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?
9
-
1$\begingroup$ Are you assuming that X is symmetric? Are you using real or complex scalars? $\endgroup$– Yemon ChoiCommented Sep 4, 2020 at 4:15
-
2$\begingroup$ 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 ChoiCommented Sep 4, 2020 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. $\endgroup$– user50394Commented Sep 4, 2020 at 4:30
-
1$\begingroup$ 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 ChoiCommented Sep 4, 2020 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. $\endgroup$– user50394Commented Sep 4, 2020 at 4:42
|
Show 4 more comments
1 Answer
$\begingroup$
$\endgroup$
2
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.
answered Sep 4, 2020 at 6:35
-
$\begingroup$ Thanks, that is helpful. Is it known whether the set of unitarily diagonalizable matrices, i.e., normal matrices, is dense in $R^{n\times n}$? $\endgroup$ Commented Sep 4, 2020 at 13:52
-
1$\begingroup$ a small perturbation of a normal matrix spoils that property, so it cannot be a dense set, can it? $\endgroup$ Commented Sep 4, 2020 at 14:57