Complementary Subspaces . How do I go about proving this? 
Two subspaces $V$ and $W$ of $\mathbb{R}^n$ are called complements
if any vector $x$ in $\mathbb{R}^n$ can be expressed uniquely as
$x = v+ w$ , where $v\in V$ and $w\in W$. Show that V
and W are complements if (and only if) $V ∩ W = {0}$
and $dim(V) + dim(W) = n.$
 A: I'll outline the proof. Fill in the blanks.
One way:
Suppose that $V$ and $W$ are complements of $X$. 
Define the map $T : X \to V \times W$ (the cartesian product), given by $T (x) = (v,w)$, where $v,w$ are the unique vectors in $V$ and $W$ respectively so that $v + w = x$. This is well defined by uniqueness.
Prove that $T$ is injective (hint : $0 = 0 + 0$ uniquely)
Prove that $T$ is surjective (hint: given $(v,w) \in V \times W$, let $x = v+w$).
Use the rank nullity theorem on $T$, noting that $\dim V + \dim W = \dim (V \times W)$.
Let $a \in V \times W$. Then, $-a \in V \times W$, but then $ 0 = 0+0 = (a) + (-a)$, so $0$ is not uniquely written as a sum of terms from $V$ and $W$, unless $a=-a \implies a=0$,so that $V \cap W = \{0\}$.
The other way:
Now, consider $S : V \times W \to X$, given by $S((v,w)) = v+w$.
Show that $S$ is the inverse of $T$. Hence, $S$ is a bijection, so every element $x$ in $X$ is uniquely the image of some $(v,w) \in V \times W$,  so that $v+w = x$, hence $V$ and $W$ are complements.
Get back if there are any doubts. 
