# unique factorization of product of linear functionals

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.

• Where $\propto$ means? Apr 9, 2021 at 6:51
• @user26857 It means "proportional to". That is, $f_1 \propto g_i$ means that some $\lambda \in \Bbb{C} \setminus \{0\}$ exists such that $f_1 = \lambda g_i$. Apr 9, 2021 at 6:53

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.