Prove that if U1 and U2 are two subspaces of V then U1 U2 is

Prove that if U1 and U2 are two subspaces of V then U1 U2 is also a subspace.

Prove that the union U1 U2 of two subspaces of V is a subspace if and only if one of the subspaces contains the other.

Solution

Let V be a vector space and U1 and U2 two subspaces of V . We show that U1 U2 is a subspace by checking the three conditions given in the text: (a) (additive identity) Since 0 U1 and 0 U2 (as they’re both subspaces), 0 U1 U2. This part’s fine (b) (closure under addition) If v + w is in U1 U2, then v + w must be in either U1 or U2 (or both). Say it’s in U1. Then since U1 is a subspace, v and w must both be in U1. But then v and w are both in U1 U2 also, so U1 U2 is closed under addition. Similarly, if v + w is in U2, then both v and w are, so they’re both in U1 U2. The whole thing is set up wrong! We should begin with two vectors in the union, and then add them. But here we started with the sum. There is another error, too. It’s not true that if v +w is in some subspace, then both v and w must be. (c) (closure under scaling) Pick any vector v U1 U2 and any scalar c. Then v U1 or v U2. If v U1 then so is cv since U1 is a subspace. So cv U1 U2. On the other hand, if v U2 instead, then cv U2 since U2 is a subspace, so again cv U1 U2