Let v_1,...,v_n be linealy independent vectors in RR^m and let w=a_1v_1+...+a_nv_n with real numbers a_1,...a_n, be linear combination of these vectors. Prove if a_1 != 0, then the vector w,v_2,...,v_n are linearly independent So far, what I have thought is forming a formula like this: x_1w+x_2v_2+...+x_nv_n=0, because v_1,...,v_n is linearly independent vectors, so x_2=...=x_n=0, then now we need to prove x_1=0 so these vector will be linearly independent. But from here, I don't know how to process further more in this problem, can anyone help me out with a way to solve this or a better alternative solution is appreciate.

pleitatsj1 2022-08-07 Answered
Let v 1 , . . . , v n be linealy independent vectors in R m and let w = a 1 v 1 + . . . + a n v n with real numbers a 1 , . . . a n , be linear combination of these vectors.
Prove if a 1 0, then the vector w , v 2 , . . . , v n are linearly independent
So far, what I have thought is forming a formula like this: x 1 w + x 2 v 2 + . . . + x n v n = 0, because v 1 , . . . , v n is linearly independent vectors, so x 2 = . . . = x n = 0, then now we need to prove x 1 = 0 so these vector will be linearly independent. But from here, I don't know how to process further more in this problem, can anyone help me out with a way to solve this or a better alternative solution is appreciate.
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Answers (1)

kunstdansvo
Answered 2022-08-08 Author has 16 answers
If w has a contribution from v 1 , then any linear combination of w , v 2 , , v n can be written as a linear combination of v 1 , , v n . That's suppose x 1 w + x 2 v 2 + + x n v n = 0. Then x 1 a 1 v 1 + ( x 1 a 2 + x 2 ) v 2 + + ( x 1 a n + x n ) v n = 0 x 1 = ( x 1 a 2 + x 2 ) = = ( x 1 a n + x n ) = 0 x 1 = x 2 = = x n = 0
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