Are These Subsets Subspaces of R^2?

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The discussion focuses on determining whether specific subsets of R^2 are subspaces by testing for closure under vector addition and scalar multiplication. The first subset, W={ (x1,0) | x1 ∈ R }, is confirmed as a subspace since it meets both criteria. The second subset, W={ (x1,0) | x1 > 0 }, is not a subspace because it does not include the zero vector, violating the closure under scalar multiplication. The third subset, W={ (2c,-3c) | c ∈ R }, is shown to be a subspace with proper demonstration of closure under addition and scalar multiplication. The fourth subset, W={ (x1,x2) | x1 > 0, x2 > 0 }, is also not a subspace for the same reason as the second subset, lacking the zero vector.
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Determine if following subsets of R^2 are subspaces of R^2. If the subset is a subspace show that it is closed under vector addition and scalar multiplication. If the subset is not a subspace show why, indicating property that fails.

1) W=\{ \left (x_1,0)\left| x_1\in\Re\} \newline
2) W=\{ \left (x_1,0)\left| x_1 > 0\} \newline
3) W=\{ \left (2c,-3c)\left| c \in\Re\} \newline
4) W=\{ \left (x_1,x_2)\left| x_1 > 0, x_2>0\} \newline

Answers:
1) (x,0) + (y,0)= (x+y,0) \in W \} \newline
c(x,0) = (c(x),0) \in W \} \newline

2) Not subspace since x1 can't be 0

3) (2c,-3c) + (x_1,x_2)= (2c+x_1,-3c+x_2) \in W \} \newline
x(2c,-3c) = (2c(x),-3c(x)) \in W \} \newline

4) Not a subspace since x1 and x2 can't be 0.

Am I on the right track with this? Thanks
 
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needhelp83 said:
Determine if following subsets of R^2 are subspaces of R^2. If the subset is a subspace show that it is closed under vector addition and scalar multiplication. If the subset is not a subspace show why, indicating property that fails.

1) W=\{ \left (x_1,0)\left| x_1\in\Re\} \newline
2) W=\{ \left (x_1,0)\left| x_1 > 0\} \newline
3) W=\{ \left (2c,-3c)\left| c \in\Re\} \newline
4) W=\{ \left (x_1,x_2)\left| x_1 > 0, x_2>0\} \newline

Answers:
1) (x,0) + (y,0)= (x+y,0) \in W \} \newline
c(x,0) = (c(x),0) \in W \} \newline
Yes, you have shown that the set is closed under addition and scalar multiplication.

2) Not subspace since x1 can't be 0
And what property of a subspace does that violate? It would be sufficient to point out that the 0 vector (0, 0) is not in that set. Since your problem specifically referred to "closed under addition and scalaar multiplication, I think it would be better to show that 0(x_1, 0)= (0, 0) is not in the set so it is not closed under scalar multiplication.

3) (2c,-3c) + (x_1,x_2)= (2c+x_1,-3c+x_2) \in W \} \newline
x(2c,-3c) = (2c(x),-3c(x)) \in W \} \newline
This is the only one where I think you have lost track of what you are doing- what are x_1 and x_2? To show "closed under addition", you need to look at (2c, -3c)+ (2d, -3d) where c and d are real numbers.

4) Not a subspace since x1 and x2 can't be 0.
Again, since closure under addition and scalar multiplication are specifically mentioned in the problem it would be better to show that 0(x1, x2)= (0, 0) is not in the set so it is not closed under scalar multiplication.

Am I on the right track with this? Thanks
 
Okay for the scalar multiplication:


3)Since c_2 \ in \ R^2
c_2(2c_1,-3c_1) = (2c_1c_2,-3c_1c_2) \in W \} \newline

Is this better?
 
needhelp83 said:
Okay for the scalar multiplication:


3)Since c_2 \ in \ R^2
c_2(2c_1,-3c_1) = (2c_1c_2,-3c_1c_2) \in W \} \newline

Is this better?
Yes, now what about addition?
 

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