# Union of two vector subspaces not a subspace?

I'm having a difficult time understanding this statement. Can someone please explain with a concrete example?

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(Let's try again, hopefully this time without typos) A nice proof of a more general result (for fields of characteristic 0): mathoverflow.net/questions/43538/… –  Andres Caicedo May 8 '13 at 19:24

The reason why this can happen is that all vector spaces, and hence subspaces too, must be closed under addition (and scalar multiplication). The union of two subspaces takes all the elements already in those spaces, and nothing more. In the union of subspaces $W_1$ and $W_2$, there are new combinations of vectors we can add together that we couldn't before, like $v_1 + w_2$ where $v_1 \in W_1$ and $w_2 \in W_2$.

For example, take $W_1$ to be the $x$-axis and $W_2$ the $y$-axis, both subspaces of $\mathbb{R}^2$.
Their union includes both $(3,0)$ and $(0,5)$, whose sum, $(3,5)$, is not in the union. Hence, the union is not a vector space.

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thanks, I wish you were my math teacher –  NSjonas Oct 12 '11 at 3:04

The union of two subspaces is a subspace if and only if one of the subspaces is contained in the other.

The "if" part should be clear: if one of the subspaces is contained in the other, then their union is just the one doing the containing, so it's a subspace.

Now suppose neither subspace is contained in the other subspace. Then there are vectors $x$ and $y$ such that $x$ is in the first subspace but not the second, and $y$ is in the second subspace but not the first. Then I claim the $x+y$ can't be in either subspace, hence, can't be in their union; hence, the union is not closed under addition, so it's not a subspace.

So, let's prove the claim. If $x+y$ is in the first subspace, well, so is $x$, so $-x$ is also there, so $(x+y)+(-x)$ is there, but that's just $y$, which we know is not there. We've reached a contradiction on the assumption that $x+y$ was in the first subspace, so it can't be. Very similar reasoning shows it can't be in the second subspace, either, and we're done.

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If $W_1$ and $W_2$ are subspaces then $W_1 \cup W_2$ is a subspace if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.

Proof:

($\Leftarrow$) This is the easy direction.

If $W_1 \subset W_2$ or $W_2 \subset W_1$ then we have $W_1 \cup W_2 = W_2$ or $W_1 \cup W_2 = W_1$, respectively. So $W_1 \cup W_2$ is a subspace as $W_1$ and $W_2$ are subspaces.

($\Rightarrow$) This is the harder direction and I give a direct proof.

Assuming $W_2 \not\subset W_1$, I'll show $W_1 \subset W_2$. Let $x \in W_1$ and $y \in W_2 - W_1$. So, by the definition of the union, we have $x \in W_1\cup W_2$ and $y \in W_1\cup W_2$. Therefore, as $W_1 \cup W_2$ is a subspace, $x + y \in W_1 \cup W_2$ which, again by the definition of the union, means that $x + y\in W_1$ or $x+y\in W_2$. If $x + y \in W_1$ then, as $W_1$ is a subspace, $y = (x + y) + (-x) \in W_1$ which is impossible as $y \in W_2-W_1$. So it must be that $x + y \in W_2$ in which case, as $W_2$ is a subspace, $x = (x + y) + (-y) \in W_2$. Therefore, as $x$ was arbitrary, $W_1 \subset W_2$ as desired. $_\Box$

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You assumed that $x+y \in W_2$. For any fixed $x$, fine, $x+y$ must lie in $W_1$ or $W_2$. But to show $W_1 \subset W_2$ you need to get the same conclusion for all $x \in W_1$. It seems to me that you have not argued for this. –  Pete L. Clark May 8 '13 at 15:47
By the way, I don't view the other answers as giving a true proof by contradiction. They are proving the contrapositive: if neither of $W_1$ and $W_2$ contains the other, then $W_1 \cup W_2$ is not a subspace. –  Pete L. Clark May 8 '13 at 15:48
About your first comment: I don't understand. I've chosen $x \in W_1$, but it's a completely arbitrary choice so why isn't that enough? I always get the same conclusion don't I? About your second comment: Yes, I suppose you're right. I'd been searching online for a direct proof and all of the ones I came across were proofs by contradiction (well, with hindsight, maybe they weren't). Re-reading Gerry's proof I definitely see he's proving the contrapositive. –  BU982T May 8 '13 at 16:17
@BryanUrízar You just proved that *if $x+y\in W_2$, then $x\in W_2$*. But it could be that $x+y\in W_1$, in which case the conclusion is that $y\in W_1$. So, you have shown: Given any $x\in W_1$ and any $y\in W_2$, either $x\in W_2$, or $y\in W_1$. This is not quite yet what you need in order to conclude $W_1\subset W_2$ or $W_2\subset W_1$. –  Andres Caicedo May 8 '13 at 16:23
@AndresCaicedo Okay, I finally understand what you've been trying to tell me. I've corrected the proof and it should be fine now. –  BU982T May 10 '13 at 14:33

This direct proof in both directions had been founded upon buri's (ie: user 70692) answer, whose comment induces me to post my edition separately for want of helping others.

If $W_1$ and $W_2$ are subspaces then $W_1 \cup W_2$ is a subspace if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.

Proof: ($\Leftarrow$) This is the easier direction.

If $W_1 \subset W_2$ or $W_2 \subset W_1$ then we have $W_1 \cup W_2 = W_2$ or $W_1 \cup W_2 = W_1$, respectively.
So $W_1 \cup W_2$ is a subspace as $W_1$ and $W_2$ are subspaces.

($\Rightarrow$) This is the harder direction. We are given that $W_1 \cup W_2$ is a subspace. Use the proof technique on P136 of Velleman's How to Prove It, 2nd Ed: break the proof into 2 cases. In each case, prove $W_2 \subset W_1$ or $W_1 \subset W_2$.

$\bbox[5px,border:2px solid green]{\text{ 1st case : } W_2 \subset W_1 \text{ is true }} \;$ Then the disjunction $W_2 \subset W_1$ OR $W_1 \subset W_2$ is trivially true.

$\bbox[5px,border:2px solid green]{\text{ 2nd case : } W_2 \not\subset W_1} \;$ Then the disjunction is true $\iff$ $W_1 \subset W_2$. Prove this directly.

Let $x \in W_1$ and $y \in W_2 - W_1$.
By the definition of the union, we have $x \in W_1\cup W_2$ and $y \in W_1\cup W_2$.
As $W_1 \cup W_2$ is a subspace, $x + y \in W_1 \cup W_2$ which, again by the definition of the union, means that $x + y\in W_1$ or $x+y\in W_2$.

If $x + y \in W_1$, then as $W_1$ is a subspace, $y = (x + y) + (-x) \in W_1$.
This is impossible because $y$ was let $\in W_2-W_1$ in the beginning.

So it must be that $x + y \in W_2$, in which case, as $W_2$ is a subspace, $x = (x + y) + (-y) \in W_2$.
As $x$ was arbitrary, $W_1 \subset W_2$ as desired. $\quad \Box$

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I don't understand why would you post a duplicate of 11E99F's answer. –  Pedro Tamaroff Oct 26 '13 at 14:22
This is not a proof by contraposition. –  BU982T Oct 26 '13 at 18:37
@PedroTamaroff: As this isn't identical but analogous, for example I essayed to detail more, and owing to buri's comments, I posted separately. –  Law Area 51 Proposal - Commit Oct 27 '13 at 9:16
@buri: Thank you. Emended. –  Law Area 51 Proposal - Commit Oct 27 '13 at 9:17