# 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?

• (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/… 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.

• thanks, I wish you were my math teacher Oct 12 '11 at 3:04
• This is a nice explanation, but I would like to add an important pointer too. One you are satisfied with the answer to the question about Union of two spaces not being a space, the next logical question would be "How do I then define the joining of two spaces?". The answer lies in what is called the direct sum: en.wikipedia.org/wiki/Direct_sum. The idea is to fill the gap. Put elements which stop the union of spaces to be a space by putting elements required into the collection. Jun 1 '18 at 10:04

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.

• But didn't a union creates a new subspace? E.g. suppose we have beforehand the three subspaces: V₁ = (x,0), V₂ = (0,y), and V₃ = (x,y). Now it seems to me that V₁ ∪ V₂ ≅ V₃, doesn't it? Nov 12 '15 at 11:19
• Do you know what a subspace is? It's not the same thing as a subset. And union isn't the same thing as addition. Nov 12 '15 at 11:48
• Wiki says, subspace is a subset of a higher dimension space. Then we could just consider my example to be subspace of three-dimension vector space. And sorry, I didn't get the point of "union vs addition" — don't the question asks about the union? Nov 12 '15 at 12:01
• Wiki says a subspace is a vector space. Yes, the question asks about union, but you seem to be doing addition to get $V_3$ from $V_1$ and $V_2$. $(x,0)+(0,y)=(x,y)$. I don't see how to make any sense out of $(x,0)\cup(0,y)=(x,y)$. Nov 12 '15 at 12:15

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. $_\Box$

• 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. 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. May 8 '13 at 15:48
• @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 $W_2\subset W_1$. 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.
– user70962
May 10 '13 at 14:33
• @Bryan: yes, it's correct now. I'm glad you took the time to work out the problem and fix it. May 10 '13 at 14:36

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. $\quad \Box$

• I don't understand why would you post a duplicate of 11E99F's answer. Oct 26 '13 at 14:22
• This is not a proof by contraposition.
– user70962
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. Oct 27 '13 at 9:16
• @buri: Thank you. Emended. Oct 27 '13 at 9:17
• I prefer this answer more because of the more detailed explaination: I can see the motivation for the proof now when you said you break it into 2 cases. Thanks very much ! Jul 15 '18 at 17:58

Take $V_1$ and $V_2$ to be the subspaces of the points on the x and y axis respectively. The union $W = V_1 \cup V_2$ is not a subspace since it is not closed under addition. Take $w_1 = (1,0)$ and $w_2 = (0,1)$. Then $w_1,w_2 \in W$, but $w_1 + w_2 \notin W$.

• What does this add to the existing 3-5 year old answers? See especially the top-voted, and accepted, answer. Nov 30 '16 at 15:58

Explained like you're 10 years old:

Here's a counter-example (where the union of two subspaces is not a subspace).

Imagine a graph with x and y axes. Each axis is a subspace; if we add stuff within each axis, we get more stuff within the axis, all the time. Nice, right?

Now take a random point $$(x_1, y_1)$$ in the middle somewhere, not on either axis. This can be written as the sum of two points, one from each axis, like $$(x_1, 0) + (0, y_1)$$.

Now consider the union of both axes - all points on either axis. Because the points we chose were from each axis, they're from the union, too. But observe (please stay awake!), their sum is that random point out in the middle somewhere, not in that union of the axes.

This violates a requirement of subspaces. While each axis is a subspace, their union isn't. We added two elements of the union and got a point outside the union. Bad news. No subspace.

The following exercise is from "Linear Algebra Done Right 3rd Edition" by Sheldon Axler. (on p.25 Exercise 12)

Prove that the union of two subspaces of $$V$$ is a subspace of $$V$$ if and only if one of the subspaces is contained in the other.

Let $$U$$ and $$W$$ be two subspaces of $$V$$.

1. If $$U\subseteq W$$, then $$U\cup W = W$$ and $$W$$ is a subspace of $$V$$ by assumption.
If $$W\subseteq U$$, then $$U\cup W = U$$ and $$U$$ is a subspace of $$V$$ by assumption.
2. Suppose $$U\cup W$$ is a subspace of $$V$$.
Assume that $$U\nsubseteq W$$ and $$W\nsubseteq U$$.
Then, there is an element $$u\in U$$ such that $$u\notin W$$ and there is an element $$w\in W$$ such that $$w\notin U$$.
Since $$U\cup W$$ is a subspace of $$V$$ and $$u\in U\subseteq U\cup W$$ and $$w\in W\subseteq U\cup W$$, $$u+w\in U\cup W$$.
So, $$u+w\in U$$ or $$u+w\in W$$.
If $$u+w\in U$$, then $$u+w=u'$$ for some $$u'\in U$$.
Since $$U$$ is a subspace of $$V$$, $$w=u'-u\in U$$.
If $$u+w\in W$$, then $$u+w=w'$$ for some $$w'\in W$$.
Since $$W$$ is a subspace of $$V$$, $$u=w'-w\in W$$.
So, $$U\subseteq W$$ or $$W\subseteq U$$.
To prove that a vector(U) is a subspace of a vector space(V). we need to prove that a+$\alpha$b,(where $\alpha$ is any scalar belonging to the field of the vector space) belongs to U.