# Proving that $(u+v,u+w,v+w)$ is linearly independent [closed]

Let $V$ be a vector space over field $f$ with characteristic not equal to 2. Prove that $(u+v,u+w,v+w)$ is linearly independent if and only if $(u,v,w)$ is linearly independent.

• What are you approaches? – Friedrich Philipp Mar 17 '16 at 2:12
• Welcome to Stackexchange. You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: What you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shahab Mar 17 '16 at 2:12
• One implication is obvious. For the other one: $u-v=u+w-(v+w)$. – user228113 Mar 17 '16 at 3:56

Suppose $\{u,v,w\}$ are linearly independent, and further that there are constants $a,b,c \in F$ such that $a(u+v)+b(u+w) + c(v+w) = 0$.
Distributing, we see $au+av + bu + bw +cv + cw = 0$, but collecting like terms, we get $(a+b)u+(a+c)v+(b+c)w = 0$, but since $u,v,w$ are linearly independent, $a+b = 0$, $a+c = 0$, and $b+c= 0$. But then $a=b=c=0$. So $\{u+v,u+w,v+w\}$ is linearly independent.
Suppose $u$, $v$, and $w$ are independent; then we can extend this list to a basis of our vector space. The determinant of a matrix with these basis vectors as its columns is nonzero iff the columns are independent, so it is indeed nonzero. Moreover, the determinant is invariant under elementary column operations, and the other list you suggested can be obtained in this way. Hence the other list is part of another basis, and is independent.