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Find a basis for the subspace of R^4 spanned by the given vectors (attempt inside) 
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#1
Nov1011, 08:57 AM

P: 18

Hello. First, I'd like to apologize because I don't know where to go ask for homework on linear algebra on the forums so if anyone could please let me know, that would be appreciated.
Here's the question: Find a basis for the subspace of R^4 spanned by the given vectors Here's the answer at the end of the textbook: so the matrix A is ( 2 4 2 3) (2 2 2 4) (1 3 1 1) and the book puts it in reduced row echelon form which comes out to ( 1 0 1 5/2 ) ( 0 1 0 1/2) (0 0 0 0 ) Thus a basis for the subspace is { (1, 0, 1, 5/2), (0, 1, 0, 1/2) } My question is: Is this the only answer you can get or can I get different answers. Say, if I used paramters "a" and "b" for column x3 and x4 respectively, then I can get x1 = a  (5/2)b x2 = (1/2)b x3 = a x4 = b then if I let a = 0 and b =1, I'd get: (5/2, 1/2, 0, 1) then if I let a = 1 and b = 0, I'd get: (1, 0, 1, 0) thus a basis for the subspace is { (5/2, 1/2, 0, 1), (1, 0, 1, 0) } Could that be another answer? Also, I thought for it to be a basis, it has to be linearly independent. Are they assuming that by inspection it's linearly independent so that's why the book isn't testing for independence? Thank you so much for any guidance. 


#2
Nov1011, 11:08 AM

Sci Advisor
P: 906

bases for vector spaces aren't unique, the only invariant is the size of the basis, that is, the dimension of the vector space.
your example isn't a valid one, however. you can't assign "parameters" to the columns, because your span is determined by row vectors. your 2nd "proposed basis" isn't even in span{(1,0,1,5/2),(0,1,0,1/2)}. suppose (5/2,1/2,0,1) = a(1,0,1,5/2) + b(0,1,0,1/2) = (a,b,a,5a/2b/2). then a must be 5/2 (1st coordinate) and a must also be 0 (3rd coordinate). similarly if (1,0,1,0) = c(1,0,1,5/2) + d(0,1,0,1/2) = (c,d,c,5c/2d/2) then c must be 1 (1st coordinate) and c must be 1 (3rd coordinate). what you can say is that {(a,0,a,5a/2),(0,b,0,b/2)} is also a basis, for nonzero a and b. the reason no check is made for linear independence for the rows of a matrix in RREF, is that the process of rowreduction ensures linear independence (only the top row has a nonzero 1st coordinate, if the first nonzero entry of row 2 is in place j, than the first j1 entries of row 2 and all subsequent rows are all 0, and the jth entry of all other rows is 0, so only the 2nd row has a nonzero entry in the jth position. continuing, (down and right), each surviving row has an entry (it's pivot entry) that no other row has, which means no linear combination of the other rows can produce a nonzero entry in that coordinate. 


#3
Nov1011, 04:28 PM

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i wonder if you are confused about whether your basis is for row space, column space, or null space?



#4
Nov1011, 08:34 PM

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Find a basis for the subspace of R^4 spanned by the given vectors (attempt inside)
You talk about a space spanned by a "given set of vectors" but give no vectors. You do, fortunately, say that you are talking about vectors in R^{4} so, since the given matrix A has four columns but only three rows, I think you mean that the given vectors are (2, 4, 2, 3), (2, 2, 2, 4), and (1, 3, 1, 1).
Rather than write matrices and try to use "rowreduction" formulas, I would prefer to use the basic definition. (w, x, y, z) would be in the span of those three vectors if and only there exist number a, b, and c such that (w, x, y, z)= a(2,4,2,3)+ b(2, 2, 2, 4)+ c(1, 2, 1, 1). That, of course, is the same as the four equations w= 2a 2b+ 2c, x= 3a 2b+ 2c, y= 2a+ 2b c, z= 3a+ 4b+ c. Now, try to eliminate a, b, and c. For example, if we subtract the first equation from the second, we eliminate both b and c: x w= a. If, instead, we add the second equation to the third, we eliminate only b: x+ y= a+ c. Similarly, we can eliminate b by adding twice the second equation to the fourth: 2x+ z= 9a+ 5c. Now subtract 5 times x+ y= a+ c to that: 2x+ z 5(x y)= 3x+ 5y+ z= 9a+ 5c 5(a+ c)= 4a. Finally, subtract 4 times x w= a from that: 3x+ 5y+ z 5(x w)= 5w 8x+ 5y+ z= 0. That is the equation any vector in the span must satisfy. 


#5
Nov1111, 10:37 PM

P: 18

To take a guess, would it be because a basis are vectors that are linearly independent, and so the author puts it in RREF and thus a basis is formed? A question about your explanation: I understand that to see if it spans, if you can find scalars a, b, c, it spans but why are you getting rid of a,b,c? Also, I've noticed that you changed the numbers. Shouldn't it be w = 2a 2b + c, and x= 4a 2b + 3c ? Thank you so much for any help. EDIT: Also, I tried testing the vectors for linear independence. c1(2,4,2,3) + c2(2,2,2,4) + c3(1,3,1,1) = (0,0,0,0) to get c1 + c3 = 0 c2+ 1/2c3=0 doesn't that mean that since c1=c2=c3 does not all equal zero, this is not linearly independent thus not a basis? Since it has to be basis it has to be linearly independent and span? Forgive me for not understanding something. The more I try to understand this problem, the more frustrated I get because it doesn't make sense. Any help would be much appreciated. 


#6
Nov1211, 04:30 AM

Sci Advisor
P: 906

ok, you're trying to find a basis for span({(2,4,2,3),(2,2,2,4),(1,3,1,1)}).
so this is the space of all vectors in R^{4} of the form: a(2,4,2,3) + b(2,2,2,4) + c(1,3,1,1), or, to put this another way, the row space of the matrix A = [tex]\begin{bmatrix}2&4&2&3\\2&2&2&4\\1&3&1&1\end{bmatrix}[/tex] what we want to do, is establish a minimal linearly independent set in our span set. now, suppose i multiply a row of A by some number (except 0). that's still in our span set, because its just a scalar multiple of one of the spanning vectors. ok, what if i switch two rows? that obviously doesn't change the span set, it just shuffles it around. fine, but what about this: what if i multiply one of the spanning vectors (rows), and then add it to another one? well, that's a diferent linear combination of our spanning vectors, but it's still a linear combination (no squaring or anything), so its still in our span set. so none of the rowreduction operations changes the spanning set (row space), so if we can get a "nicer" set of rows from rowreducing, we can be confident we haven't gone outside our spanning set. well, you might ask, ok, i see we're still in our spanning set, but maybe we "lose" some vectors along the way. well, obviously swapping rows doesn't "lose any vectors". and multiplying by a nonzero number doesn't "lose" any vectors either. so let's just concentrate on the third operation: adding a multiple of one row to another. so let's prove this: for any 2 nonzero vectors v_{1},v_{2} in a vector space V, if r ≠ 0: span({v_{1},v_{2}}) = span({v_{1}+rv_{2},v_{2}}) well, it's clear (or should be), the second set is contained in the first, because: a(v_{1}+rv_{2}) + bv_{2} = av_{1} + (ar+b)v_{2}, so we can write anything in the second spanset as linear combinations of the first. what's not immediately clear is that the first spanset is contained in the second. suppose we have av_{1}+bv_{2}. then av_{1}+bv_{2} = av_{1} + arv_{2}  arv_{2} + bv_{2} (adding and subtracting the same thing) = a(v_{1}+rv_{2}) + (bar)v_{2}, so span({v_{1},v_{2}}) ⊆ span({v_{1}+rv_{2},v_{2}}), so the two sets are equal. hopefully, this convinces you that finding rref(A) doesn't change the spanset at all: row(rref(A)) = row(A). 


#7
Nov1211, 09:21 AM

P: 18

Deveno, thank you so much for your help. It's much clearer now. Really, thank you again and Halls for really helping me out.



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