how to show a direct sum of two subspaces Let $U,V$ two subspaces of $\mathbb{R}^n$, such that:
$U = \{x \in \mathbb R^n | x_1+x_2+...+x_n=0\}$ and $V=\{x\in V |x_1=x_2=...=x_n\}$
Show that the sum of the two subspaces is a direct sum.  
What I did:
We want to show that $U\cap V = \{0\}$. 
Indeed, let $w \in U\cap V$, Hence:  
$w_1 = w_2 = ...  = w_n$
$w_1 + w_2 + ... + w_n = 0.$
it's easy to see that $\forall i.w_i = 0$, Therefore it's the zero vector.  
Now, for my understanding this isn't suffice in order to show it's a direct sum.
We also need to prove that $U+V = \mathbb{R}^n$. 
Is it right? How to do that?
 A: You don't have to do that. The task is to prove that $U+V$ is a direct sum, that is what you showed. As far as you wrote, you don't have to prove that $U+V = \mathbb R$.
If you do want to prove that, look at the dimension of $U$ and $V$. Because $U\cap V = 0$, you know that dim$(U+V) = $ dim$(U)+$dim$(V)$.
If you do not yet have the theoretical background to make the dimension argument, then take ant $x\in\mathbb R^n$ and look at $\tilde x$ defined as $\tilde x_i =x_i - \bar x$ where $\bar x$ is the average value of values $x_1,x_2,\dots, x_n$.
A: The theorem:
Let $U, W$ are subspaces of V.
Then $U + W$ is a direct sum $\iff U\cap W = \{0\}$.
The proof:
Suppose "$U + W$ is a direct sum" is true.
Then $v \in U, w \in W$ such that $0 = v+w$. And since $U + W$ is a direct sum $v = w =0$ by the theorem "Condition for a direct sum". Thus, $U\cap W = \{0\}$ is true.
Suppose that "$U\cap W = \{0\}$" is true.
Then, in order to proof that $U + W$ is a direct sum, just need to show that $v \in U, w \in W$ such that $0 = v + w$ where $v = 0$ and $w = 0$.
The equation $0 = v + w$ $\iff$ $v = -w$, where $-w \in W$ is true by property "additive inverse". And hence $v \in U\cap W$ and $v = 0$ and by the equation above $w = 0$.
