# Connection of linear equation to span.

by merry
Tags: algebra, linear combination, linear independence, span, vectors
 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
 P: 267 x, y, and z are just elements of a vector. So in the case of $$v_1=(0,1,-2)$$, x, y, and z equal 0, 1, and -2 respectively. Furthermore, your next step is to find the span of $$\{v_1, v_2\}$$.
P: 45
 Quote by epkid08 x, y, and z are just elements of a vector. So in the case of $$v_1=(0,1,-2)$$, x, y, and z equal 0, 1, and -2 respectively.
So do I take (x, y, z) as the respective components of v1 and v2 and then get 2 equations and find a, b and c? O.o

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).
P: 45
 Quote by epkid08 You won't need to worry about (x, y, z) until you find the span of (v_1, v_2).
But I dont know how to find the span of v1 and v2 O.o
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!
 P: 267 Well we know that E is the span $$\{v_1, v_2\}$$, and we also know that the span of a set of vectors is the linear combination of those vectors, so we can say that: $$E = a_1v_1 + a_2v_2=a_1(0, 1, -2) + a_2(1, 1, 1) = (0, a_1, -2a_1)+(a_2, a_2, a_2)$$
P: 45
 Quote by epkid08 Well we know that E is the span $$\{v_1, v_2\}$$, and we also know that the span of a set of vectors in the linear combination of those vectors, so we can say that: $$E = a_1v_1 + a_2v_2=a_1(0, 1, -2) + a_2(1, 1, 1) = (0, a_1, -2a_1)+(a_2, a_2, a_2)$$
Right, I did try using that but ended up with 5 variables. Since we dont know what the values of a1 and a2 are, how can we use the idea to compute a, b and c?

Thanks!
 P: 45 WOAH! THANKS A TON MAN!!! xD I finally know how to do the question! Thanks again!
 P: 267 (note: I switched a,b,c, to $$b_1,b_2,b_3$$.) E is defined by two things, $$E=\{(x, y, z) | b_1x + b_2y + b_3z = 0\}$$, as well as, $$E=(a_2, a_1 + a_2, a_2-2a_1)$$, so inserting the values of the linear combination into the constraint of E we have: $$b_1a_2 + b_2(a_1 + a_2) + b_3(a_2-2a_1) = 0$$ Rewriting it we have: $$a_2(b_1+b_2+b_3) +a_1(b_2-2b_3)= 0$$ Analyzing this, we see that in order for it to be true both of the following statements must hold: $$b_1 + b_2 + b_3 = 0$$ and $$b_2 - 2b_3=0$$ Now all you need to do is find a particular set of numbers $$\{b_1, b_2, b_3\}$$ that satisfy the above two equations. One such example could be $$\{b_1, b_2, b_3\}=\{-3, 2, 1\}$$

 Related Discussions Calculus & Beyond Homework 3 Calculus & Beyond Homework 1 Linear & Abstract Algebra 9 Calculus & Beyond Homework 4 Calculus & Beyond Homework 1