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Jordan normal form problem

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the Jordan normal form of:
    [tex]\begin{bmatrix}
    -2 & 3 & 1\\
    0 & -4 & -1\\
    1 & 2 & -3
    \end{bmatrix}[/tex]

    2. Relevant equations



    3. The attempt at a solution
    I found the eigenvalues: -3 (x3)
    I found a single eigenvector: [ 2 , -1 , 1 ]
    But now I thought of finding the nullspaces of (M + 3I)2 and (M + 3I)3 (where is the matrix we're studying) but they're empty.

    Thanks for your help.
     
  2. jcsd
  3. Jan 3, 2010 #2

    Dick

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    You've already shown that the nullspace of (M+3I) is not empty since you have an eigenvalue of -3. Then how can the nullspaces of (M+3I)^2 and (M+3I)^3 be empty?? That would be silly. In fact, (M+3I)^3=0 since it satisfies it's characteristic polynomial. I suggest you try that calculation again.
     
    Last edited: Jan 3, 2010
  4. Jan 3, 2010 #3

    HallsofIvy

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    Since -3 is a triple root, the characteristic equation must be (M+ 3)^3= 0. And, since every matrix satisfies its own characteristic equation, (M+ 3I)^3v= 0 for every vector v.
    There must exist vectors, u, such that Mu= 3u and you have found one. Since the eigenvectors do not span the space, there must exist vectors, v, such that Mv is NOT 0 but [itex](M+ 3I)^2 u= 0[/itex]. Since (M+3I)0= 0 certainly, look for a vector such that (M+ 3I)v= [2, -1, 1]. Then you will have [itex](M+ 3I)^2v= (M+ 3I)[2, -1, 1]= 0[/itex]. Once you have found that v, look for w such that [itex](M+ 3I)w= v[/itex].
     
  5. Jan 3, 2010 #4
    Thanks for your replies!

    Oh, this is bad. I was using Mathematica for my calculations, it turns out using 2 doesn't do a real matrix multiplication, it does a dot product or something similar. My bad...

    Now I took as eigenvectors:
    - (M + 3I)2 : [ 1 , 0 , 1 ]
    - (M + 3I)3 : [ 0 , 0 , 1 ]
    So that:
    [tex]P=\begin{bmatrix}
    2 & 1 & 0\\
    -1 & 0 & 0\\
    1 & 1 & 1
    \end{bmatrix}[/tex]

    But doing P·J·P-1 ≠ M
    However taking [ 1 , 0 , 0 ] instead of [ 0 , 0 , 1 ] works, why is that?
     
  6. Jan 6, 2010 #5
    Bump for help! :)
     
  7. Jan 6, 2010 #6

    Dick

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    I'm really not sure what your are doing. What's J? (M+3I)=0. Anything is an eigenvector of that.
     
  8. Jan 8, 2010 #7
    J is the Jordan normal form of M and P is the matrix so that P·J·P-1 = M.
     
  9. Jan 8, 2010 #8

    HallsofIvy

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    it's not clear to me how you got [0, 0, 1]. As I said before, you want a "v" such that [tex](M+3i)^2v= [2, -1, 1][/tex]. That means you have to solve
    [tex]\begin{bmatrix}1 & 3 & 1 \\ 0 & -1 & -1 \\ 1 & 2 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}[/tex][tex]= \begin{bmatrix}2 \\ -1 \\ 1\end{bmatrix}[/tex].
    That gives the equations x+ 3y+ z= 2 -y- z= -1, and x+ 2y= 1. The second equation is the same as z= -y+ 1 and the third is the same a x= -2y+ 1. Taking y= 0 would give v= (1, 0, 1) as you have.

    Now look for u such that (M+ 3I)u= v which means you must have
    [tex]\begin{bmatrix}1 & 3 & 1 \\ 0 & -1 & -1 \\ 1 & 2 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}[/tex].
    That gives the equations x+ 3y+ z= 1, -y-z= 0. and x+ 2y= 1. The second equation is the same as z= -y and the third equation x= -2y+1. Then the first equation is satisfied for all y. If you take y= 0, you get [1, 0, 1] the same as before, but if you take y= 1, you get u= [-1, 1, -1].

    Let P be the matrix having those vectors as columns,
    [tex]P= \begin{bmatrix}2 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & 1 & -1\end{bmatrix}[/tex]
    and you will find that
    [tex]P^{-1}AP= \begin{bmatrix}-3 & 1 & 0 \\0 & -3 & 1 \\ 0 & 0 & -3\end{bmatrix}[/tex].
     
  10. Jan 8, 2010 #9
    The thing is I was taking for the first column of P, an eigenvector of M. For the second one, a vector that is the nullspace of (M+3I)2 but is not an eigenvector of M. For the third one, a vector that is in the nullspace of (M+3I)3 (which is R3) but not in the eigenspace of (M+3I)2. Isn't that a right thing to do?
    I mean: (M+3I)3 = 0 so (M+3I)3·[ 0 , 0 , 1 ] = 0.
     
    Last edited: Jan 8, 2010
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