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Let $A$ be a $n\times n$ random matrix where every entry is i.i.d. and uniformly distributed on $[0,1]$. What is the probability that $A$ has no complex eigenvalues?

The answer cannot be 0 or 1, since the set of matrices with distinct real eigenvalues is open, and also the set with distinct, but not all real, eigenvalues is open. (the matrices with repeated eigenvalues have measure zero)

I don't see any easy transformation that links the two sets, and working on the characteristic polynomial seems quite impractical. Also, I have the feeling that $[0,1]^{n^2}$ is not a good space where to work, due to its lack of rotation invariance.

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    $\begingroup$ @AdinaGoldberg the eigenvalues of a matrix are the roots of the characteristic polynomial, and the roots are continuous functions of the coefficients $\endgroup$
    – Exodd
    Jul 27, 2020 at 20:35
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    $\begingroup$ I guess to clarify, I certainly agree that within the set of matrices with real eigenvalues, those with distinct eigenvalues is an open set. But that is relatively open, in the sense that $(0,1)$ is open in $\mathbb{R}$ but not in $\mathbb{R}^2$. $\endgroup$
    – Adina Goldberg
    Jul 27, 2020 at 20:43
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    $\begingroup$ It should be a coincidence, but the desired probability seems to be quite near the probability of an $n-1 \times n-1$ random gaussian matrix having all real eigenvalues $$\begin{array} n & & \\ 2 &1 & 1 \\ 3 &0.708 & 0.70711\\ 4 &0.346 & 0.35355\\ 5 &0.117 & 0.125\\ 6 & 0.028 & 0.03132\\ \end{array} $$ Left column: empirical. Right column: from here: core.ac.uk/download/pdf/82140233.pdf $\endgroup$
    – leonbloy
    Jul 29, 2020 at 20:22
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    $\begingroup$ @leonbloy That is quite feasible: If we pick one (probably existing) real eigenvector and take it to $e_n$ via an orthonormal base change, the other entries in the resulting matrix are essentially linear combinations of a large number of iid uniform random variables, so by the law of big numbers are gaussian. Well, the independence is a bit hand-wavy as the original eigenvector depends on all entries and vice versa all entries in the new matrix are influenced by the eigenvector. (cont.) $\endgroup$
    – Hagen von Eitzen
    Feb 5, 2021 at 7:33
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    $\begingroup$ @leonbloy (cont.) Then again, the "large amount of independence" from originally $n^2$ random values has to be "squeezed" into $(n-1)^2$ random values. So they are certainly independent indeed. Admittedly, this is still in a typical what could possible go wrong? scenario. $\endgroup$
    – Hagen von Eitzen
    Feb 5, 2021 at 7:34

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