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Eigenvalues for an Invertible Matrix

  1. May 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A is an invertible matrix, x is an eigenvector for A with an eiganvalue [tex]\lambda[/tex] [tex]\neq[/tex]0 Show that x is an eigenvector for A^-1 with eigenvalue [tex]\lambda[/tex]^-1


    2. Relevant equations
    Ax=[tex]\lambda[/tex]x
    (A - I)x

    3. The attempt at a solution

    I know that I need to find x and then apply to the inverses of my Matrix and eigenvalue, but how do I know what matrix to use for A? Do I use the inverse matrix as it is an invertible matrix? Can I use any invertible matrix to prove this?

    Thanks in advance.
     
  2. jcsd
  3. May 10, 2009 #2

    Cyosis

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    Let [itex]\mu[/itex] be the eigenvalue of [itex]A^{-1}[/itex]. Now multiply the eigenvalue equation with [itex]A^{-1}[/itex] and find [itex]\mu[/itex].
     
  4. May 10, 2009 #3
    Do you mean I should have

    A[tex]^{2}[/tex]x = A[tex]\lambda[/tex]x

    and

    A[tex]^{-2}[/tex]x = A[tex]^{-1}[/tex][tex]\mu[/tex]x

    ?

    How can I use this to show my Answer? Or do I substitute this second equation into the first?
     
  5. May 10, 2009 #4

    Cyosis

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    No with multiplying the eigenvalue equation I meant [itex]A^{-1}(Ax)=A^{-1}\lambda x[/itex]. On a side note, use [tex] brackets around your entire equation not just lambda or mu.
     
    Last edited: May 10, 2009
  6. May 10, 2009 #5
    So I can re-arrange as
    [tex]\frac{A^{-1}(A\chi)}{\lambda} = \mu\chi[/tex]
    Is that right?
     
  7. May 10, 2009 #6

    Cyosis

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    Sure, because the problem says [itex]\lambda \neq 0 [/itex].What is [itex]A^{-1}A[/itex]?
     
  8. May 10, 2009 #7
    Is A^-1*A the Inverse Matrix I?
     
  9. May 10, 2009 #8

    Cyosis

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    Yes it's the identity matrix, which has itself as an inverse. What is Ix?
     
  10. May 10, 2009 #9
    Inverse multiplied by the eigenvector will give us the original matrix?
     
  11. May 10, 2009 #10

    Cyosis

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    Are you calling I the inverse? It's the Identity matrix. If you let the Identity matrix work on a vector it yields that vector without making any changes to it. We're moving in the wrong direction now so lets start at the eigenvalue equation again. [itex] A^{-1}(Ax)=A^{-1}\lambda x \Rightarrow Ix=\lambda \mu x[/itex]. Now solve for [itex]\mu[/itex].
     
  12. May 10, 2009 #11
    Oops, my bad, I meant Identity.

    Ix gives us x again.

    So:

    [tex]\frac{I\chi}{\lambda\chi} = \mu[/tex]

    Is that right? and Ix = x?
     
  13. May 10, 2009 #12

    Cyosis

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    Well x are vectors how do you define a vector divided by another vector? While doing it in this case and letting the xs cancel you will get the correct answer, I would suggest you don't do this on your exam.

    This is what you should do, [itex]Ix=\lambda \mu x \Rightarrow x=\lambda \mu x \Rightarrow 1*x=\lambda \mu x[/itex]. So what is [itex]\mu[/itex]?
     
  14. May 10, 2009 #13
    [tex]\mu[/tex] = [tex]\frac{1}{\lambda}[/tex]

    Correct?
     
  15. May 10, 2009 #14

    Cyosis

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    Yes, perhaps it's nice if you write the full proof down now so we can see if you don't do any operations that you shouldn't really be doing.
     
  16. May 10, 2009 #15
    Alright Cool, Here Goes!

    [tex]A\chi = \lambda\chi[/tex]

    [tex]A^{-1}\chi = \lambda^{-1}\chi[/tex]

    Let [tex]\lambda^{-1} = \mu[/tex]

    [itex]
    A^{-1}(Ax)=A^{-1}\lambda x
    [/itex]

    We can re-write this as

    [tex]
    \frac{A^{-1}(A\chi)}{\lambda} = \mu\chi
    [/tex]

    [itex]
    A^{-1}A = I
    [/itex]

    Here I is the Identity Matrix

    Ix = x

    As The Identity matrix multiplied by a vector value does not change the position of the vector. So we now have:

    [tex]
    \frac{I\chi}{\lambda\chi} = \mu
    [/tex]

    This equates to:

    [tex]\frac{1}{\lambda\chi}[/tex]

    This shows that the vector [tex]\chi[/tex] is an eigen vector for the Inverse Matrix of A as it's Eigen Value is also the Inverse of [tex]\chi[/tex]
     
  17. May 10, 2009 #16

    Cyosis

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    You need to proof that [itex]\mu=\lambda^{-1}[/itex], not start with it.

    What we are given is the eigenvalue equation [itex]Ax=\lambda x[/itex] with [itex]\lambda[/itex] being an eigenvalue of A. We want to proof that x is an eigenvector of [itex]A^{-1}[/itex] with eigenvalue [itex]\lambda^{-1}[/itex].
    So lets multiply the eigenvalue equation by [itex]A^{-1}[/itex].
    This yields
    [tex]A^{-1}(Ax)=A^{-1} \lambda x \Rightarrow (A^{-1}A)x=\lambda A^{-1} x \Rightarrow Ix=\lambda A^{-1}x \Rightarrow 1*x=\lambda A^{-1} x \Rightarrow \lambda^{-1}x=A^{-1}x[/tex].
    A matrix working on a vector that yields the same vector multiplied by a scalar is an eigenvalue equation. So x must be an eigenvector of [itex]A^{-1}[/itex] and [itex]\lambda^{-1}[/itex] must be an eigenvalue of [itex]A^{-1}[/itex]. There is no dividing of any vectors happening here! If you want to divide two vectors you must first define how division with vectors works.
     
    Last edited: May 10, 2009
  18. May 10, 2009 #17
    Ohhhh!

    Thanks heaps man, is there anyway I can +rep you or something?
     
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