Thread: linear combination help?

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.

Originally Posted by bonfire09

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.

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?

Originally Posted by bonfire09

So in this problem its linearly dependent since all the scalars are nonzero?

Yes.

Originally Posted by bonfire09

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.

A set of vectors cannot have a solution; an equation can.

Originally Posted by bonfire09

Im guessing a trivial solution is meaning that x1=x2=xn=0?

Yes.

Originally Posted by bonfire09

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.

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.