Let $V,W$ be subspaces of $R^n$. Prove that $R^n = V ⊕ W$, if and only if for every $\vec{x} \in R^n$ there exists a unique $\vec{v} \in V$, $\vec{w} \in W$ such that $\vec{v} + \vec{w} = \vec{x}$.
I'm totally stumped - I'm an intro linear algebra student, and this is the first time we've been introduced to direct sums. I get the general idea behind a direct sum - for example, I can easily understand how $R ⊕ R = R^2$, but I'm having trouble figuring out where to start this proof.
Any advice/tips/answers/proofs would be greatly appreciated, thanks!
EDIT: the definition of direct sums we're given are: Let $V$ be a vector space and let $U$ and $W$ be subspaces of $V$ . We write $V = U ⊕ W$ and call $V$ the direct sum of $U$ and $W$ if $V = U + W$ and $U ∩ W = \{\vec{0}\}$.
Assume $R^n = V \oplus W$. By definition, we can write any $x\in R^n$ as $x= v+w$ with $v\in V, w \in W$. We show that this is unique:
Indeed, assume we have another decomposition
$$x= v+w= v'+w'$$
Then $v-v' = w'-w \in V \cap W = \{0\}$
and thus $v= v', w=w'$ showing uniqueness of the decomposition.
Conversely, assume that we can write any $x \in R^n$ as $x= v+w$ in a unique way. Trivially then $R^n = V+W$. Also, let $x \in V \cap W$. Then
$$x= x + 0 = 0 + x$$ are two decompositions and by uniqueness we must have $x=0$, showing $V \cap W = \{0\}$.