Let $u$ and $v$ be nonzero vectors in $R^n$ such that $Au = u$ and $Av = -v$. Prove $u$ and $v$ are linearly independent. This is my working.

Since $Au = u$ and $Av = -v$, $A$ cannot be a zero vector.

$c_1u + c_2v = 0$

$c_1Au - c_2Av = 0$

$A(c_1 - c_2v) = 0$ Since A is non zero, means $c_1u - c_2v = 0$

$c_1u + c_2v = c_1u - c_2v$

$c_2v = -c_2v$ Since v is non zero, thus $c_2 = 0$

From $c_1u - c_2v = 0$, $c_1u = c_2v$

$c_1u = 0$ Since u is non zero, thus $c_1 = 0$

So since $c_1 = 0$ and $c_2 = 0$, then u and v are linearly independent.


I have tried but is this correct? Now I question myself whether A is nonzero is even correct? Because if that is wrong, everything falls apart. Your advice is appreciated!

A continuation to this is that now $Aw = 0$ is included, and asks to prove that u, v, w are linearly independent. Do I take the same approach?
 A: You can go directly from $c_1u+c_2v=0$ to $A(c_1u+c_2v)=A0=0$ which gives $c_1u-c_2v=0$. 
A: Let $A: \Bbb R^n \to \Bbb R^n$ be given with three non-zero vectors $u$, $v$, and $w$ satisfying
$\quad Au = u$
$\quad Av = -v$
$\quad Aw = 0$
Suppose
$\tag 1 c_1u + c_2v + c_3w = 0$
Apply $A^2$ to both sides of $\text{(1)}$ and you'll be left with 
$\tag 2 c_1u + c_2v = 0$
But then substituting this result back into $\text{(1)}$ allows us to write 
$\quad 0 + c_3w = 0$
and we conclude that $c_3$ must be equal to zero.
If we apply $A$ to both sides of $\text{(2)}$ we can write as true
$\tag 3 c_1u - c_2v = 0$
Adding the lhs of $\text{(2)}$ and $\text{(3)}$ we conclude that
$\quad 2 c_1 u = 0$
and so $c_1 = 0$. Plugging this value $\text{(2)}$ we deduce that $c_2 v = 0$. But then $c_2$ must also be zero.
So whenever $\text{(1)}$ is true, $c_1 = 0$, $c_2 =0$, and $c_3=0$.
By definition, the vectors $u$, $v$, and $w$ are linearly independent.
A: Here's a much simpler approach.
Assume that $u$ and $v$ are linearly dependent.  
Then, there exist scalars $a_1$ and $a_2$, not both $0$ such that $a_1u + a_2v = 0$.
WLOG, we may assume $a_2 \neq 0$ to get $v = \lambda u$ for some scalar $\lambda$. In fact, the assumption that $v \neq 0$ also enforces $\lambda \neq 0$.
Now, we have $Av = -v.$
$\implies A(\lambda u) = -\lambda u$
$\implies \lambda u = -\lambda u \qquad (\because Au = u)$
$\implies 2\lambda u = 0$
$\implies 2\lambda = 0 \qquad (u \neq 0)$
$\implies \lambda = 0$, a contradiction.  
