Vector spaces sum issue

[Zermelo]

Hi, I have a little problem with one of the problems from my Linear Algebra 1 class.
The exercise goes like this:
Let U and W be vector subspaces of the vector space V.
Prove:
If U [MATH] \bigcup [/MATH] W is a vector subspace of V [MATH]\Longrightarrow[/MATH] U [MATH]\subseteq[/MATH] W or W [MATH]\subseteq[/MATH] U.
The proof goes like this:
Let's suppose the opposite, that U isn't a subset of W and W isn't a subset of U. That means that there exist elements u [MATH]\in[/MATH] U\W and w [MATH]\in[/MATH] W\U.
u + w [MATH]\in[/MATH] U [MATH] \bigcup [/MATH] W [MATH]\overset{don't get this} \Longrightarrow[/MATH] w [MATH]\in[/MATH] U or u [MATH]\in[/MATH] W, which is a contradiction ete etc, qed.
My question is, is it possible to have an element u from U\W and w from W\U, such that u + w is in U? I can't prove that this isn't possible, and my professor didn't prove it either, this seems trivial but I can't find a formal proof that this isn't possible.
It is clear that for all u and w, if u and w [MATH]\in[/MATH] U [MATH]\Longrightarrow[/MATH] u + w [MATH]\in[/MATH] U, because U is a vector space and that means (U, +) is an Abel group. But why does the opposite work? Here in the proof, we have u+w [MATH]\in[/MATH] U [MATH]\Longrightarrow[/MATH] u [MATH]\in[/MATH] U.
I am looking for a formal proof that it isn't possible for the sum of two elements from different vector spaces to belong to one of those vector spaces, or a formal proof of the step that I don't understand above. I hope you can understand what I mean, English isn't my first language Thanks in advance

 

[Romsek]

I think the key idea is that If [MATH]u + w \in U[/MATH] then [MATH]w \in U[/MATH] because [MATH]U[/MATH] is a vector subspace.
Same the other way if [MATH]u+w \in W[/MATH] then [MATH]u \in W[/MATH] because [MATH]W[/MATH] is a vector subspace.

Vector addition is closed in a vector subspace.

 

[Zermelo]

Well yes, but what does a "closed operation" mean? It means that [MATH]( \forall x,y \in U)x + y \in U[/MATH], which actually means [MATH](\forall x)(\forall y )(x\in U \land y \in U \Longrightarrow x+y \in U)[/MATH]. Here we have the "inverse" implication, namely, we have that [MATH]x \in U \land y\in V \land x + y \in U[/MATH] and we need to prove that [MATH]x\in U \land y \in U[/MATH] (I took only one of the cases).
The answer seems pretty logical, but I still can't find a formal, 100% clear way to prove this...

 

[Dr.Peterson]

Well yes, but what does a "closed operation" mean? It means that [MATH]( \forall x,y \in U)x + y \in U[/MATH], which actually means [MATH](\forall x)(\forall y )(x\in U \land y \in U \Longrightarrow x+y \in U)[/MATH]. Here we have the "inverse" implication, namely, we have that [MATH]x \in U \land y\in V \land x + y \in U[/MATH] and we need to prove that [MATH]x\in U \land y \in U[/MATH] (I took only one of the cases).
The answer seems pretty logical, but I still can't find a formal, 100% clear way to prove this...

If [MATH]x \in U \land y\in V \land x + y \in U[/MATH], then [MATH]y=(x+y)-x\in U[/MATH]. Right?

 

[Zermelo]

If [MATH]x \in U \land y\in V \land x + y \in U[/MATH], then [MATH]y=(x+y)-x\in U[/MATH]. Right?

Oh my God, it's so simple ? Can't believe I couldn't think of that. Thank you very much.
But I have a small follow-up question, does that mean that the following situation is impossible:
Let A and B be vector subspaces of the vector space V, and let [MATH]A\cap B = \varnothing[/MATH]. Let [MATH]a \in A,\ b\in B.[/MATH][MATH]a+b\in A\cup B \Longrightarrow a+b \in A \lor a+b \in B \\ \overset{last\ comment} \Longrightarrow a, b \in A \lor a,b\in B[/MATH], which is a contradiction with the assumption that A and B are disjoint sets. That means that two disjoint sets CANNOT BE vector subspaces of a vector space?
I'm relatively new to vector spaces which are pretty abstract and sadly my professors aren't quite communicative, so I'm really grateful for you helping me out.

 

[Dr.Peterson]

Oh my God, it's so simple ? Can't believe I couldn't think of that. Thank you very much.
But I have a small follow-up question, does that mean that the following situation is impossible:
Let A and B be vector subspaces of the vector space V, and let [MATH]A\cap B = \varnothing[/MATH]. Let [MATH]a \in A,\ b\in B.[/MATH][MATH]a+b\in A\cup B \Longrightarrow a+b \in A \lor a+b \in B \\ \overset{last\ comment} \Longrightarrow a, b \in A \lor a,b\in B[/MATH], which is a contradiction with the assumption that A and B are disjoint sets. That means that two disjoint sets CANNOT BE vector subspaces of a vector space?
I'm relatively new to vector spaces which are pretty abstract and sadly my professors aren't quite communicative, so I'm really grateful for you helping me out.

Do you see an element that all subspaces must contain? If so, then there is an "obvious" reason they can never be disjoint!

Yes, it takes a while to get used to exactly what vector spaces "feel like", but asking these sorts of questions is part of the process to get there -- especially if you take time pondering them on your own.

 

[Zermelo]

Do you see an element that all subspaces must contain? If so, then there is an "obvious" reason they can never be disjoint!

Yes, it takes a while to get used to exactly what vector spaces "feel like", but asking these sorts of questions is part of the process to get there -- especially if you take time pondering them on your own.

Oh, you mean the "0_v". Thank you, this makes a lot more sense now. Could you recommend to me some reading material? Also, a step that we didn't prove is [MATH]a+b \in A\cup B[/MATH]. Does this only work for vector subspaces?
Here's the proof that I thought of:
Let A and B be vector subspaces of some vector space V. Let a, b be arbitrary elements [MATH]a\in A,\ b\in B[/MATH].
Let's suppose the opposite, [MATH]a+b\notin A\cup B \Longrightarrow a+b \notin A \land a+b \notin B[/MATH]. Because "+" is a closed in A and B, this means that [MATH]\lnot (a,b\in A) \land \lnot (a, b \in B)[/MATH] (because if for example [MATH]a,b\in A[/MATH] that would mean [MATH]a+b \in A.[/MATH]). But because a and b are arbitrary elements from A, B, this stands for all elements in these vector spaces, and this means that A and B are disjoint, and that is a contradiction, vector subspaces can't be disjoint (last comment).

But this contradiction wouldn't happen if A and B weren't vector subspaces of the same vector space. Let's suppose X and Y are disjoint vector spaces, can the sum x+y "jump out" of [MATH]X\cup Y[/MATH]?
What I find especially weird is this idea of a sum, for example adding a matrix and a polynomial (both are elements of some vector spaces). What is the definition of "+" that we are using? What if "+" is defined differently in A and B? Does "+" only work for vector subspaces (if so, half of this post is unnecessary hahaha)

 

[Dr.Peterson]

Also, a step that we didn't prove is [MATH]a+b \in A\cup B[/MATH]. Does this only work for vector subspaces?

Why do you think that would be true?

Take an example: the x-axis and the y-axis are subspaces of R^2. But the sum of a point in each is not typically in either, so it is not in the union.

Simple examples like this are important in developing your vector-space sense.

What I find especially weird is this idea of a sum, for example adding a matrix and a polynomial (both are elements of some vector spaces). What is the definition of "+" that we are using? What if "+" is defined differently in A and B? Does "+" only work for vector subspaces (if so, half of this post is unnecessary hahaha)

When you add two elements, they must be elements of the same "something"! (You can't add 3 + chocolate, because they aren't both numbers; but you can add a natural number and an irrational number because they are both real numbers.) In this case, they should be elements of the same vector space, so the two vector spaces they are in have to be subspaces of that common space.

Clearly you are not yet accustomed to abstract math, so your confusion (trying to be too abstract!) is understandable. Again, one way to guard against this is to stick mostly to familiar examples of vector spaces, learning to walk before trying to fly.

 

[Steven G]

When did Dr Peterson say that [math]\Bbb R \subset \Bbb R^2[/math]?

 

[Zermelo]

When did Dr Peterson say that [math]\Bbb R \subset \Bbb R^2[/math]?

Yeah, I was wrong about that, figured it out a couple of minutes later and deleted it. I thought the x-axis was R, but when I thought about it, it's actually
R x {0} (when you think about an x-axis in a plane).
Either way, thank you Dr. Peterson for the counterexample, it seems that my assistant professor was wrong about this, I will let him know by showing this counter-example and I will try to find another way to prove this exercise: [MATH]A \cup B[/MATH] vector subspace of V [MATH]\Longleftrightarrow[/MATH] [MATH]A\subseteq B \lor B \subseteq A[/MATH] (A, B vector subspaces of V)

 

[Dr.Peterson]

Yeah, I was wrong about that, figured it out a couple of minutes later and deleted it. I thought the x-axis was R, but when I thought about it, it's actually
R x {0} (when you think about an x-axis in a plane).

That's right. I considered describing it formally, but chose to leave it visual and let you think through the details.

Either way, thank you Dr. Peterson for the counterexample, it seems that my assistant professor was wrong about this, I will let him know by showing this counter-example and I will try to find another way to prove this exercise: [MATH]A \cup B[/MATH] vector subspace of V [MATH]\Longleftrightarrow[/MATH] [MATH]A\subseteq B \lor B \subseteq A[/MATH] (A, B vector subspaces of V)

It may help if you show us exactly what you were told (an image if necessary), so we can see whether something wrong was actually said, or you are misinterpreting it.

 

[Zermelo]

That's right. I considered describing it formaly, but chose to leave it visual and let you think through the details.


It may help if you show us exactly what you were told (an image if necessary), so we can see whether something wrong was actually said, or you are misinterpreting it.

Here's the exercise right out of my notebook:
Let U and W be vector subspaces of the vector space V. Prove:
[MATH]U \cup W[/MATH] is a vector subspace of V if and only if [MATH]U \subseteq W \lor W \subseteq V.[/MATH]PROOF:
"->"
Let U, W, [MATH]U\cup W[/MATH] be vector subspaces of V.
Let's suppose the opposite, [MATH]\lnot (U \subseteq W) \land \lnot(W \subseteq U)[/MATH][MATH]\Longrightarrow \exists u \in U\backslash W,\ \exists w \in W\backslash U[/MATH][MATH]u+w \in U \cup W \Longrightarrow w\in U \lor u \in W[/MATH], which is a contradiction.....
Right here we can apply the counter-examaple, U is the x axis, W is the y axis, V is RxR, and this ([MATH]u + w \in U[/MATH]) doesn't hold.

 

[Zermelo]

Here's the exercise right out of my notebook:
Let U and W be vector subspaces of the vector space V. Prove:
[MATH]U \cup W[/MATH] is a vector subspace of V if and only if [MATH]U \subseteq W \lor W \subseteq V.[/MATH]PROOF:
"->"
Let U, W, [MATH]U\cup W[/MATH] be vector subspaces of V.
Let's suppose the opposite, [MATH]\lnot (U \subseteq W) \land \lnot(W \subseteq U)[/MATH][MATH]\Longrightarrow \exists u \in U\backslash W,\ \exists w \in W\backslash U[/MATH][MATH]u+w \in U \cup W \Longrightarrow w\in U \lor u \in W[/MATH], which is a contradiction.....
Right here we can apply the counter-examaple, U is the x axis, W is the y axis, V is RxR, and this ([MATH]u + w \in U[/MATH]) doesn't hold.

Oh, nevermind, I forgot about the assumption [MATH]U \cup W[/MATH] is a vector space... Now it all holds up... I was looking at the general statement [MATH]a + b \in A \cup B[/MATH] without checking what the assumptions were. Sorry for wasting your time, it's all okay, and thanks once again!