If $f$ and $g$ are two non-zero linear functionals and $\ker(f)=\ker(g)$ This question was left as an exercise in course notes on smooth manifolds and I am struck on this.
Question: Let $f: V\to \mathbb{R}$ be a non-zero linear on an $n$-dimensional vector space $V$.
(a) Prove that $\dim(\ker (f))= n-1$.
(b) If $f, g : V \to \mathbb{R}$ are two non-zero linear functionals such that $\ker(f)=\ker(g)$, prove that $g=cf$ for some constant $c\in \mathbb{R}$.
Attempt: (a) I have done.
(b) Let $P(x) = g(x)-cf(x)$ where $c\in \mathbb{R}$ is a constant. Then, if $x\in\ker(f)=\ker(g)$ then $g=cf$. But, when $x\notin\ker(f) $, I am unable to think how should I proceed.
Can you please help me with this?
 A: There is $v_0 \in V$ such that
$$ (*) \quad V= Ker(f) \oplus lin(v_0),$$
where $ lin(v_0)$ denotes the linear hull of $\{v_0\}.$
Now let $c= \frac{g(v_0)}{f(v_0)}$
If $v \in V$, there is $u_0 \in ker(f)$ and $t \in \mathbb R$ (by $(*)$),  such that
$$v=u_0 +tv_0.$$
This gives $g(v)=t g(v_0)$ and $f(v)=t f(v_0).$ Hence
$$g(v)=t c f(v_0)=c f(v).$$
A: Choose a fixed $x\notin\ker(f)$. Then $f(x)\neq0$. Consider $c=\frac{g(x)}{f(x)}$, which yields $g(x)=c\cdot f(x)$. I claim that $g(y)=c\cdot f(y)$ for any $y\in V$.
There is an $a\in\Bbb R$ such that $y-ax\in\ker(f)=\ker(g)$ (the equation $f(y)-af(x)=0$ has a solution).
Now we get
$$\begin{align}
g(y-ax)={}&cf(y-ax)\\
g(y)-g(ax)={}&cf(y-ax)\\
g(y)-ag(x)={}&cf(y-ax)\\
g(y)-acf(x)={}&cf(y-ax)\\
g(y)={}&acf(x)+cf(y-ax)\\
g(y)={}&cf(ax)+cf(y-ax)\\
g(y)={}&c(f(ax )+f(y-af))\\
g(y)={}&cf(ax+y-ax)\\
g(y)={}&cf(y)\end{align}
$$
A: Denote $W=\ker f=\ker g$ we may form the quotient map $\pi:V\to V/W\cong \mathbb{R}$. The mappings $f$ and $g$ factor through $\pi$ and we get $\bar{f},\bar{g}:\mathbb{R}\cong V/W\to \mathbb{R}$ so $\bar{f}$ and $\bar{g}$ differ by a constant of multiple. Write $\bar{f}=c\bar{g}$ we get $f=\bar f\circ \pi=c\bar g\circ \pi=cg$
