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OpenStudy (anonymous):

Find all subsets of the set that forms a basis for R^3 S= {(1,3,2),(-4,1,1),(-2,7,-3),(2,1,1)}

OpenStudy (anonymous):

Does anyone understand what to do? Cuz I don't

OpenStudy (kinggeorge):

If I remember my lin. algebra class correctly, you just have to find the vectors that are linearly independent.

OpenStudy (anonymous):

nooo they are not asking me if it is a basis

OpenStudy (anonymous):

I have to find the subsets

OpenStudy (kinggeorge):

Each set of vectors that are linearly independent with one another should form a basis of R^3.

OpenStudy (anonymous):

ohhh so i wld play around using these 4 vectors?

OpenStudy (anonymous):

Hi Samjordan, what do u want to know/ask in linear algebra???

OpenStudy (anonymous):

LOL This question

OpenStudy (kinggeorge):

I just remembered the other condition. Whatever vectors you pick to be in your subsets also need to span R^3. As far as I can tell, playing around with them would be the fastest, but you need at least 2 vectors in each subset.

OpenStudy (anonymous):

I think I am getting it

OpenStudy (anonymous):

sorry Sam, at present my knowledge on this topic is rusty !!

OpenStudy (anonymous):

hehe ok hakirat Thanks for stoppiin by

OpenStudy (anonymous):

no I think 3 vectors since it needs to be a basis

OpenStudy (kinggeorge):

I think you're right.

OpenStudy (anonymous):

I guess you wld take 3 vectors of the 4 and check if they are linearly dependant and then check another 3 and then another 3.... I am only assuming this

OpenStudy (kinggeorge):

That would get the linearly independent factor taken care of, but then you would need to show that of those that are linearly independent, they also span R^3, and that's a little trickier to show.

OpenStudy (kinggeorge):

One way you could do it is as such: Suppose you have a vector <x, y, x> in R^3. Then your set of vectors spans R^3 if you can find some combination of your vectors such that you can create the vectors <x+1, y, z>, <x, y+1, z>, and <x, y, z+1>.

OpenStudy (anonymous):

Ok just processing it

OpenStudy (anonymous):

okk wel that is liek showing the linear independance thingy

OpenStudy (kinggeorge):

almost, but not quite. I'd suggest looking here for a better explanation. http://en.wikipedia.org/wiki/Basis_(linear_algebra)#Example_of_alternative_proofs

OpenStudy (kinggeorge):

Just generalize their example to R^3

OpenStudy (anonymous):

Thanks Kinggeorge

OpenStudy (kinggeorge):

your welcome

OpenStudy (phi):

Basis requires (1) span the space (2) are independent a basis for R^3 consists of three linearly independent 3x1 vectors. Dimension of the space = number of independent vectors needed to span it. So 3 for R^3 you have to check 4choose3 = 4 different subsets. use elimination to show each set has 3 pivots (or not)

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