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Finding the Jordan canonical form of a matrix

  1. Jun 11, 2017 #1
    1. The problem statement, all variables and given/known data
    About an endomorphism ##A## over ##\mathbb{C^{11}}## the next things are know.
    $$dim\, ker\,A^{3}=10,\quad dim\, kerA^{2}=7$$
    Find the
    a) Jordan canonical form of ##A##
    b) characteristic polynomial
    c) minimal polynomial
    d) ##dim\,kerA##
    When:
    case 1: we know that ##A## is nilpotent
    case 2: we know that ##tr(A)=0##
    2. Relevant equations


    3. The attempt at a solution
    So case 1:
    ##A## is nilpotent
    therefore we know that there exists some number ##n## such that ##A^{n}=0## and since ##A^{3
    }\neq0## that must mean that ##A^{4}## must equal 0.
    ( n cannot be greater than 4 because then the dimension of its kernel would exceed the dimension of ##\mathbb{C^{11}}##
    so taking into account ##dim\, ker\,A^{3}=10,\quad dim\, kerA^{2}=7## I get .

    Slika nove bitne slike.jpg
    ##\begin{bmatrix}
    0&1&&&&&&&&&\\
    &0&1&&&&&&&&\\
    &&0&1&&&&&&&\\
    &&&0&&&&&&&\\
    &&&&0&1&&&&&\\
    &&&&&0&1&&&&\\
    &&&&&&0&&&&\\
    &&&&&&&0&1&&\\
    &&&&&&&&0&1&\\
    &&&&&&&&&0&\\
    &&&&&&&&&&0\\



    \end{bmatrix}
    ##
    characteristic polynomial ##p(\lambda)=\lambda^{11}##
    minimal polynomial ##m(\lambda)=\lambda^{4}##
    ##dim\,kerA=4##



    Case 2:
    ##tr(A)=0## here is where I get confused.
    I know that
    ##A=PJ(A)P^{-1}##
    therefore
    ##
    tr(A)=tr(PJ(A)P^{-1})\\
    tr(A)=tr(P)tr(J(A))tr(P^{-1})\\
    tr(A)=tr(PP^{-1})tr(J(A))\\
    tr(A)=tr(J(A))\\
    ##
    however the first 10 eigenvalues of J(A) are 0 so wont case 2 just be the same as case 1?
    Thanks for your help
     
  2. jcsd
  3. Jun 11, 2017 #2

    StoneTemplePython

    User Avatar
    Gold Member


    a few things. One is I'm a bit concerned that you're treating the trace operation like a determinant... They are related and important, but rather different. Trace does have a cyclic property, but in general does *not* allow you to split it apart and multiply like determinants.

    The end should read:

    ##
    tr(A)=tr(PJ(A)P^{-1})\\
    tr(A)=tr(P^{-1}PJ(A))\\
    tr(A)=tr(I J(A))\\
    tr(A)=tr(J(A))\\
    ##

    I may have mis-understood your question about part 2, but I'll give it a go anyway.

    Here is the difference between a nilpotent matrix and case 2. For n x n matrices:

    Case one:

    ##trace\big( (A^k)\big) = 0##

    for ##\{k = 1, 2, 3, ... , n\}##

    (you can actually keep counting out all natural numbers but I am happy to stop on k = n).

    At some point it is worth proving that that an n x n matrix is nilpotent if and only if

    ##trace\big( (A^k)\big) = 0##
    for ##\{k = 1, 2, 3, ... , n\}##

    Case two:

    ##trace\big( (A^1)\big) = 0##

    ##trace\big( (A^r)\big) = ?##
    for ##\{r = 2, 3, ... , n\}##

    simple example for case 2
    ## A =
    \begin{bmatrix}
    1 & 2\\
    0 & -1
    \end{bmatrix}##

    This is not nilpotent. But ##trace\big( (A^1)\big) = 0##
     
    Last edited: Jun 12, 2017
  4. Jun 12, 2017 #3

    StoneTemplePython

    User Avatar
    Gold Member

    follow-up (edit button is disabled)

    I see what you are saying now.

    ##A^3## has 10 linearly independent vectors in its nullspace, and one vector not in the nullspace.

    hence ##trace\big(A^3\big) = \lambda_1^3##

    But we know ##trace\big(A \big) = \lambda_1 + \big(\sum_{k=2}^n \lambda_k\big) = 0##

    but each ##\lambda_k## must be zero or else ##A^3## cannot be a rank one matrix. (When you look at ##A## as being similar to an upper triangular matrix, whether Jordan Form or Schur decomposition, you can lower bound its rank by the number of non-zero eigenvalues -- a basic Gaussian Elimination and pivot counting argument --, so if there were 2 non-zero eigenvalues it would be at least rank 2, but we know ##A^3## is rank one.)

    ##trace\big(A \big) = \lambda_1 + \big(0\big) = 0##, thus ##\lambda_1 = 0##

    and we have ##\lambda_1 = 0##, which brings back the nilpotent case. A bit of a trick question I suppose.
     
    Last edited: Jun 12, 2017
  5. Jun 12, 2017 #4
    thank you for point out the trace mistake I completely missed that
     
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