# Showing that $W_1\subseteq W_1+W_2$

I found this question and answer on UCLA's website:

Let $W_1$ and $W_2$ be subspaces of a vector space $V$ . Prove that $W_1 +W_2$ is a subspace of $V$ that contains both $W_1$ and $W_2$.

First, we want to show that $W_1 \subseteq W_1 +W_2$. Choose $x \in W_1$. Since $W_2$ is a subspace, $0 \in W_2$ where $0$ is the zero vector of $V$ . But $x = x + 0$ and $x \in W_1$. Thus, $x \in W_1 + W_2$ by definition. Ergo, $W_1 \subseteq W_1 + W_2$. We also must show that $W_2 \in W_1 + W_2$, but this result is completely analogous (see if you can formalize it).

My question:

Why is it enough to show that $x + 0 \in W_1 + W_2$, $0$ is just one element in $W_2$, why don't we have to show, for example, $x + y \in W_1 + W_2$?

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+1 Nice question. – Babak S. Dec 7 '12 at 7:29

You are given two subspaces $W_1$ and $W_2$ of $V$. You want to show that $W_1 + W_2$ is also a subspace of $V$. And you want to show that $W_1 + W_2$ contains $W_1$ and $W_2$. That is you need to argue that

1. $W_1 + W_2 \subseteq V$
2. $W_1 + W_2$ is a vector space.
3. $W_2 \subseteq W_1 + W_2$ and $W_2 \subseteq W_1 + W_2$.

Points number 1. and 2. aren't too hard, so it is really all about being convinced that $W_1 + W_2$ contains $W_1$ and $W_2$. Now if you have shown that it contains $W_1$, then you can apply the exact same argument to $W_2$.

So it is all reduced to proving that $W_1 \subseteq W_1 + W_2$. Let $w \in W_1$. You want to show that this given $w$ is also in $W_1 + W_2$. However, note that by definition $$W_1 + W_2 = \{w_1 + w_2 : w_1\in W_1, w_2 \in W_2\}.$$ Note that $w\in W_1$ and $0\in W_2$ (because $W_2$ is a subspace), so you have $$w = w + 0 \in W_1 + W_2.$$ And that is it.

Edit: I will try to respond to the comment below this question. First, the above is a fully formal and fine proof. Second, you can add some details to the proof. However, if you are taking a class in say linear algebra and you are studying vector spaces, I would assume that there isn't need for any more details. We are pretty much down to the definitions of things. But, I understand that it can be difficult sometimes to figure out how many details to include. So let me try to explain things a bit more.

First let us be clear about what exactly it is that we want to prove. You want to prove the third point from about. That is we want to prove that $$W_1 \subseteq W_1 + W_2.$$ So let us just focus on this one. To be clear, what are wanting to prove is that the one set $W_1$ is contained in the other set $W_1 + W_2$. The question is now: how in general do you prove that one set is contained in another set? The way that you (in general) do this is to assume that an arbitrary element from the one set ($W_1$) is given and prove that this random given element is also an element in the other set ($W_1 + W_2$). So how does this look like? Well, in the proof we will then start by saying something like: Let $w \in W_1$ be given. Ok, so now we have that random given element in $W_1$ given. Now we need to provide an argument/proof that this element is also an element of $W_1 + W_2$.

To do that we need to understand how the elements in $W_1 + W_2$ look like. How are they defined? And it is here we make the important note that by definition $W_1 + W_2 = \{x + y: x\in W_1, y\in W_2\}$. To unpack this a tiny bit more. The set $W_1 + W_2$ is defined as the collection (the set) of elements that are sums of elements $x + y$ where $x$ comes from $W_1$ and $y$ comes from $W_2$. So again, to prove that our randomly given element $w$ is an element in $W_1 + W_2$ all we need to do is to prove that $w$ can be written as a sum of an element $x$ from $W_1$ and an element $y$ from $W_2$. If we manage to write $w$ as any sum like that, then by definition $w\in W_1 + W_2$. Ok, so let us try to find these two elements. Well, this might seem a bit arbitrary, but we simply choose/define $$x = w \\ y = 0.$$ Here the key thing is that the chosen $x = w$ (remember that $w$ is the given element) is an element of $W_1$ and $y = 0$ is the zero element from $W_2$. Remember that all vector spaces (and so all subspaces) contain $0$, so indeed choosing $y = 0$ will give $y\in W_2$.

Ok, so far so good.

Now we observe/note that by these choices made $$w = x + y.$$

So again: what does this mean? This means exactly that we have written the given element from $W_1$ as the sum of an element $x$ from $W_1$ with an element $y$ from $W_2$. So by definition (as mentioned above) we have proved that $w\in W_1+ W_2$.

This this works for any such arbitrarily given element $w\in W_1$, we have (as mentioned above) that $$W_1 \subseteq W_1 + W_2.$$ Edit 2: Responding to a new comment below: I think that in all this it might be helpful for you to review what exactly it means for one set $A$ to be contained in another set $B$. If you figure this out, then what I have written above should provide all the details. However, let me try to answer your question from the comment.

The key thing is that you have to show that this given $w$ is an element in $W_1 + W_2$ (it is here that we use what it means for one set to be contained in another set). This means that you do not have to show that $w$ plus anything is contained in $W_1 + W_2$. Only $w$ needs to be in $W_1 + W_2$.

Note that $w$ doesn't depend on $y$. It isn't a function. It is just a randomly given element from $W_1$. if you add some $y\neq 0$ to $w$ (or to the $x$ from above). Then it is no longer equal to $w$ and you don't end up proving that $w$ is in $W_1 + W_2$.

We know exactly that $w\in W_1 + W_2$ because we wrote this one element as a sum of $x$ from $W_1$ and $y = 0$ from $W_2$. It doesn't really matter what $w$ and $y$ are equal to. The important thing is that the sum is equal to $w$.

So indeed we know that in general $w\in W_1 + W_2$.

If all this still doesn't make sense, you might try to start a question about what it means for one set to be contained in another set.

If this doesn't help, I probably can't help more. Sorry...

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I'm with you until the third-to-last line. I understand that $0 \in W_2$, I get that. But the definition of $W_1 + W_2$ is that for EVERY two vectors you add together, such that the first is in $W_1$ and the second is in $W_2$. You only used 1 vector - the zero vector. By that logic the definition of $W_1 + W_2$ can be $W_1 + W_2 = \{w_1 + w_2 : w_1\in W_1, w_2 \in P\}$, where $P$ is some subspace, since $0 \in P$. I hope you understand what I'm missing, because I really don't see how this is a proof. – CodyBugstein Dec 7 '12 at 1:30
@Imray: I tried to add much more detail. I hope this helps... – Thomas Dec 7 '12 at 3:52
Again, you're only proving $w \in W_1+ W_2$ when $y=0$. What about when $y \neq 0$?? Then $w$ might not be in $W_1 + W_2$ right? So in general, we don't know if $w \in W_1+ W_2$. – CodyBugstein Dec 7 '12 at 15:14
@Imray: Let me edit my question and add a last bit more. – Thomas Dec 7 '12 at 15:17
@Imray - Think this over for a bit. We can write the vector $w$ as $w = w + 0$. But $w \in W_1$ and $0 \in W_2$. So, by definition, $w$ is an element of $W_1 + W_2$. Since $w$ is an arbitrary element of $W_1$, $W_1 \subseteq W_1 + W_2$. – Chris Leary Dec 7 '12 at 16:16

$W_1+W_2=\{w_1+w_2: w_1\in W_1,w_2\in W_2\}$. To show that an element belongs to this set, we just need to show that it can be written in the form $w_1+w_2$ for some $w_1\in W_1$ and some $w_2\in W_2$.

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You kind of just repeated the answer that I'm having trouble understanding. I don't understand why saying that $0$ is in $W_2$ means that automatically you can disregard all the other things that are in $W_2$. – CodyBugstein Dec 6 '12 at 20:29
@Imray: I think you need to recall the definition of 'inclusion'. – Sugata Adhya Dec 7 '12 at 1:05

Seems to me that there is a fundamental misunderstanding of quantification here. How do I show that something is in $W_1+W_2$? All that is necessary is to find an element of $W_1$ and an element of $W_2$ that add up to the given thingie. Starting out with $w_1$ in $W_1$, the “an” elt of $W_1$ is it itself, and the “an” elt of $W_2$ is the zero elt.

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