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I'm having trouble trying to solve this question.. The question states "Let S={r,u,d,x} be a set of vectors.
If x=4r+4u+4d, determine whether or not the four vectors listed above are linearly independent or linearly dependent. If is dependent, find a non-trivial linear relation. I don't get how to get started. I know that a linear indepedant relationship is if the scalars are all 0 and a linear dependent relationship is when atleast one ore more of the scalars or weights are nonzero.
I would recommend thinking about it a bit more. This question is trivial. Also, I recommend making sure that you can write the precise definition of linear independence. "A linear indepedant relationship is if the scalars are all 0" is not a precise definition.
I dont see it still. But a more precise definition of a linear independent relationship would be like this. Let x1a1+x2a2+x3a3...xnan=0 where x1=x2=x3...=xn=0. And a linear dependent relationship where ca1+ca2+ca3...can=0 where c represents weights and it is linear dependent iff one or more of the weights c are nonzero.
Both of these definitions are wrong. It may very well be that you can't solve the problem because you don't know the correct definition.Quote:
Originally Posted by bonfire09
Note that x = 4r + 4u + 4d implies 4r + 4u + 4d + (-1)x = 0.
really its that simple? just moving over the x to the other side. Oh all along I thought I needed to find some scalar number for x. Yeah my definitions I gave are wrong. I just noticed the correct definition for linear independence is when for a set a vectors {v1,v2,...,vn} in R^n has only the trivial solution. In other wards x1v1+x2v2+...xnvn=0. Im guessing a trivial solution is meaning that x1=x2=xn=0? And then its linearly dependent when for all set of vectors {{v1,v2,...,vn} if there exists some weights or scalars c1,...,cp that are not all zero or in otherwards atleast one is nonzero c1v1+c2v2+...cnvn=0. So in this problem its linearly dependent since all the scalars are nonzero?
Yes.Quote:
Originally Posted by bonfire09
A set of vectors cannot have a solution; an equation can.Quote:
Originally Posted by bonfire09
Yes.Quote:
Originally Posted by bonfire09
What is linearly dependent, i.e., what does the bold "it" above refer to? Your definition sounds like, "It's even when for all numbers n if there exists a number k n = 2k." No, a given set of vectors v1, ..., vn is called linearly dependent if there exist numbers c1, ..., cn not all zero such that c1*v1 + ... + cn*vn = 0; otherwise this set of vectors is called linearly independent.Quote:
Originally Posted by bonfire09