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Eigenvalues question.

  1. Jun 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Let B be an invertible matrix

    a.) Verify that B cannot have zero as an eigenvalue.

    b.) Verify that if [tex]\lambda[/tex] is an eigenvalue of B, then [tex]\lambda[/tex][tex]^{-1}^[/tex] is an eigenvalue of B[tex]^{-1}[/tex].

    2. Relevant equations
    Bv = [tex]\lambda[/tex]v, where v[tex]\neq[/tex]0


    3. The attempt at a solution
    a.) I'm pretty sure that I need to manipulate the eigenvalues definition above so that I end up with v = 0, thus contradicting the definition.

    What I have so far is:
    Bv = [tex]\lambda[/tex]v
    B[tex]^{-1}[/tex]Bv = B[tex]^{-1}[/tex][tex]\lambda[/tex]v
    Iv = B[tex]^{-1}[/tex][tex]\lambda[/tex]v
    v = [tex]\lambda[/tex]B[tex]^{-1}[/tex]v
    If [tex]\lambda[/tex] = 0
    v = 0, which contradicts the definition for eigenvalues where v[tex]\neq[/tex]0
    Therefore [tex]\lambda[/tex][tex]\neq[/tex]0

    I think I am missing some crucial steps and kinda jumped ahead in my working. Am I on the right track? How can I approach this problem properly?

    b.) I'm thinking of a similar approach for b.), where I would need to use the above definition and multiply through by the inverse B, and then maybe take the reciprocal of lambda.

    However I'm not sure given that I can't do the first part yet. Any help would be much appreciated.


    Daniel
     
  2. jcsd
  3. Jun 12, 2009 #2

    rock.freak667

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    Normally, I'd say that looks like it'd work, but I am not sure as I only did this topic like 3 years ago.


    and for the first one, I'd just use what the product of the eigenvalues would give in relation to B and then show how it would contradict the statement of what B is given [itex]\lambda=0[/itex]
     
  4. Jun 12, 2009 #3

    Hurkyl

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    I suspect that it's that eerie feeling you get when you really start to understand something, and are surprised at how straightforward it all seems!

    Just to organize what you've done a little bit (and probably with too much detail), your arithmetic has shown

    Given a vector v, scalar [itex]\lambda[/itex], and matrix B:

    If [itex]Bv = \lambda v[/itex] and B is invertible
    Then [itex]v = \lambda B^{-1} v[/itex]

    If, in addition, [itex]\lambda = 0[/itex],
    Then [itex]v = 0[/itex]
    And you have correctly argued
    If B is invertible, [itex]\lambda[/itex] is an eigenvalue, and [itex]\lambda = 0[/itex]
    Then we have a contradiction​
    And correctly concluded
    If B is invertible, [itex]\lambda[/itex] is an eigenvalue
    Then [itex]\lambda \neq 0[/itex]​
     
  5. Jun 13, 2009 #4

    HallsofIvy

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    For (b), if [itex]Bv= \lamba v[/itex] then, as you have, [itex]B^{-1}Bv= B^{-1}\lambda v[/itex] or [itex]v= \lambda B^{-1}v[/itex]. Now divide on both sides by [itex]\lambda[/itex].
     
  6. Jun 13, 2009 #5
    I remember doing these problems. I think this is how I did part a:

    Bv=[tex]\lambda[/tex]v
    Bv-[tex]\lambda[/tex]v=0
    (B-[tex]\lambda[/tex])v=0

    If [tex]\lambda[/tex]=0 then Bv=0 which can only happen if B is not invertible, which is a contradiction.


    As for part b: If Bv= [tex]\lambda[/tex]v then B-1v=[tex]\lambda[/tex]-1v
    Then B-1v-[tex]\lambda[/tex]-1v=0
    (B-1-[tex]\lambda[/tex]-1)v=0 and you now have the definition of an eigenvalue, where [tex]\lambda[/tex]-1 is an eigenvalue for B-1.
     
    Last edited: Jun 13, 2009
  7. Jun 13, 2009 #6

    Hurkyl

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    You know, you guys shouldn't be doing his homework for him. :grumpy:

    Fortunately, from the opening poster's comments, I'm sure he had worked it out already.
     
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