| written 7.8 years ago by | • modified 3.9 years ago |
(i) W = (a , 1 , 1) / a ϵ R
(ii) W = (a , b , c) / b = a + c + 1 , a , b ,c ϵ R
Subject: Applied Mathematics 4
Topic: Vector Spaces
Difficulty: Medium
| written 7.8 years ago by | • modified 3.9 years ago |
(i) W = (a , 1 , 1) / a ϵ R
(ii) W = (a , b , c) / b = a + c + 1 , a , b ,c ϵ R
Subject: Applied Mathematics 4
Topic: Vector Spaces
Difficulty: Medium
| written 7.5 years ago by |
Let u = (a,1,1) and v = (b,1,1) where a,b $ \epsilon $ IR
u + v = (a+b,2,2) $ \notin $ W
Therefore, W is not closed under addition.
Therefore, W is not a subspace.
(ii) W = {(a,b,c) | b = a+c+1, a,b,c $ \epsilon $ IR}
Let,
$ u = (a_1,b_1,c_1) \hspace{0.20cm} \epsilon \hspace{0.20cm} W \\ v = (a_2,b_2,c_2) \hspace{0.20cm} \epsilon \hspace{0.20cm} W $
$ b_1 = a_1 + c_1 + 1 \\ b_2 = a_2 + c_2 + 1 $
$ u + v = (a_1 + a_2,b_1 + b_2, c_1 + c_2) $
Consider,
$ b_1 + b_2 = a_1 + c_1 + 1 + a_2 + c_2 + 1 \\ = (a_1 + a_2) + (c_1 + c_2) + 2 \\ \neq (a_1 + a_2) + (c_1 + c_2) + 1 $
u,v $ \epsilon $ W but u + v $\notin$ W
Therefore, W is not in subspace of IR$^3$