Just to clarify, a sum of subspaces $U_{1},...,U_{n}$ of the vector space $V$ is said to be direct if
$U_{j} \cap (U_{1}+...+U_{j}^*+...+U_{n})={0}$
In which the $U_{j}^*$ term is ommited from the sum. Also, the symbol used in this case is the $\oplus$.
I want to prove that $V=U_{1}\oplus ... \oplus U_{n} \iff \forall \ v \in V, \ \forall \ j = 1,...,n \ \exists! \ u_{j} \in U_{j}$ such that $v=u_{1}+...+u_{n}$.
What I thought to do first for the direct implication $(\Rightarrow)$ was to use contradiction, i.e. Assume $V=U_{1}\oplus ... \oplus U_{n} \ and \ \exists \ v \in V, \ \exists \ j \in \{1,...,n\}$ such that $\nexists \ or \ \exists^{>1}$ $u_{j} \in U_{j}$ such that $v \neq u_{1}+...+u_{n}$. I found this statement a little bit confusing. Does anyone could prove this by using contradiction or other method? If I incorrectly stated the negative please correct me.
For this direction $\implies$, the existence of a representation $$v=u_1+\cdots +u_n$$ is clear. The issue is to prove the uniqueness. Now suppose you have two representations of the same vector: $$v=u_1+\cdots +u_n\quad\hbox{and}\quad v=w_1+\cdots +w_n$$ Then $$(u_1-w_1)+\cdots +(u_n-w_n)=0$$ Assume that for some $1\leq i \leq n$, $(u_i-w_i)\neq 0$. Then this non-zero vector is equal to the vector $$-\sum_{j\neq i} (u_j-w_j)$$ which means that $U_i$ intersects non-trivialy the sum of the rest of the $U_j$'s, contradictory to the definition of the direct sum. Therefore, if $V$ is a direct sum of the $U_i$'s, then the representation is unique.