Kernel of Linear Functionals Problem: Prove that for all non zero linear functionials $f:M\to\mathbb{K}$ where $M$ is a vector space over field $\mathbb{K}$, subspace $(f^{-1}(0))$ is of co-dimension one.
Could someone solve this for me?
 A: The following is a proof in the finite dimensional case: The dimension of the image of $f$ is 1 because $\textrm{im} f$ is a subspace of $\Bbb{K}$ that has dimension 1 over itself. Since $\textrm{im} f \neq 0$ it must be the whole of $\Bbb{K}$. By rank nullity, 
$$\begin{eqnarray*} 1 &=& \dim \textrm{im} f \\
&=&   \dim_\Bbb{K} M- \dim \ker f\end{eqnarray*}$$
showing that $\ker f$ has codimension 1.
A: Take $u \in M$ so that $f(u) = 1$.  Then for any $v \in M$ you can write $v = f(v) u + w$ where $w \in f^{-1}(0)$ since $f(w) = f(v - f(v) u) = 0$.  That says $M = {\mathbb K} u + f^{-1}(0)$, so $f^{-1}(0)$ has codimension $1$.
A: If $V$ and $W$ are vector spaces and $T:V\rightarrow W$ is linear, the isomorphism theorem says that
$${V\over \ker T} \cong {\rm Im}(T) $$
In the case that $W$ is the base field and $T\not= 0$, $T$ must be onto, so 
the dimension of $V/{\ker(T)}$ is 1.  This does not depend upon finite-dimensionality.
A: First, we show that $Ker(f)$ is a hyperplane in $M$. Let $W_{1} \subset M$ a subspace, it's suffice shows that if $Ker(f) \neq W_{1}$ then $W_{1}=M$. Since $f \neq 0$ we have $Ker(f) \neq M$. 
Suppose $Ker(f) \neq W_{1}$ and choose $v_{0} \in W_{1} \setminus Ker(f)$. Let $v \in M$ and taking $u= v - \frac{f(v)}{f(v_{0})}v_{0}$ we have that $u \in Ker(f) \subset W_{1}$, therefore $v \in W_{1}$, thus $W_{1}= M$.
Thereby, since $Ker(f)$ is a hyperplane in $M$, it's the maximal subspace in M. Write $W' = Lin(x_{0} \cup ker(f))$. Since that $W \subset W'$, we have that $W' = M$. Then,
$M = Lin(x_{0}) + ker(f)$
So, $ker(f)$ has codimesion 1
