Understanding Direct Sum The definition of direct sum that I learned is: $U$ is the direct sum of $U_1,U_2$ if $\forall u\in U,\exists !(u_1,u_2)\in U_1\times U_2:u=u_1+u_2$.
I want to understand what $U=U_1\oplus U_2$ means.
From the definition I know that every element of $U$ gets decomposed in a unique way on terms of $U_1,U_2$. However, does that also mean that every $u_1\in U_1,u_2\in U_2$ can be associated to a unique $u\in U$?
Thanks for helping! :D
 A: I think a characterization of the definition might be useful: 
$$U_1 \oplus U_2 \iff U_1+U_2=U \text{ and } U_1 \cap U_2 = \{0\}$$
So let's see what each condition means:
$U_1+U_2=U$ means that given an element $u \in U$ you can always find elements $u_1 \in U_1$ and $u_2 \in U_2$ such that $u_1+u_2 = u$.
But the uniqueness of this way of writing $U$ comes from the condition that $U_1 \cap U_2=\{0\}$. If you have $U_1 +U_2=U$ and $U_1 \cap U_2 \neq \{0\}$, then there may be more than one choice for $u_1$ and $u_2$. Here's an example of a sum that is not a direct sum:
Take $U=\Bbb R^3$ and suppose $U_1=\langle e_1, e_2\rangle$ and $U_2=\langle e_2,e_3\rangle$. (Here $e_i$ is the vector with a $1$ in the $i$th coordinate and a $0$ in the other two.)
Then $U_1 \cap U_2 = \langle e_2\rangle$, and indeed elements of $U$ don't have a unique decomposition. For example, consider
$$(1,2,3) = e_1+2e_2+3e_3$$ 
This can be interpreted in many different ways; here are three of them:
$$(e_1)+(2e_2+3e_3) \text{ with $e_1 \in U_1$ and $2e_2+3e_3 \in U_2$} \\ 
(e_1+2e_2)+(3e_3) \text{ with $e_1+2e_2 \in U_1$ and $3e_3 \in U_2$} \\
(e_1+e_2)+(e_2+3e_3) \text{ with $e_1+e_2 \in U_1$ and $e_2+3e_3 \in U_2$}$$
Hence, in the context of your definition, in this example you could take $(u_1,u_2)$ to be any of $$(e_1, 2e_2+3e_3), (e_1+2e_2, 3e_3),(e_1+e_2, e_2+ 3e_3),$$ which means that $U$ is not the direct sum of $U_1$ and $U_2$.
A: We take vector spaces $U_1$ and $U_2$ and form $U_1 \times U_2$ by the natural component-wise addition and scalar multiplication:
$$ (a,b)+(x,y) = (a+x,b+y) \qquad \& \qquad c(a,b) = (ca,cb)$$
we do have:
$$ u=(u_1,u_2) = (u_1,0)+(0,u_2)  $$
but, I wouldn't write $u=u_1+u_2$ here... unless I was identifying $u_1$ with $(u_1,0)$ and $u_2$ with $(0,u_2)$.
In contrast, I would write $U = U_1 \oplus U_2$ if both $U_1$ and $U_2$ were subspaces of $U$ for which every $u \in U$ is uniquely decomposed as $u=u_1+u_2$ for $u_1 \in U_1$ and $u_2 \in U_2$.
Of course, there is an isomorphism between these and often it is abused, wait, no, used.
