Why are two characters of a commutative Banach algebra with the same kernel equal?

Question

Let $A$ be a commutative Banach algebra. Let $\chi_1$ and $\chi_2$ be characters of $A$.

I am having some difficulty seeing why the following statement is true:

If $\ker \chi_1 = \ker\chi_2$, then since $\chi_1(\textbf{1})=\chi_2(\textbf{1})=\textbf{1}$, we have that $\chi_1 = \chi_2$.

Answer 1

Suppose $a$ is an element of the algebra, and $\chi_1(a) = \lambda$. Then $a - \lambda \mathbf{1} \in \ker \chi_1 = \ker \chi_2$ so

$$ 0 = \chi_2(a - \lambda \mathbf{1}) = \chi_2(a) - \lambda $$