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Let $A$ be an $n\times n$ random matrix where every entry is i.i.d. and uniformly distributed on $[0,1]$. What is the probability that $A$ has only real 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 to work in, due to its lack of rotational invariance.

Exodd
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  • From here: http://web.math.princeton.edu/mathlab/projects/ranmatrices/yl/randmtx.PDF following the proof of Theorem 1, you might be able to find the distribution in terms of the distribution of traces. If you can't follow their method with complex numbers directly, maybe defining $A_+=A+A^$ and $A_- = i(A - A^)$ (both of which have real eigenvalues) could work. – Sam Jaques Jul 27 '20 at 11:21
  • @SamJaques they are dealing with symmetric matrices, and they work for big $n$.. Are you sure? – Exodd Jul 27 '20 at 11:34
  • I'm not sure it will work, no. On a second look it seems like a key component of the argument for higher moments is that $\sum_{j=1}^N(A^2){jj}=\sum{j,k=1}^N(A_{jk})^2$ which requires symmetric matrices. – Sam Jaques Jul 27 '20 at 11:37
  • Perhaps you could try to say something about the distribution of the discriminant of the characteristic polynomial – Ben Grossmann Jul 27 '20 at 11:41
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    I'm curious why the set of matrices with distinct real eigenvalues is open (I'm assuming as a subset of the $n \times n$ real matrices.) – Adina Goldberg Jul 27 '20 at 20:19
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    @AdinaGoldberg the eigenvalues of a matrix are the roots of the characteristic polynomial, and the roots are continuous functions of the coefficients – Exodd Jul 27 '20 at 20:35
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    @Exodd Yes, indeed! But $\mathbb{R}$ is not open as a subset of $\mathbb{C}$. How can you complete your argument? – Adina Goldberg Jul 27 '20 at 20:37
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    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 $\mathbb{R}^2$. – Adina Goldberg Jul 27 '20 at 20:43
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    @AdinaGoldberg We are on the set of real matrices, so the eigenvalues are either real or come in couples of conjugated complex numbers. If you have distinct real eigenvalues, then you can always find a small enough neighbourhood where the eigenvalues are not complex due to the continuity of the eigenvalues – Exodd Jul 27 '20 at 21:03
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    Empirically : $3$: $0.708$, $4$: $0.346$, $5$: $0.117$ – leonbloy Jul 27 '20 at 22:31
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    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: https://core.ac.uk/download/pdf/82140233.pdf     – leonbloy Jul 29 '20 at 20:22
  • @leonbloy how many examples have you generated for the empirical evaluation? – Exodd Jul 29 '20 at 22:58
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    about one million (trusting Octave's random generator) – leonbloy Jul 29 '20 at 23:26
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    with 10kk the results do not change much... – Exodd Jul 29 '20 at 23:27
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    This paper: https://shub.ccny.cuny.edu/articles/1994-how_many_eigenvalues_of_a_random_matrix_are_real.pdf says:Mehta [24, Conjectures 1.2.1 and 1.2.21 conjectures from extensive numerical experience that the statistical properties of matrices with independent identically distributed entries behave as if they were normally distributed as n -+ rn . Mehta focuses on the symmetric or Hermitian cases, but surely the idea is quite general. – Jap88 Jan 26 '21 at 22:34
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    @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.) – Hagen von Eitzen Feb 05 '21 at 07:33
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    @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. – Hagen von Eitzen Feb 05 '21 at 07:34
  • just a quick thought. Assume $n=2$. If $A$ eigenvalue $\lambda_2$ is complex, then $A-\lambda_2 I $ cannot be drawn as a parallelogram in, say $XY$ plane i.e. its area (in $XY$ plane) does not exist. Since the number of eigenvalues increases as $n\rightarrow \infty$, it feels for sure that at least for one of them, say $\lambda_n$, the $n$-dimenional area $det(A-\lambda_n I)$ does not exist. Thus the sought probability is $0$. – rrv Oct 29 '22 at 14:38
  • Since this question is getting a lot of attention, I ask you to reformulate it. Every square matrix has at least one complex eigenvalue, since every polynomial of degree $\geq 1$ has at least one complex root (note that real numbers are also complex numbers, by definition). What you ask is the probability for all eigenvalues to be real. Right? – IljaKlebanov Feb 26 '23 at 00:01
  • My intuition is that the probability is zero, and that intuition is based on a slightly simpler problem: Consider the set of n x n matrices where all eigenvalues satisfy $|\lambda| < 1$. The subset of these matrices where all eigenvalues is real is a set of subset of measure zero relative to the entire set in eigenspace. I would expect something similar if you change the constraint from the one I suggested to the original problem, but I'm not sure what the mapping from n x n matrices with entries all in [0, 1] "looks like" in eigenspace, so that's where my intuition might fall apart. – Adam Cataldo Dec 08 '23 at 03:47
  • @AdamCataldo the probability Cannot be zero, since the set of all-real eigenvalues matrices has non-empty interior part. Your reasoning holds if you consider complex-entries matrices, while here we have only real entries – Exodd Dec 08 '23 at 04:05
  • Not sure I follow this, because what I'm saying is that in the unit circle in the complex plane, the real line, that is the real values between -1 and 1, has measure zero. It does in fact have an empty interior, because if you draw a ball of any radius around any of the points on the real line, the ball will include complex numbers, not just real values. – Adam Cataldo Dec 10 '23 at 19:22
  • @AdamCataldo if the eigenvalues of the matrices were "continuously" distributed in the unit circle, then you would be correct. But if you take real matrices, the distribution of their eigenvalues is biased on certain areas of the unit circle, and in particular they greatly concentrate on the real line, so much that even if $[-1,1]$ is negligible in the unit circle, the probability of having real eigenvalues is not zero – Exodd Dec 10 '23 at 20:46
  • I think what I'm stuck on is measure space you're trying to work in, since you're suggesting in the problem it shouldn't be $[0, 1]^{n^2}$ (presumably with Lebesgue measure). It sounds like it's also not the eigenspace (the subset of $\mathbb{C}^n$ that is projected on by the set of matrices in $[0, 1]^{n \times n}$). Or do you think the projection into the eigenspace is such that the eigenvalues on the real line have non-zero measure? – Adam Cataldo Dec 11 '23 at 00:13
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    @AdamCataldo yes, if you only consider real matrices then the eigenvalues are very biased towards the real line. Maybe finding a bigger space where to work may not be a bad idea, but the choice of the biger space has to reflect/expand somehow consistently the properties of the spectra of real matrices – Exodd Dec 11 '23 at 00:39
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    The probability the matrix is symmetric is zero, but maybe one can argue that a certain class of “almost symmetric” matrices (via a perturbation) have real eigenvalues and with positive probability, to get a lower bound on the desired probability in question. – jdods Mar 29 '24 at 13:54
  • Why don’t you just use the characteristic polynomial? All the eigenvalues are real if $det(xI-A)$ only has real solution. For example, for 2-by-2, $(a_{11}-a_{22})^2+4a_{12}a_{21}>0$ if and only if all the eigenvalues are real. Thus the probability is the volume of the set of inequality intersection $[0,1]^4$. – J1U Nov 25 '24 at 17:19
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    @J1U for the 2x2 it is easy indeed (actually the proobability is 1) but for larger degrees how do you ensure the polynomial has only real solutions? – Exodd Nov 26 '24 at 07:10
  • @Exodd See this: https://en.m.wikipedia.org/wiki/Discriminant For degree 3 and 4, the discriminant is explicitly computed. For even higher degrees, the discriminant is still expressed as the algebraic relation of the coefficients of the polynomial, although it would be messier. – J1U Nov 26 '24 at 07:35
  • @J1U already for degree 4, "If the coefficients are real numbers and the discriminant is positive, then the roots are either all real or all non-real" so it cannot predict if the roots are all real or not – Exodd Nov 26 '24 at 11:50
  • @EXodd Ok. Then we can try this way for $n\geq 4$ case. Let $f(x)=det(xI−A)$ be the characteristic polynomial. Let $a_1 >⋯>a_{n−1}$ be roots of $f′(x)$. Then the roots of f are all real if $f(a_1)<0,f(a_3)<0,…$ and $f(a2)>0,f(a4)>0,…$. For $n=4$, this seem to give the explicit algebraic relation. For $n>4$, I am not sure. – J1U Nov 26 '24 at 12:23

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