Connection of linear equation to span.by merry Tags: algebra, linear combination, linear independence, span, vectors 

#1
Nov809, 07:22 PM

P: 45

Let E = span{v1, v2} be the linear subspace of R3 spanned by the vectors v1 = (0,1,2) and v2 = (1, 1, 1). Find numbers a, b, c so that
E = {(x, y, z) of R3 : ax + by + cz = 0} I tried doing this question, but I am totally lost. I know that any vector in E can be represented as a linear combination of v1 and v2, but then how do I interpret (x, y, z) in terms of v1 and v2? @.@ If someone could please give me an idea of how I should be going about with the solution, I'd really appreciate it! I dont need a full solution, just an idea of how to do the question. THANKS!!! and please help! T.T 



#2
Nov809, 07:51 PM

P: 267

x, y, and z are just elements of a vector. So in the case of [tex]v_1=(0,1,2)[/tex], x, y, and z equal 0, 1, and 2 respectively.
Furthermore, your next step is to find the span of [tex]\{v_1, v_2\}[/tex]. 



#3
Nov809, 07:52 PM

P: 45





#4
Nov809, 07:55 PM

P: 267

Connection of linear equation to span.
You won't need to worry about (x, y, z) until you find the span of (v_1, v_2).




#5
Nov809, 07:58 PM

P: 45

I know that if I take v1 and v2 and put them in the form of a linear combination: av1 + bv2 = (m, n) where m and n are arbitrary, I can solve for a and b in terms of m and n. Is that what you call solving for the span? O.o Thanks! 



#6
Nov809, 08:07 PM

P: 267

Well we know that E is the span [tex]\{v_1, v_2\}[/tex], and we also know that the span of a set of vectors is the linear combination of those vectors, so we can say that:
[tex]E = a_1v_1 + a_2v_2=a_1(0, 1, 2) + a_2(1, 1, 1) = (0, a_1, 2a_1)+(a_2, a_2, a_2)[/tex] 



#7
Nov809, 08:11 PM

P: 45

Thanks! 



#8
Nov809, 08:46 PM

P: 45

WOAH! THANKS A TON MAN!!! xD I finally know how to do the question!
Thanks again! 



#9
Nov809, 09:12 PM

P: 267

(note: I switched a,b,c, to [tex]b_1,b_2,b_3[/tex].)
E is defined by two things, [tex]E=\{(x, y, z)  b_1x + b_2y + b_3z = 0\}[/tex], as well as, [tex]E=(a_2, a_1 + a_2, a_22a_1)[/tex], so inserting the values of the linear combination into the constraint of E we have: [tex]b_1a_2 + b_2(a_1 + a_2) + b_3(a_22a_1) = 0[/tex] Rewriting it we have: [tex]a_2(b_1+b_2+b_3) +a_1(b_22b_3)= 0[/tex] Analyzing this, we see that in order for it to be true both of the following statements must hold: [tex]b_1 + b_2 + b_3 = 0[/tex] and [tex]b_2  2b_3=0[/tex] Now all you need to do is find a particular set of numbers [tex]\{b_1, b_2, b_3\}[/tex] that satisfy the above two equations. One such example could be [tex]\{b_1, b_2, b_3\}=\{3, 2, 1\}[/tex] 


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