Prove that $\operatorname{Ann}(U \cap W) = \operatorname{Ann}(U) + \operatorname{Ann}(W)$ if $\dim V < \infty$ Prove that $\operatorname{Ann}(U \cap W) = \operatorname{Ann}(U) + \operatorname{Ann}(W)$ if $\dim  V < \infty$ for $U$ and $W$ subspaces of $V$.
Annihilator of U = Ann($U$) = $\{ \phi \in V^* | \phi(u) = 0 \text{ for } u \in U\}$. 
Here $V^*$ is the dual space of $V$. 
 A: Here's a proof. Since linear forms that vanish on all of $U$ certainly vanish on all of $U\cap W$, one has $\mathrm{Ann}(U\cap W)\supset \mathrm{Ann}(U)$. Similarly $\mathrm{Ann}(U\cap W)\supset \mathrm{Ann}(W)$, so 
$$\mathrm{Ann}(U\cap W)\supset \mathrm{Ann}(U)+\mathrm{Ann}(W)$$
We dualize this to get an inclusion in the double dual space:
$$\mathrm{Ann}\Big(\mathrm{Ann}(U\cap W)\Big)\subset \mathrm{Ann}\Big(\mathrm{Ann}(U)+\mathrm{Ann}(W)\Big)$$

On the other hand, if a linear form vanishes on all of $U + W$ , then it surely vanishes on both $U$ and $W$, so
$$\mathrm{Ann}(U+ W)\subset \mathrm{Ann}(U)\cap\mathrm{Ann}(W)$$
Use this second fact with the subspaces $U'=\mathrm{Ann}(U)\subset V^*$ and $W'=\mathrm{Ann}(W)\subset V^*$. This gives you
$$\mathrm{Ann}\Big(\mathrm{Ann}(U)+ \mathrm{Ann}(W)\Big)\subset \mathrm{Ann}\Big(\mathrm{Ann}(U)\Big)\cap\mathrm{Ann}\Big(\mathrm{Ann}(W)\Big)$$

Combine the two facts and you get the inclusions
$$\mathrm{Ann}\Big(\mathrm{Ann}(U\cap W)\Big)\subset \mathrm{Ann}\Big(\mathrm{Ann}(U)+\mathrm{Ann}(W)\Big)\subset \mathrm{Ann}\Big(\mathrm{Ann}(U)\Big)\cap\mathrm{Ann}\Big(\mathrm{Ann}(W)\Big)$$

Under the canoncal isomorphism $\varphi:V\simeq V^{**},~x\mapsto\text{ (evaluation at }x)$ , this is where the finite dimension of $V$ comes into play, one has $\mathrm{Ann}(\mathrm{Ann}(U))=\varphi(U)$. Apply $\varphi^{-1}$ to the term on the left and the one on the right, and you get that they are in fact the same, so 
$$\mathrm{Ann}\Big(\mathrm{Ann}(U\cap W)\Big)= \mathrm{Ann}\Big(\mathrm{Ann}(U)+\mathrm{Ann}(W)\Big)=\mathrm{Ann}\Big(\mathrm{Ann}(U)\Big)\cap\mathrm{Ann}\Big(\mathrm{Ann}(W)\Big)$$

As a last step, dualize one more time:
$$\mathrm{Ann}\bigg( \mathrm{Ann}\Big(\mathrm{Ann}(U\cap W)\Big)\bigg) = \mathrm{Ann} \bigg(\mathrm{Ann}\Big(\mathrm{Ann}(U)+\mathrm{Ann}(W)\Big)\bigg)$$
using the canonical isomorphism between $V^*$ and $(V^*)^{**}$, we get
$$\mathrm{Ann}(U\cap W)=\mathrm{Ann}(U)+\mathrm{Ann}(W).$$
A: It's pretty clear that some functional $\Lambda$ annihilates $U+V$ iff it annihilates $U$ and $V$. So we have 
$$\operatorname{Ann}(U+W) = \operatorname{Ann}(U) \cap \operatorname{Ann}(W)$$
You didn't want this, but it's an easy and related result. 
If $\Lambda = u+w \in \operatorname{Ann}(U)+\operatorname{Ann}(W)$, with $u$ coming from $\operatorname{Ann}(U)$ and $w$ from $\operatorname{Ann}(W)$, then both $s$ and $t$, and therefore $\Lambda$ are in $\operatorname{Ann}(U\cap W)$. We therefore have the inclusion:
$$\operatorname{Ann}(U)+\operatorname{Ann}(W) \subseteq \operatorname{Ann}(U\cap W)$$
We now need to prove inclusion the the reverse direction, so we show that an element $\Lambda \in \operatorname{Ann}(U\cap W)$ is in $\operatorname{Ann}(U)+\operatorname{Ann}(W)$. We do this as follows. If $U = U'\oplus(U \cap W),W= (U \cap W)\oplus W'$ we have $$V = U' \oplus (U \cap W) \oplus W' \oplus Q$$
for some $Q$. We then define two functionals $\sigma,\tau \in V^*$.
$$\sigma|U' = \Lambda,~~~~~ \sigma|(U\cap W) = 0~~~~~ \sigma|W' = 0, ~~~~~ \sigma|Q = \Lambda\\
\tau|U' = 0,~~~~~ \tau|(U\cap W) = 0~~~~~ \tau|W' = \Lambda, ~~~~~ \tau|Q = 0$$ 
And it is easy to see that $\tau \in \operatorname{Ann}(U), \sigma \in \operatorname{Ann}(W)$, as well as that $\sigma + \tau = \Lambda$, which gives the other inclusion 
$$\operatorname{Ann}(U)+\operatorname{Ann}(W) \supseteq \operatorname{Ann}(U\cap W)$$
and proves the theorem. 
