# Thread: Another linear dependence problem

1. ## Another linear dependence problem

if $\displaystyle \{u,v,w,\}$ is a linearly independent set of vectors, then $\displaystyle S = \{u+v-2w,u-v-w,u+w\}$ is also linearly independent.

True or false?

thanks

2. Originally Posted by Bucephalus
if $\displaystyle \{u,v,w,\}$ is a linearly independent set of vectors, then $\displaystyle S = \{u+v-2w,u-v-w,u+w\}$ is also linearly independent.

True or false?

thanks
Form the determinant of those coefficients and show it is non-zero.

3. Originally Posted by ThePerfectHacker
Form the determinant of those coefficients and show it is non-zero.
At first I set up the matrix with each vector in a row like this:

$\displaystyle \left( \begin{matrix} u & v & -2w\\ u & -v & -w\\ u & 0 & w\\ \end{matrix} \right)$

But then I couldn't see how this would work. Then I realised that you meant the following.

$\displaystyle \left( \begin{matrix} u & u & u\\ v & -v & 0\\ -2w & -w & w\\ \end{matrix} \right)$

So I formed the determinant of this matrix and got a non-zero result.
Thanks for your help.