$\mathbf{Question:}$ Let $V$ be a vector space over a field $F$. Let $f$ and $g$ be two non zero functionals defined on $V$ such that $f(x)=0 \implies g(x)=0$ for all $x \in V$. Show that $g= \lambda f$ for some $\lambda \in F$.
$\mathbf{Attempt:}$ $f,g :V \to F$ are both linear mappings and $dim(F)=1$ over $F$. Thereby, $0\leq dim(Im(f,g)) \leq 1$. Now, $f$ and $g$ both being non zero functionals, $dim(Im(f,g))=1$.
Now, $dim (Ker f)+dim (Im f)=dim V \implies dim(Ker f)=dim V-1$. Same goes for $g$.
Let $\{x_1, x_2, ..., x_n\}$ be such a basis of $V$ that for only $x_1$ (say), $f(x_1) \neq 0$ and $g(x_1) \neq 0$ [ We can find such a basis by extending the basis of kernel of $f$ to a basis of $V$].
Let $f(x_1)=p$ and $g(x_1)=q$. Then, for $p^{-1}q= \lambda \in F$, $q=\lambda p$.
Now, $\lambda f(x_1)=g(x_1)$. Hence, $\forall x= c_1x_1+...+c_nx_n \in V$, $g(x)=c_1g(x_1)=\lambda c_1 f(x_1)=\lambda f(x)$.
Hence $g= \lambda f$.
Is this proof correct? Kindly verify.