Re: linear combination help?

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.

Re: linear combination help?

I dont see it still. But a more precise definition of a linear independent relationship would be like this. Let x1**a1**+x2**a2**+x3**a3**...xn**an**=0 where x1=x2=x3...=xn=0. And a linear dependent relationship where c**a1**+c**a2**+c**a3.**..c**an**=0 where c represents weights and it is linear dependent iff one or more of the weights c are nonzero.

Re: linear combination help?

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Originally Posted by

**bonfire09** But a more precise definition of a linear independent relationship would be like this. Let x1**a1**+x2**a2**+x3**a3**...xn**an**=0 where x1=x2=x3...=xn=0. And a linear dependent relationship where c**a1**+c**a2**+c**a3.**..c**an**=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.

Re: linear combination help?

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 x1**v1**+x2**v2**+...xn**vn**=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 c1**v1**+c2**v2**+...cn**vn**=0. So in this problem its linearly dependent since all the scalars are nonzero?

Re: linear combination help?

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Originally Posted by

**bonfire09** So in this problem its linearly dependent since all the scalars are nonzero?

Yes.

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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.

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Originally Posted by

**bonfire09** Im guessing a trivial solution is meaning that x1=x2=xn=0?

Yes.

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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 c1**v1**+c2**v2**+...cn**vn**=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.