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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....

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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.

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  • $\begingroup$ Ow.. I really did not think of this. Thank you! $\endgroup$
    – Shashi
    Dec 8, 2017 at 14:14

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