# If S= {$v_1$, $v_2$, $v_3$, $v_4$} is linearly independent, then so is T= {$v_1$ − $v_2$, $v_2$ − $v_3$, $v_3$ − $v_4$, $v_4$ − $v_1$}

If S= {$v_1$, $v_2$, $v_3$, $v_4$} is linearly independent, then so is T= {$v_1$ − $v_2$, $v_2$ − $v_3$, $v_3$ − $v_4$, $v_4$ − $v_1$}

I can't find an example as to why this is false

• Hint: add them all up. – dxiv Nov 8 '16 at 3:52
• Sometimes it is better to look at the problem on a smaller scale. Consider only two vectors S= {$v_1$, $v_2$,} where the problem becomes to consider {$v_1-v_2,v_2-v_1$}. and take for the vectors unit vectors. It becomes rather obvious then... – imranfat Nov 8 '16 at 4:27

$(\nu_1-\nu_2)+(\nu_2-\nu_3)+(\nu_3-\nu_4)+(\nu_4-\nu_1)=0$

Therefore they cannot be linear independent.