Q:

Linear functionals, complex vector spaces

I'm stuck with this problem, could somebody help me?
"Let $V$ be a complex vector space and assume that $f, g \colon V \to \mathbb{C}$ are two linear functionals such that $\ker f \subseteq \ker g$. Prove that there exists $c \in \mathbb{C}$ such that $g(x) = cf(x)$ for all $x \in V$."

A:

Define $f_{0}:V/\ker f\rightarrow{\bf{C}}$, $f_{0}(x+\ker f)=f(x)$, it is routine to show that $f_{0}$ is an one-to-one onto linear map. So $V/\ker f$ is algebraic isomorphic to ${\bf{C}}$, which is one-dimensional. 
Hence $V/\ker f=\left<a+\ker f\right>$ for some $a\in V$.
On the other hand, let $g_{0}:V/\ker f\rightarrow{\bf{C}}$, $g_{0}(x+\ker f)=g(x)$. Since $\ker f\subseteq \ker g$, such a $g_{0}$ is well-defined.
Assume that $f=0$, then $\ker f=V$, so is $\ker g=V$, then $g=0$, take $c=0$.
Assume now that $f\ne 0$, as $f_{0}(a+\ker f)=f(a)$ and $a+\ker f$ generates $V+\ker f$, if $f(a)=0$, then $f_{0}=0$, then $f=0$ by construction. So $f(a)\ne 0$.
Let $c=\dfrac{g(a)}{f(a)}$. For any $x\in V$, let $k_{x}\in{\bf{C}}$ be its coefficient, that is, $x+\ker f=k_{x}(a+\ker f)$, then 
\begin{align*}
g(x)&=g_{0}(x+\ker f)\\
&=g_{0}(k_{x}(a+\ker f))\\
&=k_{x}g_{0}(a+\ker f)\\
&=k_{x}g(a)\\
&=k_{x}cf(a)\\
&=c(k_{x}f(a))\\
&=c(k_{x}f_{0}(a+\ker f))\\
&=cf_{0}(k_{x}(a+\ker f))\\
&=cf_{0}(x+\ker f)\\
&=cf(x).
\end{align*} 

