- #1

dmatador

- 120

- 1

I did this:

a_1( v_1 - 2v_2 + v_3) + a_2(v_2) + a_3(v_3) + ... + a_m(v_m)

= a_1(v_1) - a_1(2v_2) + a_1(v_3) + a_2(v_2) + a_3(v_3) + ... +a_m(v_m)

= -a_1(2v_2) + a_1(v_3) + a_1(v_1) + a_2(v_2) + a_3(v_3) + ... +a_m(v_m)

and since we know the first part, we know that a_1(v_1) + a_2(v_2) + a_3(v_3) + ... +a_m(v_m) is Linearly independent so we know that a_1 = a_2 = a_3 = ... = a_m = 0. so we can see now that the first two terms have scalars a_1 and a_1 which equal zero to the second set of vectors is also Linearly independent.