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span what tho? or does it matter
lol here it is
if the vectors are linearly D they will span either an R^1 or R^2 if they are linearly INd they will span R^3
if det=0; they are D, otherwise they are INd
ohhh i see. wish u were my teacher
lol it was so not explained clearly
2(5-8) = -6 --1(10-0) = 10 0(....) = 0 --------- add: not= 0 so linearly INd
you can do a rref on the vectors to see if there is a pivot point in each row too
Thanks that was a big help. I was kind of feeling lost
http://www.wolframalpha.com/input/?i=rref%7B%7B2%2C0%2C1%7D%2C%7B-1%2C1%2C2%7D%2C%7B0%2C4%2C5%7D%7D pivot in every row; independant http://www3.wolframalpha.com/input/?i=det%7B%7B2%2C0%2C1%7D%2C%7B-1%2C1%2C2%7D%2C%7B0%2C4%2C5%7D%7D det not= 0 ; so NOT dependant :)
ok thanks amistre. haha wish u cld be my tutor
@amistre, wow how did you figure out rref ?
i typed it into the wolf and it gave me back the rref; lazy math i know
I would actually disagree. I think it is linearly dependant
lol whoops i am wrong :D
c1 = c2 = c3 = 0 is the trival solution and as such is not a good "dep, ind" indicator
if c1v1 + c2v2 = v3 or some other mix of cs and vs then we know that at least one vector can be made from a linear combo of the others
one vector would DEPEND on the others and not need to be included in a "most effieciant" span
probably the easiest way to show independent vectors is to use elimination -1 1 2 0 4 5 2 0 1 <-- zero out the 2 by multiplying the top row by 2 and adding to this row -1 1 2 0 4 5 0 -2 4<-- now get rid of the -2 by multiplying the 2nd row by 1/2 and adding -1 1 2 0 4 5 0 0 2.5 you have 3 pivots (the leading number in each row) so you have 3 independent vectors
Well actually we are trying to prove that it spans
the three pivots show that your set has dimension 3, so you know it spans R^3 or, if you want to go to first principles, the span of a set of vectors is all possible linear combinations of the vectors. So, if we show that there is a linear combination that forms (1,0,0), another combination that forms (0,1,0), and a third combination that forms (0,0,1), then you know that you can create any linear combination in R^3. to find the linear combinations of v1,v2,v3 that form 1,0,0 etc. Find the inverse of the matrix [v1,v2,v3]. The entries in the inverse tell you the linear combination needed to create the natural basis for R^3 ie. A*Ainv = I so the first column of Ainv tell you how to add the columns (v1,v2,v3) of A to get (1,0,0)transpose
Thanks phi I really appreciate it. I dont know a thing abt spanning and you are definitely helping me there. My textbook always explains things backwards
linear albagra tends to be more on the pure side of math to me. It kinda cuts the imbilical cord betwen needing pictures and something tangible to rely upon and just uses the world of math.
span itself is a rather basic concept tho; just stretch out your arms as wide as they wil go and everything inbetween them is within their span of influence.
the arms of a vector space are the vectors themselves that can be streched out and away into infinity. the vectors for the usual xy plane are then: <1,0> and <0,1> with these vectors we can combine them into any combonation to created every other vector; or rather, reach every single point within the xy plane
if we had say: <1,0> <0,1> and <2,3> as our span for the xy plane , that would be fine except for we want to be as efficient as possible; only use the least number of vector to reach every single point. So, in this case, we could omit one of them and still be able to span R^2
Thanks :D That was helpful