# What is $\ \overline{\bigcup_{p≥ 1}\ \{A\in M_n(\mathbb C), \ A^p = I_n\}} \$?

Let $\Gamma_p = \{A\in M_n(\mathbb C), A^p = I_n\}$ and let $\Gamma = \bigcup_{p≥ 1}\ \Gamma_p$.

What is the closure of $\Gamma$ ? (This is from an oral exam).

Let $B \in M_n(\mathbb C)$ such that there exists a sequence $(A_k)_k$ of elements of $\Gamma$ such that $A_k \to B$.

$\chi_{A_k} \to_{k\to\infty} \chi_B$ ($\chi$ is the characteristic polynomial). We will prove that each eigenvalue $\lambda$ of $B$ satisfies $|\lambda | =1$. Writing $\chi_{A_k} = \prod_{1 ≤ i ≤ n}(X-a_{i,k})$, with $|a_{i,k}|=1$ for all $i$ and $k$ ;

all $(a_{i,k})_k, 1≤i≤n,$ are bounded so we can apply Bolzano-Weierstrass theorem in $\mathbb C^n$ to prove there exists $\phi$ an extraction such that $a_{i, \phi(k)} \to_{k\to\infty} l_i$ for all $i$, $1≤i≤n$.

$\chi_{A_{\phi(k)}} \to_{k \to \infty} P$ and the roots of $P$ are the limits $l_i$ of $a_{i, \phi(k)}$, hence $|l_i|=1$ for all $i$. Otherwise $\chi_{A_{\phi(k)}} \to \chi_B$, hence $\chi_B = P$.

Reciprocally, suppose $B$ is such that for all eigenvalue $\lambda,$ $|\lambda| =1$.

$B = P^{-1}TP$ with $T$ being upper triangular. For every diagonal coefficient $t_i$ of $T$, $t_i$ is one of the eigenvalue of $B$, hence $|t_i|=1$.

For all $t_i$, we consider a sequence of roots of $1$ $(a_{i,k})_k$ such that $a_{i,k} \to_{k\to\infty} t_i$ and we choose these sequences such that for all $k$ and for all $(i,j), \ 1 ≤ i < j≤ n$, $a_{i,k} \neq a_{j,k}$ (using the density of the roots of $1$ in the unit circle).

Let $A_k$ be the upper triangular matrix with diagonal coefficient $i$ equal to $a_{i,k}$ for $1 ≤ i ≤ n$ and the other coefficients equal to the ones of $T$.

We will prove that for all $k$, there exists $p_k ≥ 1$ such that $A_k^{p_k} = I_n$ :

For all $i$ there exists $p_{i,k}$ such that $a_{i,k}^{p_{i,k}} = 1$.

Let $p_k$ be a common multiple of the $p_{i,k}, 1 ≤ i ≤ n$.

$\chi_{A_k} = \prod_{1 ≤ i ≤ n}(X-a_{i,k})$ vanishes $A_k$ and the $a_{i,k}, \ 1 ≤ i ≤ n$ are distinct, hence $X^{p_k} - 1$ vanishes $A_k$ i.e. $A_k^{p_k} = I_n$.

Finally, $(P^{-1}A_kP)_k$ is a sequence of $\Gamma$ (because $(P^{-1}A_kP)^{p_k} = I_n$) that converges towards $B$ since $A_k \to T$.

Is that correct?

• Your argument seems correct to me, for what that's worth. – Omnomnomnom May 17 '18 at 13:04
• Does the bar over $\Gamma$ refer to the complex conjugate? – Doubt May 17 '18 at 15:02
• @Doubt: It is the closure in topology. – user371663 May 17 '18 at 15:05