# Have I proved W is a subspace of R3?

## Homework Statement

Let W = { (x, y, z) | x - 2y + z = 0 }

Is W a subspace of R3?

## The Attempt at a Solution

Using another post here I tried the following to show W is closed under addition:

1. Let u = (x, y, z) and v = (i, j , k). u and v are both in W.
2. So, x - 2y + z = 0, and i - 2j + k = 0.
3. Then u + v = (x+i, y+j, z+k).
4. Does (x+i) - 2(y+j) + (z+k) = 0?

Now we're in the land of real numbers with no vectors, I can rearrange all the terms as I wish, so I'll try to reconstruct the original equations in (2).

(x+i) - 2(y+j) + (z+k) = 0
x + i -2y -2j + z + k = 0
(x -2y +z) +( i - 2j + k) = 0

From (2) we know both those groups = 0, so adding them together still gives 0, so u+v is closed under addition. I'm pretty sure we didn't do it this way in class, but it does seem to satisfy the condition of the set W.

Then to show it is closed under scalar multiplication I have:

1. Let u = (x, y, z) and k be a real scalar constant.
2. Is ku $$\in$$ W?
3. ku = (kx, ky, kz)
4. Does kx -2(ky) + kz = 0?

Dividing both sides by k gives x -2y + z = 0/k = 0. So ku $$\in$$ W, so scalar multiplication is closed.

So W is a subspace of R3.

Is anyone convinced by this proof? What have I missed?

Regards,
David Taylor.

HallsofIvy
Homework Helper

## Homework Statement

Let W = { (x, y, z) | x - 2y + z = 0 }

Is W a subspace of R3?

## The Attempt at a Solution

Using another post here I tried the following to show W is closed under addition:

1. Let u = (x, y, z) and v = (i, j , k). u and v are both in W.
2. So, x - 2y + z = 0, and i - 2j + k = 0.
3. Then u + v = (x+i, y+j, z+k).
4. Does (x+i) - 2(y+j) + (z+k) = 0?

Now we're in the land of real numbers with no vectors, I can rearrange all the terms as I wish, so I'll try to reconstruct the original equations in (2).

(x+i) - 2(y+j) + (z+k) = 0
x + i -2y -2j + z + k = 0
(x -2y +z) +( i - 2j + k) = 0

From (2) we know both those groups = 0, so adding them together still gives 0, so u+v is closed under addition. I'm pretty sure we didn't do it this way in class, but it does seem to satisfy the condition of the set W.
Yes that is perfectly correct. I'd be surprised if that was NOT how you did it in class.

Then to show it is closed under scalar multiplication I have:

1. Let u = (x, y, z) and k be a real scalar constant.
2. Is ku $$\in$$ W?
3. ku = (kx, ky, kz)
4. Does kx -2(ky) + kz = 0?

Dividing both sides by k gives x -2y + z = 0/k = 0. So ku $$\in$$ W, so scalar multiplication is closed.
there is no reason to divide by k- you don't want to show that x- 2y+ z= 0, you already know that. Instead, kx- 2ky+ kz= k(x- 2y+ z)= k(0)= 0.

So W is a subspace of R3.

Is anyone convinced by this proof? What have I missed?

Regards,
David Taylor.

You also may or may not have to show that W contains the zero vector. If you didn't have to in class, then don't worry about it.

... there is no reason to divide by k- you don't want to show that x- 2y+ z= 0, you already know that. Instead, kx- 2ky+ kz= k(x- 2y+ z)= k(0)= 0.

OK, thanks. I see your point that I just restated what is known. Is my answer valid and provides proof by showing you can ignore k, or doesn't it prove anything?

Regards,
David.

You also may or may not have to show that W contains the zero vector. If you didn't have to in class, then don't worry about it.

Thanks. We only have to show closure under + and . We did cover that, but for some reason the lecturer isn't asking for it in the proof.

To show the zero vector is part of W, I could just show something like:

u = (0, 0, 0). 0 - 2(0) + 0 = 0, so u is an element of W. Is that right?

That would also mean any W where ax + by + cz != 0 can't have the zero vector in it, so isn't a subset or R3?

Regards,
David.

HallsofIvy
Essentially, yes. Every subspace has to satisfy some "homogenous" equation. You can think of "lines through the origin" and "planes through the origin" as one and two dimensional subspaces of $$\displaystyle R^3$$. (x, y, z) such that ax+ by+ cz= a non-zero number would be a plane that does NOT contain the origin. Those are "linear manifolds" but not subspaces. Of course, that would also fail to be closed under addition or scalar multiplication. If ax+ by+ cz= p where p is not 0, then (x, y, z)+ (x', y', z') = (x+ x', y+ y', z+ z') where ax+ by+ cz= p and ax'+ by'+ cz'= p satifies a(x+ x')+ b(y+ y')+ c(z+ z')= (ax+ by+ cz)+ (ax'+ by'+ cz')= p+ p= 2p which is NOT p. Similarly, k(x, y, z)= (kx, ky, kz) satifies a(kx)+ b(ky)+ c(kz)= k(ax+ by+ cz)= kp, not p.