# Linear Independence - Prove or Give Counterexample

Problem:

Let $u, v, w \in V$ be three nonzero vectors in a vector space over $C$ such that any list of two of the three vectors is linearly independent. With these conditions, prove or give a counterexample that $u, v, w$ is linearly independent.

Work:

My guess is that $u,v,w$ cannot be assumed to be linearly independent with the given conditions that each list of two vectors is linearly independent.

For $u,v,w$ to be linearly independent, there must exist some $a,b,c$ such that

$au+bv+cw=0$

which means that $a=b=c=0$.

Can I assume that since each set of 2 vectors are linearly independant, that these $a,b,c$ are automatically $0$ for $u,v,w$?

• You should be able to come up with a counter-example in $\mathbb R^2$. – John Douma Oct 7 '15 at 2:42

A slightly wilder example for 3-dimensional case: If $u:= (1,0,1)$, if $v:= (-1,0,0)$, and if $w := (0,0,-1)$, then $u,v,w$ are pairwisely linearly independent; but $u + v + w = (0,0,0)$.
Consider $V = \mathbb{C}^2$ and $u = (1,0), v = (0,1), w = (1,1)$. Then $u,v,w$ are pairwise linearly independent and but the whole set it not.