Show that the vectors
$v_1=(1,1,1)$, $v_2=(1,1,0)$, $v_3=(1,0,0)$, $v_4=(3,2,0)$ span $\mathbb{R}^3$ but do not form a basis for $\mathbb{R}^3$:
I understand how to find different find the general solution through RREF using the augmented matrix,
but I do not understand how the arbitrary vector $x=(x,y,z)$ can be written as
$x=zv_1+(y-z)v_2+(x-y)v_3+0v_4$
Best Answer
We have $ x=zv_{1}+(y-z)v_{2}+(x-y)v_{3}+0v_{4}=z(1,1,1)+(y-z)(1,1,0)+(x-y)(1,0,0)=(z,z,z)+(y-z,y-z,0)+(x-y,0,0)=(z+y-z+x-y,z+y-z,z)=(x,y,z) $.
To obtain this you need to find a, b, c and d such that $av_{1}+bv_{2}+cv_{2}+dv_{3}=(x,y,z)$
Clearly the vectors span $\mathbb{R}^3$ since any arbitrary vector from $\mathbb{R}^3$ can be generated from these four vectors, however they do not form a basis for $\mathbb{R}^3$ since they are linearly dependent.