## A fact about linearly independent triples

Let $F$ be a field and let $V$ be an $F$-vector space. Suppose $T = \{x,y,z\} \subseteq V$ is linearly independent. Show that $S = \{x+y,x-z,y-z\}$ is also linearly independent.

Suppose $a(x+y) + b(x-z) + c(y-z) = 0$; then $(a+b)x + (a+c)y - (b+c)z = 0$. Since $T$ is linearly independent, $a+b = a+c = b+c = 0$. In particular, $b = c$ and $b = -c$; thus $b = c = 0$, and so $a = 0$ as well. Thus $S$ is linearly independent.