Why do elements of the Gel'fand spectrum map self-adjoint elements to points in the usual spectrum of the element?

Question

A. Döring says in the book "Deep Beauty Understanding the Quantum World Through Mathematical Innovation", that, for the Gelfand spectrum defined as:

Definition: A Gelfand spectrum $\Sigma_V$ consists of the positive linear functionals from a abelian subalgebra $V$ of a encompassing von Neumann algebra $\mathfrak{M}$, such that they have norm $1$ and are algebra homomorphisms from $V$ to $\mathbb{C}$.

He then says that for any $\lambda \in \Sigma_V$, it can be shown that for any self-adjoint $A \in V_{sa}$, the following relation holds: $$\lambda(A) \in \sigma(A)$$ where $\sigma(A) \subset \mathbb{R}$ is the usual spectrum of $A$.

How does one prove this? Is it a consequence of the Gelfand–Mazur theorem? In which case, is this then implied because $\lambda$ sends the self-adjoint $A$ to an arbitrary complex number and, by the Gelfand–Mazur theorem(?), the entirety of $\mathbb{C}$ is isomorphic to $\sigma(A)$, making that arbitrary complex number be somewhere in $\sigma(A)$?

(Not sure if that is what the Gelfand–Mazur theorem implies, I kinda stumbled upon it when looking for the reason why $\lambda(A) \in \sigma(A)$, and from a surface reading of the statement of the theorem, he seems to be exactly what needs to be used to prove that type of relation.).

Answer 1

It is just a direct observation: $$ \lambda(A-\lambda(A)1)=\lambda(A)-\lambda(A)=0, $$ so $A-\lambda(A)\in\ker\lambda$. Because $\lambda$ is not zero, one easily check that $\lambda(1)$ has to be a projection; and the only nonzero projection in $\mathbb C$ is $1$. So $\lambda$ is unital. If $A-\lambda(A)1$ were invertible, then its image would have to be nonzero by the multiplicativity of $\lambda$. Hence $A-\lambda(A)1$ is not invertible, and $\lambda(A)\in\sigma(A)$.

With a bit more work one can show that on any unital abelian Banach algebra you have $$ \sigma(A)=\{\lambda(A):\ \lambda:V\to\mathbb C\ \text{ nonzero homomorphism}\}. $$

Answer 2

I strongly suspect this is unavoidably circular. See Martin's comment below.

If $\lambda(A)=0$, then you can show that $A$ is not invertible, so zero is in the spectrum.

Suppose $\lambda(A)$ is non-zero then consider $\mathrm{C}^*(A)$ the unital $\mathrm{C}^*$-algebra generated by $A\in\mathfrak{M}$. This is isomorphic to $C(\sigma(A))$ where $A$ in $C(\sigma(A))$ is identified with the function $t\mapsto t$.

With $\lambda(A)$ non-zero, it restricted to $\mathrm{C}^*(A)$ is a character. And characters on $\mathrm{C}^*(A)$ are exactly evaluation functionals $\mathrm{ev}_x$ for some $x\in\sigma(A)$. So in this picture $\lambda(A)$ is nothing but $\mathrm{ev}_x(t\mapsto t)=x$.