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

For a quick hint to a speedier solution, notice you have the following relation:
$$\begin{pmatrix} u'\\v' \\w' \end{pmatrix}=\begin{pmatrix} u+v+w \\ v-w \\ 2w \end{pmatrix}=\begin{pmatrix}1_K & 1_K & 1_K \\ 0_K & 1_K & -1_K \\ 0_K & 0_K & 2_K \end{pmatrix} \begin{pmatrix} u\\ v\\ w \end{pmatrix}$$
where $K$ is the unspecified field of scalars of the unspecified vector space $V$ you are considering and where for any integer $n \in \mathbb{Z}$ we denote $n_K=n1_K$.
The square matrix of order $3$ that has appeared --let us call it $A$-- is upper triangular and as long as $\mathrm{char}\ K \neq 2$ (in other words as long as $2_K \neq 0_K$) all the diagonal entries will be nonzero hence invertible hence the whole matrix will be invertible.
