[Math] If $\ker f\subset \ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $c\in F$.

Question

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?

Answer 1

There is no need for finite dimensionality and no need to use bases. Let $f(x) \neq 0$, $y$ be arbitrary and consider $y-\frac {f(y)} {f(x)} x$. By linearity we get $f(y-\frac {f(y)} {f(x)} x)=0$. By hypothesis this implies $g(y-\frac {f(y)} {f(x)} x)=0$. Hence $g(y)=cf(y)$ where $c=\frac {g(x)} {f(x)}$. Hypothesis implies that $c \neq 0$ so we can write $f=\frac 1 c g$.