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

Question

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

Answer 1

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 $u,v,w$.

Answer 2

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.