Connection of linear equation to span.


by merry
Tags: algebra, linear combination, linear independence, span, vectors
merry
merry is offline
#1
Nov8-09, 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
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epkid08
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#2
Nov8-09, 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].
merry
merry is offline
#3
Nov8-09, 07:52 PM
P: 45
Quote Quote by epkid08 View Post
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.
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

epkid08
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#4
Nov8-09, 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).
merry
merry is offline
#5
Nov8-09, 07:58 PM
P: 45
Quote Quote by epkid08 View Post
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!
epkid08
epkid08 is offline
#6
Nov8-09, 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]
merry
merry is offline
#7
Nov8-09, 08:11 PM
P: 45
Quote Quote by epkid08 View Post
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 in 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]
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!
merry
merry is offline
#8
Nov8-09, 08:46 PM
P: 45
WOAH! THANKS A TON MAN!!! xD I finally know how to do the question!
Thanks again!
epkid08
epkid08 is offline
#9
Nov8-09, 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_2-2a_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_2-2a_1) = 0[/tex]

Rewriting it we have:

[tex]a_2(b_1+b_2+b_3) +a_1(b_2-2b_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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