Proving a subspace

  • #1

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.

Homework Equations



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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

  • #2
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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?
 
  • #3
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?
 
  • #4
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If [itex]U \subset V[/itex] and [itex]u \in U[/itex], is [itex]u \in V[/itex]?
 
  • #5
I'm not sure what you are asking? Sorry...
 
  • #6
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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.
 
  • #7
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.
 
  • #8
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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.
 
  • #9
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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 [itex]x_1, x_2 \in U+W[/itex], then

1. [itex]x_1+x_2 \in U+W[/itex];

2. [itex]c x_1 \in U+W[/itex], where c is a constant; and

3. [itex]0 \in U+W[/itex]

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.
 
  • #10
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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.
 
  • #11
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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.
 
  • #12
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?
 
  • #13
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You don't need to know that level of detail for U and W. You only need to know they are subspaces.

Let [itex]x_1 \in U+W[/itex]. By definition of U+W, you can say [itex]x_1 = u_1 + w_1[/itex] where [itex]u_1 \in U[/itex] and [itex]w_1 \in W[/itex]. You can say the same sort of thing for [itex]x_2 \in U+W[/itex]. Now look at the sum
[tex]x = x_1 + x_2 = (u_1 + w_1) + (u_2+w_2) = (u_1+u_2) + (w_1+w_2)[/tex]
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 [itex]u \in U[/itex] and [itex]w \in W[/itex]. How can you do this using the fact that U and W are subspaces of V?
 
  • #14
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?
 
  • #15
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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?
 
  • #16
It means that x1 = u1 + w1, where u1 is an element of U and w1 is an element of W.
 
  • #17
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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?
 
  • #18
I would need to show that u1 + w1 + u2 + w2 is in U+W... But I'm not sure how I would do this?
 
  • #19
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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:
  • #20
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?
 
  • #21
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Yes.
 
  • #22
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?
 
  • #23
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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.
 

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