Trying to understand how the Gelfand spectrum relates to the usual notion of spectrum of an operator? I am trying to understand how the Gelfand spectrum relates to the usual notion of spectrum of an operator. I have tried to construct a simple artificial example to help with understanding but I am stuck. First I will give the definitions and then I will state my example.


*

*Gelfand spectrum: Let $A$ be a commutative Banach algebra with unit $1$ such that $||1|| = 1$. The Gelfand spectrum of $A$ is defined as the set
$$
\text{sp}(A) = \{\omega \in \text{hom}(A,\mathbb{C}) : \omega \neq 0 \}
$$
of all nontrivial complex homomorphisms of $A$.

*Gelfand map: Any $x\in A$ gives rise to $\hat{x}:\text{sp}(A) \to \mathbb{C}$ such that $\hat{x}(\omega) = \omega(x)$, for $\omega \in \text{sp}(A)$. The map $x \mapsto \hat{x}$ is known as the Gelfand map.

*Gelfand spectrum correspondence with spectrum of elements of $A$: For any $x \in A$ it holds that
$$
\sigma(x) = \{\hat{x}(\omega):\omega \in \text{sp}(A)\}
$$


Explicit example
So here is my explicit example. Take $A$ to be the set of diagonal $2\times 2$ matrices with sub-multiplicative matrix norm. So this seems to be a valid explicit form for $A$ (i.e. a commutative Banach algebra with unit $1$).
Take the element $x\in A$ given by
$$
x =
\begin{bmatrix}
3 & 0 \\
0 & 2
\end{bmatrix},
$$
and clearly $\sigma(x) = \{3,2\}$. So we should be able to recover this from definition 3. above. So using our explicit element $x\in A$ we can write
$$
\sigma(x) = \{\hat{x}(\omega):\omega \in \text{sp}(A)\}
$$
as
$$
\begin{align}
\sigma(\begin{bmatrix}
3 & 0 \\
0 & 2
\end{bmatrix}) & = \{\hat{\begin{bmatrix}
3 & 0 \\
0 & 2
\end{bmatrix}}(\omega):\omega \in \text{sp}(A)\} \\
& = \{\omega(\begin{bmatrix}
3 & 0 \\
0 & 2
\end{bmatrix}):\omega \in \text{sp}(A)\}
\end{align}
$$
Now I don't see how to get $\sigma(x) = \{3,2\}$ from this? It seems I can go no further. Is there an invalid assumption in my example? What am I doing wrong?
 A: Say $D(a,b)$ is the diagonal matrix with $a$ and $b$ on the diagonal. In fact there are exactly two complex homomorphisms of $A$, one with $\omega(D(a,b))=a$ and one with $\omega(D(a,b))=b$:
Say $I$ is an ideal in $A$. If there exists $D(a,b)\in I$ with $a\ne0$ and $b\ne0$ then $I=A$. So if $I$ is the kernel of a complex homomorphism $\omega$ then either $a=0$ for all $D(a,b)\in I$, in which case $\omega(D(a,b))=a$, or ..., in which case $\omega(D(a,b))=b$.
In detail: For every $D(a,b)\in I$ we must have $a=0$ or $b=0$, since if $a\ne0$ and $b\ne0$ then the fact that $I$ is an ideal shows that $I=A$. But if $D(a,0)\in I$ and $D(0,b)\in I$ with $a\ne0$ and $b\ne0$ then $D(a,b)\in I$ again implies that $I=A$. So either $a=0$ for all $D(a,b)\in I$ or $b=0$ for all $D(a,b)\in I$. Wlog $a=0$ for all $D(a,b)
\in I$.
Now for any $x=D(a,b)\in A$, let $\alpha=\omega(x)$. Let $e=D(1,1)$, the identity. Then $\omega(x-\alpha e)=0$, so $D(a-\alpha,b-\alpha)=x-\alpha e\in I$; hence $a-\alpha=0$. So $\omega(D(a,b))=a$, qed.
Note In fact $A$ is just $C(K)$, where $K$ is a two-point set; the above is a special case of the standard proof that any complex homomorphism of $C(K)$ is given by evaluation at some point of $K$.
