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Jordan form of a matrix

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data
    I need to find the Jordan Normal (canonical) form of the following matrix:

    {(3, 1, 0, 0); (-4, -1, 0, 0); (7, 1, 2, 1); (-17, -6, -1 0)}

    Where each ( , , , ) is a row.


    2. Relevant equations

    There are not really any equations involved, but rather theorems involving Jordan form.

    3. The attempt at a solution

    I have found that the characteristic polynomial is (x - 1)^4. So therefore I my four eigen values are all 1.

    I am a little confused with how to proceed. It seems my Jordan form will now have all 1's on the main diagonal but I am not sure how to find out what my blocks will look like.

    Now I went ahead and reduced my matrix (after taking 1 away from all entries of the main diagonal) to rref. I got

    {(1, 0 , 0 , 0); (0, 1, 0, 0); (0, 0, 1, 1); (0, 0, 0, 0)}

    I think that means that my eigen space has dimension 3. But I am not sure what that tells me about my Jordan form.

    Someone please help me out.

    Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 11, 2008 #2

    HallsofIvy

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    Assuming that x= 1 is an eigenvalue, then the matrix equation is
    [tex]\left[\begin{array}{cccc}3 & 1 & 0 & 0 \\ -4 & 1 & 0 & 0 \\ 7 & 1 & 2 & 1 \\ -16 & -6 & -1 & 0\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \\ u\end{array}\right]= \left[\begin{array}{c} x \\ y \\ z \\ u \end{array}\right][/tex]
    which gives the four equations
    3x+ y= x, -4x+ y= y, 7x+ y+ 2z+ u= z, and -16x- 6y- z= u which reduce to
    2x+ y= 0, -4x = 0, 7x+ y+ z+ u= 0, and -16x- 6y- z- u= 0.

    From the second equation x= 0 so the first equation becomes y= 0. Putting x= y= 0 in the last two equations, z+ u= 0 and -z- u= 0, which are satisfied by z= -u for any u. That is, the eigenspace is spanned by (0, 0, -1, 1) and is one dimensional.

    Actually, if I have done the determinant correctly, 1 is a double root of the eigenvalue equation but there are two other eigenvalues.
     
    Last edited: Dec 11, 2008
  4. Dec 11, 2008 #3

    Dick

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    Hi, Halls. You've got a sign error in the second row. It's supposed to be [-4,-1,0,0]. The characteristic polynomial IS (x-1)^4.
     
  5. Dec 11, 2008 #4
    I just reworked calculating the characteristic polynomial, and I again got (x-1)^4

    I can't understand what I am doing wrong.

    Det(A-xI) =
    (3-x)(-1-x)((2-x)(-x) + 1) - ((-4)((2-x)(-x) + 1)) =

    (3 - x)(-1-x)(x^2 - 2x + 1) + 4(x^2 - 2x + 1) =

    (x^2 -2x - 3)(x^2 - 2x + 1) + 4(x^2 - 2x + 1) =

    ((x^2 -2x - 3) + 4)(x^2 - 2x + 1) =
    (x^2 - 2x + 1)(x^2 - 2x + 1) =

    (x-1)^4

    Right?
     
  6. Dec 11, 2008 #5
    OK so assuming that I have the right characteristic, am I any closer to finding Jordan form?
     
  7. Dec 11, 2008 #6

    Dick

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    I think your next step is to figure out how many linearly independent eigenvectors there are, if I recall correctly. Fix Halls' mistake and solve that system.
     
  8. Dec 11, 2008 #7
    Alright

    I get that x & y can be any value,

    z = -10x + 7y

    and

    u = 7x + y

    So I guess that gives only two eigen vectors

    (1,0, 7, -10) and (0, 1, 1, 7). Now what?
     
  9. Dec 11, 2008 #8

    Dick

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    You probably know more about this than I do. I think you want to complete those two eigenvectors to a basis by finding generalized eigenvectors. Better check the book. And you'd better check those eigenvectors. I don't think they work.
     
  10. Dec 11, 2008 #9
    Woops should vectors should be

    (1,0,-10,7) and (0,1,7,1)
     
  11. Dec 11, 2008 #10

    HallsofIvy

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    Corrected version. Thanks for catching that, Dick. I also has "-16" where it should have been "-17".

    Assuming that x= 1 is an eigenvalue, then the matrix equation is
    [tex]\left[\begin{array}{cccc}3 & 1 & 0 & 0 \\ -4 & -1 & 0 & 0 \\ 7 & 1 & 2 & 1 \\ -17 & -6 & -1 & 0\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \\ u\end{array}\right]= \left[\begin{array}{c} x \\ y \\ z \\ u \end{array}\right][/tex]
    which gives the four equations
    3x+ y= x, -4x- y= y, 7x+ y+ 2z+ u= z, and -17x- 6y- z= u which reduce to
    2x+ y= 0, -4x-2y = 0, 7x+ y+ z+ u= 0, and -17x- 6y- z- u= 0.

    The first two equations both reduce to y= -2x. Putting y= -2x in the last two equations, 7x- 2x+ z+ u= 5x+ z+ u= 0 and -17x+ 12x- z- u= -5x- z- u= 0, both of which reduce to z+ u= -5x. We can then write z= -5x- u and y= -2x. Taking x= 1, u= 0, y= -2 and z= -5. One basis vector for the eigenspace is (1, -2, -5, 0). Taking x= 0, u= 1, y= 0 and z= -1. The other basis vector for the eigenspace, which is two dimensional, is (0, 0, -1, 1).

    The Jordan Normal form can be written as
    [tex]\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\end{array}\right][/tex]

    Essentially, the top and last rows represent the two dimensional eigenspace while the middle two rows are the "Jordan block" for the missing dimensions.
     
  12. Dec 11, 2008 #11
    Thanks Halls. I think I almost understand.

    But I don't get how "take" x= 1, u= 0, y= -2 and z= -5 and then take x= 0, u= 1, y= 0 and z= -1.

    And then I also don't quite see how the basis leads to the Jordan form. Can you explain a bit?

    Again I really appreciate your help.
     
  13. Dec 11, 2008 #12

    Dick

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    Ok, I took some time to actually review the Jordan form stuff. Halls' answer is actually wrong. If you let M be your original matrix and J be the matrix Halls gave, then you can check (M-I)^2=0, but (J-I)^2 is not zero ((J-I)^3 is zero). You need two 2x2 Jordan blocks to make it work. To actually construct the Jordan basis, find the two eigenvectors e1 and e2. Now find two vectors f1 and f2 such that (M-I)f1=e1 and (M-I)f2=e2. These are the generalized eigenvectors. Now write M in the basis {e1,f1,e2,f2} to see the Jordan form.
     
  14. Dec 11, 2008 #13

    HallsofIvy

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    Unfortunate how often that happens. I should have been more careful. Thanks for the correction.

     
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