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Eigenvalue question

  1. Oct 31, 2012 #1
    If ##\hat{A}\vec{X}=\lambda\vec{X}## then ##\hat{A}^{-1}\vec{X}=\frac{1}{\lambda}\vec{X}##

    And what if ##\lambda=0##?
     
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  3. Oct 31, 2012 #2

    micromass

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    0 can never be an eigenvalue of an invertible matrix. So if [itex]\lambda=0[/itex], then [itex]\hat{A}[/itex] is not invertible, so [itex]\hat{A}^{-1}[/itex] makes no sense.
     
  4. Nov 1, 2012 #3
    Is there some easy way to see that?
     
  5. Nov 1, 2012 #4

    AlephZero

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    If ##X## is an eigenvector of ##A## with eigenvalue zero, then ##AX = 0## and ##X \ne 0##.

    But if ##A## is non-singular, then ##X = A^{-1}0 = 0##.

    For any matrix ##A##, only one of the above can be true.
     
  6. Nov 1, 2012 #5
    To put it less formally (but perhaps less transparently if you haven't gone far in linear algebra yet), in order for a matrix to have an inverse, it has to be associated to a one to one linear transformation. If you can hit a vector, v, with a linear map T and kill it (make it zero), then you hit λv with T and kill it for any scalar λ, so the map is not one to one because the pre-image of 0 (or any other vector) contains at least one parameter worth of stuff.
     
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