Proof that the sum of two subspaces $A$ and $B$ when $A\cap B = \{0\}$ If 
$$A\cap B = \{0\}$$ and $A$ and $B$ are subspaces, then the sum of the subspaces is defined by
$$A+B = \{a+b |a\in A, b\in B\}$$
If their intersection is $\{0\}$ then its natural to think that $w$ in $A+B$ has na unique decomposition in terms of $A$ and $B$. 
My book proves this by this argument:
$$w = u+v = u_1 + v_1\implies u-u_1 = v_1-v$$But $v_1-v\in V$, then $u-u_1\in U\cap V = \{0\}$. Then $u-u_1 = 0\implies u_1 = u$
Can somebody explain to me why 
$$v_1-v\in V \implies u-u_1\in U\cap V = \{0\}$$?
After this, the book follows by proving the converse. He picks $w\in U\cap V$ and $u\in U$, $v\in V$. Then he writes the sum:
$$u+v = (u+w) + (v-w)$$
Then, by the unicity of the hypotesis, he says that
$u = u+w$ and $v = v-w$
Then $w = 0$, then $U\cap V = \{0\}$
Why the unicity implies $u = u+w$ and $v = v-w$?
 A: When your textbook writes $w=u+v=u_1+v_1$ it (implicitly or explicitly) assumes that $u\in U$, $v\in V$, $u_1\in U$ and $v_1$. (I don't really understand why you change from $A$ and $B$ to $U$ and $V$, but it's not so relevant).
Since $U$ is a subspace, $u-u_1\in U$; similarly, $v_1-v\in V$. As $u-u_1=v_1-v$, this vector is in the intersection, so it's $0$.
The converse amounts to proving that

if every vector $x\in U+V$ can be written in one and only one way as $x=u+v$, with $u\in U$ and $v\in V$, then $U\cap V=\{0\}$.

So, let $w\in U\cap V$; we wish to prove that $w=0$. Let $u\in U$ and $v\in V$; then set $x=u+v$. By simple algebra, we have also
$$
x=(u+w)+(v-w)
$$
and $u+w\in U$, $v-w\in V$. By assumption, there is only one way to write $x$ as sum of a vector in $U$ with a vector in $V$, but we have found two:
$$
x=u+v,\qquad x=(u+w)+(v-w)
$$
It follows that these ways are not different, so
$$
u=u+w\quad\text{and}v=v-w
$$
which means $w=0$.
A: Not only is $v - v_1 \in V$, but also $u - u_1 \in U$ (by the same argument). Therefore $u-u_1 = v-v_1 \in U \cap V = \{0\}$.
