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Jordan Form- Please confirm my solution

  1. May 31, 2010 #1
    1. The problem statement, all variables and given/known data
    edit, sorry its not jordan form, I misplaced words.
    Hey,
    Let A be an nxn matrix over some field F. [tex] A \neg 0 [/tex] and [tex] A^{2009}+A^{2007}=0 [/tex]. Is a diagnizable?

    Attempt at solution
    we can notice that [tex] A^{2009}+A^{2007}=0 \iff A^{2007}(A^2+I)=0 \iff A^{2007}=0 or A^2=-I [/tex]
    If [tex] A^2 = -I [/tex] then a is a diagnol matrix with eigenvalues i and -i (i is defined as the sqaure root of -1 in the field. if the field does not have sucha number then A is not diagnizable).
    If [/tex] A^2 \neq -I [/tex] then[tex] A^{2007}=0 [/tex] but the A is nillpotent and so has less the n eigenvectors, so A is not diagnizable. q.e.d.
    Is this correct?
    Thanks
    Tal
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: May 31, 2010
  2. jcsd
  3. May 31, 2010 #2

    Mark44

    Staff: Mentor

    You are omitting a possibility. For square matrices A and B, it is possible for AB = 0 where neither A nor B is the 0 matrix.

    Here's a simple example.
    [tex]\left[
    \begin{array}{c c} 0 & 1 \\ 0 & 0 \end{array} \right]
    \left[
    \begin{array}{c c} 0 & 2 \\ 0 & 0 \end{array} \right] =
    \left[
    \begin{array}{c c} 0 & 0 \\ 0 & 0 \end{array} \right][/tex]
     
  4. May 31, 2010 #3
    Hi Mark,
    thanks for your input. I think I'm missing your point.
    As I see it, I am only talking about AA, and not Ab, two different matrices. if B= A^2006 abnd AB=-0 tehn AB=A*A*A&2006=A^2007=0 which still leaves A nilpotent.
    If that is not the case then we need A^2=-I which still means that A is diagnozable iff sqrt(-1) exists in the field.
    What am I missing?
    Thanks
    Tal
     
  5. May 31, 2010 #4

    Mark44

    Staff: Mentor

    You are talking about two different matrices: A2007 and A2 + I. You said that since A2007(A2 + I) = 0, then it must be that A2700 = 0 or A2 + I = 0.

    What I said was that it is possible for the product of two matrices to be zero, where neither matrix is the zero matrix.

    It is true that if A = 0 or B = 0, then AB = 0, but the converse is not necessarily true, as my example shows.
     
  6. Jun 1, 2010 #5
    Ok, heres an improvment:
    Assume A is diagnizable. Let v be an eigenvector of A with an eigenvalue, c, that is not 0. then we have
    [tex] (A^{2009}+A^{2007})\vec(v) =A^{2009}\vec(v) + A^{2007}\vec(v) = c^{2009}\vec(v) + c^{2007}\vec(v)=0 \iff c=i or c=-i[/tex]
    If A is a 2x2 matrix over a field where i exists then the matrix is diagnizable. otherwise it i not.
    if RkA> 3 then for A to be diagnizable it would need other eigenvalues but none would atisfy the condition.
    Is that correct?
    Thanks
    Tal
     
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