# How does $\operatorname{codim}(\mathcal{N}(f))=1$ holds if $\mathcal{N}(f)=\{0\}$

I'm studying Functional Analysis of Kreyszig and in problem 2.8.10 I must show that for a linear functional $f\neq 0$ we have $\operatorname{codim}\mathcal{N}(f)=1$.

I am not asking for a solution to how to show this. The only question that I have is shouldn't the author of the book state that $\mathcal{N}(f)\neq \{0\}$, or am I missing something? otherwise if we would have $\mathcal{N}(f)=\{0\}$ then $\operatorname{codim}\mathcal{N}(f)=\operatorname{dim}(X/\{0\})=\operatorname{dim}(X)$ , where $X$ is the normed space we are working on. But $\operatorname{dim}(X)$ is not $1$ in general....

If $\mathcal N(f) = \{0\}$, then $f : X \to \mathbb{K}$ is injective, where $\mathbb K$ is your field. Hence, $f$ is a bijection and $X$ is one-dimensional.