# $\ker T\subset \ker S\Rightarrow S=rT$ 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.

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There are two cases: either the kernels are equal or $S = 0$. – t.b. Aug 29 '11 at 5:21
In the infinite-dimensional case you can use codimensions instead of dimensions. – Mark Aug 29 '11 at 6:32
The codimension of kernel of $S$ is at most as large as that of $T$, so is either 1 or 0. What is the codimension of the intersection between two subspaces of codimension 1? – Willie Wong Aug 29 '11 at 13:15
I would argue as follows: If $T = 0$ then $S = 0$. If $T \neq 0$ choose a basis $\{v_i\}_{i \in I}$ of $\ker{T}$ and choose a vector $v$ not in $\ker{T}$. Then $\{v\} \cup \{v_i\}_{i \in I}$ is a ... of $V$. What can you say about $T(v_i)$ and $T(v)$? What about $S(v_i)$ and $S(v)$? Can you fill in the dots and the rest of the argument? – t.b. Aug 29 '11 at 14:17
By the way: When you've figured it out, please do post your solution here and ping me, I'll then have a look at it. – t.b. Aug 29 '11 at 14:30

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 $x'\in{\rm ker}\>T$.

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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.

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