How to get the spectrum of an element of $\mathbb{C}^{2}$?

Question

I know the definition of the spectrum, which is not invertible in an algebra $\mathbf{A}$.

For a complex vector space $\mathbf{A}=\mathbb{C}^2$, define the norm: $$\| (a_1,a_2)\| = \max\{|a_{1}|,\, |a_{2}|\}. $$

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

Now, $\mathbf{A}$ is a $C^{*}$ algebra. Problem about the maximal ideals I have asked here: The Gelfand transform of a example: a complex vector space $\mathbf{A}=\mathbb{C}^2$

But about the spectrum of an element $\lambda \in \sigma(a)$ means that $\lambda e-a=\langle \lambda-a_{1}, \lambda-a_{2}\rangle $ is not invertible. $\textbf{How to write more specifically about the $\sigma(a)$?}$

Answer 1

A useful point of view is to note that $\mathbb C^2=C(\{1,2\})$, the algebra of continuous functions on the discrete set with two points.

So we might as well find the spectrum of $f\in C(X)$, where $X$ is any Hausdorff compact space. A function will be invertible in $C(X)$ when $1/f$ makes sense: so we need $f\ne0$, or $|f|>0$. Because of continuity and compactness, this condition is also sufficient:

$f\in C(X)$ is invertible if and only if $f(x)\ne0$ for all $x$.

Now, $\lambda\not\in\sigma(f)$ precisely when $f-\lambda$ is invertible. By the above, when $f(x)-\lambda\ne0$ for all $x$. That is to say, when $\lambda$ is not in the image of $f$. It follows that $\lambda\in \sigma(f)$ if and only if $\lambda$ is in the image of $f$. In other words, $$ \sigma(f)=\{f(x):\ x\in X\}. $$