
#1
Apr310, 04:18 PM

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1. The problem statement, all variables and given/known data
How to determine if a set of vectors span a space in general? say, V=R^n and you're given a few vectors and asked to determine if they span the space.. how do you do that? 2. Relevant equations 3. The attempt at a solution 



#2
Apr310, 04:21 PM

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Just to make this a little less abstract, suppose V = R^{3}, and that S = {<1, 0, 1>, <0, 2, 5>}. Does this set of vectors span V? 



#3
Apr310, 04:32 PM

P: 50

I'm not quite sure with my way of finding the answer, which is NO i.e. do not span
would it be okay if you show me your working based on this example? :) 



#4
Apr310, 04:45 PM

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how to determine if a set of vectors span a space
You tell me why you think this set doesn't span R^{3}.




#5
Apr310, 04:52 PM

P: 50

erm, I wrote it as augmented matrix
1 0  1 0 0 0 2  0 1 0 1 5  0 0 1 and start to reduce it to "reduced row echelon form" (that's why I started another thread before this asking about RRE form because I'm unsure how this works) and on the 3rd row, i get 0 0  0 1/2 1/5 (inconsistent, so do not span?) :( I'm looking for another way of determining the spanning set this is what I get from googleing O_O 



#6
Apr310, 07:22 PM

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There's a much simpler way to answer the question, that can be done with no computation. Answer the following questions and it will be obvious to you.
What's the dimension of R^{3} (i.e., dim(R^{3}))? How many vectors does it take to span R^{3}? How many vectors are there in S? Does S span R^{3}? Going back to the work you did, you have a lot of extra stuff that doesn't make any sense to me. A given set of vectors spans R^{3} if any vector in R^{3} is some linear combination of the vectors in the set. IOW, for any vector <x, y, z>, there is a solution for the constants a and b in this equation: a<1, 0, 1> + b<0, 2, 5> = <x, y, z> Setting this up as an augmented matrix gives you this: 1 0  x 0 2  y 1 5  z After row reduction, I get j 1 0  x 0 1  y/2 0 0  zx 5y/2 The first two rows say that a = x and b = y/2, but the bottom row says that 0a + 0b = z  x  5y/2. This last equation is saying that the system of equations has a solution only if z  x 5y/2 = 0. IOW, for some vectors <x, y, z> there is no solution. 


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