4 Subspaces Recall that a nonempty subset W of a vector spac

4. [Subspaces]
Recall that, a nonempty subset W of a vector space V is a subspace of V if and onlyif the following two conditions are satised for any vectors u, v in W and any scalars c
in R:
(I) If u and v are vectors in W, then the vector u + v is also in W.(II) If u is a vector in W and c is a scalar, then the vector cu is also in W.
Are the following set of vectors W a subspace of V ? Give either a proof (if you thinkit is true) or a counterexample (if you think it is false). Counterexamples withoutexplanation/justication will receive zero credit.
(a) V = R3; W is the set of vectors (, , )T in R3 such that 0.

(b) V = R3; W is the set of vectors (, , ) in R3 such that 0.

c) V = R4; W is the set of vectors (, , ) in R3 such that 100 37 + = 0.

(d) V = R3, W is the set of vectors (, , ) in R3 such that = , + 2 = .

Solution

(a)

False

Consider the vector: (0,2,0). This belongs to W. Now consider scalar multiplication by -1.

We get: (-1)*(0,2,0)=(0,-2,0) which is not in W hence W is not a subspace of V

(b)

False

Consider the vectors in W: a=(-4,-1,0) and b=(1,5,0) ,c=a+b=(-3,4,0)

-3*4<0

Hence c does not belong to W. Hence W is not a subspace of V

(c)

False

Let two vectors in W,a=(a1,a2,a3),b=(b1,b2,b3)

100a1-37a2+pi=0,100b1-37b2+pi=0

Adding the two we get: 100(a1+b1)-37(a2+b2)+2*pi=0

But we need: 100(a1+b1)-37(a2+b2)+pi=0 which means pi=0 which is not possible. Hence W is not a subspace.

(d)

Yes.

Let two vectors in W:a=(a1,a2,a3),b=(b1,b2,b3)

Consider, c=a+b=(a1+b1,a2+b2,a3+b3)

(a1+b1)-(a2+b2)=(a1-a2)+(b1-b2)=-a3-b3=-(a3+b3)

(a1+b1)+2(a2+b2)=a1+2a2+b1+2b2=a3+b3

Let k be a scalar:

Consider: d=ka=(ka1,ka2,ka3)

ka1-ka2=k(a1-a2)=ka3

ka1+2ka2=k(a1+2a2)=ka3

Hence W is a subspace.