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Let $X$ be an $n\times n$ matrix whose elements are i.i.d. sampled from a normal distribution of zero mean and unit variance. Is $X$ diagonalizable over $\mathbb{C}$ with probability 1? Is there a good reference for diagonalizability of random matrices?

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    $\begingroup$ Are you assuming that X is symmetric? Are you using real or complex scalars? $\endgroup$
    – Yemon Choi
    Sep 4 '20 at 4:15
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    $\begingroup$ If you work in $M_n({\bf C})$ then the subset of matrices that are not diagonalizable over ${\bf C}$ has Lebesgue measure zero, and hence for any probability density on $M_n({\bf C})$ that is absolutely continuous with respect to Lebesgue measure, a random element of $M_n({\bf C})$ will be almost surely diagonalizable over ${\bf C}$ $\endgroup$
    – Yemon Choi
    Sep 4 '20 at 4:16
  • $\begingroup$ @YemonChoi X is not symmetric; every element is sampled independently from a normal distribution. Thanks for the result over C. I'd be more curious to know the density when X is real-valued. $\endgroup$
    – user50394
    Sep 4 '20 at 4:30
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    $\begingroup$ But why would you expect a "typical" real matrix to be diagonalizable over R? (The particular Gaussian i.i.d. model you are asking about is the Ginibre ensemble, so we know that the spectrum is almost surely a disc (in the asymptotic sense as n tends to infinity. However, my point is that even ignoring probability theory, it is very easy for real matrices to have complex eigenvalues) $\endgroup$
    – Yemon Choi
    Sep 4 '20 at 4:36
  • $\begingroup$ Ah I see, thanks. I am very new to random matrix theory. I think I am curious to know the density of real-valued diagonalizable matrices whose eigenvalues can be complex numbers. $\endgroup$
    – user50394
    Sep 4 '20 at 4:42
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The measure of real matrices that are not diagonalizable over $\mathbb{C}$ equals to 0, see for example On the computation of Jordan canonical form, so the probability for a random matrix with a continuous probability distribution to be non-diagonalizable vanishes. To put it differently, the set of real matrices without multiple eigenvalues is dense, and a matrix without multiple eigenvalues is definitely diagonalizable.

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  • $\begingroup$ Thanks, that is helpful. Is it known whether the set of unitarily diagonalizable matrices, i.e., normal matrices, is dense in $R^{n\times n}$? $\endgroup$
    – user50394
    Sep 4 '20 at 13:52
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    $\begingroup$ a small perturbation of a normal matrix spoils that property, so it cannot be a dense set, can it? $\endgroup$ Sep 4 '20 at 14:57

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