Suppose $\varphi_1$ and $\varphi_2$ are linear maps from $V$ to $\mathbf F$ that have the same null space. Show that there exists a constant $c \in \mathbf F$ such that $\varphi_1 = c \varphi_2$.

Notice $\dim \text{range }\varphi_1 \le \dim \mathbf F = 1$ by 2.38 as it's a subspace. Assume $\dim \text{range }\varphi_1 = 1$ as zero dimension is the trivial $\varphi_1 = \varphi_2 = 0$ case.

Let $v \in V \setminus \text{null }\varphi_1$ this spans $\text{range }\varphi_1$ since $\dim \text{range }\varphi_1 = 1$. Thus after picking $c$ so $\varphi_1 v = c\varphi_2 v$ we automatically have $\varphi_1 = c\varphi_2$ since every $w \in V$ has $\varphi_1 w = \varphi_1 \lambda v = c \varphi_2 \lambda v$.

TODO: Make the last part more rigorous, currently I'm relying on my intuition from 3b/16, which doesn't directly apply here since $V$ could be infinite dimensional.