How to prove that two non-zero linear functionals defined on the same vector space and having the same null-space are proportional? Let $f$ and $g$ be two non-zero linear functionals defined on a vector space $X$ such that the null-space of $f$ is equal to that of $g$. How to prove that $f$ and $g$ are proportional (i.e. one is a scalar multiple of the other)?
 A: Let $H$ be the null space and take a vector $v$ outside $H$. The point is that $H+\langle v\rangle$ is the whole vector space, this I assume you know (i.e. $H$ has codimension 1). Then $f(v)$ and $g(v)$ uniquely determine the functions $f$ and $v$ and all $x\in X$ can be written as $x=h+tv$ with $h\in H$ so:
$$
  f(x) / g(x) = f(tv)/g(tv) = f(v)/g(v). 
$$
A: If $N$ is the common nullspace, then $f$ and $g$ both push forward to well-defined linear functionals on $X/N$.  Since $\dim X/N=1$, any pair of nonzero linear functionals on it are proportional to each other. 
Addendum: Here is a proof that does not use the quotient space.  Since $f\ne 0$, there is some $v_0\in X$ such that $f(v_0)\ne 0$.  Let $v_1:=v_0/f(v_0)$; then, since $f$ is linear, $f(v_1)=1$.  Now, for any $w\in X$, again using the linearity of $f$,
$$
f(w - f(w)v_1) = f(w) - f(w) f(v_1) = 0.
$$
Then, by assumption, $g(w-f(w)v_1)$ is also zero, so, using the linearity of $g$,
$$
g(w)=g(w-f(w)v_1) + g(f(w)v_1) = g(f(w)v_1)=f(w) g(v_1).
$$
Since $w$ was arbitrary, this proves that $f$ and $g$ are proportional.
