I have a question regarding linear functionals. In "Finite-dimensional vector spaces" from Paul Halmos there is an exercise which I would like to visualize better. Given two functionals $y$ and $z$ on the same vector space $V$ such that $y(x)=0$ whenever $z(x)=0$ for $x \in V$ prove that there exists a scalar $\alpha$ such that $y=\alpha z$.
The way I interpret this is that $ker(z) \subseteq ker(y)$. So geometrically speaking the hyperplane which is described by the kernel of $z$ is a subset of the hyperplane of which is described by the kernel $y$. Let's say I have $ker(z) = ker(y)$ then in $\mathbb{R}^3$ I would have two planes that describe the same linear functional, so it makes sense that $y=\alpha z$ holds. I wonder though what would be an example for $ker(z) \subset ker(y)$ geometrically. First I thought about a line and a plane in $\mathbb{R}^3$ which go through the origin and where the line lies in the plane. That would an example for $ker(z) \subset ker(y)$, but $z$ wouldn't be a linear functional in that case so my exampe is flawed.
My question is: Could someone give me a non-trivial visualized example for the exercise.
Thanks for any help!