There are 4 vectors: $v_1 = (1,4,2,-3) , v_2 = (-2,4,-4,27) , v_3 = (1,8,2,4) , v_4 = (2,-12,4,-41)$

Are there any scalars $a_i, b_i, c_i ( 1 \le i \le 4)$ that the vectors

$u_1 = \sum\limits_{i=1}^4 a_iv_i$

$u_2 = \sum\limits_{i=1}^4 b_iv_i$

$u_3 = \sum\limits_{i=1}^4 c_iv_i$

be independent?

Well, I put $v_1,v_2,v_3,v_4$ on a matrix and got 3 vectors which are linearly independent.

Which means that if I want $u_1,u_2,u_3$ to be linearly independent I will need to put for every $u_i$ vector two scalars to be zero.

for example:

$u_1 = a_1v_1$

$u_2 = b_2v_2$

$u_3 = c_3v_3$

and like that I will have about 6 options I guess to make linearly independency. What do you think? Did I solve this all wrong and there's a better way to solve it?

  • $\begingroup$ solved it. thanks! $\endgroup$ Jan 10 '15 at 16:44

After putting v1, v2, v3, v4 into a matrix and obtain its row echelon form, obtain the linear independent set from the corresponding pivotal ones. Suppose v1, v2 and v3 are linearly independent. Then just choose u1=v1, u2=v2 and u3=v3 by assigning appropriate values for ai's.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.