Orthonormals, v1,v2. Given: $2$ vectors $v1,v2 \ne 0$ and $v1 \ne v2$ and $v1,v2 \in R^n$
Prove: $\{v1\}^\bot = \{v2\}^\bot$ if and only if ${v_1,v_2}$ are linearly dependent.
Well, I do have a solution for this but I didn't understand it well and I'll be really glad if you guys can explain how to prove it. (both ways because of "if and only if")
If needed, I can write the given solution, although I didn't understand it so I prefer to see if you can help me sort out what I didn't understand by giving your solution with some explanation if possible.
 A: If $v_1$ and $v_2$ are linearly dependent and are both nonzero, then it is easy to see that $v_1=\lambda v_2$ with $\lambda\ne0$. Thus, all vectors orthogonal to $v_2$ will be orthogonal to $v_1$ and vice versa. If $v_1$ and $v_2$ are linearly independent, then the vector $\xi=v_1-\frac{(v_1,v_2)}{||v_2||^2}v_2$ is nonzero and orthogonal to $v_2$ but not to $v_1$.
Indeed, $(\xi,v_2)=0$ by straightforward computation. $(\xi,v_1)=||v_1||^2-\frac{(v_1,v_2)^2}{||v_2||^2}$. If $(\xi,v_1)=0$, then $||v_1||^2||v_2||^2=(v_1,v_2)^2$ and hence
$$
||v_1-\lambda v_2||^2=||v_1^2||+\lambda^2||v_2||^2-2\lambda(v_1,v_2)
$$
is zero for $\lambda=(v_1,v_2)/||v_2||^2$; i.e., $v_1=\lambda v_2$, a contradiction. QED
A: 
If $\{v_2,v_1\}$ are linearly dependent then: $v_2 = \Lambda v_1$
$ v \in \{v_1\}^\bot \iff v * v_1 =0 \iff v * \Lambda v_1 = 0 \iff v * v_2 = 0 \iff v \in \{v_2\}^\bot$
Which means $ \{v_1\}^\bot = \{v_2\}^\bot $.

Now, the other way:

If $ \{v_1\}^\bot = \{v_2\}^\bot $ then,
We know that every sub-set of A holds $(A^\bot)^\bot = Sp(A)$
So we can say that $ \{v_1\} = \{v_2\} $ Thus,
We can say that $v_1 = Sp(v_1)$ and $v_2 = Sp(v_2)$ $\rightarrow$ $Sp(v_1) = Sp(v_2)$ Thus,
$\{v1,v2\}$ is linearly dependent.

