# Proof of Gelfand's formula without using $\rho(A) < 1$ iff $\lim A^n = 0$

Gelfand's formula states that the spectral radius $\rho(A)$ of a square matrix $A$ satisfies

$$\rho(A) = \lim_{n \to \infty} \|A^n\|^{\frac{1}{n}}$$

The standard proof relies on knowing that $\rho(A) < 1$ iff $\lim_{n \rightarrow \infty} A^n = 0$. Is there a proof of Gelfand's formula without using this result, whose proof happens to rely on a burdensome use of Jordan Normal form.

• uniform boundness principle Oct 26, 2017 at 23:27
• There is a proof of this in the more general setting of Banach algebras (where the Jordan normal form makes no sense), but it requires a fair amount of complex and functional analysis. I think the Jordan normal form approach is probably the cleanest approach you'll find. Oct 28, 2017 at 3:22
• Remark: Jordan form is not required to prove the statement "$\rho(A)<1$ iff $\lim_{n\to\infty}A^n=0$". See this answer for instance. Nov 16, 2021 at 21:32

Since all norms are equivalent on a finite-dimensional vector space, it suffices to prove Gelfand's formula using Frobenius norm. Suppose first that $$A$$ can be diagonalised as $$PDP^{-1}$$. Then $$\rho(A)=\rho(A^n)^{1/n}\le\|A^n\|_F^{1/n}=\|PD^nP^{-1}\|_F^{1/n} \le\|P\|^{1/n}\|D^n\|_F^{1/n}\|P^{-1}\|_F^{1/n}.$$ Pass $$n$$ to the limit, the result follows.

Next, suppose $$A$$ is not diagonalisable. Since Frobenius norm is unitarily invariant, we may assume that $$A$$ is triangular. Denote by $$|A|$$ its entrywise absolute value. Take any sequence of nonnegative diagonalisable triangular matrices $$\{B_m\}_{m\in\mathbb N}$$ such that $$|A|\le B_m$$ entrywise and $$\lim_{m\to\infty}B_m=|A|$$. (E.g. we may obtain $$B_m$$ by adding small positive amounts to the diagonal entries of $$|A|$$ to make them distinct.) Since each $$B_m$$ is diagonalisable, Gelfand's formula holds for it. Therefore $$\limsup_{n\to\infty}\|A^n\|_F^{1/n} \le\limsup_{n\to\infty}\|\,|A|^n\,\|_F^{1/n} \le\limsup_{n\to\infty}\|B_m^n\|_F^{1/n}=\rho(B_m)$$ and hence by passing $$m$$ to the limit, $$\limsup_{n\to\infty}\|A^n\|_F^{1/n}\le\lim_{m\to\infty}\rho(B_m)=\rho(|A|)=\rho(A).$$ Yet we also have $$\rho(A)\le\liminf_{n\to\infty}\|A^n\|_F^{1/n}$$ because $$\rho(A)=\rho(A^n)^{1/n}\le\|A^n\|_F^{1/n}$$ for every $$n$$. Hence the result follows.

Here's another elementary proof (which also works for general Banach algebras and nontheless avoids complex analysis) based on Rickart's proof of the Gelfand formula

*First things first: the limit of $$\|A^n\|^{1/n}$$ exists by virtue of the subadditivityof the sequence $$a_n=\log(\|A^n\|^{1/n})$$ and Fekete's lemma. The limit equals $$\inf_{n\in \mathbb{N}}\|A^n\|^{1/n}=:\nu$$

*Next, it is easy to see that $$\rho(A) \leq \nu$$. Indeed, for any $$\lambda>\nu$$, one can check that $$B:=\frac{1}{\lambda}\sum_{j=0}^\infty \left(\frac{A}{\lambda}\right)^j$$ is well-defined and $$(\lambda I-A)B = I = B(\lambda I -A)$$, hence $$\lambda\notin \sigma(A)$$.

*So we still have to prove that $$\nu \leq \rho(A)$$.

First suppose that $$\nu=0$$. I will show that $$0$$ must be an eigenvalue then, which suffices for our proof in this case. If $$0$$ were not an eigenvalue of $$A$$, then $$A$$ is invertible and taking the norm of $$I=A^n (A^{-1})^n$$ yields $$1 \leq \|A^n\|^{1/n} \|(A^{-1})^n\|^{1/n} \leq \|A^n\|^{1/n}\|A^{-1}\|\to \nu \|A^{-1}\|$$ which results in a contradiction against the assumption $$\nu=0$$.

From now on suppose that $$\nu>0$$. Assume that $$\rho(A)<\nu$$ (anticipating a contradiction). The following identity, which holds for any $$q>\rho(A)$$, will come in handy, $$\left(\frac{A^n}{q^n}-I\right)^{-1}=\frac{1}{n}\sum_{j=0}^{n-1} \left(\frac{A}{\omega_n^jq}-I\right)^{-1}\qquad (1)$$ where $$\omega_n\in \mathbb{C}$$ is a primitive n'th root of unity. This identity can be verified by multiplying both sides by $$\frac{A^n}{q^n}-I=\frac{A^n}{(\omega_n^j q)^n}-I$$. Fix $$\varepsilon>0$$ (we will later send it to 0). By our assumption that $$\rho(A) <\nu$$, both $$\nu$$ and $$\nu_\varepsilon:=\frac{\nu}{1-\varepsilon}$$ can substitute $$q$$ in the identity (1). Hence $$\left\|\left(\frac{A^n}{\nu_\varepsilon^n}-I\right)^{-1}-\left(\frac{A^n}{\nu^n}-I\right)^{-1}\right\| \leq \frac{1}{n}\sum_{j=0}^{n-1} \left\|\left(\frac{A}{\omega_n^j\nu_\varepsilon}-I\right)^{-1}-\left(\frac{A}{\omega_n^j\nu}-I\right)^{-1}\right\|$$ $$=\frac{1}{n}\sum_{j=0}^{n-1}|\nu_\varepsilon^{-1}-\nu^{-1}|\left\|\left(\frac{A}{\omega_n^j\nu_\varepsilon}-I\right)^{-1}A\left(\frac{A}{\omega_n^j\nu}-I\right)^{-1}\right\|$$ $$\leq \frac{\varepsilon}{n\nu}\sum_{j=0}^{n-1}\left\|\left(\frac{A}{\omega_n^j\nu_\varepsilon}-I\right)^{-1}\right\|\|A\|\left\|\left(\frac{A}{\omega_n^j\nu}-I\right)^{-1}\right\|$$ where I used a variant of the resolvent identity in the step in the middle. The key now is to find an $$M>0$$ which bounds the individual terms in the final sum from above. To find $$M$$ one establishes the continuity of the map $$\varphi: {\cal A}\subset \mathbb{C} \to M^{m \times m}: z \mapsto \left(\frac{A}{z}-I\right)^{-1}$$ where $${\cal A}$$ is the (compact) annulus $$\{z \in \mathbb{C}:\,\nu\leq |z|\leq 2\nu\}$$ (easy and therefore omitted). Compactness of $${\cal A}$$ means that $$\|\varphi\|$$ reaches a maximum $$N<+\infty$$. Wrapping everything together, we then find $$\left\|\left(\frac{A^n}{\nu_\varepsilon^n}-I\right)^{-1}-\left(\frac{A^n}{\nu^n}-I\right)^{-1}\right\| \leq \frac{\varepsilon N^2 \|A\|}{\nu} = C \varepsilon\qquad (2)$$ where it is important that $$C$$ does not depend on $$n$$. Letting $$n \to \infty$$, we have $$\frac{\|A^n\|^{1/n}}{\nu_{\varepsilon}} \to 1-\varepsilon$$ and therefore $$\frac{\|A^n\|}{\nu_{\varepsilon}^n} \to 0$$ and therefore $$\frac{A^n}{\nu_{\varepsilon}^n} \to 0$$ and therefore $$\left(\frac{A^n}{\nu_{\varepsilon}^n}-I\right)^{-1} \to -I$$ and therefore (since the right hand side of (2) can be made arbitrarily small) necessarily $$\left(\frac{A^n}{\nu^n}-I\right)^{-1}\to -I+O(\varepsilon)$$ which requires that $$\frac{A^n}{\nu^n}=O(\varepsilon)$$. But that is impossible since $$\nu=\inf_{n\in \mathbb{N}}\|A^n\|^{1/n}$$. So we arrive at the anticipated contradiction.