Show that there exists a constant a such that $f = ag$ where $f,g$ are linear applications from $W$ to $V*$ (vector spaces)

Question

So, I couldn't insert everything in my title, but I'm currently struggling with this linear algebra exercise, and I need your help:

Let $V, W$ be vector spaces over a field $K$ and let $V^*$ be the dual of V. Let $f,g:W \rightarrow V^*$ be injective linear applications. Show that if $Ker(f(x)) = Ker(g(x))$ then there exists a constant a in $K$ such that $f=ag$

Answer 1

Hint: Show that for $\gamma,\delta \in V^*$, if $\ker \gamma = \ker \delta$, then there exists an $a$ for which $\gamma = a \delta$.

Then, show that if there exist vectors $x_1,x_2 \in V$ for which $f(x_1) = a_1 g(x_1)$ and $f(x_2) = a_2 g(x_2)$ with $a_1 \neq a_2$, then there exists an $x \in V$ for which $g(x)$ fails to be a multiple of $f(x)$.