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Let L and T be two linear functionals on a real vector space $V$ such that $L(v) = 0$ implies $T (v) = 0$. Show that $T = cL$ for some real number $c$.

how can i prove the above problem. clearly converse part is true.but how can i proceed in this case.

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@did: How is a linear functional supposed to be invertible unless $V$ is one-dimensional? –  wj32 Nov 2 '12 at 21:49
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@wj32 Right. Sorry about the noise. –  Did Nov 3 '12 at 6:35
    
Consider this approach. Given $N(L) \subset N(T)\,,$ where $N$ is the null space of the linear functional. Assume $v\in N(L)\,,$ then we have $$ T(v)=L(v)=0=L(cv)=cL(v) \implies Tv-cLv=0 \implies (T-cL)(v)=0 \implies T=cL \,. $$ –  Mhenni Benghorbal Jan 6 '13 at 7:30
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1 Answer 1

1) Show that a linear functional is non-zero iff it is onto, and in this last case we always have that its kernel is a maximal proper subspace of $\,V\,$

2) So

$$\ker T=\ker L\Longrightarrow \,\exists v\in V\,\,\,s.t.\,\,\,V=\langle\,\ker T\,,\,v\,\rangle=\langle\,\ker L\,,\,v\,\rangle\,\,,\,\,v\notin\ker T=\ker L$$

Suppose $\,Tv=k\,\,,\,\,Lv=r$ and let $\,c\,$ be a solution to $\,xr=k\,$ ,thus

$$Tv=k=cr=cLv=L(cv)$$

and since obviously $\,T(u)=cL(u)=L(cu)=0\,\,\,,\,\,\forall\,u\in\ker T=\ker L\,$ , we are thus done.

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