# 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:

1. Fundamental theorem of linear maps
2. $\dim V/U = \dim V-\dim U$
3. $V/U = \{\vec{v} + U : \vec{v} \in V\}$
4. $\widetilde{T} \in \mathcal{L}(V/(\operatorname{null}T),W)$ 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.

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 $\text{dim}(U)\geq\text{dim}(V)-\text{dim}(W)$.

• Sir, I've been working on this suggestion. Wouldn't the extension of the basis of need n+1 additional vectors in order for the hypothesis criteria to hold? Or does that not matter for your suggestion to work? In any case, I am not creative, so I imagine I'll be peeking at that link pretty soon. Apr 3, 2017 at 23:33
• @rocksNwaves We have $1=dim(V/U)=dim(V)-dim(U)$. Hence, we need only 1 additional basis vector, when we extend the basis of $U$ to a basis of $V$. Apr 4, 2017 at 9:35
• @SeverinSchraven If we are given $\dim(V/U)=1$, I am not sure if this implies that $V$ is finite-dimensional. According to a result in the textbook Linear Algebra Done Right by Sheldon Axler, we require that $V$ be finite-dimensional and $U$ be a subspace of $V$ in order to use the formula $\dim V/U=\dim V-\dim U$. Jul 21, 2019 at 18:32
• @Cookie Of course you are right. But the tag says Linear Algebra, thus the implicite assumption is that our spaces are finite-dimensional. However, the same argument can be done in a way such that it works in any dimension. Jul 21, 2019 at 20:19

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.