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Estimating eigenvalue of perturbed matrix

  1. Feb 22, 2012 #1

    julian

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    Say M_{ij} = A_{ij} + s B_{ij}/2, where the matrices are 3 by3 and A_{ij} symmetric, s \in [0,s^*], and the smallest eigenvalue of A is lambda \leq -(1/2). Given that |M_{ij} - A_{ij}| \leq to C_{s^*} s/2 and |A_{ij}| \leq 1, plus that the cubic equation determining the eigenvalues has an explicit fomrula to solve, how do you show that there is some s_0 \in [0,s^*] such that M_{ij} has an eigenvalue \leq -(1/4)?
     
    Last edited: Feb 22, 2012
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  3. Feb 23, 2012 #2

    julian

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    Actually it's quite easy isn't it... The min eigenvalue of A is <= -(1/2). The deviation of M from A can be made arbitrarily small by choice of s_0, the min eigenvalue is a continuous function of the entries of the matrix hence by choosing s_0 smal enough we can gaurantee the min eigenvalue be <= to say -(1/4).
     
  4. Feb 23, 2012 #3

    julian

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    In our case, we know for 3 by 3 matrices that the eigenvalues are continuous functions of the matrix entrie because we have an explicit formula for roots of a cubic polynomial. Remind me,do higher order polynomials have analytic solutions? Could we still say the roots depend continuously on the matrix elements?
     
  5. Feb 23, 2012 #4

    julian

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    Is this where we would use the implicit function theorem or somerthing to estaablish continuity of eigenvalue on matrix elements - I'm a physicist, not completely familiar with this stuff.
     
  6. Feb 23, 2012 #5

    julian

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    Plus a part of the original problem was to show s_0 can be chosen in a way independent of A_{ij}. Forgot to mention that -what makes the problem a bit more difficult.
     
  7. Feb 24, 2012 #6

    julian

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    So the min eigenvalue is a continuous function over the space given by |A_{ij}| <=1. Because this spce is compact does this mean that the continuous functions l_s (A_{ij}) converge uniformally to l_0 A_{ij} in the parameter s? This sounds familiar to me.
     
  8. Feb 24, 2012 #7

    julian

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    If I had to use the implicit function theorem would it work only locally? Would this cause me any problems? If not why not? If a maths person could help fill in details for a poor physicist that would be great.
     
    Last edited: Feb 24, 2012
  9. Feb 25, 2012 #8

    morphism

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    I'm having a hard time understanding what you want to accomplish. Are you just looking for the result that the eigenvalues of a matrix vary continuously with the entries of the matrix? For this you don't need the implicit function theorem. You just need to observe that the eigenvalues vary continuously with the coefficients of the characteristic polynomial, which in turn varies continuously with the entries of the matrix. Note that for this to make sense, you need to be working over the complex numbers (or some algebraically closed field), to guarantee that your matrix and perturbations of it have eigenvalues.
     
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