I would like only a hint to the following exercise:
Let $V$ be a vector space over the field $K$, and $T$, $S$ linear functionals on V such that $Tv=0\Rightarrow Sv=0$. Prove that there exists $r\in K$ such that $S=rT$.
I know how to prove this when $V$ is finite dimensional. I show that if there is no such constant $r$ then $n-2=\operatorname{dim}\textrm{ }(\ker\textrm{ }T\textrm{ }\cap \ker\textrm{ }S)=\operatorname{dim}\ker T=n-1$, a contradiction. But this approach doesn't seem to help at all for the stated problem.
Best Answer
Following Theo Buehler's suggestion:
$T=0 \Rightarrow \ker T = V \Rightarrow \ker S = V \Rightarrow S = 0$, and the existence of $r$ is trival.
If $T\neq 0$, let $\{v_i\}_{i\in I}$ be a basis of $\ker T$. If $v\notin\ker T$, then $\{v_i\}_{i\in I}\cup\{v\}$ is a basis of $V$: say $Tv=k\neq 0$. If $Tw=0$ then $w\in\left<v_i\right>_{i\in I}$; if $Tw=h\neq 0$, then $Tw=\frac{h}{k}k=\frac{h}{k}Tv=T\frac{h}{k}v$, therefore $w-\frac{h}{k}v\in\ker T$, and we're done.
$Tv_i=0$ implies $Sv_i=0$. If $Tv=k\neq 0$, then $Sv=h\neq 0$ unless $S=0$ in which case $r=0$.
But then $S=\frac{h}{k}T$, and the verification is immediate on the basis elements.