If $R$ is a unital ring and $x \in R$ a nilpotent element, then $1-x$ is a unit, and so is $1-x^s$ for any $s \geq 1 $ since powers of $x$ are again nilpotent.
My question concerns the converse of this statement in the case in which $R = M_n(\mathbb{C})$, that is:
Question 1: if $A \in M_n(\mathbb{C})$ is a square matrix with complex entries and $I-A^s$ is invertible for all $s \geq 1$, what conditions can we impose on $A$ to guarantee that it is nilpotent?
Some condition has to be in place, as the general statement is already false in dimension $1$. In that case, the matrices $1-a^n$ will be always invertible whenever $a$ is not a root of unity, and the only nilpotent element is zero. More generally, any diagonal matrix $\mathrm{diag}(a_1, \ldots, a_n)$ of non-zero elements that are not roots of unity will do the trick.
Here's a partial result,
If $\{\det(I-A^n)\}_n \subset \mathbb{R} \setminus \{0\}$ converges, then either the spectrum of $A$ is contained in $B_1(0)$. In particular, if $A$ has integer coefficients then it is nilpotent.
A proof follows: let $c_n := \det(I-A^s)$ and $\lambda_1, \ldots, \lambda_n \in \mathbb{C}$ be the eigenvalues of $A$, counted with multiplicity. Then
$$ (1-\lambda_1^s) \cdots (1-\lambda_n^s) = c_s $$
is a convergent subsequence. Noting $d_s = \prod_{i \ : \ |\lambda_i| \neq 1}(1-\lambda_i^s)$, we then have
$$ \prod_{i \ : \ |\lambda_i| = 1}(1-\lambda_i^s) = c_s/d_s \to L. $$ with either $L = 0$ or $L = \lim_s c_s$, depending on whether there are eigenvalues with absolute value greater than $1$. If has some eigenvalue outside the unit disc, the right hand side converges and moreover the left hand side has some non-trival factor.
Observe that since $c_n$ is never zero, no $\lambda_i$ can be a root of unity; and so we are left with showing that if $k \geq 1$ and $\alpha_1, \ldots, \alpha_k \in [0,1)$ are irrational then
$$ a_s := \prod_{j = 1}^k(1-e^{i2\pi\alpha_j s}) $$ does not converge to $0$ nor $\lim_s c_s$.
By Dirichlet's approximation theorem, for each $k \geq 1$ we can find a natural number $s_k$ and an integer $p_k$ for which $|s_k\alpha_j - p_k| < 1/k$, and thus there is a subsequence $(s_k)_k$ for which $e^{i2\pi\alpha_j s_k} \to 1$ for all $j$, and $a_{s_k}\to 0$. But then taking $s'_k = s_k+1$ yields $e^{i2\pi\alpha_j s_k'} \to e^{i2\pi\alpha_j s}$ and $a_{s_k'} \not \to 0$. Hence $(a_s)$ never converges.
Are there results in this direction whose proof is purely algebraic? In particular, can we recover the former partial results without doing analysis?
It seems to me like there is some relation between the spectrum of $A$ and the fact that $I-A^s$ being invertible implies the invertibility of $\sum_{i = 0}^s A^i$, the limit of the latter - were to exist- being the inverse of $I-A$.
Is there a reference for this statement, at least in the case in which $\det(I-A^s) = \pm 1$ for each $s$ and $A$ has integer coefficients?
I suspect this specific case can be proved in a purely algebraic fashion but so far I haven't been able to come up with a proof. Here's what I've done so far:
- Suppose that $f = \chi_A = X^m g, g = (X-\lambda_1) \cdots (X-\lambda_k)$ with $X \not \mid g$. Note $p^{(m)}$ the polynomial whose roots are the $m$-th powers of $p$'s roots. Then our hypothesis is precisely that $f^{(m)}(1)$ is always a unit in $\mathbb{Z}$. Thus, the same holds for $g$, write $g = p_1 \cdots p_r$ where $p_j \in \mathbb{Z}[X]$ are irreducible. Since no eigenvalue $\lambda_i$ is a root of unity, once again by Kronecker's theorem each $p_j$ has a - possibly complex - root of absolute value greater than $1$. Moreover the proof notes that in general if $p \in \mathbb{Z}[X]$ then $p^{(m)} \in \mathbb{Z}[X]$, so in any case, we have $$ \pm 1 = g^{(m)}(1) = p_1^{(m)}(1) \cdots p_r^{(m)}(1) $$ and so $p_j^{(m)}(1) = \pm 1$. This allows us to reduce the original problem to:
Attempted Subproblem 1. Let $f = \prod_{i=1}^k (X-\alpha_i) \in \mathbb{Z}[X]$ be an irreducible monic polynomial. If $f^{(m)}(1) = \pm 1$ for each $m \geq 1$, where $f^{(m)} = \prod_{i=1}^k (X-\alpha_i^m)$, then $f = X$.
- Since $\det(I-A^s) = \pm 1$ for all $s$, dividing by $\det(I-A)$ gives $\det(\sum_{i=0}^s A^i) = \pm 1$ for all $s$. This yields a pretty large sum of all monomials $x_{i_1}^{j_1} \ldots x_{i_r}^{j_r}$ where $\{i_1, \ldots, i_r\} \subset \{1,\ldots, k\}$ and $j_t \in \{0,\ldots, s\}$, being equal to $\pm 1$. Maybe there's some underlying interpretation in terms of symmetric polynomials.
You can definitely say that if $I-A\cdot P(A)$ is invertible for every $P$ polynomial in $A$, then $P$ is nilpotent. Or: if $I - Q(A)$ is invertible for every polynomial $Q$ with $Q(0) = 0$, then $A$ is nilpotent. Since otherwise, let $\lambda\ne 0$ an eigenvalue of $A$, and $Q$ polynomial, $Q(0) = 0$, $Q(\lambda) = 1$. We get $I- Q(A)$ has an eigenvalue $1- Q(\lambda) = 1-1=0$.