Suppose a finite dimensional vector space V is the direct sum of two subspaces $U_1,U_2$. Then $(V-U_1 )\cup$ $\{$ $\textbf{0}$ $\}$ $=$ $U_2$. How do I show this?
I tried breaking it into the following cases:
(1)$U_1=U_2$ $=$ $\textbf{0}$
(2) Without loss of generality, $U_1$ is the zero vector, while $U_2$ contains atleast one non-zero vector.
(3) Both, $U_1$ and $U_2$ are non-zero. The difficulty im encountering is with the third case. If both are non-zero then $U_1$ $\neq U_2$. If I take an arbitrary element in $(V-U_1) \cup$ $\{$ 0 $\}$ then I can break it into cases:
$x \in V- U_1$ and $x$ is the zero vector. The latter case is trivial. If $x\in V-U_1$ then $x\in V$, which is the direct sum of $U_1$ and $U_2$ and so $x=u_1+u_2$ for some $u_1$ in $U_1$ and $u_2$ in $U_2$. How do I proceed with this case?
Cases 1 and 2 are easy. I'm only wondering about the third case. By $V-U_1$ I mean the set difference.