Linear Functional over $\mathbb{R}$ Vector Space If f and g are two non-zero linear functionals on $\mathbb{R}$ vector space $V$[finite dimensional]. Such that whenever $f(x)\geq 0$  we have  $g(x)\geq 0$. The claim is $Ker(f)=Ker(g)$ and $f=\alpha g$ for some $\alpha > 0.$
My Thinking:
Every linear functional over from $V$ to $\mathbb{R}$ is of the form $f(x)=\sum_{i=1}^{n} a_ix_i$ and $g(x)=\sum_{i=1}^{n} b_ix_i$ where $x=\sum_{i=1}^{n} x_ie_i$. 
To conclude $Ker(f)=Ker(g)$. 
$f(x)=0$ $\implies g(x) \geq 0.$ Now if $g(x)=0$ then nothing to show. If $g(x) > 0$ then considering
$g(x)-f(x)=g(x)$ we get $$\sum_{i=1}^{n}(b_i-a_i)x_i=\sum_{i=1}^{n}b_ix_i$$ We cannot conclude from here that $b_i-a_i=a_i$ for all $1\leq i\leq n$. Right? If no, then we can arrive at a contradiction that $f$ is $0$.
One Observation:
If $(g(x) \geq 0$ and $f(x)<0)$ then $f(-x)>0\implies g(-x)\geq 0 \implies g(x)=0. $
Hence if $g(x) >0$ then $f(x) \geq 0.$
 A: Let us assume to the contrary that there exists $x\in\ker(f)$ such that $x\not\in\ker(g)$. Now $f(x)=0$, so we must have $g(x)\ge0$, or more precisely $g(x)>0$ as we have assumed $g(x)\ne0$. But then $g(-x)=-g(x)<0$. Although $f(-x)=0$. This contradicts our condition. So $\ker(f)\subseteq\ker(g)$.
To prove the reverse inclusion, let us assume there exists $x\in\ker(g)$ such that $x\not\in\ker(f)$. Now take any $y$ such that $g(y)<0$. Such an $y$ must exist, because for any $z$ with $g(z)>0$, $g(-z)<0$, and we have assumed that $g$ is not the zero function. Now let $k>|f(y)|/|f(x)|$. Then if $f(x)>0$, then $f(y+kx)>0$, otherwise if $f(x)<0$, then $f(y-kx)>0$. But $g(y+kx)=g(y-kx)=g(y)<0$. Hence contradicts our condition. So $\ker(g)\subseteq\ker(f)$.
Take any $x\not\in\ker(g)$. Let $g(x)=r$. Let $y=r^{-1}x$. Then $g(y)=1$. Also let $\alpha=f(y)$. Now as $f$ and $g$ is linear non-zero functionals, so dimension of their image is $1$. By rank nullity theorem we have $V=\ker(g)+\mathrm{span}(y)$. So any $v\in V$ can be written as $v=z+ky$, where $z\in\ker(g)$ and $k\in\mathbb{R}$. Then $$f(v)=f(z+ky)=f(z)+kf(y)\stackrel{(1)}{=}k\alpha=\alpha(g(z)+kg(y))=\alpha g(v)$$
where $(1)$ follows from the fact that $z\in\ker(g)=\ker(f)$. Hence $f=\alpha g$.
