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.
Best Answer
For any $v,w\in V$, \begin{eqnarray} f(v)g(v)+f(w)g(w)&=&h(v)+h(w)=h(v+w)=f(v+w)g(v+w)\\\\ &=&f(v)g(v)+f(w)g(w)+f(v)g(w)+f(w)g(v). \end{eqnarray} We conclude that $$ f(v)g(w)+f(w)g(v)=0,\ \ v,w\in V. $$ In particular, $f(v)g(v)=0$ for all $v\in V$. Now let $e_1,\ldots,e_n$ be a basis of $V$. If $f\ne0$, then there exists $i$ with $f(e_i)\ne0$.
We have $0=f(e_i)g(e_i)$, so $g(e_i)=0$. For any other $j$, $$ 0=f(e_i)g(e_j)+f(e_j)g(e_i)=f(e_i)g(e_j); $$ as $f(e_i)\ne0$, we get that $g(e_j)=0$ for all $j$, i.e. $g=0$.