Prove $U^0\cap W^0=(U+W)^0$ $V$ is a finite dimensional vector space, $U$ and $W$ are subspaces and $U^0,W^0, (U+W)^0$ are the relevant annihilators. I would please appreciate help proving:


$U^0\cap W^0=(U+W)^0$


I would think that anything that annihilates $(U+W)$  would annihilate $U$ and annihilate $W$. But I am even confused about that in general, given that $\dim V=\dim U+\dim U^0$, the larger the dimension of $U$, the smaller the dimension of $U^0$. 
But in that $\dim (U+W)\ge \dim U$, and similarly $\ge \dim W$, I am not even sure of that.
I would appreciate help in proving the highlighted equality as an inclusion of sets in both directions, and also, if possible, using dimensions.
Thanks
 A: Let $f \in U^0 \cap W^0$. Then $f \in U^0$ and $f \in W^0$. Therefore, $f$ is a linear functional defined on $V$ such that $f(u) = f(w) = 0$ for all $u \in U$ and for all $w \in W$. 
So, for all $x \in U+W = \left\{ \ u+w \colon \ u\in U, \ w \in W \ \right\}$, we can write $x$ as $x = u+w$ for some $u \in U$ and $w \in W$ and thus 
$$ f(x) = f(u+w) = f(u)+f(w) = 0+0=0,$$
showing that $f \in \left( U+W \right)^0$. 
Therefore, $$U^0 \cap W^0 \subset \left( U+W \right)^0.$$
Conversely, suppose that $g \in \left( U+W \right)^0$. Then $g$ is, by definition, a linear functional defined on $V$ such that $$g(x) = 0 \ \mbox{ for all } x \in U+W. \ \tag{1}$$
Now let $\theta$ denote the zero vector in $V$, and let $u \in U$ and $w \in W$ be arbitrary. 
We note that, as both $U$ and $W$ are subspaces of $V$, so $\theta \in U$ and $\theta \in W$, and therefore 
$$u = u+\theta \in U+W, \ \mbox{ and } \ w = \theta+w \in U+W,$$
and so by (1), we can conclude that $$g(u) = 0 = g(w),$$
and since $u \in U$ and $w\in W$ are arbitrary, we can conclude that 
$$g \in U^0 \cap W^0,$$
and therefore $$\left( U+W \right)^0 \subset U^0 \cap W^0.$$
Hope this helps!!
A: Suppose $\phi\in U^0\cap W^0$, and let $v = u + w\in U + W$ be arbitrary. Then $\phi(v) = \phi(u + w) = \phi(u) + \phi(w) = 0$, so $\phi\in (U + W)^0$.
Conversely, let $\phi\in(U + W)^0$. Then since $U\subseteq U + W$ and $W\subseteq U + W$, $\phi$ annihilates both $U$ and $W$, so $\phi\in U^0\cap W^0$.
