The Gelfand transform of a example: a complex vector space $\mathbf{A}=\mathbb{C}^2$

Question

Consider a complex vector space $\mathbf{A}=\mathbb{C}^2$, define the norm: $$\parallel a\parallel = max\{\mid a_{1}\mid, \mid a_{2}\mid\}, \forall a=<a_{1},a_{2}> \in \mathbb{C}^2.$$

$\mathbf{A}$ is a Banach space, and define the product law and involution: $$ab=<a_{1}b_{1},a_{2}b_{2}>, a^{*}=<\bar{a_{1}}, \bar{a_{2}}>.$$

Obviously, it is a C* algebra, but I do not understand how to find the maximal ideals?

My questions is:

Consider the Gelfand transform: $$ \mathbf{A} \rightarrow C(\mathbf{M})$$ $$a \mapsto \widehat{a}(J)$$ where $\mathbf{M}$ is the maximal ideals of $ \mathbf{A}$.

What is the $\widehat{a}(J)$?

Answer 1

$\mathbf A$ is isometrically *-isomorphic to $C(K)$, where $K=\{1,2\}$. (Not because of anything to do with Banach algebra theory, but trivially: Define $T:\mathbf A\to C(K)$ by $T(a)=f$, where $f(1)=a_1$ and $f(2)=a_2$. Then $T$ is an isometric $*$-isomorphism.)

Edit: Comments from "From Review" notwithstanding, that really does answer the question, for anyone who knnows the first thing about Banach algebras. A review of the basic fact about $C(K)$, for readers unfamiliar with the topic: Note that $K$ will always denote a compact Hausdorff space. If $I$ is an ideal in $C(K)$ we will let $Z(I)$ denote the set of $x\in K$ such that $f(x)=0$ for every $f\in I$.

Lemma If $I$ is a proper ideal in $C(K)$ then $Z(I)\ne\emptyset$.

Suppose otoh that $Z(I)=\emptyset$. For every $x\in K$ there exists $f\in I$ with $f(x)\ne0$. Since $K$ is compact there exist $f_1,\dots,f_n\in I$ such that for every $x\in K$ there exists $j$ with $f_j(x)\ne0$.

Since $I$ is an ideal it follows that $g\in I$, where $$g=\sum|f_j|^2=\sum f_j\overline f_j.$$But our choice of $f_1,\dots f_n$ shows that $g(x)>0$ for every $x\in K$. In particular $1/g\in C(K)$, so $1\in I$, hence $I=C(K)$. QED.

Note that a "complex homomorphism" of $C(K)$ is a nonzero homomorphism $\phi:C(K)\to\Bbb C$; in particular if $\phi$ is a complex homomorphism of $C(K)$ then $\phi(1)=1$.

The Main Theorem About $C(K)$: If $\phi$ is a complex homomorphism of $C(K)$ then there exists $x\in K$ such that $\phi(f)=f(x)$ for every $f\in C(K)$.

Let $I$ be the kernel of $\phi$; then $I$ is a proper ideal, so $Z(I)\ne\emptyset$. Say $x\in Z(I)$. Then $I\subset J$, where $$J=\{f\in C(K):f(x)=0\}.$$Since $I$ is a maximal ideal, $I=J$.

Now suppose $f\in C(K)$. Let $a=f(x)$ and define $$g=f-a=f-a \cdot 1.$$Then $g(x)=0$, so $g\in I$. Hence $\phi(g)=0$, which says that $$\phi(f)=\phi(a\cdot 1)=a\phi(1)=a=f(x).$$