TITLE: $\ker T\subset \ker S\Rightarrow S=rT$ when $S$ and $T$ are linear functionals
QUESTION [14 upvotes]: 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.

REPLY [4 votes]: This is an exercise in Halmos' books and he has a very nice and simple basis-free solution, which only requires the definition of linear functional:

If $T=0$, then $S=0$ (by hypothesis) and thus the desired result holds trivially.

If $T\neq 0$, then there exists $v_0\in V$ such that $T(v_0)\neq 0$. As
$$T(T(v_0)v-T(v)v_0)=T(v_0)T(v)-T(v)T(v_0)=0,\quad\forall \ v\in V$$
we conlcude (by hypothesis) that
$$0=S(T(v_0)v-T(v)v_0)=T(v_0)S(v)-T(v)S(v_0),\quad\forall \ v\in V.$$
Then,
$$S(v)=\frac{S(v_0)}{T(v_0)}T(v),\quad\forall \ v\in V$$
and thus the result holds with $r=\frac{S(v_0)}{T(v_0)}$.


Motivation for this solution: If such a constant $r$ does exist, then it has to be $r=\frac{S(v)}{T(v)}$ for all $v$  such that $T(v)\neq 0$.