For the \"if\" direction, V is contained in W, then V U W = W, so it\'s obviousl
ID: 2987645 • Letter: F
Question
For the "if" direction, V is contained in W, then V U W = W, so it's obviously is a subspace i get that part.
The "only if" direction is a little harder:
If V is contained in W, then we're done, so let's assume that it's not... that is, there's an element v in V such that v isn't in W.
For any element w in W, we get:
v+w is in V U W
so v+w is in one of V or W.
Case 1: v+w is in W.
then since w is in W, (v+w) - w = v is in W (contradiction)
Case 2: v+w is in V
then since v is in V, (v+w) - v = w is in V.
Can anyone show a picture for the second part with vectors and subspaces and v+w. i would really appreciate it
a picture would help me understand the concept better thanks.
Explanation / Answer
Call your subspaces V and W. We want to see when V U W is a vector space.
For the "if" direction, note that if V is contained in W, then V U W = W, so it's obviously a subspace.
The "only if" direction is a little harder:
If V is contained in W, then we're done, so let's assume that it's not... that is, there's an element v in V such that v isn't in W.
For any element w in W, we get:
v+w is in V U W
so v+w is in one of V or W.
Case 1: v+w is in W.
then since w is in W, (v+w) - w = v is in W (contradiction)
Case 2: v+w is in V
then since v is in V, (v+w) - v = w is in V.
Since this was for any arbitrary w in W, this shows that W is contained in V, and we're done.