Figuring out if subset of R2 is a subspace

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In summary: Then av1 + bv2 is an element of span(E U B), and hence span(E U B) is a subspace of V.In summary, we are trying to determine if the union of two subspaces, E and B, is a subspace of the space V. We do this by taking the span of E U B, which is the set of all linear combinations of elements from E U B, and see if it satisfies the conditions for a subspace. In this case, E U B is not a subspace of V because there exists a linear combination that is not in either E or B.
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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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  • #2
What are the conditions for a space to be a subspace of some other space? Can you apply these to find a counterexample, perhaps?
 
  • #3
ahh.. So E is not closed under scalar multiplication since -k(x,y)=(-kx, -ky) and this is not an element of E
 
  • #4
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.
 
  • #5
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?
 
  • #6
Yes. Also, it's even easier to see that (0, 0) is not in E in this case.
 
  • #7
Oh that is much easier! Thank you so much!
 
  • #8
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?
 
  • #9
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).
 
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FAQ: Figuring out if subset of R2 is a subspace

What is a subspace in R2?

A subspace in R2 is a subset of the two-dimensional coordinate plane that satisfies three conditions: it contains the origin (0,0), it is closed under vector addition, and it is closed under scalar multiplication. Essentially, it is a set of vectors that can be added and multiplied by a scalar without leaving the two-dimensional plane.

How do I determine if a subset of R2 is a subspace?

To determine if a subset of R2 is a subspace, you must check if it satisfies the three conditions mentioned above. If it contains the origin, and any two vectors in the subset can be added and multiplied by a scalar to produce a vector that is also in the subset, then it is a subspace. If any of these conditions are not met, then it is not a subspace.

Can a subspace in R2 be a line?

Yes, a subspace in R2 can be a line as long as it satisfies the three conditions for a subspace. For example, the x-axis or y-axis can be considered as subspaces in R2 because they contain the origin and are closed under vector addition and scalar multiplication.

Is the empty set considered a subspace in R2?

No, the empty set is not considered a subspace in R2. This is because it does not contain the origin, and therefore cannot satisfy the first condition for a subspace. Additionally, there are no vectors in the empty set, making it impossible to satisfy the other two conditions.

Can a subspace in R2 be a circle?

No, a subspace in R2 cannot be a circle. This is because a circle does not contain the origin, and therefore cannot satisfy the first condition for a subspace. Additionally, a circle is not closed under vector addition and scalar multiplication, as adding or multiplying by a scalar would result in a point outside of the circle.

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