$U\times W$ isomorphic to $V$? 
Let $V$ be a direct sum of two subspaces $U,W$. Prove that $U\times W$ is isomorphic to $V$?

My problem lies in the fact the book has not provided any formal definition or theorem about $U\times W$, I mean about the product of Vector Spaces, on this case subspaces. As far as I can see the $ker(T)=0$, because $0\times 0$ is preserved  but I cannot talk about dimensions.
Questions:
Could anyone provide me a definition of $U\times W$?
How can I prove $U\times W$ is isomorphic to V?
Thanks in advance!
 A: The product $U \times W$ for any sets $U$ and $V$ is defined as $U \times W = \{(u,w) \ | \ u\in U, w \in W\}$. If $U$ and $V$ are vector spaces over the same field, their Cartesian product is always a vector space by defining
$$
(I).\ \  (u_1,w_1) + (u_2,w_2) = (u_1+u_2,w_1+w_2),
\ \ (II). \ \ a(u,w) = (au,aw),
$$
where $a$ is a scalar. If $V = U \oplus W$ then (as you point out) $U \cap W = 0$. You can easily show $\dim (U\times W) = \dim U + \dim W$. It is also true that by identifying $U$ with $\{(u,0) \ | \ u \in U\}$ and $W$ with $\{(0,w) \ | \ w \in W\}$ we may regard $U$ and $W$ as subspaces of $U \times W.$ 
Does this help?
A: Hint:
$U\times W$  is by definition the set of pairs $(u,w)$, $u\in U,\;w\in W$, endowed with operations componentwise
$$(u,w)+(u',w')=(u+u', w+w')\quad \lambda (u,w)= (\lambda u,\lambda w),\qquad\text{
(external direct sum)}$$
You can define a linear map
\begin{align}U\times W&\longrightarrow V\\
(u,w)&\longmapsto u-w\end{align}
and show it is an isomorphism.
A: The space $U\times W$ consists of all ordered pairs $(u,w)$, where $u\in U$ and $w\in W$. The operations are given by coordinate wise addition and scalar multiplication. 
Now define $T:U\times W\to V$ by $T(u,w)=u+w$. We have $$T(\lambda u_1+u_2,\lambda w_1+w_2)=\lambda(u_1+w_1)+u_2+w_2=\lambda T(u_1,w_1)+T(u_2,w_2).$$ So $T$ is linear. 
Since $V=U+W$, given $v\in V$ there exist $U\in U$ and $w\in W$ with $v=u+w=T(u,w)$. So $T$ is onto. 
If $T(u,w)=0$, then $u+w=0$. Thus $u=-w$.  Because $U\cap W=\{0\}$, $u=w=0$. So $T$ is one-to-one. 
