Prove that $\ker f=\ker g$ implies $f=cg$.

Question

Let $V$ be a vector space of dimension $n$; let $f: V \to \Bbb K$, $g :V \to \Bbb K$ be linear maps, so that $f,g \in V^*$. Prove that if $$\ker f=\ker g,$$ then there is some $c \in \Bbb{K}$ such that $$f= cg$$ on $V$.

I know that $f= a_1f_1^*+......a_nf_n^*$ where {${f_1^*,....,f_n^*}$} represents a basis of $V^*$.

The same thing for $g= b_1f_1^*+......ab_nf_n^*$ and I know that $f(v)=g(v)$ How can I continue?

Answer 1

Hint: If $f(v) = 0$ for all $v$, there is nothing to prove. Otherwise, pick $v_0$ such that $f(v_{0}) \neq 0$, and put $c = f(v_{0})/g(v_{0})$. (Why is $g(v_{0}) \neq 0$?)

Now look at the kernel of $f - cg$.