# Basis for the indicated subspace

by 164694605
Tags: basis, span subspace
 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.
 HW Helper 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?
P: 4
 Quote by radou 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?

 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?
P: 4
 Quote by VeeEight 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.
 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!
 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.
Math
Emeritus