If two linear functionals are such that the kernel of one is contained in the kernel of the other, then they are proportional Let $V$ be a vector space over $K$ and let $f,g \in V^*$ and satisfy $\ker f \subseteq \ker g$. Show there exist such $c \in K$ so that $c \cdot f =g$ 
How to approach this problem ? 
 A: Here is an elementary proof, with your notations. If one among $f,g$ is $0$, then choose $\lambda = 0$. If $f\not = 0$ and $g\not = 0$, then you have in fact equality of the kernels as both are hyperplanes, note $H$ this hyperplane. Let $e\in H$, and $D$ the line generated by $e$, which is a supplementary subspace of $H$ in $V$ as $f$ and $g$ are non zero linear forms. Now look at $\varphi = f(e) g - g(e) f$. This $\varphi$ is equal to zero, as it is zero on $H$ and on $D$ (being zero on $e$.) Then $\lambda = \frac{g(e)}{f(e)}$ will do the job. (Note that $f(e)$ is non zero as $e\in H$.)
Remark 0. Caracterization of non zero linear forms on $V$. The following are equivalent, for a subspace $H$ of $V$ :


*

*there is a non zero linear form $l$ on $V$ such that $H$ is $l$'s kernel

*$H$ has a supplementary subspace in $V$ that is a line (meaning by that : that is of dimension $1$

*the quotient $V/H$ is on dimension $1$

*$H$ is maximal (for the inclusion) in the set of strict subspaces of $V$


More conceptual proof, as this has nothing to do with the dual $V^{*}$ of $V$, as it is quite general and it has to do with factorization. What you want to prove results from the following more general fact :
Let $E,F,G$ be three vector spaces over $k$, and $f : E\to F$ and $g : E \to G$ linear such that $\textrm{Ker}(f)$ contains $\textrm{Ker}(g)$. Then, there exist $h : G \to F$ linear such that $h\circ g = f$. (The inverse is trivial.)
Take $y\in\textrm{Im}(g)$ and write it $y = g(x)$ for a $x\in E$. Set $h_1 (y) = f(x)$. This is well defined, as if $x'$ is another element of $E$ such that $y=g(x')$, then $x-x'$ is in the kernel of $g$ which is included in the kernel of $f$ so that $f(x) = f(x')$. So we have defined a map $h_1 : \textrm{Im}(g)\to F$ such that $h_1 \circ g = f$ on $\textrm{Im}(g)$, and $h_1$ is linear by construction. (Same argument than the one used to check that it is well-defined.)
Now, take a subspace $H$ of $G$ such that $\textrm{Im}(g)\oplus H = G$, extend $h_1$ to $G = \textrm{Im}(g)\oplus H$ in a linear map equal to $h_1$ on $\textrm{Im}(g)$ and $0$ on $H$. We have obviously $h\circ g = f$. $\square$
Now, in your case, $h$ will be a linear map from $k$ to $k$, so that it is an homothety of ratio $\lambda$ for some $\lambda \in k$, and this $\lambda$ is the one you are looking for.
Remark 1. In this above construction, I switched notation, as kernel of my $g$ is included in the kernel of my $f$, whereas in your notation, it's the opposite. But that's not an issue. Note that the existence of $H$ is unsured by Zorn's lemma, which is not used in the elementary proof my answer is beginning with.
Remark 2. You have a version of this with images, namely : if $\varphi : A \to B$ and $\psi : C\to B$ are linear such that image of $\varphi$ is included in the image $\psi$, then there exists a linear $\delta : A \to C$ such that $\psi \circ \delta = \varphi$. Here also in general, you need Zorn's lemma.
