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Eigenvalues & Similar Matrices

  1. Oct 21, 2007 #1
    Q: Suppose the only eigenvalues of A are 1 and -1, and A is similar to a diagonal matrix. Prove that A^-1 = A

    My Attempt:
    Suppose the only eigenvalues of A are 1 and -1, and A is similar to a diagonal matrix.
    =>A is invertible (since 0 is not an eigenvalue of A)
    and there exists invertible P s.t.
    (P^-1) A P = D is diagonal
    => A= P D (P^-1)
    => A^-1 = P (D^-1) (P^-1)

    Now if I can prove that D = D^-1, then I am done. But I am stuck right here. The trouble is that the eignevalues of A can have any number of multiplicities, so D can be diag{1,1,-1,-1,-1}, diag{1,-1,-1,-1,-1,-1}, etc., there are infinite number of possible D's, how can I prove that D = D^-1 is always true for these infinite number of different settings?


    Can someone please help me?
    Thanks a lot!
     
  2. jcsd
  3. Oct 21, 2007 #2

    Dick

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    The inverse of adiagonal matrix is formed by inverting the diagonal elements. 1^(-1)=1 and (-1)^(-1)=-1. What's the problem?
     
  4. Oct 21, 2007 #3

    JasonRox

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    Whenever you're stuck, write out an example and solve it. If you have done so, you would have noticed what Dick is saying.
     
  5. Oct 21, 2007 #4
    How can I prove this statement? I haven't learnt this before...
     
  6. Oct 21, 2007 #5

    Dick

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    Follow JasonRox's advice and just do it. Multiplying diagonal matrices just involves multiplying the diagonal elements. c_ij=a_ik*b_kj sum over k. If they are diagonal, c_ii=a_ii*b_ii.
     
  7. Oct 22, 2007 #6

    HallsofIvy

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    You don't really need to calculate A-1 at all. Just showing that A2= A[itex]\cdot[/itex]A= I is sufficient.
     
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