Show that $f=cg$ for some scalar $c$. This question is in connection with this one: Show that $\langle v,w\rangle _1=c\langle v,w\rangle _2$ for some scalar $c$..

Let $V$ be a vector space. Let $f,g:V\to F$ such that $\ker f =\ker g$. Show that $f=cg$ for some scalar $c$.

If $f\equiv 0\implies g\equiv 0$, take $c=1$ we are done.
If $f\neq 0$ then $f(v_0)\neq 0\implies g(v_0)\neq 0$.
How can I prove $f=cg$ from here?
 A: Well, let's see . . . a smallish lemma might kick this off:
Smallish Lemma: Let $V$ be a vector space over the field $F$, and let
$\lambda:V \to F \tag 1$
be linear over $F$; if $\lambda \ne 0$, then
a.)  There exists $w \in V$ with
$\lambda(w) = 1; \tag 2$
b.)  For any $v \in V$, 
$v = aw + y, \; a = \lambda(v) \in F, \; y \in \ker \lambda; \tag 3$
c.) $a$ and $y$ are unique.
Proof of Smallish Lemma:  For (a), since $\lambda \ne 0$, there is some $x \in V$ with 
$\lambda(x) \ne 0; \tag 4$
set
$w = (\lambda(x))^{-1} x; \tag 5$
then
$\lambda(w) = \lambda( (\lambda(x))^{-1} x) =  (\lambda(x))^{-1} \lambda(x) = 1; \tag 6$ 
For (b):  if $v \in V$, set $a = \lambda(v)$ and $y = v - aw$, with $w \in V$ as in (a); then
$\lambda(y) = \lambda(v) - a \lambda(w) = \lambda(v) - \lambda(v) \cdot 1 = 0; \tag 7$
thus
$y \in \ker \lambda, \tag 8$
and
$v = aw + y = \lambda(v) w + y; \tag 9$
For (c), if $a_1, a_2 \in F$, $y_1, y_2 \in \ker \lambda \subset V$, and
$a_1 w + y_1 = v = a_2 w + y_2, \tag{10}$
then
$(a_1 - a_2)w = y_2 - y_1 \in \ker \lambda; \tag{11}$
so that
$a_1 - a_2 = (a_1 - a_2) \lambda(w) = \lambda((a_1 - a_2)w) = \lambda(y_2 - y_1) = 0; \tag{12}$
thus $a_1 = a_2$ from which (10) yields $y_1 = y_2$.  End:  Proof of Smallish Lemma.
We may apply the Smallish Lemma to the present situation as follows: if $f = 0$, take $c = 0$ and we are done.  If $f \ne 0$, by Smallish there exists $w \in V$ with $f(w) = 1$; since for any $v \in V$,
$v = f(v)w + y, \; y \in \ker f = \ker g, \tag{13}$
$g(v) = g(f(v)w + y) = f(v)g(w) + g(y) = f(v)g(w) = g(w)f(v); \tag{14}$
now taking $c = g(w)$ we are done!
