orthogonals and linear dependence Well, I did solve this but I want to see if you guys have another way to solve this:
Given: $v_1,v_2$ are two vectors that are not zero and they are in R^n.
Prove that ${v_1}^{\bot} = {v_2}^{\bot}$ if and only if ${v_1,v_2}$ are linearly dependent.
 A: The following might be a bit overkill, but I think the arguments are worth understanding.
Lemma 1. Let $V$ be a finite-dimensional inner-product space and let $W$ be a subspace of $V$. Then $\dim W+\dim W^\bot=\dim V$.
Proof. Let $\{w_1,\dotsc,w_k\}$ be a basis of $W$ and let $T:V\to V$ be the linear map
$$
T(v)= \langle v,w_1\rangle\cdot w_1+\dotsb+\langle v,w_k\rangle\cdot w_k
$$
Then $\DeclareMathOperator{im}{im}\im T=W$ and $\ker T=W^\bot$ (can you prove this?). Hence rank-nullity implies the lemma. $\Box$
Lemma 2. Let $V$ be a finite-dimensional inner-product space and let $W$ be a subspace of $V$. Then $(W^\bot)^\bot=W$.
Proof. Showing that $W\subseteq (W^\bot)^\bot$ is an easy exercise (do this yourself!).
Moreover, Lemma 1 implies that $$\dim(W^\bot)^\bot=\dim V-\dim W^\bot=\dim V-(\dim V-\dim W)=\dim W$$ Hence $W$ has full dimension in $(W^\bot)^\bot$ so $(W^\bot)^\bot=W$. $\Box$
Your problem now easily follows from Lemma 2. Let $v_1,v_2\in\Bbb R^n$ and let $W_1=\DeclareMathOperator{Span}{Span}\Span\{v_1\}$ and $W_2=\Span\{v_2\}$. Note that $v_1$ and $v_2$ are linearly independent if and only if $W_1=W_2$.
Suppose that $W_1^\bot=W_2^\bot$. Then Lemma 2 implies $W_1=W_2$ (can you see why?). Hence $v_1$ and $v_2$ are linearly independent.
Can you prove the converse?
