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Jordan Basis question.

  1. May 3, 2007 #1
    i have this matrix:
    Code (Text):

    011
    001
    000
     
    and i need to find the matrix jordan basis, and jordan form.
    for the jordan bassis i need first to find the eigen value which is zero.
    i also found that: Ker(A)={e_1} Ker(A^2)={e_1,e_2} Ker(A^3)=R^3
    now Ae1=0, Ae2=e1, Ae3=e1+e2.
    now the basis must include e2 and e1, cause we have a transformation from e_2 in Ker(A^2) to e_1 in ker(A), but we don't have a transformation from Ker(A^3) to Ker(A^2), Ae3 should be one of either e2 or e1, can someone help me on this?
    thanks in advance.
     
  2. jcsd
  3. May 3, 2007 #2

    mathwonk

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    thid matrix has minimal polynomial X^3 hence the jordan form is the given matrix but with the upper right "1" removed.

    cant you just replace e3 by e3-e2?
     
    Last edited: May 3, 2007
  4. May 3, 2007 #3
    yes i got this polynomial, my question is how do i find the jordan basis?

    as i stated so far e1 and e2 are in this basis but e3 isn't.
    there's an algorithm which was shown to me in class, that in order to find the JB, you need first to find the cyclic bases i.e the kernels, and then you find a vector in the basis of R^3 which isn't in Ker(A^2), and then you multiply by A and find the next vector in Ker(A^2) and now you choose a vector which is in Ker(A^2) and not in Ker(A) and do it again.

    does it mean that the Jordan basis is {e_2,e_1}?
     
  5. May 3, 2007 #4

    mathwonk

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    re read my post, i added the basis.
     
  6. May 3, 2007 #5
    ok, that will do.
    and that means that the basis is {e3-e2,e1,e2} right?

    ok, i can proceed from here.
    thanks.
     
  7. May 3, 2007 #6

    mathwonk

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    well it depends on whetehr you want upper or lower, i gave you upper.


    didnt i? and yours seems to be neither.

    but as you said you can take it from here.
     
  8. May 3, 2007 #7
    another question with finding JB:
    this time i have this matrix:
    Code (Text):

    0100
    0001
    0000
    0000
     
    let's denote it A, then A^2 equals:
    Code (Text):

    0001
    0000
    0000
    0000
     
    and A^3=0.
    i got that Ker(A)={e1,e3} Ker(A^2)={e1,e2,e3}
    KerA^3=R^4
    now Ae4=e2 and Ae2=e1. so iv'e got 3 vectors and i need four, i tried your way by substracting with knowing that Ae3=Ae1=0.
    but it doesnt seem to work, i tried with replacing e3 with e2-e3 but it doesnt work as i said already.
    if it matters the the char polynomial is x^4 and minimal polynomial is x^3.
     
  9. May 3, 2007 #8
    btw, for the first question what is the lower JB?
    i know that JB isn't unique, but i havent heard of upper lower basis.
     
  10. May 3, 2007 #9

    mathwonk

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    well the upper one gives the 1's above the diagonal,....
     
  11. May 3, 2007 #10
    isnt Jordan Form is with ones above the diagonal? i guess you can also define it with 1's below, but i havent learned this part, and i think i wasnt about to find this anyway.
    about the second matrix, do you have another simple trick down your sleeve? (im not sure you spell it this way).
     
  12. May 4, 2007 #11
    ok i figured out this second matrix.

    i have another question, let A be a matrix of nxn, i need to prove that the char polynomial of A divides m(x)^n where m(x) is the minimal polynomial.
    well iv'e proven it by stating that m(x) and p(x) have the same linear factors,
    m(x)=(x-a1)^l1...(x-a_m)^l_m and p(x)=(x-a1)^k1...(x-a_m)^k_m where k1+...+k_m=n and 1<=l1,..,l_m<=k1,..k_m<=n so by raising m(x) we achieve higher degress of the factors and thus m(x)^n is divisble by p(x).
    now iv'e checked in the book and it has a different proof, is my proof valid here?
     
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