Are These Subsets Subspaces of R^2?

[needhelp83]

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

 

[HallsofIvy]

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

 

[needhelp83]

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?

 

[HallsofIvy]

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?