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)?
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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.
David Moews
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1For those who may not know why dim$X/N=1$ : for any element $v \in X$ s.t. $f(v)\neq 0$, we have $\mathbb R=\mathbb R f(v)=f(\mathbb R v)$. Noting that for any other element $u \in X$, we will have $f(u)\in \mathbb R.$ So we will have that, $f(tv-u)=0$ for some constant $t\in \mathbb R$ s.t. $tf(v)=f(u).$ And so $u\in \langle t\rangle+N,$ i.e. dim$X/N=1.$ – Sam Wong Feb 02 '20 at 15:14
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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: $$ f(x) / g(x) = f(tv)/g(tv) = f(v)/g(v). $$
Emanuele Paolini
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1@SaaqibMahmuud: Clearly $f=g$ on $\ker(f)=\ker(g)$. Let $c=f(v)/g(v)$. If $x\in\langle v \rangle$ then $x=rv$, so $f(x)=rcg(v)=cg(x)$. Therefore $f=cg$ on $X$. – wj32 Feb 23 '13 at 10:13