Definition of direct sum of linear subspaces Suppose $V$ is vector space and let $U,W$ are subspaces of $V$. 
Define $U+W:=\{u+w: u\in U,w\in W\}$. We say that the sum of subspaces $U$ and $W$ is direct if any vector $x\in U+W$ has a unique representation $x=u+w$ with $u\in U$ and $w\in W$.
Question 1: if the vector $x$ can be written as the sum $x=u+w$ and we also know that $x=w+u$. Could I conclude that $u=w$? This uniqueness from definition confuses me a bit!
Question 2: I want to show  the following: if the sum of $U$ and $W$ is direct then $U\cap W=\{0\}$.
Let $z\in U\cap W$ and $z\neq 0$ then one can write $z=z+0=0+z$. Since the sum is direct hence there is a unique representation could I conclude that $z=0$?
I would be very grateful if anyone can in detail explain that confusing moment with uniqueness.
 A: Perhaps an example is helpful. Let $V = \mathbb{R}^2$, let $U$ be the $x$-axis, and let $W$ be the $y$-axis. Then consider the element $x := (2, 1) \in V$. The unique representation of $x$ as a sum of elements from $U$ and $W$ is
$$
(2, 1) = (2, 0) + (0, 1),
$$
where $u = (2, 0) \in U$ and $w = (0, 1) \in V$. Now, you could also write the sum in the order:
$$
(2, 1) = (0, 1) + (2, 0).
$$
But what you can't do is use any other element from $U$ than $(2, 0)$. Like, you couldn't use $(7, 0)$ instead; no matter how hard you look, you'll never find a $w \in W$ such that
$$
(2, 1) = (7, 0) + w.
$$
The element that comes from $U$, whether you write it first or second in the sum, must be $(2, 0)$. And similarly, the element from $V$ must be $(0, 1)$. That's what uniqueness of the representation as $u + w$ means.
A bit more formally, what uniqueness means is that if $u + w = u' + w'$, where $u, u' \in U$ and $w, w' \in W$, then it must be the case that $u = u'$ and $w = w'$.
For the second question, since $z$ and $0$ are in both $U$ and $W$, we can write $z$ as
$$
z + 0 \qquad\text{(with $u = z$ and $w = 0$)}
$$
or as
$$
0 + z \qquad\text{(with $u' = 0$ and $w' = z$)}.
$$
But uniqueness means that $u = u'$, which is to say $z = 0$ (and also $w = w'$, which again tells us that $z = 0$).
A: Answer 1: the addition of vectors is a commutative operation, so $u+w=w+u$, however $u\in U$ and $w\in W$, doesn't matter in what order you add them. If the representation is unique and $u+w\neq 0$ then it is not possible that $u=w$, otherwise we will had the representation $2u+0=u+w$ choosing $2u\in U$ and $0\in W$.
Answer 2: if $z\in U\cap W$ then $z\in U$ and $z\in W$, so the representation of $z$, as the addition of two elements of $U$ and $W$ respectively, is not unique except in the case that $z=0$. That is: if $z\neq 0$ then we can represent $z=z+0$ choosing $z\in U$ and $0\in W$, but we also can represent $z$ as the same addition but this time choosing $z\in W$ and $0\in U$.

Another way to say that $U+W$ is a direct sum is to say that for any chosen $z\in U+W$ there exists a unique $(u,w)\in U\times W$ such that $z=w+u$. This is also equivalent to say that the map
$$
S:U\times W\to U+W,\quad (u,w)\mapsto u+w
$$
is a bijection.
A: Defininition: The sum of two subspaces is said to be direct if the union of basis of the spaces are going to be basis of the sum. V $\cap$ W = {0} 
Proof: let u = v1 + w2  this implies that  u $\in$ V + W because  v1 $\in$ V and w1 $\in$ W. 
Lets try to rewrite this in a different manner u = v2 + w2;  By combining both ways 
of expressing this we obtain the following 
v1+w1 = v2+ w2 
0 = (v2+w2) - (v1+w1) 
0 = (v2-v1) + (w2-w1)
v2 - v1 = w2-w1 
Recall (v2 is not equal v1 and w2 is not equal w1) 
then v2-v1 $\in$ V, therefore w2-w1 $\in$ W, however, we claimed that V$\cap$W = {0} 
Therefore v2-v1 = 0 and w2 - w1 = 0
v2 = v1 and w2 = w1. 
This is a contradiction since we assumed that v2 is not equal v1 (same applies for w1 and w2). Therefore we can conclude that every vector u $\in$ V+W has an unique way of decomposing it as the sum of one vector in V and one vector in W. 
