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(distinct eigenvalues) help!

  1. Dec 5, 2005 #1
    Let A be an nxn mx with n distinct eigenvalues and let B be an nxn mx with AB=BA. if X is an eigenvector of A, show that BX is zero or is an eigenvector of A with the same eigenvalue. Conclude that X is also an eigenvector of B.



    I could show BX is zero or is an eigenvector of A with the same eigenvalue, but i dont know how to Conclude that X is also an eigenvector of B. Does anyone know how to do it? Thanks!
     
  2. jcsd
  3. Dec 5, 2005 #2

    JasonRox

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    Why don't we ever get neat questions like that?

    I have an idea how to tackle this, but I can't help you at this moment.
     
  4. Dec 5, 2005 #3
    why you cannot help me?
     
  5. Dec 6, 2005 #4

    matt grime

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    The answer follows from all the definitions given in one line:

    If BX=0 we're done, if not to show BX is an eigenvalue of A we consider ABX and use the information in the question.
     
  6. Dec 6, 2005 #5
    But that was what JerryKelly had already shown...
     
  7. Dec 6, 2005 #6

    HallsofIvy

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    Yes, that part he said he could do. The remaining problem is to show that X is in fact an eigenvector of B.

    You haven't used the fact that A has n distince eigenvalues. If that is true then there exist a basis for the vector space consisting of eigenvectors of A.
     
  8. Dec 6, 2005 #7

    Hurkyl

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    How many eigenvectors does A have for any particular eigenvalue?
     
  9. Dec 6, 2005 #8

    matt grime

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    But it also proves the rest of the question (admittedly it is a trivial observation since any question that is doable is doable because of the information in the question, but here it is a case of follow your nose). We have proved B preserves (generalized) eigenspaces (of A) which are all 1-d according to the information in the question, thus answering the last part.
     
  10. Dec 6, 2005 #9

    HallsofIvy

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    Since matt grime is about to burst trying to give clues without giving the whole thing, I'm going to give up and spell it out in "donkey steps".

    Given A and B are n by n matrices with AB= BA.
    1. If x is an eigenvector of A with eigenvalue [itex]\lambda[/itex] then Bx is also also an eigenvector of A with eigenvalue [itex]\lambda[/itex].
    (A(Bx))= (AB)x= (BA)x= B(Ax)= B([itex]\lambda[/itex]x)= [itex]\lambda[/itex](Bx))

    2. If lambda is an eigenvalue of A then the set of all vectors x such that Ax= [itex]\lambda[/itex]x forms a subspace (often called the "eigenspace" of [itex]\lambda[/itex]). Further the eigenspaces of two distinct eigenvalues have only 0 in common.

    3. Since A is an n by n matrix it is on an n dimensional vector space.

    4. Since A has n distinct eigenvalues, each eigenspace has dimension 1.

    5. If two vectors are in the same one-dimensional subspace then one is a multiple of the other.

    6. Since x and Bx are in the same one-dimensional eigenspace, Bx is a multiple of x.
     
    Last edited: Dec 7, 2005
  11. Dec 6, 2005 #10

    Hurkyl

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    Hey, no doing the whole problem for the OP!
     
  12. Dec 6, 2005 #11
    Thanks for help,HallsofIvy! It made perfert sence! for the first step, it is so similar what i was doing. I was doing ABx=BAx=B[itex]\lambda[/itex]x=[itex]\lambda[/itex]Bx.
    since A(Bx)=[itex]\lambda[/itex](Bx)
    Bx=0 or Bx is eigenvector.
     
  13. Dec 6, 2005 #12
    your way is seeing better than my way. Thanks,agian! It is very helpful.
     
  14. Dec 7, 2005 #13

    HallsofIvy

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    Mea Culpa, Mea Culpa, Mea Maxima Culpa.
     
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