unique factorization of product of linear functionals

Question

Let $V$ be a vector space over the complex numbers. Let $f_{1\leq i \leq n}$ and $g_{1\leq i \leq n}$ be two sets of nonzero linear functionals on $V$.

Suppose we have

$$f_1(v) f_2 (v)\ldots f_n (v) = g_1(v) g_2 (v)\ldots g_n (v) $$

for arbitrary $v\in V $. It is conjectured that $f_i \propto g_{\sigma(i)}$ for some permutation $\sigma \in S_n$. How to prove it?

I can show that $f_i $ appears among the $g$'s. For example, consider $f_1$. Take $v\in Ker(f_1)$. Then $v\in \bigcup_i Ker(g_i) . $ As this holds for all $v\in Ker(f_1)$, we have

$$Ker(f_1) \subseteq \bigcup_i Ker(g_i) $$

Or

$$ Ker(f_1) = \bigcup_i \left(Ker(f_1) \bigcap Ker(g_i ) \right) . $$

By the well known fact that a vector space cannot be the union of a finite number of proper subspaces, we know $Ker(f_1) = Ker(g_i )$ for some $i$, which means $f_1 \propto g_i $.

The difficulty is to handle the case of repeated factors. Can anyone provide a way out? I thought of using the unique factorization property of multi-variable polynomials. But $V$ could be an infinite dimensional space.

Answer 1

Up to a permutation of the $g_i$ and a scalar factor, we can assume $f_1=g_1$. So divide both terms by $f_1$. You still have equality of $g_2\ldots g_n$ and $f_2 \ldots f_n$ outside a proper subspace $W$ of $V$ (the kernel of $f_1$). Now, take any $z \in W$, $y \notin W$. Then $t \longmapsto (g_2\ldots g_n)(z+ty)-(f_2\ldots f_n)(z+ty)$ is a complex polynomial that vanishes at every nonzero $t$, so it is zero and $(g_2\ldots g_n)(z)=(f_2 \ldots f_n)(s)$ for each $z \in W$ too. Thus, for any $v \in V$, $(g_2\ldots g_n)(v)=(f_2\ldots f_n)(v)$ and thus you can prove your statement by induction.