$C^*$-subalgebra of $M_2(\mathbb{C})$

Question

Consider the following subalgebra of $M_2(\mathbb{C})$: $$A= \left\{\begin{bmatrix} a & b \\ b & a \end{bmatrix} : a,b\in \mathbb{C}\right\}.$$

One method I know that $A$ is isometrically *-isomorphic to $C(K)$, where $K=\{1,2\}$. (Because we can assume function $f$ such that $f(1)= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $f(2)=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, then we use the Gelfand transform between $C(K)$ and $C(\mathscr{M})$, where $\mathscr{M}$ is the maximal ideals of $C(K)$. Hence, using the Gelfand transform $\widehat{f}(J)=f(x), J\in \mathscr{M}, x\in K $ to get the maximal ideals of $A$.

But this problem is asked to show that $\Delta=\{\phi_{1}, \phi_{2}\}$, where $\phi_{1}(y)=a+b, \phi_{2}(y)=a-b$. Here, $\Delta$ is the set of all complex homomorphism $A \to \mathbb{C}$.

$\textbf{How to solve this problem in this way? }$

Answer 1

Suppose that $\pi:A\to\mathbb C$ is a homorphism.

If $\pi(I)=0$, then $\pi=0$. Otherwise, from $\pi(I)=\pi(I^2)=\pi(I)^2$, we deduce that $\pi(I)=1$. In any case, $\pi(aI)=a$. Also, since $$ x=\begin{bmatrix} 0&1\\1&0\end{bmatrix} $$ satisfies $x^2=I$, we deduce that $\pi(x)^2=1$. So $\pi(x)=1$ or $\pi(x)=-1$. And those are all the choice we have: when $\pi(x)=1$, we obtain the homomorphism $$ \pi\left(\begin{bmatrix} a&b\\b&a\end{bmatrix} \right)=\pi(aI+bx)=a+b. $$ When $\pi(x)=-1$, we obtain $$ \pi\left(\begin{bmatrix} a&b\\b&a\end{bmatrix} \right)=\pi(aI+bx)=a-b. $$

Answer 2

We can explicitly construct an isomorphism $\Psi:A\to C(K)$ by $$\Psi\begin{pmatrix}a&b\\b&a\end{pmatrix}=(a+b,a-b).$$ Then any complex homomorphism of $A$ is of the form $\phi\circ\Psi,$ where $\phi$ is a complex homomorphism of $C(K)$.