SOLUTION: let{u1,u2....uk}be a linearly independent set of vector in R^n and c1,c2,....ck are scalars. prove that {c1u1, c2u2....., ckuk}is also limearly independent.

Algebra ->  College  -> Linear Algebra -> SOLUTION: let{u1,u2....uk}be a linearly independent set of vector in R^n and c1,c2,....ck are scalars. prove that {c1u1, c2u2....., ckuk}is also limearly independent.      Log On


   



Question 400630: let{u1,u2....uk}be a linearly independent set of vector in R^n and c1,c2,....ck are scalars. prove that {c1u1, c2u2....., ckuk}is also limearly independent.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Since { u%5B1%5D,+u%5B2%5D, u%5B3%5D, ..., u%5Bk%5D } are linearly independent, then the equation d%5B1%5Du%5B1%5D+%2B+d%5B2%5Du%5B2%5D+%2B+d%5B3%5Du%5B3%5D+ ...+ d%5Bk%5Du%5Bk%5D+=+theta has only the trivial solution d%5B1%5D=+d%5B2%5D+=+d%5B3%5D= ...= d%5Bk%5D+=+0. (theta+is the zero vector, and d%5Bn%5D's are scalars.)
We have to assume that none of the c%5Bn%5D's are equal to zero, otherwise,
{ c%5B1%5Du%5B1%5D,+c%5B2%5Du%5B2%5D, c%5B3%5Du%5B3%5D, ..., c%5Bk%5Du%5Bk%5D } would contain the zero vector, automatically making the set linearly dependent. Hence
+ ...+ %28d%5Bk%5D%2Fc%5Bk%5D%29%2A%28c%5Bk%5Du%5Bk%5D%29+=+theta
==> d%5Bn%5D%2Fc%5Bn%5D+=+0 for every n = 1,2,3,...,k, and { c%5B1%5Du%5B1%5D,+c%5B2%5Du%5B2%5D, c%5B3%5Du%5B3%5D, ..., c%5Bk%5Du%5Bk%5D } is a linearly independent set.