Let $\ u_1, u_2, u_3 $ be linearly independent vectors in $\ V$. and
$\ v_1 = u_1 + u_2+ u_3, \ \ v_2 = u_1 – u_2 + u_3 , \ v_3 = -u_1 + 3u_2 – u_3 $
Does (1) $\operatorname{span}\{ u_1,u_2,u_3 \} =\operatorname{span}\{ v_1 , v_2 , v_3 \} $ ?
It is clear that $\operatorname{span}\{ v_1, v_2, v_3 \}\subseteq\operatorname{span}\{ u_1,u_2,u_3 \} $ So need to prove that $\operatorname{span}\{u_1,u_2,u_3 \} \subseteq\operatorname{span}\{ v_1, v_2, v_3 \} $
So I chose random vector $\ w_1 \in\operatorname{span}\{u_1,u_2,u_3\} $ , $\ w_1 = \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3$ and then trying to express it using $\ v_1, v_2, v_3 $
Now trying to solve it using the equations I only get
$\ u_2 = \frac{v_2 + v_3}{2} $ and can't express $\ u_1, u_3 $ using the v's. So I guess it means those vectors are linearly dependent and therefore equation (1) is wrong?
Best Answer
Simply note that
$$u_2=\frac{v_1+v_3}4=\frac{v_1-v_2}2$$
therefore $v_1,v_2,v_3$ are not linearly independent and
$$\ Sp \{ u_1,u_2,u_3 \} \neq Sp\{ v_1 , v_2 , v_3 \}$$