I came across this problem in a Linear Algebra text today:
Let $u$ and $v$ be distinct vectors in a vector space $V$ over a field $F$. Prove that $\{u,v\}$ is linearly independent if and only if $\{u+v,u-v\}$ is linearly independent.
Working on ($\Rightarrow$), I must show that
$$c_1(u+v) + c_2(u-v) = 0 \implies c_1 = c_2 = 0 \text{, where } c_1,c_2 \in F.$$
Using linear independence of $\{u,v\}$ I arrive at the equation $c_1 + c_1 = 0$. However, if $F = \mathbb{Z}_2$ then $1 + 1 = 0$. Also, if $c_1$ and $c_2$ were "integers" then in a field of characteristic $2$ I have
$$c_1 + c_1 = \underbrace{(1 + \cdots + 1)}_{c_1 \text{ times}} + \underbrace{(1 + \cdots + 1)}_{c_1 \text{ times}} = (1 + 1) + \cdots + (1 + 1) = 0 + \cdots + 0 = 0$$
However, I'm assuming that they're integers and I'm not sure (as for example $\mathbb{R}$ is a field which isn't only integers) that $x + x = 0$ in any field with characteristic $2$. Can someone clarify?
Also, I see often the restriction that $F$ be of characteristic not equal to $2$. Why is that?
My knowledge of fields is very limited so any references would also be appreciated.
Thank-you.
Answer
The statement you are trying to prove is, indeed, not true if the field has characteristic $2$. If $u=(1,0)$ and $v=(0,1)$ then $u$ and $v$ are linearly independent but $u+v=u-v=(1,1)$.
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