Show that the statements "$v_1 + v_2 = 0 \Rightarrow v_1 = v_2 = 0$" and "if $v \in V_1 \cap V_2$, then $v = 0$" are equivalent. Let $V_1, V_2 \leq V$ and $\vec{v}_1 \in V_1$ and $\vec{v}_2 \in V_2$ Show that the following conditions are equivalent:


*

*$\vec{v}_1 + \vec{v}_2 = \vec{0} \Rightarrow \vec{v}_1 = \vec{v}_2 = \vec{0}$

*if $\vec{v} \in V_1 \cap V_2$, then $\vec{v} = 0$. (That is, $V_1 \cap V_2$ = $\{\vec{0}\}$).


My main issue is that I don't know how to structure an argument with the two expressions.  Basically the logical form of the argument can be stripped down to proving the logical statement:
$$(A \Rightarrow B) = (C \Rightarrow D)$$
For the forward implication can I simply assume that A, B, and C are true, and then prove D?
Can I do something like the following for the forward implication:
Assume $\vec{v}_1 + \vec{v}_2 = \vec{0}$ and that $\vec{v}_1 = \vec{v}_2 = \vec{0}$.
Assume $\vec{v}_1, \vec{v}_2 \in V_1 \cap V_2$.  Let $\vec{u} = \vec{v}_1 + \vec{v}_2$.  Therefore $\vec{u} \in  V_1 \cap V_2$ because $V_1 \cap V_2$ is a subspace of $V$.  Furthermore, because $\vec{v}_1 + \vec{v}_2 = \vec{0}$, $\vec{u} = \vec{0}$.  Since $\vec{v}_1, \vec{v}_2$, and $\vec{u}$ were arbitrary $V_1 \cap V_2$ = $\{\vec{0}\}$.
Thank you,
ICG
 A: You can assume C is true and try to prove D using the fact that $A⇒B$.
In your case, we can just assume $A⇒B$ and prove the equvalent statement "$V_1 \cap V_2$ = $\{\vec{0}\}$".
Suppose $V=V_1 \cap V_2 \neq \{\vec{0}\}$, then there exists $v\neq 0\in V$ and also $-v\neq 0\in V$ such that $v+(-v)=0$, but since that implies $v=-v=0$, a contradiction.
For the other direction, assume $C⇒D$ then suppose A and prove B.
Or in this case, we prove the contrapositive. 
Suppose $\vec{v}_1 , \vec{v}_2 \neq \vec{0}$ (it can't be only one of them is 0 for sure), then suppose $\vec{v}_1 + \vec{v}_2 = \vec{0}$, we see that $v_1=-v_2$, hence $v_2\in V_1$, so $v_2\in V_1 \cap V_2$
But by our assumption "$C⇒D$", $D$ must be true and $v_2=v_1=0$, a contradiction, hence $A$ is false when $B$ is false, and we've shown $A$ implies $B$.
A: Both implications can be proven directly. 
Suppose that if $v_1+v_2=0$, then $v_1=v_2=0$.  If $v\in V_1\cap V_2$, Then $v+(-v)=0$. So $v=-v=0$. 
Conversely, if $V_1\cap V_2=\{0\}$ and $v_1+v_2=0$, then $v_1=-v_2\in V_1\cap V_2$, so $v_1=0$, and then $v_2=0$. 
A: Hint for the forward implication:
$$\Big(\vec{v}_1 + \vec{v}_2 = \vec{0} \;\;\Rightarrow\;\; \vec{v}_1 = \vec{v}_2 = \vec{0}\Big) \quad\implies\quad \Big( \,\text{if}\;\;\vec{v} \in V_1 \cap V_2 \,\text{,}\;\text{ then}\;\vec{v} = 0\,\Big)$$
Suppose $\,\vec{v} \in V_1 \cap V_2\,$, then $\,\vec v \in V_1\,$ and $\,\vec v \in V_2\,$, and also $\,-\vec v \in V_2\,$ since $\,V_2\,$ is a subspace. We can then choose the two vectors $\,\vec v_1= \vec v \in V_1\,$ and $\vec v_2 = -\vec v \in V_2$ so that $\,\vec v_1+\vec v_2=\vec v - \vec v = \vec 0\,$.
But $\,\vec{v}_1 + \vec{v}_2 = \vec{0} \;\Rightarrow\; \vec{v}_1 = \vec{v}_2 = \vec{0}\,$ by the given premise, so it follows that $\,\vec v = \vec 0\,$.
