Spectral radius formula

Remarks

Proof idea: if $\lambda \in \sigma(T)$, then $\lambda^n \in \sigma(T^n)$, and the Neumann series shows $\abs{\lambda}^n \le \norm{T^n}$; this gives $r(T) \le \inf_n \norm{T^n}^{1/n}$. For $\abs{\lambda} > \norm{T}$, the resolvent has the expansion $R(\lambda, T) = -\sum_{n \ge 0} T^n / \lambda^{n+1}$. The resolvent is analytic on $\abs{\lambda} > r(T)$, so this Laurent series converges there, and the uniform boundedness principle with the Cauchy estimates gives $\limsup_n \norm{T^n}^{1/n} \le r(T)$. Nonemptiness of $\sigma(T)$ follows from Liouville's theorem applied to $\lambda \mapsto f(R(\lambda, T))$ for continuous linear functionals $f$.

Example: the Volterra operator $(Vf)(x) = \int_0^x f(t) \, dt$ on $C[0, 1]$ with the supremum norm satisfies $\norm{V^n} = 1/n!$, attained at $f = 1$. So $r(V) = \lim (1/n!)^{1/n} = 0$ and $\sigma(V) = \set{0}$, although $V \ne 0$.

The field must be $\C$. The rotation $T(x, y) = (-y, x)$ of the real plane has $\norm{T^n} = 1$ for all $n$, but $T - \lambda I$ is invertible for every real $\lambda$, so its real spectrum is empty.

For matrices this is Gelfand's formula for the spectral radius. The same statement, with the same proof, holds for every element of a unital complex Banach algebra.

Going deeper

For a student

The formula shows that $T^n \to 0$ in norm if and only if $r(T) < 1$, and that $\sum T^n$ converges in that case, which sharpens the Neumann series. It also shows that $r(T)$ depends only on the algebra, not on the choice of an equivalent norm. A quasinilpotent operator, with $r(T) = 0$, need not be nilpotent, as the Volterra operator shows. For normal operators on a Hilbert space, $\norm{T^{2^k}} = \norm{T}^{2^k}$, so $r(T) = \norm{T}$; this is a key step in the theory of C*-algebras.