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.

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    $\begingroup$ uniform boundness principle $\endgroup$ Oct 26, 2017 at 23:27
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    $\begingroup$ 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. $\endgroup$
    – Aweygan
    Oct 28, 2017 at 3:22
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    $\begingroup$ 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. $\endgroup$
    – user1551
    Nov 16, 2021 at 21:32

2 Answers 2


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.


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