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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.

Davide Giraudo
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Weltschmerz
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    There are two cases: either the kernels are equal or $S = 0$. – t.b. Aug 29 '11 at 05:21
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    In the infinite-dimensional case you can use codimensions instead of dimensions. – Mark Aug 29 '11 at 06:32
  • @Theo: Sorry, I've been thinking, and I need a little more. – Weltschmerz Aug 29 '11 at 12:40
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    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
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    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
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    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
  • @Willie: I understand why the codimension of $\ker T$ or $\ker S$ is at most 1, and that the codimension of $\ker S$ is not greater than the codimension of $\ker T$. However I don't know about the codimension of the intersection between two subspaces (in general, because in this case it's either 0 or 1). I just read the definition of codimension. – Weltschmerz Aug 29 '11 at 17:02
  • @Theo: Answer posted, thank you. – Weltschmerz Aug 29 '11 at 17:03
  • This looks very good! I hope I've not given too much away... – t.b. Aug 29 '11 at 18:50
  • @Theo: would you mind pointing me more explicity towards Wille's suggestion? – Weltschmerz Aug 31 '11 at 23:21
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    Honestly, I didn't understand Mark's and @Willie 's point either. I guess they wanted to argue as follows: If $S \neq 0 \neq T$ then the kernels have both codimension $1$. Their intersection has then dimension $1$ or $2$ (but that's something you need to prove...) If it is two get a contradiction to the fact that $\ker{T} = \ker{T} \cap \ker{S}$ has codimension $1$ by hypothesis (the same contradiction as you found in finite dimensions). So it must be $1$. Now... well, now what? Do the argument Christian Blatter suggested (or what you did). Doesn't seem to really help to me. – t.b. Sep 01 '11 at 00:17
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    @Theo's right. That line of argument is toward showing that if $S\neq 0 \neq T$ then $\ker S = \ker T$. Then you look at the action of $S$ and $T$ on the 1-dimensional quotient $V / \ker T$ (which is essentially the second half of Christian Blatter's answer). But as Christian noted, $\ker T\subset \ker S$ is enough for the argument. (In my mind I was taking the quotient by $\ker S$, which would not be good for the operator $T$ unless $\ker S \subset \ker T$; I was making the solution more complicated than necessary.) – Willie Wong Sep 01 '11 at 00:38

3 Answers3

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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.

Weltschmerz
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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 $T(v)\neq 0$.

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

Christian Blatter
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