Two linear functionals are equal Let, $f$ and $g$ be two linear functionals such that ker$f$=ker $g$ and $f(a)$=$g(a)$. Then to prove $f(x)$=$g(x)$.
 A: 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 
$$f(x)=\alpha f(a)=\alpha g(a)=g(x)$$
A: 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 $x\in V$
