Linear Algebra - Orthogonal problem (from exam can I appeal this?) I have this problem :
Let $A=\{v_1,v_2....,v_k\}$ in $R^n$ while $2 \leq k$.
Prove if $A^\perp=(A-\{v_1\})^\perp$, then A is not linear independant.
Please take a look at my solution since this is my answer from the exam and I think to appeal on this question, I'd like to hear your thoughts if I could appeal this.
My solution
Lets assume that $A$ is linear independant.
We know that $$A^\perp=(A-\{v_1\})^\perp \rightarrow A^\perp = \{v_2,v_3,...,v_k\}^\perp$$
Using the definition we know that $A^\perp+A=R^n$ Therefore we can conclude:
$$A^\perp = \{v_2,v_3,...,v_k\}^\perp \rightarrow A^\perp=\{v_1,v_{k+1},v_{k+2}...v_n\}$$
It a known fact that for any $x \in A$ and $y \in A^\perp \rightarrow$ $x \bullet y = 0$
But we found that $v_1 \in A$ and also $v1 \in A^\perp$ which means that $v_1 \bullet v_1$ =0, and by theorem is can only be true if $v_1=0$ therefore $A$ is linear independant.
After reading my answer I found nothing wrong with it, I'll be glad if someone could tell me what's wrong.
Thanks!
 A: It looks to me as if you've assumed that all the $v_i$'s (the $v_1,\dots,v_k$ in the problem and also the $v_{k+1},v_{k+2},\dots$ that came from nowhere) are orthogonal to each other.  There's no justification for this assumption.
A: I assume that all $v_i$ are non-zero.
Here is one example for the assumption that $A$ is not linearly independent: $v_1 = v_2 + v_3$.
In this case, obviously, $v_1 \in [ \{ v_2, v_3 \} ]$, i.e., $v_1 \not \in A^\perp$. Therefore, your first step is wrong.
Further, I'm quite positive that the original statement is wrong. I think that $A^\perp = (A - \{v_1\})^\perp$ implies that $v_1$ is a linear combination of other vectors in $A$ (specifically, that $A$ is not linearly independent).
The reason (not exactly a proof, but not far from it either) is simple. Assume that $v_2,\dots,v_k$ are linearly independent (if they are not, we can just remove some of those that depend on some others and nothing will change). So,
$$\dim A^\perp = n - \dim [A] = n - k$$
if and only if $A$ is linearly independent. However,
$$\dim (A - \{v_1\})^\perp = n - \dim [\{v_2,\dots,v_k\}] = n - (k - 1) = n - k + 1.$$
