Clarification of linear functionals Let $V$ be a vector space of dimension $n$.
Let $f: V \to \Bbb K$, $g :V \to \Bbb K$ be linear maps, so that $f,g \in V^*$. Prove that if 
$$\ker f=\ker g,$$
then there is some $c \in \Bbb{K}$
such that 
$$f= cg$$
on $V$.
I'm having a first approach to linear algebra and I can't understand how I should proceed to prove the following
 A: Consider subspace $W$ of $V$ such that $\mathrm{codim}(W) = 1$. Then you can write 
$$V = W \oplus Kv_0$$
for some nonzero vector $v_0\in V$. Now if 
$$\mathrm{ker}(f) = W = \mathrm{ker}(g)$$
and $f(v_0) = cg(v_0)$ for some $c\in K$. Then you can show that $f = cg$ i.e. $f(v) = cg(v)$ for all $v\in V$
If you are more advanced, then you can use some argument based on isomorphism theorem.
Edit (version with isomorphism theorem).
Isomorphism theorem (for vector spaces).
If $f:V\rightarrow W$ is a surjective linear map between arbitrary vector spaces then there exists an isomorphism of vector spaces $\phi:W\rightarrow V/\mathrm{ker}(f) $ such that $q = \phi \circ f$, where $q:V\rightarrow V/\mathrm{ker}(f)$ is the canonical quotient linear map.
Corollary. Suppose that $f, g:V\rightarrow W$ are linear surjections such that $\mathrm{ker}(f) = \mathrm{ker}(g)$. Then it follows by Theorem that there exists an isomorphism $\phi:W\rightarrow W$ such that  $\phi \circ f = g$.
Next suppose that two linear maps $f,g:V\rightarrow K$ have the same kernel. If one of $f, g$ is zero, then $g = 0 \cdot f$. So assume that none of them is zero. Then they are both surjective. Applying Corollary we derive that there exists an isomorphism $\phi:K\rightarrow K$ such that $\phi\circ f = g$. Since $\phi$ is linear automorphism of one dimensional vector space, we derive that $\phi$ is multiplication by some $c\in K$. Thus
$$g = cf$$
for some $c\in K$.
A: $$V=\ker(f)\oplus Span\{x_0\}=\ker(g)\oplus Span\{x_0\}.$$
 Set $c=\frac{f(x_0)}{g(x_0)}$. Let $x=tx_0\in Span\{x_0\}$ where $t\in\mathbb K$. Then,
$$f(x)=f(tx_0)=tf(x_0)=tcg(x_0)=cg(tx_0)=cg(x).$$
A: If $g\equiv 0$, there's nothing to prove. So assume  $g\not\equiv 0$ and let $a\in V $ be such that  $g (a)\neq 0$. For $x\in V $, define  $x_1=x-\frac {g (x)}{g (a)}a $. Then $x_1\in \ker g=\ker f $. So $f (x_1)=0$. Rewriting, $f (x)=\frac {f(a)}{g (a)}g (x)$.
