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Polynomial of Jordan Decomposition

  1. Apr 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Let A be a real or complex nxn matrix with Jordan decomposition A = [tex]X \Lambda X^{-1}[/tex] where [tex]\Lambda[/tex] is a diagonal matrix with diagonal elements [tex]\lambda_1,...,[/tex] [tex]\lambda_n[/tex]. Show that for any polynomial p(x):
    p(A)=[tex]Xp(\Lambda)X^{-1}[/tex]

    [tex]p(\Lambda)[/tex] should really be the matrix with p([tex]\lambda_j[/tex]) on its diagonal for j=1,...,n but I couldn't figure out how to make that matrix in latex.

    3. The attempt at a solution
    I'm guessing there should be a way to take p of both sides and somehow extract the X and X inverse, but I can't seem to figure it out. Does anyone see anything? Thank you.
     
    Last edited: Apr 2, 2009
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  3. Apr 2, 2009 #2

    tiny-tim

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    Hi azdang! :wink:

    (for matrices etc in LaTeX, see http://www.physics.udel.edu/~dubois/lshort2e/node56.html#SECTION00850000000000000000 [Broken])

    Hint: if the polynomial is A2, then A2= XΛX-1XΛX-1 = XΛ2X-1 = … ? :smile:
     
    Last edited by a moderator: May 4, 2017
  4. Apr 2, 2009 #3
    Hey tiny-tim. Thank you for your response. Quick question: Is it okay to assume p(A) = [tex]A^2[/tex]? (This would be the polynomial p(x)=[tex]x^2[/tex] right?) Can we say that because the problem say 'for any polynomial'? Or is there a more general way to prove this?

    Also, I do see then that [tex]\Lambda^2=p(\Lambda)[/tex], so that is all clear. Thank you very much. I'm just wondering if we should find a way to represent p(x) more generally?
     
  5. Apr 2, 2009 #4

    tiny-tim

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    Yes, we need to deal with a general p(x).

    So you do still need to prove it for p(A) = An, for any n …

    my n = 2 was just an example for you. :wink:
     
  6. Apr 2, 2009 #5
    Alright, cool.

    So, we could say p(x) = xn.

    Then, p(A)=An=[tex](X \Lambda X^{-1})^n[/tex]=[tex]X \Lambda X^{-1}X \Lambda X^{-1}...X \Lambda X^{-1}[/tex].

    All of the X's and X-1 will cancel out except for the ones on the edges so we are left with: p(A)=[tex]X \Lambda^n X^{-1}[/tex].

    So, we know p([tex]\Lambda[/tex])=[tex]\Lambda^n[/tex]. Can we say that that is equivalent to the matrix with p([tex]\lambda_j[/tex]) on its diagonal where j=1,...,n because it is a diagonal matrix?
     
  7. Apr 3, 2009 #6

    tiny-tim

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    just got up :zzz: …

    Yes, of course (for Λm). :wink:

    (I don't think the examiner would even expect a reason for that :wink:)

    And then prove that it's "additive", in the sense that if B = XΛBX-1 and C = xΛCX-1 then ΛB+C = … ? :smile:
     
  8. Apr 3, 2009 #7
    I'm kind of confused by this last part. I don't really understand the subscripts on Lambda or where B and C are coming from. Thank you again, tiny-tim, for helping me get through this :)
     
  9. Apr 3, 2009 #8

    tiny-tim

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    I mean, if B and C are real or complex nxn matrices with Jordan decomposition B = XΛBX-1 and C = XΛCX-1 and B + C = XΛX-1 where Λ ΛB and ΛC are diagonal matrices, what is the equation relating Λ ΛB and ΛC ? :smile:
     
  10. Apr 3, 2009 #9
    But we cannot be guaranteed that the X's in the Jordan decompositions for B and C are the same, can we?
     
  11. Apr 3, 2009 #10

    tiny-tim

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    ?? :confused:

    Assume they are … then prove the proposition … and then use it to answer the original question.
     
  12. Apr 3, 2009 #11
    Well then, does [tex]X\Lambda_B X^{-1} + X\Lambda_C X^{-1}=X(\Lambda_B + \Lambda_C)x^{-1}[/tex]?

    Then [tex]p(\Lambda_B + \Lambda_C)=p(\Lambda_B)+p(\Lambda_C)[/tex], so the matrix would be the diagonal matrix with [tex]\lambda_B_j + \lambda_C_j[/tex] on the diagonal, j=1,...n, right? I'm not sure if I'm understanding all this, or why we are showing it is additive. Sorry!!
     
  13. Apr 3, 2009 #12

    tiny-tim

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    That's right! :smile:
    Because it's easy to prove the question for p(A) = Am, but not for p(A) = Am + Al

    so we just prove it for Am, show that it's the same X for any m, and then use the theorem above to prove that it's true for any sum of powers of A, ie for any polynomial. :wink:
     
  14. Apr 3, 2009 #13
    Okay, I think I'm understanding, though I might have a hard time explaining it. One thing in what you just said that confused me, what do you mean 'show that it's the same X for any m'. I really can't thank you enough for walking me through this btw.
     
  15. Apr 3, 2009 #14

    tiny-tim

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    I mean, for example, A2 = XΛ2X-1 and A3 = XΛ3X-1 (and it's the same X, which was worrying you earlier :wink:)
     
  16. Apr 3, 2009 #15
    Oooh, okay. Yes, I can see right away that they are the same X!

    So, I should really just show this for:
    p(x)=[tex]\alpha_0+\alpha_1x+\alpha_2x^2+...+\alpha_nx^n[/tex] for scalars alpha.

    P.S. I totally just did this ^ and I TOTALLY get it. Again, thank you very very much!!
     
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