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Continuous eigenvalue of n by n matrix

  1. May 7, 2006 #1
    Hello, all,
    This is my fisrt time post on this forum, I have this question for long time but people around me couldn't really answer it, hopes I can get the answer from you guys...

    Given a complex n by n matrix A, Under what restriction, its eigenvalue(s) is the continuous function of A?
  2. jcsd
  3. May 8, 2006 #2


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    What do you mean, "continuous function"?

    A general condition for an eigenvalue to be complex?
  4. May 8, 2006 #3
    There are no restrictions whatsoever. For example, in two dimensions the trace and determinant are continuous functions in the matrix topology and the eigenvalue polynomial is:
    x^2 - tr(A)x + det(A) with roots (tr(A) + sqrt(tr(A)^2 - 4 det(A) ) )/2 (here sqrt means `all possible roots´)
    The latter are continuous functions (obviously, you have to be careful with the sqrt function).


    Last edited: May 8, 2006
  5. May 8, 2006 #4


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    well this is correct except the answer may be misleading, i.e. since their ordering is undetermined, the eigenvalues are not functions of the matrix at all ["be careful of the square root"], but the elementary symmetric functions of the eigenvalues, e.g. the trace and determinant, are.

    it is the unordered set of eigenvalues that is a continuous function of the matrix, and this is equivalent to saying the elementary symmetric functions of them are continuous.

    this is the same question as asking whether the roots of a polynomial are continuous functions of the coefficients.
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