Figuring out if subset of R2 is a subspace

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Homework Help Overview

The discussion revolves around determining whether specific subsets of R² qualify as subspaces. The original poster presents a subset defined by E={(x,y)|x≥0,y∈ℝ} and questions its status as a subspace. Other participants explore the necessary conditions for subspaces and provide examples related to closure under scalar multiplication and addition.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the conditions required for a subset to be a subspace, including closure properties. There are attempts to apply these conditions to the given subsets, with some questioning the implications of specific elements within the subsets.

Discussion Status

The conversation includes various perspectives on the subsets in question, with some participants providing insights into closure under scalar multiplication and addition. There is acknowledgment of the need to demonstrate specific cases to support claims about the subsets' properties.

Contextual Notes

Some participants express uncertainty about the definitions and properties of subspaces, while others reference specific examples that illustrate the concepts being discussed. There is a mention of the implications of including the zero vector in the subsets.

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Homework Statement


Let E be the subset of R2 defined by E={(x,y)|x≥0,y[itex]\in[/itex]ℝ}. Is E a subspace of R2?


Homework Equations


E={(x,y)|x≥0,y[itex]\in[/itex]ℝ}


The Attempt at a Solution


I honestly have no idea where to start. Please help ! I'm not asking for the answer per se, just a starting point
 
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What are the conditions for a space to be a subspace of some other space? Can you apply these to find a counterexample, perhaps?
 
ahh.. So E is not closed under scalar multiplication since -k(x,y)=(-kx, -ky) and this is not an element of E
 
csc2iffy said:
ahh.. So E is not closed under scalar multiplication since -k(x,y)=(-kx, -ky) and this is not an element of E

Exactly. Note only that you have to take an element (x, y) of E where x > 0 to demonstrate that it works, since if you take an element of the form (0, y) in E, then for some k < 0 you have k(0, y) = (0, ky), which is still in E.
 
Thanks! I was also struggling a bit with this problem:
Let E={(x,2x+1|x∈ℝ}, so E is a subset of R2. Is E a subspace of R2?

I said no because it is not closed under addition:
(x,2x+1)+(y,2y+1)=(x+y,2x+1+2y+1)=(x+y,2(x+y)+2) which isn't equal to (x+y,2(x+y)+1)
Is this correct?
 
Yes. Also, it's even easier to see that (0, 0) is not in E in this case.
 
Oh that is much easier! Thank you so much!
 
Sorry, no one is replying to my other thread so I figured I would ask you one more thing here.

Let E={(2a,a)|a∈ℝ}. Let B={(b,b)|b∈ℝ}.
Is E∪B a subspace of R2?
What is E+B

My solution:
E∪B={(2a,a),(b,b)|a,b∈ℝ}
I don't know how to show if tis is a subspace of R2 or if that is the correct union

For the second part, I know that E+B = span (E∪B)
but I don't know if I have the right union?
 
A union of subspaces of a given space need not be a subspace of that space. For example, take a non-zero a in R, and let (2a, a) and (a, a) be elements of E U B. Then (2a, a) + (a, a) = (3a, 2a) is neither in E nor in B.

The sum of two subspaces is again a subspace of that space.
 
  • #10
So, is E+B= span(EUB)=
a(2,1)+b(1,1)=(2a,a)+(b,b)=(2a+b, a+b)?
 
  • #11
You have two subspaces of a space V, E and B. The span of E U B is defined to be the set of all linear combinations of elements from E U B. Let a and b be scalars, and let v1 and v2 be elements of E U B. Then av1 + bv2 is an element of E U B, and hence E U B is a subspace of V.

Correction: let v1 and v2 be elements of span(E U B) (note that v1 and v2 are linear combinations).
 
Last edited:

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