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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.

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2 Answers 2

up vote 3 down vote accepted

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 $x+x'$?

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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 $f(e_i)\ne0$. Let us assume (just for notational simplicity) that $i=1$.

We have $0=f(e_1)g(e_1)$, so $g(e_1)=0$. For any other $j$, $$ 0=f(e_1)g(e_j)+f(e_j)g(e_1)=f(e_1)g(e_j); $$ as $f(e_1)\ne0$, we get that $g(e_j)=0$ for all $j$, i.e. $g=0$.

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