The following proof was confusing me so any help would be appreciated!
Let V be a vector space and let W_1 and W_2 be subspaces of V. Prove that W_1 (insert upside down U symbol here) W_2 is a subspace of V. Remember to show that W_1 (insert upside down U symbol here) W_2 is nonempty.)
Apr 8th 2007, 07:18 PM
Okay...so I think I came up with a solution, but I'm not sure if it's right. Here it is:
Let a and b be vectors in the intersection. This means that a is in W_1 and a is in W_2, and b is in W_1 and b is in W_2. So this means that a + b has to be in W_1, as W_1 under the Closure Property of Addition. The same goes for W_2. Therefore, a + b is in W1 and a + b is in W_2. So a + b must be in the intersection.
Then for scalar multiplication...
Let a be some vector in the intersection. This means that a is in W_1 and a is in W_2. Therefore, a times a scalar x will be in W_1 as well as W_2 by the closure property of scalar multiplication. So a*x must be in the intersection.