If $\phi \in W^*$, show that we can find a $\widetilde{\phi} \in V^*$ such that $\widetilde{\phi}\Bigr|_{W} = \phi$

Question

Let W be a subspace of the vector space $V$. If $\phi \in W^*$, show that we can find a $\widetilde{\phi} \in V^*$ such that $\widetilde{\phi}\Bigr|_{W} = \phi$

So, this is what I know:

For $W$ to be a subspace of $V$ it is closed under addition and scalar multiplication on $V$.

$V^*$ means the dual space of $V$ which means the set of all linear maps $\phi: V \rightarrow k$

$\phi \in W^*$ means the linear map $\phi: W \rightarrow k$ with $k$ being the field that $V$ is over

And I'm given a hint: If $f: V \rightarrow k$ is a linear map, then $f\Bigr|_{W}$ means the linear map $W \rightarrow k$ given by the same formula as $f$ (in other words, just shrink the domain of $f$).

I can see intuitively that this makes sense but I am unsure of how to approach proving it.

Answer 1

I will assume $V $ is finite-dimensional. Let $w_1,...w_m$ be a basis of $W$, extend to a basis of $V$ by adding $u_1,...u_n $ and define $U = span (u_1,...u_n) $.

From bases of $W $ and $U$, obtain dual bases $w_1^*,...,w_m^*$ and $u_1^*,...,u_n^*$ of $W^*$ and $U^*$ respectively. Note $w_1^*,...,w_m^*,u_1^*,...,u_n^*$ is a basis of $V^*$

Suppose $\phi = \lambda_1 w_1^* +...+ \lambda_m w_m^* $ and extend the functional to $V^*$ by defining $\widetilde{\phi} = \lambda_1 w_1^* +...+ \lambda_m w_m^* +0 u_1^* +...+0 u_n^*$.