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Put $s={\gamma }^{-1/2}$. ThenFor an eigenvalue $\lambda$, the corresponding $2\times 2$ block has characteristic polynomialFor $\lambda =1$,so the polynomial isFor $\lambda =\gamma$, it isIts discriminant is $\beta (\beta -4)<0$, so its conjugate roots have product $\beta$ and modulus $\sqrt{\beta }$. Both blocks therefore have the same spectral radius, andThis is the heavy-ball rate for a two-eigenvalue diagonal quadratic.

By contrast, the steepest-descent factor isThe heavy-ball factor is $1-{\gamma }^{-1/2}$, which is substantially smaller for $\gamma \gg 1$: it gives an iteration scale of order $\sqrt{\gamma }$ rather than order $\gamma$.

A character of an algebra is a nonzero multiplicative complex-linear functional $\phi :A\to \mathbb{C}$, and the character space of an algebra is the set ${\Phi }_A$ of all such characters. Since $A$ is unital, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. The spectral radius estimate therefore yieldsThus every character is continuous and has norm one.

Let $M$ be a maximal ideal. Its norm closure is again an ideal. It cannot equal $A$, because then some $m\in M$ would satisfy $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Hence $M$ is closed. The quotient $A/M$ is a complex unital Banach division algebra, so the Gelfand-Mazur theorem identifies it with $\mathbb{C}$. Composing the quotient map with this isomorphism gives a character with kernel $M$. Conversely, a character kernel is maximal because its quotient is $\mathbb{C}$.

Now $\lambda \in {\sigma }_A(x)$ exactly when $x-\lambda 1$ is not invertible, equivalently when it lies in some maximal ideal. The preceding result turns that ideal into $\ker \phi$, giving $\phi (x)=\lambda$. The reverse implication follows from the first paragraph, so

The Gelfand topology is the weak-star topology on ${\Phi }_A\subseteq A^{\ast }$. The Gelfand transform isIts values are continuous by the definition of the topology, and multiplicativity and linearity of characters show that it is a unital algebra homomorphism. Finallyso it is continuous.

Assume for contradiction that any algebra norm exists. Part iii makes $K$ compact, so choose an interval disjoint from $K$ and a nonzero $a\in C(\mathbb{R})$ supported in that interval. Choose $b\in C(\mathbb{R})$ with $b=1$ on $K$ and $b=0$ on the support of $a$. Then $a,b$ are nonzero and $ab=0$.

In the completion $B$, every character is evaluation at a point of $K$, so ${\sigma }_B(b)=\{1\}$. Thus $q=1-b$ has spectral radius zero. The equality $ab=0$ givesfor every $n$. The spectral radius formula supplies an $n$ with $\Vert q^n\Vert <1$, and submultiplicativity then givesan impossibility because $a\ne 0$. Hence $C(\mathbb{R})$ admits no algebra norm at all.

A stationary iteration has the formwith a fixed iteration matrix $H$. It converges for every initial vector exactly when the spectral radius satisfies $\rho (H)<1$.