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Linear Algebra: Similar matrices *Prove*

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data
    if A is similar to B with P^-1 A P = B and x is an eigenvector of A, prove that P^-1x is an eigenvector of B


    2. Relevant equations



    3. The attempt at a solution

    so far I did this:

    A = PBP^-1
    Ax = PBP^-1x
    P^-1Ax = BP^-1x
    B^-1P^-1Ax = P^-1x

    is that sufficient to prove that P^-1x is the eigenvector of B? ??

    I don't think so .. if yes please explain how.. or else please give a proper "proof" .

    I guess the reason I don't think this is right is becaues I am not able to bring it to the format of Ax = tx where t is an eigenvalue and x is the eigenvector. Any help would be appreciated.


    Final exam on tuesday ... :(
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 9, 2009 #2

    jbunniii

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    No, but you're on the right track. You got as far as

    [tex]P^{-1}Ax = BP^{-1}x[/tex]

    Now you need to use the fact that x is an eigenvector of A, so you can replace Ax in this equation with something else, and then you're essentially done.
     
  4. May 9, 2009 #3
    so that would mean:
    Ax = tx Where t is the eigenvalue:

    so [tex]P^{-1}tx = BP^{-1}x[/tex]
    [tex] tx = PBP^{-1}x [/tex]
    and we know [tex]PBP^{-1} [/tex] is A
    so we essentially have [tex]tx = Ax [/tex]

    but how does this show that [tex] P^{-1}x [/tex] is the eigenvector of B?
     
  5. May 9, 2009 #4

    jbunniii

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    You can see it from the first equation that you wrote:

    [tex]P^{-1}tx = BP^{-1}x[/tex]

    Rearrange the left hand slightly, then add some parentheses in appropriate places on both sides, until you see something that looks like an eigenvalue/eigenvector relationship. Notice that this shows something interesting beyond what was asked for: in particular, compare the eigenVALUE of B corresponding to [tex]P^{-1}x[/tex] to the eigenvalue of A corresponding to x.
     
  6. May 9, 2009 #5
    I SEE IT!!!

    [tex] t(P^{-1}x) = B(P^{-1}x) [/tex]

    cool!
    Thanks a lot.
     
  7. May 9, 2009 #6

    jbunniii

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    Isn't that "ah ha!" moment great?

    Cheers!
     
  8. May 9, 2009 #7
    absolutely!
     
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