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Eigenvalue proof

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Proof: Prove that if A is an nxn (square mtx) such that A^2=A, then A has 0 or 1 as an eigenvalue.




    3. The attempt at a solution
    A=A^2
    A^2-A=0
    A(A-I)=0
    A=0 or A=1
    and then plugging the A solutions in to the characteristic equation and solving for λ
     
  2. jcsd
  3. Oct 24, 2011 #2

    lanedance

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    start with eigenvalue [itex] lambda [/itex] with a correpsonding eigenvector u.

    Is u also an eigenvector of A^2?

    if so what is the corresponding eigenvalue?
     
    Last edited: Oct 24, 2011
  4. Oct 24, 2011 #3
    can we assume, for the proof, the eigenvalues are both equal to λ?
     
  5. Oct 24, 2011 #4

    lanedance

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    no, don't assume, but can you show it?

    Also though I get your meaning please be explicit in you question (eg. what do you mean by "both")
     
  6. Oct 24, 2011 #5
    Sorry about that, what I meant was could we associate λ as an eigenvector for A and A^2. For example:

    If Au=λu
    then (A^2)u=λu, where u=/=0
     
  7. Oct 24, 2011 #6

    lurflurf

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    ^What do you mean both there are n eigenvalues.
    Are you working over a splitting field?
    Sketch of proof
    1)since A=A2
    A and A2 have the same eigenvalues
    2)find out when A and A2 have the same eigenvalues
     
  8. Oct 24, 2011 #7

    HallsofIvy

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    If v is an eigenvector of A with eigenvalue [itex]\lambda[/itex], then [itex]A^2v= A(Av)= A(\lambda v)= \lambda Av= \text{what?}[/itex]
     
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