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anonymous
 4 years ago
Prove that the vectors form a spanning set
anonymous
 4 years ago
Prove that the vectors form a spanning set

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1span what tho? or does it matter

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328808886645:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if det=0; they are D, otherwise they are INd

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhh i see. wish u were my teacher

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol it was so not explained clearly

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.12(58) = 6 1(100) = 10 0(....) = 0  add: not= 0 so linearly INd

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1you can do a rref on the vectors to see if there is a pivot point in each row too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks that was a big help. I was kind of feeling lost

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=rref%7B%7B2%2C0%2C1%7D%2C%7B1%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%7B1%2C1%2C2%7D%2C%7B0%2C4%2C5%7D%7D det not= 0 ; so NOT dependant :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok thanks amistre. haha wish u cld be my tutor

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@amistre, wow how did you figure out rref ?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i typed it into the wolf and it gave me back the rref; lazy math i know

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would actually disagree. I think it is linearly dependant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328810589623:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol whoops i am wrong :D

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1c1 = c2 = c3 = 0 is the trival solution and as such is not a good "dep, ind" indicator

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1one vector would DEPEND on the others and not need to be included in a "most effieciant" span

phi
 4 years ago
Best ResponseYou've already chosen the best response.2probably 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well actually we are trying to prove that it spans

phi
 4 years ago
Best ResponseYou've already chosen the best response.2the 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1linear 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.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1span 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.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks :D That was helpful
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