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## Homework Statement

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

Is W a subspace of R

^{3}?

## Homework Equations

## 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 [tex]\in[/tex] 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 [tex]\in[/tex] W, so scalar multiplication is closed.

So W is a subspace of R

^{3}.

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

Regards,

David Taylor.