Let $V$ be a vector space with $\dim V=n$ .
If $\ker f\subset \ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $c\in F$.
Now let $\mathcal B=\{v_1,v_2,\ldots ,v_n\}$ be a basis of $V$,
since $f,g$ are non-zero linear functionals then $\exists v_i\in \mathcal B $ such that $g(v_i)\neq 0\implies f(v_i)\neq 0$
Take $i=1$ without any loss of generality so take $g(v_1)\neq 0,f(v_1)\neq 0$.
Now take $c=\dfrac{f(v_1)}{g(v_1)}$
Then we need to show that $(f-cg)(v_i)=0\forall i$
Now $(f-cg)(v_1)=0$
How to show that $(f-cg)(v_i)=0\forall i\ge 2$
Can someone please help?
Note::Another Question Why do we need the dimension of the vector space to be finite?