TITLE: How does one prove that if $f$ and $g$ are linear functionals on $V$ such that $h=fg$ is also a linear functional, then either $f=0$ or $g=0?$
QUESTION [6 upvotes]: I am self-studying Hoffman and Kunze's book Linear Algebra. This is the exercise 13 from page 106.

Let $\mathbb{F}$ be a subfield of the field of complex numbers and let $V$ be any vector space over $\mathbb{F}.$ Suppose that $f$ and $g$ are linear functionals on $V$ such that the function $h$ defined by $h(v)=f(v)g(v)$ is also a linear functional on $V$. Prove that either $f=0$ or $g=0.$

I was able to show that $h=0$. Therefore $V=\operatorname {Ker} (f)\cup \operatorname{Ker}(g)$. I am assuming that $f\neq 0$ and I would like to show that $\operatorname {Ker} (f)\subset \operatorname {Ker} (g)$, but I wasn't able to acomplish that.
I would appreciate your help.

REPLY [4 votes]: Another way to complete the story (using your starting point that $h=0$, and therefore $V=\mathrm{Ker}\,f\cup \mathrm{Ker}\,g$), is to prove that the union of two proper subspaces of $V$ cannot ever be all of $V$. This is true over any field. Here is a (rather big) hint for how to prove this:
Let $W,W'$ be proper subspaces of $V$. If one is contained in the other, their union is not all of $V$ because they are both assumed proper. If neither is contained in the other, then there exists $x$ in $W$ but not in $W'$ and $x'$ in $W'$ but not in $W$. What can you say about $x+x'$?