Let $p_n$ the probability that a random matrix $M\in\mathcal{M}_n(\mathbb{R})$ such that its entries $(a_{i,j})_{1\leqslant i,j\leqslant n}$ are independant and following an uniform distribution over $[-1,1]$, is diagonalizable. I was wondering how to calculate $p_n$ and maybe how to find its limit or an equivalent.
Diagonalization in $\mathbb{C}$ : I proved that $p_n=1$ for all $n\in\mathbb{N}$ if we talk about diagonalization in $\mathbb{C}$ :
Let $$ \Phi_n : \left|\begin{aligned} &\ \ \ \ \ \ \ _ \ \ \ \ \mathbb{R}_{=n}[X] &\longrightarrow &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbb{C} \\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P &\longmapsto &\prod_{1\leqslant i<j\leqslant n}\left(\lambda_i(P)-\lambda_j(P)\right) \end{aligned}\right. $$ where $(\lambda_i(P))_{1\leqslant i\leqslant n}$ are the roots of $P$ ordered in the lexicographical order. For any $P\in\mathbb{R}_{=n}[X]$, $\Phi_n(P)$ is, by a factor, the discriminant of $P$ and thus is a polynomial function of the coefficients of $P$. Moreover, $\Phi_n(P)=0$ if and only if $P$ has a multiple root so that $$ p_n \geqslant \mathbb{P}(\Phi_n(\chi_M)\neq 0)=1-\mathbb{P}(\Phi_n(\chi_M)=0) $$ because $M$ is diagonalizable in $\mathbb{C}$ if $\chi_M$ has no multiple root.
Moreover, if we denote $\lambda_n$ the Lebesgue measure on $\mathbb{R}^n$, one can show that for any non-constant $P\in\mathbb{R}[X_1,\ldots,X_n]$, if $$ \zeta(P) := \{ x\in\mathbb{R}^n\ |\ P(x)=0 \} $$ then $\lambda_n(\zeta(P))=0$. We show it by induction on $n$ : if $n=1$ then $\zeta(P)$ is finite so that $\lambda_1(\zeta(P))=0$. If $n\geqslant 2$, we write $$ \zeta(P)=\bigcup_{t\in\mathbb{R}}\zeta(P(\cdot,t)) $$ where $P(\cdot,t):(x_1,\ldots,x_{n-1})\mapsto P(x_1,\ldots,x_{n-1},t)$. By hypothesis $\lambda_{n-1}(\zeta(P(\cdot,t)))=0$ for all $t\in\mathbb{R}$, thus using Fubini's theorem we have $$ \lambda_n(\zeta(P))=\int_{-\infty}^{+\infty}\lambda_{n-1}(\zeta(P(\cdot,t)))dt=0 $$
Finally, since $M\mapsto\Phi_n(\chi_M)$ is a polynomial function of the coefficients of $M$ (because $\Phi_n$ and $M\mapsto\chi_M$ are), the measure of the set $$ \{M\in\mathcal{M}_n(\mathbb{R})\ |\ \Phi_n(\chi_M)=0\} $$ is $0$ and $\mathbb{P}(\Phi_n(\chi_M)=0)=0$ and thus $p_n=1$.
Diagonalization in $\mathbb{R}$ : Because of what said above, for any $M\in\mathcal{M}_n(\mathbb{R})$, $\chi_M$ has no multiple root almost surely so that $$ p_n=\mathbb{P}(\text{Sp}(M)\subset\mathbb{R}) $$ I believe that $\lim\limits_{n\rightarrow +\infty}p_n=0$ but I don't know how to prove it, and even less how to find an equivalent of $p_n$.
from Hot Weekly Questions - Mathematics Stack Exchange
Post a Comment