TITLE: Proof of Gelfand's formula without using $\rho(A) < 1$ iff $\lim A^n = 0$
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: 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.