# Another vector problem

1. Nov 28, 2007

### 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.

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

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

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

2. Nov 28, 2007

### HallsofIvy

Staff Emeritus
Yes, you have shown that the set is closed under addition and scalar multiplication.

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.

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.

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.

3. Nov 29, 2007

### 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?

4. Nov 29, 2007

### HallsofIvy

Staff Emeritus