Let be $V$ a vector space over $F$ and let $u,v \in V$. $u$ and $v$ have the follow property:
''whenever $f(u)=0, $ we have $f(v)=0, \forall f \in V ^{*}$''. Show that $u = \alpha v$, fo some $\alpha \in F$.
I am trying to solve this exercise by contradiction, but I can solve if $\dim V < \infty$.
Suppose that $u$ and $v$ are linearly independent, then call $u = v_1, v=v_2$ and complete a basis for $V$, that is, $B = \{v_1,v_2, v_3,\cdots,v_n \}$.
Now define $f_i(v_j)= \delta_{ij}$, then $B' = \{f_1,\cdots, f_n \}$ is a basis for $V^{*}$. But we have $f_1(u) =1 $ and $f_1(v) = 0$. It is a contradiction!
Therefore $\{u,v\}$ is linearly dependent.
But if $\dim V = \infty$, how can I solve that?
Assume that $u$ and $v$ are linearly independent,and extend it to a basis $B$ for $V$.This can be done in infinite dimensional spaces too,you would need Zorn's lemma,though.Now define a functional $f$ by $f(u)=0$,$f(v)=1$ and $f(w)=0$ whenever $w$ is an element of $B$,not equal to $u$ or $v$.In case you want the same thing with continuous linear functionals on normed spaces over $\mathbb{R}$ or $\mathbb{C}$,then like above,assuming linear independence of $u$ and $v$,define a functional(bounded,of course) on the subspace $S$ spanned by $u$ and $v$ by $f(u)=0$ and $f(v)=1$.Using Hahn-Banach theorem,extend this to a bounded functional on $V$,that will do it.