Question

Determine whether or not W is a subspace of R³ where W consists of all vectors (a,b,c) in R³ such that

1. a =3b

2. a <= b <= c

3. ab=0

4. a+b+c=0

5. b=a²

6. a=2b=3c

Answer 1

Let X = (a,b,c) and Y = (d,e,f) be 2 arbitrary vectors in W and let k be an arbitrary scalar.

1. We have a = 3b and d = 3e so that X+Y = (3b,b,c)+(3e,e,f) = (3(b+e),(b+e),c+f). This implies that X+Y ∈ W so that W is closed under vector addition. Also, kX = k (3b,b,c) = (3kb,kb,kc). This implies that kX ∈ W so that W is closed under scalar multiplication.

Further, since 3*0 = 0, hence the zero vector(0,0,0) ∈ W. Therefore, W is a vector space and, hence, a subspace of R³ .

2. Since, kX = (ka,kb,kc) and Since a ≤ b ≤ c , hence, when k is a negative scalar, we have ka≥ kb ≥ kc. This implies that kX ∉ W so that W is not closed under scalar multiplication. Therefore, W is not a vector space and, hence, not a subspace of R³ .

3. We have ab= 0 and de = 0 . Now, X+Y = (a,b,c)+(d,e,f) = = (a+d,b+e,c+f). Further, (a+d)(b+e) = (ab+ de) + (ae+bd) =0+0+ ae+bd = ae+bd. Since ae+bd need not be 0, hence X+Y ∉

W so that W is not closed under vector addition. Therefore, W is not a vector space and, hence, not a subspace of R³ .

4. We have a+b+c= 0 and d+e+f = 0 . Now, X+Y = (a,b,c)+(d,e,f) = = (a+d,b+e,c+f). Further, (a+d)+(b+e)+(c+f) = (a+b+c)+(d+e+f) = 0+0 = 0 . This implies that X+Y ∈ W so that W is closed under vector addition. Also, kX = k(a,b,c) = (ka,kb,kc) and since ka+kb+kc = k(a+b+c)= k.0 = 0, hence kX ∈ W so that W is closed under scalar multiplication.

Further, since 0+0+0 = 0, hence the zero vector(0,0,0) ∈ W. Therefore, W is a vector space and, hence, a subspace of R³ .

5. We have b = a² and e = d². Now, X+Y = (a,b,c)+(d,e,f) = = (a+d,b+e,c+f). Further, b+e = a² + d² while (a+d)² = a² + d² +2ad ≠ a² + d² . Thus, b+e ≠(a+d)². Hence X+Y ∉

W so that W is not closed under vector addition. Therefore, W is not a vector space and, hence, not a subspace of R³ .

6. We have a = 2b = 3c so that c = (2b/3) and d = 2e = 3f so that f = (2e/3). Now, X+Y = (2b,b,2b/3)+(2e,e,2e/3) = (2(b+e), b+e, (2/3)(b+e)) . This This implies that X+Y ∈ W so that W is closed under vector addition. Also, kX = k(2b,b,2b/3) = (2kb,kb,2kb/3). This implies that kX ∈ W so that W is closed under scalar multiplication.

Further, since 2*0 = 0 and 3*0 = 0, hence the zero vector(0,0,0) ∈ W. Therefore, W is a vector space and, hence, a subspace of R³ .