I must proof the following:
- Prop.: Let $E_1,E_2$ two vector subspace of $V$ then $$E_1+E_2\doteq E_1 \oplus E_2 \leftrightarrow E_1 \cap E_2=\{0_V\}$$
- Proof: I must show $$1)E_1+E_2\doteq E_1 \oplus E_2 \to E_1 \cap E_2=\{0_V\}$$ $$and$$ $$2)E_1 \cap E_2=\{0_V\} \to E_1+E_2\doteq E_1 \oplus E_2$$
$1)$ I proof by contradiction, therefore $E_1 \cap E_2 \neq \{0_V\}$ then $ E_1 \cap E_2 \nsubseteq \{0_v\} \vee \{0_V\}\nsubseteq E_1 \cap E_2$, if $E_1 \cap E_2 \nsubseteq \{0\}$ then $ \exists x \in E_1 \cap E_2 (x \notin \{0_V\})$ therefore $x \in E_1 \cap E_2 \wedge x \neq 0_V$, if $x \in E_1 \cap E_2$ then $-x \in E_2$ therefore $x+(-x) \in E_1 +E_2$, but $x+(-x)=0_v$ and by hypothesis $x+(-x)=0 \to x=-x=0_V$ and it is an absurd with $x\neq 0_V$; if $\{0_V\} \nsubseteq E_1 \cap E_2$ then $0_V \notin E_1 \vee 0_V \notin E_2$ and it is an absurd beacuse $E_1$ and $E_2$ are vector subspaces of $V$.
$2)$ I proof by contradiction, therefore $E_1 +E_2$ is not $E_1 \oplus E_2$, therefore $\exists x \in E_1, y \in E_2(x+y=0_V \wedge x\neq 0_V \vee y \neq 0_V)$, if $x \neq 0_V$ therefore $x=-y$ but $x \in E_1$ and $-y \in E_2$ then $x \in E_2$ therefore $0_V\neq x=-y \in E_1 \cap E_2$ but it is an absurd because $E_1 \cap E_2=\{0_V\}$ therefore if $x \in E_1 \cap E_2 \to 0_V \neq x=0_V$ (an absurd!!); if $y \neq 0_V$ I have the result $0_V \neq y = 0_V$ (an absurd!)
It is correct?