Show that vectors $u$ and $v$ are linearly independent iff $u+v$ and $u-v$ are linearly independent. 
Show that vectors $u$ and $v$ are linearly independent iff $u+v$ and $u-v$ are linearly independent.

We know that $v_1=u+v$ and $v_2=u-v$ are linearly independent which implies the only solution to $\alpha v_1+ \beta v_2 = 0 $ is $\alpha = \beta = 0$.
Thus we are required to prove that $u$ and $v$ are linearly independent, how do we use the first part to prove the second?
 A: This is not true if the scalar field has characteristic 2.  In the 2 dimensional vector space of row vectors over $\mathbb F_2$, for instance, $u=(1,0)$ and $v=(0,1)$ are linearly independent, for the usual reason. But $x=u+v$ and $y=u-v$ are not linearly independent: the non trivial linear realation $x+y=0$ holds.
One might as well ask, when is the matrix $\pmatrix{1&1\\1&-1}$ invertible?
Answer: when its determinant, which is $-2$, is invertible.  That happens in only in those  fields whose characteristic is not $2$.
A: The comments and several of the other answers direct you to consider the linear combination
$$ au + bv = \frac{a+b}{2} (u+v) + \frac{a-b}{2}(u-v), $$
though they don't give much intuition as to why or how this is the right thing to consider.  It is something of a rabbit drawn from a hat.  However, it does come up quite naturally.  The goal is to rewrite $u$ and $v$ in terms of the vectors $(u+v)$ and $(u-v)$.  Notice that if we add these together we get $2u$, and if we subtract them we get $2v$.  That is,
$$ 2u = (u+v) + (u-v)
\qquad\text{and}\qquad
2v = (u+v) - (u-v).
$$
But then if there are scalars $a$ and $b$ such that $au + bv = 0$, then we can use the above to obtain
\begin{align}
0 &= au + bv \\
&= \frac{a}{2} [(u+v) + (u-v)] + \frac{b}{2}[(u+v) - (u-v)] \\
&= \underbrace{\frac{a+b}{2}}_{=\alpha}(u+v) + \underbrace{\frac{a-b}{2}}_{=\beta} (u-v).\end{align}
By the linear independence of $(u+v)$ and $(u-v)$, this implies that
$$\alpha = \beta = 0 \implies a+b = a-b = 0, $$
which is only possible if $a = b = 0$.  Therefore $u$ and $v$ must be linearly independent.
A: The approach that occurs to me is to show that a non-trivial solution to $zu+bv=0$ implies a non-trivial solution to $av_1+bv_2=0$. Can you put this idea together with the hint provided in a comment?
A: Suppose $u$ and $v$ are linearly independent.
Let $\beta_1$, $\beta_2$ be scalars such that
$$\beta_1(u-v)+\beta_2(u+v)=0\text{.}$$
Write
the above as
$$(\beta_1+\beta_2)u+(\beta_1-\beta_2)v=0\text{.}$$
Since $u$ and $v$ are linearly independent, it follows that $\beta_1 + \beta_2 = 0$ and $\beta_1 - \beta_2 = 0$. The only way this can occur is if $\beta_1 = \beta_2 = 0$. Hence, it follows that $u-v$ and $u+v$ are linearly independent.
A: We show this by proving the contrapositive:

$u,v$ are linearly dependent $\iff$ $u+v,u-v$ are linearly dependent.

Let $u,v$ be linearly dependent. Then there exists constant $a$ and $b$, not both equal to $0$, such that $au+bv=0$, but this implies 
$$a\left(\frac{(u+v)+(u-v)}{2}\right)+b\left(\frac{(u+v)-(u-v)}{2}\right)=\frac{a+b}{2}(u+v)+\frac{a-b}2(u-v)=0$$
But $\frac{a+b}2$ and $\frac{a-b}2$ can't be both equal to zero as that would imply $a=b=0$, cintai to our assumption.
We can show the converse similarly.
