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Eigenvector proof

  1. Aug 12, 2014 #1
    Hello PF, brand new member here.

    A question about a proof:

    If A*v=λ*v, then w = c*v is also an eigenvector of A.

    This seems really simple to me, but perhaps I am doing it incorrectly:

    A*c*v=λ*c*v, divide both sides by c and you are left with your original eigenvector of A. Am I missing something here?
     
  2. jcsd
  3. Aug 12, 2014 #2
    What you've shown is that if cv is an eigenvector, then v is an eigenvector. You need to show the other way. In addition, you can't always divide by c.

    Also, this should be in homework help.
     
  4. Aug 13, 2014 #3

    WWGD

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    notice that subspaces are closed under scaling. Or just factor out the c as the matrix cId.
     
  5. Aug 13, 2014 #4

    ehild

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    Welcome to PF!:smile:
    Do not use * when applying an operator on a vector. The operator A assigns a vector u to vector v. Write u=A(v). In case v is an eigenvector of operator A, A(v)= λ*v. The right side is a product - a vector multiplied by a scalar, but the left-hand side is not.

    You can use the property of linear operators that A(cv)=c A(v) (c is a scalar).

    ehild
     
    Last edited: Aug 13, 2014
  6. Aug 13, 2014 #5

    HallsofIvy

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    As you have been told, what you shown is that "If cv is an eigenvector of A with eigenvalue λ then so is v". To prove the other way, reverse your proof: from Av= λv, multiply both sides by c.
     
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