How to prove that this kernel has codimension 1? Let $\phi\in X'$ , where X is a Hilbert space and $X'$ its dual. Then I want to check that $\ker\phi$ is a closed subspace of $X$ of codimension 1 ($\phi \ne 0$).
So to see the closeness we pick a sequence $(x_n) \in \ker \phi$ that converges to $x$, then, since $\phi$ is a bounded functional it is continuous, thus $\phi((x_n)) \to \phi(x)=0$ therefore $\ker \phi$ is closed.
The thing is that I don't know how to attached the socond part, because I think I need to prove that the space $X/ \ker \phi=\{x+\ker \phi : x \in X\}$ has dimension 1.
Can someone help me to prove this assertion please?
Thanks a lot in advance.
 A: This is a purely algebraic result that you can prove without using the inner product or any other additional structure on your vector space:

Proposition. Let $V$ be a vector space over some field $\mathbb{F}$, and let $\phi\colon V\to\mathbb{F}$ be a nontrivial linear functional. Then $\ker(\phi)$ has codimension 1.

The proof will consist of the following three steps:

*

*Note that, by the first isomorphism theorem, $V/\ker(\phi)\cong\operatorname{im}(\phi)$ (as $\mathbb{F}$-vector spaces).


*Note that $\operatorname{im}(\phi)$ is a subspace of $\mathbb{F}$, and that the only subspaces of $\mathbb{F}$ are $\{0\}$ and $\mathbb{F}$ itself. Since $\phi$ is nontrivial, the only possibility is $\operatorname{im}(\phi)=\mathbb{F}$.


*Use the fact that isomorphic spaces have the same dimension to conclude that $\operatorname{codim}_\mathbb{F}(\ker(\phi))\overset{\mathrm{def}}{=}\dim_\mathbb{F}(V/\ker(\phi))=\dim_\mathbb{F}(\mathbb{F})=1$.
A: Let $z\not\in \ker\phi$, the observe that for each $x\in X$, 
$$x= \frac{\phi(x)}{\phi(z)} z - \frac{\phi(x)}{\phi(z)} z + x\\
= \frac{\phi(x)}{\phi(z)} z + \bigg[x - \frac{\phi(x)}{\phi(z)} z \bigg]$$
and $\bigg[x - \frac{\phi(x)}{\phi(z)} z \bigg] \in ker \phi$. This means that $X$ can be written as the direct sum of its two subspaces that is 
$$X = \text{span} \{z \}\oplus \ker \phi.$$
Therefore the $\ker \phi$ has codim $1$.
