"Let $W_1, W_2$ be subspaces of a vector space $V$. Then $V=W_1 \oplus W_2$ if and only if $V=W_1 + W_2$ and $W_1 \cap W_2=\{0\}$."
I've looked over the proof of this statement and I'm interested about the "$\implies$" direction. The proof states the following : "Let $u$ be an arbitrary $u\in W_1 \cap W_2$. Now $u\in V$ and it has the unique representation $u=u_1 + u_2$ when $u_1 \in W_1$ and $u_2 \in W_2$. Now $$u_1=u_1+0_2$$ $$u_2=0_1 + u_2$$
where $0_1 \in W_1$ and $0_2 \in W_2$ being the zero vector. This is only possible when $u=0$ and thus $W_1 \cap W_2 = \{0\}$." First off, the zero vector is certainly just the same vector of $V$, so $0_2 = 0_1$ right? Second, if we rewrite these two equations we get $u_1 - u_1 = 0$ and $u_2 - u_2 = 0$. Now if these two representations are unique, it must mean that $0=u$ is the only solution. Otherwise any $u_1,u_2$ would be a solution. Have I understood this correctly?
I think that there is a notation's problem. I try to rewrite the proof in a similiar way: "Let $u$ be an arbitrary $u\in W_1\cap W_2$. Then you can write $u$ in two different way: \begin{gather} u=u+0\\ u=0+u \end{gather} In the first equality you have $u\in W_1$ and $0\in W_2$; in the second one you have $0\in W_1$ and $u\in W_2$. So you found two different represetation of $u\in W_1\oplus W_2$, but the representation is unique in $W_1\oplus W_2$ so the two decomposition have to be same compontentwise (i.e. $u=0$ and $0=u$). Then $u=0$."