Let, $f$ and $g$ be two linear functionals such that ker$f$=ker $g$ and $f(a)$=$g(a)$. Then to prove $f(x)$=$g(x)$.
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2I think you miscopied the question. The point $a$ should be outside the kernel of $f$. – Andreas Blass Aug 06 '14 at 17:16
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1This may help. – David Mitra Aug 06 '14 at 17:17
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See also: http://math.stackexchange.com/questions/890101/if-two-functionals-have-the-same-kernel-then-one-is-a-multiple-of-the-other – Martin Sleziak Aug 07 '14 at 15:58
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We need to add the assumption: $a\not\in \ker f=\ker g$. Notice that $\ker f$ is an hyper-plan so it's easy to see that $$E=\ker f\oplus \operatorname{span}(a)$$ so let $x=x_1+\alpha a\in E$ where $x_1\in \ker f=\ker g$ and we have $$f(x)=\alpha f(a)=\alpha g(a)=g(x)$$
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Suppose $T$ is the zero map then we have nothing to prove, we can consider $c=0$..
Suppose $T$ is not zero map then $L$ can not be zero map as $L(v)=0$ for all $v\in V$ then we would have $T(v)=0$ for all $v\in V$ concluding that $T$ is a zero map.
So, we have assumed that $T$ is not a zero map and $L$ is also not a zero map.
As $T:V\rightarrow \mathbb{R}$ linear, we have $\dim(V)=\dim(Ker(T))+\dim(Im(T))$ as $T$ is non zero, we have $\dim(Im(T))=1$ and so $\dim(Ker(T))=n-1$ assuming that dimension of $V$ is $n$..
Same rule applies for $L$ and we have $\dim(Ker(L))=n-1$..
As $Ker(L)\subset Ker(T)$ and dimensions are same, we have $Ker(T)=Ker(L)$..
I do not know how to get that $c$ such that $T=cL$...
As $T$ is non zero, there exists $a\in V$ such that $T(a)=1$..
Clearly $L(a)\neq 0$ as $T(a)\neq 0$...
Let $x\in V$ then $x=ka+x'$ for $x'\in Ker(T)$
So, $L(x)=kL(a)$ and $T(x)=k$ so, $L(x)=T(x)L(a)$
Thus, $L(x)=L(a)T(x)$ for all $x\in V$