Linear Independence for different fields

Question

I have a statement for a space over $R^n$:

{x, y, z} is linearly ind. $\implies$ {x + y, x + z, y + z} is linearly independent

Quick proof:

a(x+y) + b(x+z) + c(y+z) = 0 $\implies$ (a+b)x + (a+c)y + (b+c)z = 0

a+b = 0, a+c = 0, b+c = 0 (assuming x,y,z are LI) $\implies$ a,b,c = 0

The question is, would that be true for an arbitrary field?

I suppose, that it is wrong for some $Z_p$ field because of mod rules which make addition weird.

But I don't have an idea how to show this in general way.

Answer 1

The reasoning up to \begin{cases} a+b=0\\ a+c=0\\ b+c=0 \end{cases} can be performed in any field.

From $b+c=0$ you get $c=-b$, so $a-b=0$ and $a=b$. Therefore $2a=0$ and you can conclude that $a=0$ only if the characteristic of the base field is not $2$.

On the other hand, if the characteristic is $2$, the assertion is false, because $$ 1(x+y)+1(x+z)+1(y+z)=0. $$

Answer 2

It is not true for $\mathbb Z_2$: Take $a=b=c=1$, then $a+b=a+c=b+c=0$. You need $1+1\ne 0$ to get linear independent vectors.

Answer 3

Nothing wrong with the (+1) already posted solutions. I present a way that is perhaps a bit more easily generalizable. We observe that the coordinates of the new set $\{x+y,x+z,y+z\}$ in terms of the known linearly dependent set $\{x,y,z\}$ form the matrix $$ A=\left(\begin{array}{rrr}1&1&0\\1&0&1\\0&1&1\end{array}\right). $$ We have $\det A=-2$, so by the determinant test the new set is linearly independent, if and only if $-2\neq0$. Thus characteristic two is the only exception.