subspace

(1) U is not a subspace.



The vector (1,0,-1) belongs to U but 2 x (1,0,-1) = (2,0,-2) is not, therefore U is not a subspace.



(2) V is not a subspace.



The vector (1,1,0) belongs to V but -1 x (1,1,0) = (-1,-1,0) is not, therefore V is not a subspace.



(3) W is a subspace.



Let x = t, y = -t, z = 2t. The subset W is actually a line through origin with the direction (1,-1,2). Therefore it is a subspace.

I agree with the first two answers from the above replier. For number 3, I want to specify below:
Let u=(a,b, c) and v=(A, B, C) be the vectors in W. And let r be the scalar.

Vector addition is closed:
1) (a, b, c ) (A, B, C) = (a A, b B, c C) = (a A) - (B b) (c C)
= (a-b c) (A-B C) = 0 0 = 0.
2) (a A) (b B) = (a b) (A B) = 0 0 = 0

Scalar multiplication is also closed:
1) r*(a, b, c) = (ra, rb, rc)
ra- rb rc = r*(a-b c) = r*0= 0
2) r*a r*b = r*(a b) = r*0 = 0

Thus, W is a subspace

2008-08-12 06:26:53 補充：
I don't know why the "+"sign doesn't show. I will do it one more time:
1) (a, b, c) + (A, B, C) = (a+A, b+B, c+C)
i) (a+A) - (B + b) + (c + C) = (a-b+c) + (A-B+C) = 0 + 0 = 0
ii) (a+A) + (b+B) = 0 + 0 = 0

2008-08-12 06:28:13 補充：
2) r(a, b, c) = (ra,rb, rc)
i) ra-rb+rc = r(a-b+c) = r*0 = 0
ii) r*a + r*b = r*(a+b) = r*0 = 0
So scalar multiplication also holds.