Answer to Question #105461 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).
Suppose v1 = (5,−5,0,0), v2 = (10,5,−10,−10), and v3 =(−5,0,−5,5).
Find u1, u2, and u3.

Answer 1

We known that every element of $V$ is written as linear combination of element of $S$ .

Therefore, According to question

$v_1=1.u_1-2.u_2+0.u_3$ ................(1)

$v_2=2.u_1-7.u_2+4.u_3$ ..................(2)

$v_3=-3.u_1+8.u_2-1.u_3$ ...................(3)

Now ,we have to solving the above 3 equation.

Multiply equation (3) by 4 and then add with (2), we get

$v_2+4v_3= -10u_1+25u_2$ .............(4)

Multiplying 10 with equation (1) and then add with (4),we get

$10v_1+v_2+4v_3=5u_2$

$\implies$ $u_2=2v_1+\frac{1}{5}v_2+\frac{4}{5}v_3$

$=2[5,-5,0,0]^t+\frac{1}{5}[10,5,-10,-10]^t+\frac{4}{5}[-5,0,-5,5]^t$

$=$$[8,-9,-6,2]^t$

Now ,from (1)

$u_1=v_1+2u_2$

$=[5,-5,0,0]^t+2[8,-9,-6,2]^t$

$=[21,-23,-12,4]^t$

Now, from (3)

$u_3=v_3+3u_1-8u_2$

$=[5,0,-5,5]^t+3[21,-23,-12,4]^t-$ $8[8,-9,-6,2]^t$

$=[4,3,7,1]^t.$