# Proving a subspace

## Homework Statement

Let U and W be subspaces of a vector space V.

Show that the set

U + W = {v (element symbol) V : v = u + w, where u (element symbol) U and w (element symbol) W}

Is a subspace of V.

-

## The Attempt at a Solution

I really don't know where to start here? What would I work with? How can I add these vectors to show that their sum is in fact forming a subspace of V? I just don't know where to start?

## Answers and Replies

This is equivalent to showing that V is closed under addition. So, what's the definition of V, and how is the addition operator defined?

Well, we are given that V is a vector space, therefore, by our axiom (1), if u and w are in V, then u + w is in V. But we are not given that u and w are in V, we are given that u and w are in U and W, which are both subspaces of V.

So, is it possible to show that u + w is a subspace of V?

If $U \subset V$ and $u \in U$, is $u \in V$?

I'm not sure what you are asking? Sorry...

If u is an element in the space U, and if U is a subspace of V, does this imply that u is an element of V? What is the definition of a subspace?

Answering that should tell you how to answer the question you asked in #3.

A subspace of a vector space V is a subset of V which itself is a vector space under the addition and scalar multiplication defined on V.

Ok, this makes sense, I suppose I just was not looking at it properly.

So this kind of proof, it would mainly be in words as I can imagine it. I would write:

u and w are both elements of V, since they are elements of subspaces of V. And, by axiom (1), which states that if u and w are elements of V, then u + w is also an element of V. Thus, since u and w are elements of a subspace (which are of course vector spaces), u + w is itself a subspace of V.

??? Is this okay?

Thanks.

That's the general logic. The last step (proving that u+w is a valid subspace) may or may not need more rigor depending on the assumptions of the problem.

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
A subspace of a vector space V is a subset of V which itself is a vector space under the addition and scalar multiplication defined on V.

Ok, this makes sense, I suppose I just was not looking at it properly.

So this kind of proof, it would mainly be in words as I can imagine it. I would write:

u and w are both elements of V, since they are elements of subspaces of V. And, by axiom (1), which states that if u and w are elements of V, then u + w is also an element of V. Thus, since u and w are elements of a subspace (which are of course vector spaces), u + w is itself a subspace of V.
U+W and u+w are not the same thing. u+w is an element of V; it is not a subspace of V. U+W, on the other hand, is what you're trying to prove is a subspace.
??? Is this okay?

Thanks.
No, it's not. You're trying to prove that U+W is a subspace. That means you need to show that if $x_1, x_2 \in U+W$, then

1. $x_1+x_2 \in U+W$;

2. $c x_1 \in U+W$, where c is a constant; and

3. $0 \in U+W$

Note that you're trying to show that the various elements are in U+W, not V. You're proving U+W is non-empty and is closed under addition and scalar multiplication.

No, it's not.

But it is the first step. As long as he understands the logic behind the construction of U+W (which is where it seems he was stuck) then proving that U+W satisfies the properties of a subspace is relatively easy.

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
I'll have to disagree. Showing V is closed under addition is axiomatically true since V is a vector space, and the OP seemed to be under the impression that it was all he needed to show to prove U+W is a subspace. It misses the whole point that you have to show it is U+W, not V, that is closed under both addition and scalar multiplication.

Ok you see. This is my problem. I am not given any constructs of what U and W are, just that they are both subspaces of V. How do I go about showing that if they are both subspaces, then U+W is itself a subspace of V? Thinking about it, I can only conclude that my worded answer is correct?

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
You don't need to know that level of detail for U and W. You only need to know they are subspaces.

Let $x_1 \in U+W$. By definition of U+W, you can say $x_1 = u_1 + w_1$ where $u_1 \in U$ and $w_1 \in W$. You can say the same sort of thing for $x_2 \in U+W$. Now look at the sum
$$x = x_1 + x_2 = (u_1 + w_1) + (u_2+w_2) = (u_1+u_2) + (w_1+w_2)$$
You want to show that this vector x is in U+W. In other words, prove it's equal to the sum of two vectors u and w where $u \in U$ and $w \in W$. How can you do this using the fact that U and W are subspaces of V?

Ok. I would imagine that would involve using the fact that we are given that

velement of V = u + w, where uelement of U and welement of W...??

Using this piece of information, I could say that:

By definition of U+W, u = u1 + w1, where u1 is an element of U and w1 is an element of W. And w = w2 + u2 where w2 is an element of W and u2 is an element of U.

Now, v = u + w
= (u1 + w1) + (w2 + u2
= (u1+u2) + (w1 + w2)
= u + w

...look, I don't know...just what exactly am I doing here?

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
Nooooo...

You're trying to show x1, x2 ∈ U+W implies that x1+x2 ∈ U+W.

When you say x1 ∈ U+W, what does that mean given the definition of U+W?

It means that x1 = u1 + w1, where u1 is an element of U and w1 is an element of W.

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
OK. You can say the analogous thing for x2. So now you form their sum x = x1+x2. What do you need to show to prove that x is in U+W?

I would need to show that u1 + w1 + u2 + w2 is in U+W... But I'm not sure how I would do this?

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
Use the definition of U+W. In other words, you have to show that x is the sum of two vectors, one of which is in U and one of which is in W.

Hint: To do this, you'll need to rely on the fact that U and W are subspaces so that they are closed under addition.

Last edited:
In other words, use axiom (1), which states that if u and v are both elements of a vector space/subspace, then u+v is itself in that space or subspace...correct?

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
Yes.

So my solution goes something like this, but I lose the plot...:

Let x1 be an element of U+W, therefore x1 = u1 + w1 where u1 is an element of U and w1 is an element of W. Now let x2 be another element of U+W, therefore x2 = u2 + w2 where u2 is an element of U and w2 is an element of W.

Now, let us look at the sum x1 + x2 = (u1 + w1) + (u2 + w2)
= (u1 + u2) + (w1 + w2)
But, by axiom (1), u1 + u2 is in U, and (w1 + w2) is in W, therefore x1 + x2 is in U+W...what do I do from here? Is this even correct?

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
Yes, that's correct so far. You've shown U+W is closed under vector addition.

There's a theorem that says to prove a subset of a vector space is a subspace, you have to show three things, which I enumerated in post #9. You've just shown the first one holds for U+W. Once you show the other two conditions hold, you can conclude U+W is indeed a subspace.