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Eigenvector Proof

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Let B = S^-1 * A * S and x be an eigenvector of B belonging to an eigenvalue [tex]\lambda[/tex]. Show S*x is an eigenvector of A belonging to [tex]\lambda[/tex].


    2. Relevant equations



    3. The attempt at a solution
    The only place I can think of to start, is that B*x = [tex]\lambda[/tex]*x.
    However, even starting with that, I can't figure out where to go next.
    Could someone point me in the right direction?
     
  2. jcsd
  3. Apr 5, 2010 #2

    Mark44

    Staff: Mentor

    That's a decent start. Next, show that A(Sx) = [itex]\lambda[/itex]x. That's what it means to say that Sx is an eigenvector of A corresponding to [itex]\lambda[/itex].
     
  4. Apr 5, 2010 #3
    What do I use to show that?

    The only new information I've got that might be helpful is that A = S * B * S^-1

    Multiplying on the left by S gives A*S = S*B

    After doing that, I'm stuck again. I feel like this is the right track, but I don't know how to relate this back to what I'm trying to prove.
     
  5. Apr 5, 2010 #4

    vela

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    Staff Emeritus
    Science Advisor
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    Slight correction: You want to show that A(Sx) = [itex]\lambda[/itex](Sx)
    Multiply by x now.
     
  6. Apr 5, 2010 #5
    I think I got it now.

    After multiplying by x, I have ASx = SBx.

    Bx has already been shown equal to [tex]\lambda[/tex]x, so I substitute that in, giving

    ASx = S[tex]\lambda[/tex]x

    [tex]\lambda[/tex] can be moved to the other side of S since it's a scalar, giving ASx = [tex]\lambda[/tex]Sx.
     
  7. Apr 5, 2010 #6

    Mark44

    Staff: Mentor

    Right.
     
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