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?

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 $f=\bar f\circ \pi=c\bar g\circ \pi=cg$
