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Basis for the indicated subspace 
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#1
Apr2410, 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. 


#2
Apr2410, 07:27 AM

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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 (yz, y, z) = y(1, 1, 0) + z(1, 0, 1) tell you about the subspace?



#3
Apr2410, 11:06 PM

P: 4

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=c2c1 u3=c2 can this prove span? Find really confused on this question, please help 


#4
Apr2410, 11:59 PM

P: 610

Basis for the indicated subspace
To show it spans, consider (yz, 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? 


#5
Apr2510, 12:29 AM

P: 4

I can't just say because one is the linear combination of another, thus one set spans another, can i. 


#6
Apr2510, 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=c2c1 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! 


#7
Apr2510, 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. 


#8
Apr2510, 10:59 AM

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PF Gold
P: 39,552

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? 


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