1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear Algebra - Jordan form basis

  1. Jun 29, 2012 #1
    Hi all,
    I'm having trouble finding jordan basis for matrix A, e.g. the P matrix of: [itex]J=P^{-1}AP[/itex]
    Given [itex]A = \begin{pmatrix} 4 & 1 & 1 & 1 \\ -1 & 2 & -1 & -1 \\ 6 & 1 & -1 & 1 \\ -6 & -1 & 4 & 2 \end{pmatrix}[/itex]

    I found Jordan form to be: [itex]J = \begin{pmatrix} -2 & & & \\ & 3 & 1 & \\ & & 3 & \\ & & & 3 \end{pmatrix}[/itex]

    Now wer'e looking for [itex]v_1, v_2, v_3, v_4[/itex] such that:

    [itex] Av_1 = -2v_1 → (A+2I)v_1=0[/itex]
    [itex]Av_2 = 3v_2 → (A-3I)v_2=0[/itex]
    [itex]Av_3 = v_2+3v_3 → (A-3I)v_3=v_2[/itex]
    [itex]Av_4 = 3v_4 → (A-3I)v_4=0 [/itex]

    So now I find: [itex]v_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \end{pmatrix} \hspace{10mm} v_2,v_4 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}[/itex]

    Now I try to solve [itex] (A-3I)v_3=v_2[/itex] for each of the possible v2's I just found above, but there's no solution for any of em'...

    [itex]A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix}\hspace{5mm} OR \hspace{5mm} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}[/itex]


    Where am I going wrong? Thanks in advance!
     
    Last edited: Jun 29, 2012
  2. jcsd
  3. Jun 29, 2012 #2

    I like Serena

    User Avatar
    Homework Helper

    Hi oferon! :smile:

    Did you consider that the proper v2 could be a linear combination of your current v2 and v4?
    What if you try ##\lambda v_2 + \mu v_4## to find v3?
     
  4. Jul 10, 2012 #3
    If it was a linear combination of other vectors then V1-4 would not be a basis.. Am I wrong?

    Plus, another student told me the method I tried was completly wrong and that the correct method is finding more vectors through
    [itex] Ker (A-λI)^j[/itex] where j=2,3,... depends on how many more vectors I need for my basis.

    Which of the methods should I use? Any why? I'm lost :confused:
     
    Last edited: Jul 10, 2012
  5. Jul 11, 2012 #4

    I like Serena

    User Avatar
    Homework Helper

    You need to find a ##v_3## that satisfies ##(A-3I)v_3=λv_2+μv_4##.
    When you find it, v1-v4 will form a basis.


    That would work too, but it seems to me that it is a lot more work.
    (Short story: that student is wrong. Your method is fine. You just did not finish it.)


    If you're wondering... try both?
     
  6. Jul 12, 2012 #5
    Hi, thanks for your kind replies.

    Ok, first I try what you suggested.. I take [itex] (A-3I)v_3 = λv_2+μv_4 [/itex] I get:

    [itex]\begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} λ \\ μ \\ λ \\ -2λ-μ \end{pmatrix} ----> \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 6 & 1 & -4 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} λ \\ λ+μ \\ λ \\ -λ-μ \end{pmatrix}[/itex]

    Now I see it must satisfy [itex] μ = -λ[/itex] so I pick [itex]λ=1, μ=-1[/itex] thus [itex] v_2-v_4=\begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}[/itex] so now I solve: [itex]\begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}[/itex]
    But the solutions I get are exactly [itex] \begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix} [/itex] The same v2,v4... So where am I wrong now?


    Second thing, I've searched all over the net, and found this method. Yet the method the other student told me is what was taught in class. Can I be 100% sure both methods are equivalent and can be used both in all cases?
    I thank you again for your time.
     
  7. Jul 12, 2012 #6
    Ok, so I asked our instructor about the second question and yes, both methods are good.
    I prefer "my" method, but as you can see I still get stucked with it.. So how do I move on with this [itex] (A-3I)v_3 = λv_2+μv_4 [/itex] ?
    Thanks again
     
  8. Jul 12, 2012 #7

    I like Serena

    User Avatar
    Homework Helper

    Can you find a 3rd solution that is independent of v2 and v4?
    (Let's say with the first 2 entries set to zero. ;)
     
  9. Jul 13, 2012 #8
    Hmm, ok I see what you say..
    So now I have 3 final questions to close this case for good:

    1) I thought all solutions were given by span of [itex] \begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix} [/itex]
    So where did this [itex] \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}[/itex] (tho I agree it IS a solution for this system) come from??

    2) How is it possible that [itex]v_2 , v_4[/itex] are solutions of both homogeneous and non-homogeneous
    [itex](A−3I)v_3=0[/itex] and [itex](A−3I)v_3=v_2-v_4[/itex]. I doubt if it was just by accident..

    3) Final question is how come I'm allowed to go from the equation I got by comparing columns of PJ and AP: [itex](A−3I)v_3=v_2[/itex], ,
    to the equation [itex](A−3I)v_3=λv_2+μv_4[/itex]?
    The third column in J matrix [itex] \begin{pmatrix} 0 \\ 1 \\ 3 \\ 0 \end{pmatrix}[/itex] clearly shows I should find [itex]Av_3=v_2+3v_3[/itex] , not [itex]Av_3=v_2+3v_3-v_4[/itex]

    I appreciate your help alot! Thank you.
     
  10. Jul 13, 2012 #9
    Oh ok, I discard my 3rd question... The answer is that I pick v2 to be [itex]\begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}[/itex]

    Now I remain only with questions 1, and 2.. More related to equations system rather than J form I suppose
     
  11. Jul 14, 2012 #10
    OK, please discard all of my question, I'm an idiot :)
    Everything is clear now, I thank you very much for the last time :)
     
  12. Jul 14, 2012 #11

    I like Serena

    User Avatar
    Homework Helper

    Okay... I just got around to looking at your thread again.
    But it seems you've already answered your own questions.

    Good! :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear Algebra - Jordan form basis
Loading...