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1. What the problem is asking

We have an (n\times n) matrix

[ A=\bigl(a_{ij}\bigr){1\le i,j\le n}, \qquad a{ij}\stackrel{\text{i.i.d.}}{\sim}\text{Uniform}[0,1], ]

and we would like to know

[ \boxed{\;P_n:=\Pr\bigl(\text{all eigenvalues of }A\text{ are real}\bigr)\;} ]

for a given size (n).
In words: pick a matrix at random from the unit cube ([0,1]^{n^{2}}) (all entries are independent and uniformly distributed); what is the chance that the matrix has no complex eigenvalues?

The question is non‑trivial because the eigenvalues are complicated polynomial functions of the entries, and the set

[ \mathcal R_n={A\in\mathbb R^{n\times n} : \text{all eigenvalues of }A\text{ are real}} ]

is neither open nor closed in the usual Euclidean topology.

2. Step‑by‑step analysis

2.1 Small dimensions

(n) Reasoning (P_n)
1 A (1\times1) matrix is just a real number, so its (single) eigenvalue is always real. (1)
2 For a (2\times2) matrix (\begin{pmatrix}a&b\c&d\end{pmatrix}) the eigenvalues are real iff the discriminant ((a-d)^2+4bc) is non‑negative. Because (b,c\ge 0) (the entries lie in ([0,1])), the term (4bc) is never negative, hence the discriminant is always (\ge0). Consequently every such matrix has real eigenvalues. (1)

Thus the problem becomes interesting only from (n\ge 3).

2.2 The set of “real‑eigenvalue” matrices

For (n\ge 3) let

[ p_A(\lambda)=\det(\lambda I -A)=\lambda^{n}+c_{1}(A)\lambda^{n-1}+ \dots +c_{n}(A) ]

be the characteristic polynomial. All coefficients (c_k(A)) are polynomial functions of the entries of (A) (they are (up to sign) elementary symmetric functions of the eigenvalues).

A real‑coefficient polynomial of degree (n) has only real roots iff its discriminant (\Delta(p_A)) is non‑negative (for (n\ge 3) the discriminant is a polynomial of degree (2n-2) in the coefficients). Hence

[ \mathcal R_n = {A\in[0,1]^{n^{2}} \; :\; \Delta(p_A)\ge 0}. ]

The map

[ \Phi : [0,1]^{n^{2}} \longrightarrow \mathbb R^{n},\qquad \Phi(A)=(c_{1}(A),\dots ,c_{n}(A)) ]

is polynomial, thus absolutely continuous with respect to Lebesgue measure on (\mathbb R^{n}). Consequently the probability we are looking for can be written as the volume of the image region defined by the inequality (\Delta\ge0):

[ P_n = \frac{\operatorname{Vol}{n^{2}}\bigl({A\in[0,1]^{n^{2}}:\Delta(p_A)\ge0}\bigr)} {\operatorname{Vol}{n^{2}}\bigl([0,1]^{n^{2}}\bigr)} . ]

Because (\Delta(p_A)) is a non‑constant polynomial, the set ({A:\Delta(p_A)=0}) has Lebesgue measure zero. Therefore the probability of hitting the boundary (multiple eigenvalues) is zero, and we may replace “(\ge0)” by “(>0)’’ without changing the value of (P_n).

2.3 Why a closed‑form answer is not known

For the Gaussian (real Ginibre) ensemble the joint density of the eigenvalues is known explicitly, and one can integrate it to obtain the exact probability

[ P_n^{\text{Ginibre}} = 2^{-\,n(n-1)/4}\qquad (n\ge 1). ]

The crucial point is the rotational symmetry of the Gaussian law; the eigenvalue density factorises in a way that makes the integration possible.

The uniform law on ([0,1]) lacks rotational invariance; the joint eigenvalue density is not known in a tractable form. Consequently no closed elementary expression for (P_n) is available at present. The problem reduces to evaluating a high‑dimensional integral of the indicator of the set ({\Delta>0}), which is computationally intractable for symbolic calculation beyond the smallest sizes.

2.4 What is known about the magnitude of (P_n)

Even though an exact formula is missing, a number of results give the asymptotic behaviour of (P_n) for any continuous distribution with a density (including the uniform distribution). The key facts are:

Fact Statement
Universality of the leading exponent For i.i.d

Original question: Probability for an $n\times n$ matrix to have only real eigenvalues on Mathematics Stack Exchange, licensed CC BY-SA.