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Eigenvalue question

  1. May 22, 2009 #1
    Hello,

    I was reading something in my text/wikipedia, and they both said that "...the eigenvalues of a matrix are the zeros of its characteristic polynomial." Do they mean that λ in the characteristic polynomial causes det (A - λI) = 0 (in particular A = λI)?

    JL
     
    Last edited: May 22, 2009
  2. jcsd
  3. May 23, 2009 #2

    dx

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    The characteristic polynomial is p(x) = det(A - xI). The zeroes of the characteristic polynomial are the x values that satisfy det(A-xI) = 0.
     
  4. May 23, 2009 #3

    HallsofIvy

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    Yes, that's exactly what they mean.

    If the determinant of [itex]A- \lambda I[/itex] is not 0, then [itex]A- \lambda I[/itex] has an inverse and so the equation [itex](A- \lambda I)v= 0[/itex] has a unique solution [itex](A- \lambda I)^{-1}(A- \lambda I)v= v= (A- \lambda I)^{-1}0= 0[/itex] which contradicts the definition of "eigenvalue" which is that the equation [itex]Av= \lambda v[/itex], equivalent to [itex](A- \lambda I)v= 0[/itex], has "non-trivial" (non-zero) solutions.
     
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