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Eigen value proof

  1. Dec 10, 2012 #1
    Let A be an invertible matrix. Show that if λ is
    an eigenvalue of A, then 1/λ is an eigenvalue of
    A^−1.


    det((A-λI))
    det((A-λI)^-1)
    =det(A^-1 - λ^-1 * I)
    =det(A-1-1/λ*I)

    Is this enough to show that?



    Another question I have is:
    Let A be an n × n matrix. Show that A is not
    invertible if and only if λ = 0 is an eigenvalue
    of A.

    Not sure how to approach this prob.
    det(A-λI)=0
    det(A-0I)=0
    det(A)=0

    But idk how to show what the problem is asking.
     
  2. jcsd
  3. Dec 10, 2012 #2

    Mark44

    Staff: Mentor

    (Above) No... Is this some kind of new distributive property for exponents? Kind of like (a+ b)2 = a2 + b2?
    No. Start with what it means for λ to be an eigenvalue of A.
    And what it means for λ-1 to be an eigenvalue of A-1.

    IOW, start with some definitions.
     
    Last edited: Dec 10, 2012
  4. Dec 10, 2012 #3

    Mark44

    Staff: Mentor

    This is an "if and only if" proof, so you have to prove two propositions:
    1. If λ = 0 is an eigenvalue of A, then A is not invertible.
    2. If A is not invertible, then λ = 0 is an eigenvalue of A.

    For the first one, start with what it means for a number to be an eigenvalue of a matrix.
    For the second one, what can you say about a matrix that does not have an inverse?
     
  5. Dec 10, 2012 #4
    Let A be an invertible matrix. Show that if λ is
    an eigenvalue of A, then 1/λ is an eigenvalue of
    A^−1.
    det((A-λI))
    det((A-λI)^-1)
    =det(A^-1 - λ^-1 * I)
    =det(A-1-1/λ*I)

    if λ is an eigenvalue of A then the following is true:
    Av=λv where v=eigenvec
    A^-1*Av=A^-1*(λv)=A^-1*Av=Iv
    A^-1*Av=Iv=v
    A^-1*λv=Iv=v
    A^-1*v=(1/λ)*v
     
  6. Dec 10, 2012 #5

    Mark44

    Staff: Mentor

    You can get rid of this stuff - it's wrong.
    And v has to be nonzero.
    It seems like you're going around in a circle here.
    Just from the left side above, we have A-1Av = Iv = v
    From where you started, you want to end up with this statement:

    A-1u = (1/λ)u
     
  7. Dec 10, 2012 #6
    1. If λ = 0 then det(A-λI)=0=det(A-0)=det(A) so A is not invertible
    2. det(A)=0 if it doesn't have an inverse. so that means λ = 0
     
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