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?

  • $\begingroup$ Your reasoning is not clear. How did you suddenly get $c_1Au - c_2Av = 0$? $\endgroup$ – Marc van Leeuwen Mar 12 '16 at 9:20

You can go directly from $c_1u+c_2v=0$ to $A(c_1u+c_2v)=A0=0$ which gives $c_1u-c_2v=0$.

  • $\begingroup$ thanks a lot! that's help circumvent that needless assumption $\endgroup$ – matthew.j Mar 12 '16 at 9:12

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