$V=V^*$? the dual space 
Let V be a finite dimensional vector space over the field $K$. Let $U,W$ be the subspaces, and assume that $V$ is the direct sum $U\oplus W$. Show that $V^*$ is equal to the direct sum  $U^{\perp}\oplus W^{\perp}$. 

I am aware the dual space has the same dimension. "Theorem:The map of $v\to L_v$ of $V$ into $V^*$ is an isomorphism." 
I guess $U^{\perp}=W$ and $W^{\perp}=U$, so $U^{\perp}\oplus W^{\perp}=U^{\perp}+W^{\perp}$. 
But unless $K=1$ 
I am not seeing how $U^{\perp}\oplus W^{\perp}\in V^{*}$ because that would imply V*=V, right?
Questions:
1) Is $V=V^{*}$?
2) How can I complete the proof?
Thanks in advance!
 A: 1: You are not wrong with the statement $V=V^*$, but they are only congorphic, so i think it would be better to write as $V \cong V^*$. For the purpose of the second question, this statement is not useful and i suggest you ignore it.
2:
Here i will use the definition
$$
 U^\perp = \{f\in V^*: f(u)=0 \forall u\in U\}.
$$
Using this definition (which is very common!) we can see that $U^\perp\subset V^*$.
For the direct sum you have to show two things: First, that $U^\perp \cap W^\perp=\{0\}$.
Second, that $V^*=U^\perp+W^\perp$.
The first property is easier to show (try it).
For the second property, let $f\in V^*$ be given.
Note that $f$ is uniquely defined by its values on $U$ and $W$.
So we can define $f_U,f_W \in V^*$ as
$$
 f_U(u+w) = f(u) \quad,\quad f_W(u+w) = f(w).
$$
where $u \in U$ and $w\in W$.
(it might be necessary to show that these are well-defined).
Then $f_U \in W^\perp$ and $f_W\in U^\perp$ (why?).
Finally $f = f_U+f_W \in W^\perp +U^\perp$, thus completing the proof of question 2.
There might be easier proofs, depending on what tools you have available, but the above is rather elemtary and you can get maybe a better feeling for dual spaces and $U^\perp$.
