Intuition behind Theorem pertaining to Direct Sums We are all aware of the Theorem that says that Given any two subspaces $U_1$ and $U_2$ of the vector spaces $V$ there sum $U_1+U_2=U_1\oplus U_2$ if and only if $U_1\cap U_2=\{0\}$.
I am able to understand the proof behind this but what is the intuition behind this result a geometric explanation would quite appreciated.
 A: In a (general) direct sum $U_1 \oplus U_2$ of $U_1$ and $U_2$ (where $U_1$ and $U_2$ could be totally unrelated vector spaces in their own right) we take all formal sums $x+y$ with $x \in U_1$ and $y \in U_2$ and compute with them as if they were real sums, and for two such sums  $x+y = u +v$ iff $x=u$ and $y=v$; we could just see them as pairs $(x,y) \in U_1 \times U_2$ really, added coordinatewise, and scalar multiplication as well.
But in the special case where $U_1, U_2 \subseteq V$, some vector space in which they are both a subspace, we actually can do additions between $x\in U_1$ and $y\in U_2$, so it makes sense to look at $U_1 + U_2  =\{z \in V: \exists x \in U_1, \exists y \in U_2: z = x+y\}$, and wonder whether this subspace of $V$ is isomorphic to $U_1 \oplus U_2$. For this we need to be able to write every such $z \in U_1 + U_2$ to be written as $z=x+y, x \in U_1, y \in U_2$ in a unique way and then we have a well-defined linear map mapping $z$ to $(x,y)$. Its' clearly necessary to have $U_1 \cap U_2 = \{0\}$ because if $v \neq 0$ lies in both, we have $v = v +0 = 0 + v$ in two ways. The theorem you studied says that this in fact a sufficient condition: if the subspaces of $V$ intersect only in $\{0\}$ then we have unique decomposition for $U_1 + U_2$ and we have a concrete representation for $U_1 \oplus U_2$ inside $V$.
