Eigenvalue question

  • #1
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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
 
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Answers and Replies

  • #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.
 
  • #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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