# Help Understanding Fields

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.

• Yes, characteristic 2 means $x+x=0$ for all $x$ in the field. Commented May 24, 2013 at 10:00
• ... as $x+x = x(1+1) = x\cdot 0$. Commented May 24, 2013 at 10:01

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 $u+v=u-v=(1,1)$.
• I have always been uneasy about statements such as (the set) $\{ u, v \}$ is linearly independent. This is of course because if $u$ is a nonzero vector, then $\{ u \} = \{ u, u \}$, so is this set independent or not? In this case, even in characteristic two, one might thus claim that $\{ u+v, u-v \} = \{ u+ v \}$, with $u + v \ne 0$ if $u \ne v$, so the set $\{ u+ v \}$ is linearly independent. Commented May 24, 2013 at 10:11
• @AndreasCaranti Why is that a problem? Linear dependence is defined as $S$ is linear dependent if there exists a finite number of distinct vectors and scalars $a_1,\ldots,a_n$, not all zero, such that $a_1v_1 + \cdots + a_nv_n = 0$. Linear independence is then defined as not linear dependent. So linear combinations should be taken over distinct vectors of $S$ in which case if $u \neq 0$ $\{u,u\}$ is linearly independent.
• @Andreas, you raise a valid point. Perhaps I should have phrased my answer in these terms: in characteristic 2, there are examples where $u$ and $v$ are independent, but $u+v$ and $u-v$ are dependent. Commented May 24, 2013 at 10:50