To determine whether a subset is a subspace, we need to check two conditions.
1) The set is nonempty (it's usually easiest to see if the zero vector is in the subset).
2) The set is closed under addition and scalar multiplication.
1) `(0,0)in U,` so `U` is nonempty.
2) `(0,y_1)+(0,y_2)=(0,y_1+y_2),` and since `y_1, y_2<=0,` `y_1+y_2<=0` and `U` is closed under addition.
`c(0,y)=(0,cy)` for all `cinRR` , but if `c` is negative, `cy` may be positive. For example, `-1(0,-1)=(0,1),` which is not an element of `U` because its second entry is not less than or equal to zero. Therefore `U` is not a subset of `RR^2` .
1) The third entry in `(0,0,0,0)` is the sum of the second and fourth entries, so `W` is nonempty.
2) Take two elements `(a,b,b+d,d)` and `(s,t,t+v,v)` in `W.` Their sum is `(a+s,b+t,b+t+d+v,d+v)` and since the third entry is the sum of the second and fourth entries, the sum is in `W`.
For scalar multiplication, we get `c(a,b,b+d,d)=(ca,cb,cb+cd,cd),` and since again the third entry is the sum of the second and fourth entries, the product is in `W`. Therefore `W` is nonempty and closed under the two vector space operations and is a subspace of `RR^4.`
`U` is not a subspace of `RR^2`, and `W` is a subspace of `RR^4.`