Prove if $u,v,w$ linearly independent then ${\{u+v+w,v-w,2w}\}$ linearly independent 
Prove if $u,v,w$ linearly independent then ${\{u+v+w,v-w,2w}\}$
  linearly independent

what I did is : 
I need to Prove that for  $x,y,z  \in F$ 
$x(u+v+w) + y(v-w) + z(2w) = 0 $
implies that $x=y=z=0$
$x(u+v+w) + y(v-w) + z(2w) = xu  + xv + xw + yv - yw + 2zw =  $
$= (x)u + (x+y)v + (x-y+2z)w =0$
$ u,v,w $    linearly independent $\rightarrow   x=0  $  and $x+y=0$ and $(x-y+2z)=0$ $\rightarrow x=y=z=0$
is that enough to prove the question ? 
thanks

The systematic way is to consider matrices:
$$
\begin{pmatrix} u' \\ v' \\ w' \end{pmatrix}
=
\begin{pmatrix} 1 & 1 & \hphantom{-}1 \\ 0 & 1 & -1 \\ 0 & 0 & \hphantom{-}2 \end{pmatrix}
\begin{pmatrix} u \\ v \\ w \end{pmatrix}
$$
The matrix is triangular with nonzero diagonal entries. Therefore, it is invertible.
Thus, the subspace generated by $u',v',w'$ is the same as the subspace generated by $u,v,w$.
