Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

perhaps this question was asked before, but if not give me a hand for it. My question is how to prove that if $W_1$ and $W_2$ be sub-spaces of vector space $V$ with the union of $W_1$ and $W_2$ is also a subspace, then one of the subspace $W_i$ is contained in other ?

Please give me a hint for start. Where should I look here?

share|cite|improve this question

marked as duplicate by Gerry Myerson, Babak S., Zev Chonoles Feb 21 '13 at 12:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Duplicate of – Zev Chonoles Feb 21 '13 at 11:46
@BasilR More specifically, see the answer of Gerry Myerson to the above-linked question. – user1551 Feb 21 '13 at 12:13
Thaak you. ok i will... – Basil R Feb 21 '13 at 14:31
up vote 2 down vote accepted

Suppose that $W_1$ has a basis $v_1, \ldots , v_n$, and that $W_2$ has a basis $u_1, \ldots, u_m$.

If neither space is a subspace of the other then there is a $u_i$ not in $W_1$, and a $v_j$ not in $W_2$. But since the union is a subspace that contains $u_i$ and $v_j$ then it must contain $v_j + u_i$. But this implies the sum is in either $W_1$ or $W_2$ and hence that either $u_i \in W_1$ or $v_j \in W_2$, a contradiction.

share|cite|improve this answer

Hint: What would be happen if for example $W_1$ is not contained in $W_2$ and $W_2$ is not contained in $W_1$? Think of the point that with this assumption the set $W_1\cup W_2$ would not be a subspace of whole space $V$. This is a nice contradicton!

share|cite|improve this answer
Sharp observation! – amWhy Feb 21 '13 at 13:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.