Suppose $U$ is a subspace of $V$ such that dim$V/U=1$. Prove that there exists $\varphi \in \mathcal{L} (V, \mathbb{F}) $ such that null$ \varphi = U$ Thus far my exploration has centered around the following definitions and theorems from my book, but I have not been able to link them together in a proof:


*

*Fundamental theorem of linear maps

*$\dim V/U = \dim V-\dim U$

*$V/U = \{\vec{v} + U : \vec{v} \in V\}$

*$\widetilde{T} \in \mathcal{L}(V/(\operatorname{null}T),W)$ defined by $\widetilde{T}(\vec{v}+\operatorname{null} T)=T\vec{v}$


I think this exercise is meant to foreshadow "Linear Functionals" which we have not learned about yet, and so I want to show this without using them. 
 A: Hint: As usual in linear algebra, everything becomes easier when we have chosen an appropriate basis. Choose one of $U$ (say $u_1, \dots, u_n$) and extend it to a basis of $V$ (say $u_1, \dots, u_n, v_1$). Try to define a linear map $\varphi: V \rightarrow \mathbb{F}$ that has the desired properties in terms of this basis.
If you can't figure it out yourself, it is a special case of this Prove $\exists T\in\mathfrak{L}(V,W)$ s.t. $\text{null}(T)=U$ iff $\text{dim}(U)\geq\text{dim}(V)-\text{dim}(W)$.
A: This answer is what I came up with in response to the hint given me by Severin Schraven: 
Take a basis in $U$:
$$B_{U} = \{u_{1},...,u_{n}\}$$
Extend $B_{U}$ to a basis in $V$:
$$B_{V} = \{u_{1},...,u_{n},v_{1}\}$$
Then
$$\forall v \in V, v= \lambda_{1}u_{1} +\,...\,+\lambda_{n}u_{n}+\lambda_{n+1}v_{1}$$
Then define $\varphi :V\rightarrow \mathbb{F}$:
$$
\forall v \in V, \,\varphi(v) = \lambda_{n+1}$$
So, $$ v \in U \rightarrow\lambda_{n+1}=0\\ v\notin U \rightarrow \lambda_{n+1} \not = 0$$
Thus null $\varphi$ = $U$.
And this mapping is linear, completing the proof.
