Example 1

Suppose that there exists $\lambda \in \mathbf{C}$ such that Sp$_A(x)$ is reduced to $\lambda$. Then one has $x=\lambda \cdot 1$. In fact, the identity function of Sp$_A(x)$ is equal to $\lambda$, therefore its image under the functional calculus mapping, that is to say $x$, is equal to $\lambda \cdot 1$.