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Eigenvalues and Eigenvectors

  1. Dec 14, 2006 #1
    Suppose that B is the inverse of A. Show that if |psi> is an eigenvector of A with eigenvalue a not equal to 0, then |psi> is an eigenvector of B with eigenvalue 1/a.


    So I know that A|psi> = a|psi>, and I'm trying to prove that A^(-1)|psi> = 1/a|psi>. I tried simplifying A as a 2x2 matrix and then doing the inverse of that. And then I assumed that the inverse of A has an eigenvalue b. So then I did the determinant of A^(-1)-b = 0 in the hopes to find b and see that it's equal to 1/a. But that became really messy.

    Any suggestions on how to solve this problem? Thank you so much!
     
  2. jcsd
  3. Dec 14, 2006 #2

    marcusl

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    Start with A|psi> = a|psi> and solve for (isolate) |psi> in terms of inv(A). You should be able to take it from there.
     
  4. Dec 18, 2006 #3
    well it is not too difficult:

    you supposed A invertible and B=A^(-1).

    since AB=BA=1---->[A,B]=0.

    (1)|psi>=BA|psi>=Ba|psi>=AB|psi>=ab|psi>.


    but ab=1 so the eigenvalue b must be b=1/a
     
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