Character on Banach Algebra is Surjective/Proof 1

Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.

Let $\phi : A \to \C$ be a character on $A$.

Then $\phi$ is surjective.

Proof

From Image of Submodule under Linear Transformation is Submodule, $\phi \sqbrk A$ is a vector subspace of $\C$.

From Dimension of Proper Subspace is Less Than its Superspace, we have:

$\dim \phi \sqbrk A \le \dim \C = 1$

and so we either have $\phi \sqbrk A = \set 0$ or $\phi \sqbrk A = \C$.

Since $\phi \ne 0$ by the definition of a character, we have $\phi \sqbrk A = \C$.

$\blacksquare$