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I have read that if a random matrix is hermitian then its eigenvalues are continuous, hence also measurable.

If the random matrix is not hermitian, the eigenvalues are not continuous in some cases. However, is it possible to show that the eigenvalues are still measurable functions?

Thank you

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  • $\begingroup$ Measurable with respect to what $\sigma$-algebra on (say) $\mathbb R^{n\times n}$? Also, in what sense do you mean "the eigenvalues are [...] measurable"? For each matrix $A$, we can only talk about the set of its eigenvalues. So, you also need some $\sigma$-algebra on the set of such eigenvalue sets. $\endgroup$
    – Iosif Pinelis
    Commented Jun 7, 2023 at 17:31
  • $\begingroup$ Here is a related question where the asker observed that the function mapping complex matrices to the sets of its eigenvalues is Holder continuous with exponent $1/d$. mathoverflow.net/q/433834/22277 $\endgroup$
    – Joseph Van Name
    Commented Jun 7, 2023 at 17:43
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    $\begingroup$ As a rule of thumb, if one can write down an algorithm in, say, MATLAB, to compute some statistic of some mathematical object to arbitrarily specified precision, then that statistic is a measurable function of the object with respect to reasonable choices of sigma-algebras, and the proof basically consists of running the algorithm a countable number of times to get increasing levels of precision and then taking a limit. "If it's computable, it's measurable." $\endgroup$
    – Terry Tao
    Commented Jun 8, 2023 at 0:17

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