Given the subset $W$ of the vector space $V$, call $A(W)$ = {$\phi\in V^* | \phi$ annihilates $W$} the annihilator of $W$. Show that $A(W)$ is a subspace of $V^*$. Show also that if $W' \subseteq W$ then $A(W) \subseteq A(W')$.
This is what I know:
For $A(W)$ to be a subspace of $V$ it must be closed under addition and scalar multiplication over $V$.
$V^*$ means the dual space of $V$ which means the set of all linear maps $\phi: V\rightarrow k$.
$\phi\in V^*$ annihilates $W$ if $\phi(w) = 0$ for every $w\in W$.
I'm understanding this intuitively, but I'm having trouble understanding how to write a proof.
To see that $A (W) $ is a subspace take arbitrary elements $\phi$ and $\psi \in A (W)$ and let $\lambda$ be an element of the underlying field.
It needs to be shown that for all $w \in W $, $(\phi + \lambda \psi )(w) =0$. (The space is non empty as it contains the zero map.)
But $(\phi + \lambda \psi )(w) = \phi(w) + \lambda \psi(w)$. Which is the sum of two zero elements because $\phi$ and $\psi$ annihilate $W$. Hence $(\phi + \lambda \psi )(w) =0$.
A proof of the containment:
Suppose $\phi \in A (W)$.
Then from the definition of annihilator, $\phi(w) = 0$ for every $w\in W$.
Now for all $w' \in W' \subseteq W$ , $\phi(w') = 0$, since $w' \in W $.
Hence $\phi \in A (W')$.