$\ker T\subset \ker S\Rightarrow S=rT$ when $S$ and $T$ are linear functionals 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.
 A: You have to bring the assumption ${\rm ker}\>T\subset {\rm ker}\>S$ to bear without talking about a basis.
If $T=0$ everything is easy.
If there is a vector $a\in V$ with $Ta=1$ make an educated guess what $S$ would have to be and prove that guess. To this end you should (prove and) use the fact that any vector $x\in V$ can be written in the form $x=\xi a + x'$ for some $\xi\in K$ and $x'\in{\rm ker}\>T$.
A: 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 $T(v)\neq 0$.
A: 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.
