Register to reply 
Linear Algebra (showing if a vector spans a vector space) 
Share this thread: 
#1
Nov2311, 08:46 PM

P: 25

1. The problem statement, all variables and given/known data
I was wondering if someone could explain the easiest way to determine if a set S spans V? some example questions would be: show that S = {v1, v2, v3, v4} spans R4 where v1 = [1 0 +1 0] v2 = [0 1 1 2] v3 = [0 2 +2 1] v4 = [1 0 0 1] 2. Relevant equations 3. The attempt at a solution I know that you need to let x = [a, b, c, d] be any vector in R4 and form an eqn like a1v1 + a2v2 + a3v3 + a4v4 = x but now I'm lost... please help! 


#2
Nov2311, 09:33 PM

Mentor
P: 21,307




#3
Nov2311, 10:04 PM

P: 25




#4
Nov2311, 10:06 PM

Mentor
P: 21,307

Linear Algebra (showing if a vector spans a vector space)
Start with that and tell me what you get.



#5
Nov2311, 10:16 PM

P: 25

I get
a1v1 + a2v2 + a3v3 + a4v4 = x => a1 00 00 a4 = a 00 a2 2a3 00 = b a1 a2 2a3 00 = c 00 2a2 00 a4 = d => 1 0 0 1  a 0 1 2 0  b 1 1 2 0  c 0 2 0 1  d => RRE form 1 0 0 .... OOooohhh I see!! Now when I turn it into RREF, I'll get something like a1 = a+bc etc.. correct?! => RREF 1 0 0 1  a 0 1 2 0  b 0 0 1 (1/4)  d2b/4 0 0 0 0  ca+b+(d2b) Does that look like I'm on the right track? 


#6
Nov2311, 11:49 PM

P: 771




#7
Nov2311, 11:54 PM

P: 25




#8
Nov2411, 12:20 AM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,773




#9
Nov2411, 12:24 AM

P: 2,568

Now @OP, what does this mean? Having no nonpivot columns? How does this relate to Span? 


#10
Nov2411, 01:08 AM

Mentor
P: 21,307

And to continue with what vela and flyingpig said, think about what system of equations you're reduced augmented matrix represents. If <a, b, c, d> is any arbitrary vector in R^{4}, is there a specific linear combination of the v_{i} vectors that forms that vector <a, b, c, d>? If so, those v_{i} vectors span R^{4}.



#11
Nov2411, 01:27 AM

P: 25

Umm I don't remember learning about pivots.
and I did the thing again and I still got 0 0 0 0 for the last row :O Here are my steps: r1 + r3 > r3 r2 + r3 > r3 r42r2 > r4 1/4 r3 > r3 r4 + 4r3 > r4 this gave me a final matrix of 1 0 0 1 a 0 1 2 0 b 0 0 1 1/4 [(ca+b)/4] 0 0 0 0 d2b+4c Now what do I do? Or even better, could someone please explain to me the general steps of finding if a vector spans the v.s. so I can complete these problems? Thank you! EDIT: Mark I just saw your response, how would I know if there is a linear combination that forms <a, b, c, d>? 


#12
Nov2411, 03:37 AM

P: 771

It's not clear to me whether you want to demonstrate a knowledge of basic methods of determining whether a set of vectors spans a space, or whether you just want to know how to use functions available in modern mathematical software such as Maple, Mathematica or Matlab to make the determination.
If it's the latter, proceed as follows: If you have a bunch of N element vectors, let's say you have M of them, such that M≥N, let the vectors form the rows of an MxN matrix. Then you don't need to augment that matrix to determine if there are at least N linearly independent vectors in the set. Simply row reduce the MxN matrix. The number of 1's on the main diagonal of the result is the number of linearly independent vectors in the set. Only if that number is N does the set span the space R(N). The number of linearly independent rows (your vectors) is also equal to the rank of the MxN matrix of vectors, and you can use the "rank" function to determine this. You could also calculate the singular values of the MxN matrix; the rank is equal to the number of nonzero singular values (that's what the "rank" function does). See the attached image. 


#13
Nov2411, 04:11 AM

P: 25

Sorry, I don't mean to be rude but I'm just kinda lost. This is my first course in linear algebra and that's why I'm pretty confused.
What I was trying to do was to find a basis. The theorem states that vectors are said to form basis for V if a) if they span V b) they are linearly independent I know that to prove b) I need to put it in a matrix, reduce and if I get a matrix with trivial solution meaning everything equals to 0 then it's a trivial solution and it's linearly independant. But I'm having trouble with checking if they span or not... that's the whole point of this post :) The examples in my book don't show step by step solution so that's why I'm lost. e.g. Show that S = {v1, v2, v3, v4} where v1 = [1 0 0 1], v2 = [0 1 1 2], v3 = [0 2 2 1], v4 = [1 0 0 1] is a basis for R4. (R4 is written as R subscript 4 meaning it's referring to the row spaces?...) to show that S is linearly independant, I formed the eqn a1v1 + a2v2 + a3v3 + a4v4 = 0 turned it into a augmented matrix and got its RREF. This means that it's linearly independent. NOW, to show that S spans R4, I let x = [a, b, c, d] be a vector in R4 what should I do next? thanks! 


#14
Nov2411, 05:56 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,564

In either case, showing linear independence or spanning, you are, in effect, solving a system of equations. You are either trying to solve [itex]a_1v_1+ a_2v_2+ a_3v_3= 0[/itex] (for linear independence) or [itex]a_1v_1+ a_2v_2+ a_3v_3+ a_4v_4= y[/itex] for any y in the space (for spanning). You could do either by setting up the "augmented" matrix with the right side as the "fifth column". Of course, with all "0"s in the fifth column, no row operations will change those: [itex]a_1= a_2= a_3= a_4= 0[/itex] is an obvious solution. Showing that you can reduce the first four columns to the identity matrix is enough to show that is the only solution. Similarly, to show that the set spans the space you add the components of the vector y as the fifth column. You would rowreduce the matrix in exactly the same way, just doing the row operations on the fifth column on the y components rather than "0"s. But the actual value of the coefficients is not important, only that you can find them. So the values you wind up with in the fifth column is not important, only that you can reduce the first four columns to the identity matrix exactly what you did to show independence. They really are exactly the same thing. When someone said you could look at the determinant of the matrix, earlier, you asked what to do if the matrix is not square. In that case, there is not work to do such a set of vectors cannot be a basis (though it may still span the space or be independent). The number of columns of the matrix, the number of components in each vector, is the dimension of the vector space. The number of rows is the number of vectors in the set. And those must be equal in order to have a basis. If there are fewer vectors than the dimension if there are more columns than row they might be independent but cannot span the space. If there are more vectors than the dimension if there are more rows than columns, they might span the space but cannot be independent. 


#15
Nov2411, 10:17 AM

P: 25

Thank you all for being patient and helping me, I do understand it now! I really appreciate your help guys! :)
and as a side note to anyone who comes across this, yes I did make a mistake in the vectors (late night mistakes...) v1 was actually [1 0 1 0] 


Register to reply 
Related Discussions  
Linear algebra what is a vector space?  Calculus & Beyond Homework  3  
(Linear Algebra) Vector Space  Calculus & Beyond Homework  5  
Linear algebra vector space  Calculus & Beyond Homework  2  
Linear Algebra: Vector space axioms  Calculus & Beyond Homework  4  
Linear Algebra: The vector space R and Rank  Calculus & Beyond Homework  1 