1
$\begingroup$

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 $u,v,w$?

$\endgroup$
  • 1
    $\begingroup$ You should be able to come up with a counter-example in $\mathbb R^2$. $\endgroup$ – John Douma Oct 7 '15 at 2:42
3
$\begingroup$

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 $u + v + w = (0,0,0)$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.