Register to reply

Basis for the indicated subspace

by 164694605
Tags: basis, span subspace
Share this thread:
164694605
#1
Apr24-10, 04:43 AM
P: 4
hi guys, i have no idea of how to do the following question, could u give some ideas?

Q:determine whether or not the given set forms a basis for the indicated subspace
{(1,-1,0),(0,1,-1)}for the subspace of R^3 consisting of all (x,y,z) such that x+y+z=0

how should i start?
i know the vectors are linearly independent, and then i think i need to show they span. but doesn't it require 3 vectors to prove span in R^3?

or maybe the (x,y,z) such that x+y+z=0 can be used somewhere? I know this will form a plane, but how should i say this?

Thank you.
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
radou
#2
Apr24-10, 07:27 AM
HW Helper
radou's Avatar
P: 3,220
You are dealing with a subspace of R^3 here. Its dimension is not 3, although this doesn't mean that 3 or more vectors can't span it, of course. What does (-y-z, y, z) = y(-1, 1, 0) + z(-1, 0, 1) tell you about the subspace?
164694605
#3
Apr24-10, 11:06 PM
P: 4
Quote Quote by radou View Post
You are dealing with a subspace of R^3 here. Its dimension is not 3, although this doesn't mean that 3 or more vectors can't span it, of course. What does (-y-z, y, z) = y(-1, 1, 0) + z(-1, 0, 1) tell you about the subspace?
thanks for reply. I get the last part, but i have trouble to interpret it.

this is what i have done so far:
because x+y+z=0 so y=-(x+z)
because the standard basis for R^3 is (1,0,0) (0,1,0) (0,0,1)
so (x,y,z)=(x,-(x+z),z)=(1,0,0)x-(0,1,0)(x+z)+(0,0,1)z=(1,-1,0)x+(0,-1,1)z
but the numbers don't match, the question given is (1,-1,0) (0,1,-1), and I get (1,-1,0) (0,-1,1).

also in order to show a basis, we need to show linearly independent and span.
In this case, how should i prove the set of vectors span the subspace?
I tried this way, c1(1,-1,0)+c2(0,1,-1)=(u1,u2,u3)
and i get u1=c1 u2=c2-c1 u3=-c2 can this prove span?

Find really confused on this question, please help

VeeEight
#4
Apr24-10, 11:59 PM
P: 610
Basis for the indicated subspace

To show it spans, consider (-y-z, y, z) = y(-1, 1, 0) + z(-1, 0, 1), where y and z are in R. Thus, B = {(-1, 1, 0), (-1, 0, 1)} generates the subspace.

You have your original set A = {(1,-1,0), (0,1,-1)}. How does A relate to B? Can you write elements of B as a linear combination of elements from A?
164694605
#5
Apr25-10, 12:29 AM
P: 4
Quote Quote by VeeEight View Post
To show it spans, consider (-y-z, y, z) = y(-1, 1, 0) + z(-1, 0, 1), where y and z are in R. Thus, B = {(-1, 1, 0), (-1, 0, 1)} generates the subspace.

You have your original set A = {(1,-1,0), (0,1,-1)}. How does A relate to B? Can you write elements of B as a linear combination of elements from A?
Exactly, as my previous post, I consider (x,-(x+z),z)=(1,-1,0)x+(0,-1,1)z which is more obvious that elements of B can be write as a linear combination of elements from A, but the question is how should I link these things into a proof?
I can't just say because one is the linear combination of another, thus one set spans another, can i.
164694605
#6
Apr25-10, 12:32 AM
P: 4
Here is my thinking, please help me to finish it, thank you!
If these two vectors (1,-1,0) and (0,1,-1)will span R^3 then for each u=(u1,u2,u3) in R^3 there must be scalars c1 and c2, so that
c1(1,-1,0)+c2(0,1,-1)=(u1,u2,u3)
then i get u1=c1 u2=c2-c1 u3=-c2, which means there will be exactly one solution to the system for each u, so span is proved?
Then should I related it to the x+y+z=0, to show it satisfy for the given subspace, or what?

Again, thank you for help me out!
VeeEight
#7
Apr25-10, 01:07 AM
P: 610
No, your proof does not follow any order. You are asked to check that A is a basis

See what to do in post #4. If B spans X and elements in B are a linear combination of elements of A, then A also spans X.
HallsofIvy
#8
Apr25-10, 10:59 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,552
Quote Quote by 164694605 View Post
Exactly, as my previous post, I consider (x,-(x+z),z)=(1,-1,0)x+(0,-1,1)z which is more obvious that elements of B can be write as a linear combination of elements from A, but the question is how should I link these things into a proof?
I can't just say because one is the linear combination of another, thus one set spans another, can i.
Well, that's pretty much what "spans" means, isn't it?

Because any vector in this subspace can be written as x(1, -1, 0)+ z(0, -1, 1) {(1, -1, 0), (0, -1, 1) spans the given subspace. To prove it is a basis, all you need to do is show that they are independent. Suppose a(1, -1, 0)+ b(0, -1, 1)= (0, 0, 0). What must a and b equal?


Register to reply

Related Discussions
Basis of a Subspace Linear & Abstract Algebra 4
Basis and subspace help Calculus & Beyond Homework 6
Basis for a subspace in R^4 Calculus & Beyond Homework 1
Basis of a subspace? Calculus & Beyond Homework 4
Basis for a Subspace. Precalculus Mathematics Homework 6