If $W$ is a subspace of the finite-dimensional vector space $V$, show that $W^* \cong V^*/A(W)$. Conclude that $\dim A(W) = \dim V - \dim W$.
Hint: Define a map $\psi: V^* \rightarrow W^*$ by $\psi(\phi) = \phi\vert_{W}$ and show that $\psi$ is surjective with ker$\psi$ = $A(W)$. Now apply the theorem $V / \ker \psi \cong \psi(V)$.
What I know:
For $W$ to be a subspace of $V$ it must be closed under addition and scalar multiplication over $V$.
$V^* / A(W)$ is the quotient space of $V^*$ and $A(W)$ where $V^*/A(W)$ = {$\bar{v};v\in V$} where $\bar{v} = v + A(W) = $ {$v+a; a\in A(W)$}.
$V^*$ means the dual space of $V$ which means that it is the set of all linear maps $\phi: V\rightarrow k$
If $f: V → k$ is a linear map, then $f\vert_{W}$ means the linear map $W → k$ given by the same formula as $f$. (in other words, just shrink the domain of $f$)
$A(W)$ means the set of $\phi \in V^*$ such that $\phi(w) = 0$ for every $w \in W$. (The annihilator of $W$)
I'm just having trouble putting all of this information into a cohesive proof.