Union of two vector subspaces not a subspace?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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. $_\Box$

Answer 4

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.

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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 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. $\quad \Box$

Answer 5

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.

Answer 6

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 $w_1 + w_2 \notin W$.

Answer 7

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 $V$, $w=u'-u\in U$.
   This is a contradiction.
   If $u+w\in W$, then $u+w=w'$ for some $w'\in W$.
   Since $W$ is a subspace of $V$, $u=w'-w\in W$.
   This is a contradiction.
   So, $U\subseteq W$ or $W\subseteq U$.