec{p}_1##. That'll give you a more useful equation to solve.
     
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