Answer to Question #105458 in Linear Algebra for michael

Question

Let V be a subspace of R^4 and S = {u1,u2,u3} be a basis for V. Suppose v1,v2,v3 are vectors in V such that(v1)S = (1,−2,0), (v2)S = (2,−7,4), and (v3)S = (−3,8,−1).
Show that {v1, v2, v3} is also a basis for V.

Answer 1

{u1,u2,u3} is basis of V.

$c_1u_1+c_2u_2+c_3u_3=0$

$v_1=u_1-2u_2$

$v_2=2u_1-7u_2+4u_3$

$v_3=-3u_1+8u_2-u_3$

$v_1+v_2+v_3=3u_3-u_2$

$u_3=\frac{1}{3}(u_2+(v_1+v_2+v_3))$

$u_1=v_1+2u_2$

$v_2=2( v_1+2u_2 )-7u_2+\frac{4}{3}(u_2+(v_1+v_2+v_3)) )$

$u_2=(10v_1+v_2+4v_3)/3$

$v_1,v_2,v_3$ in terms of $u_1,u_2,u_3$ .

$u_1=(23v_1+2v_2+8v_3)/3$

$u_2=(10v_1+v_2+4v_3)/3$

$u_3=(11v_1+2v_2+5v_3)/3$

Putting these in $c_1u_1+c_2u_2+c_3u_3=0$

$c_1(23v_1+2v_2+8v_3)/3+c_2(10v_1+v_2+4v_3)/3+c_3(11v_1+2v_2+5v_3)/3=0$

$v_1(23c_1+10c_2+11c_3)/3+v_2(2c_1+c_2+2c_3)/3+v_3(8c_1+4c_2+5c_3)/3=0$

$d_1v_1+d_2v_2+d_3v_3=0$

$d_1=(23c_1+10c_2+11c_3)/3$

$d_2=(2c_1+c_2+2c_3)/3$

$d_3=(8c_1+4c_2+5c_3)/3$

$c_1=c_2=c_3=0$

So, $d_1=d_2=d_3=0$

Thus, {$v_1,v_2,v_3$ } spans and is linearly independent.

{$v_1 , v_2 , v_3$} is also a basis of V.