I'm having a difficult time understanding this statement. Can someone please explain with a concrete example?
Union of two vector subspaces not a subspace?
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0(Let's try again, hopefully this time without typos) A nice proof of a more general result (for fields of characteristic 0): http://mathoverflow.net/questions/43538/wonderful-applications-of-the-vandermonde-determinant – 2013-05-08
6 Answers
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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0This 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: https://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. – 2018-06-01
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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1Wiki 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 $(x,0)\cup(0,y)=(x,y)$. – 2015-11-12
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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0I find it rather funny how people are down voting my answer now :) – 2013-10-28
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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0I 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 ! – 2018-07-15
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 $w_1 + w_2 \notin W$.
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0What does this add to the existing 3-5 year old answers? See especially the top-voted, and accepted, answer. – 2016-11-30
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.
so we will make that the two vectors make a single vector. And that is possible only when one of the vector space is a subset of the other.
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0thanks for helping – 2018-06-29