# Homework Help: How many elements does the set A union B have?

1. Oct 31, 2015

### ZenchiT

1. The problem statement, all variables and given/known data
A and B are both spans. A = span(a1,a2,a3) B = span(b1,b2)
Then how many elements would AUB have?

2. Relevant equations

N/A

3. The attempt at a solution

I'm almost certain the answer is 2, with the two elements being span(a1,a2,a3) and span(b1,b2) but I'm unsure as to whether a span can be an element? Can someone verify this for me please.

Thank you!

2. Oct 31, 2015

### Krylov

Where do the elements $a_i$ and $b_j$ live? And what, according to you, is the definition of "span"?

3. Oct 31, 2015

### ZenchiT

They live inside the span? A span is a set of all linear combinations of all the elements inside it?

4. Oct 31, 2015

### Krylov

For this it is necessary that the $a_i$ are vectors in a linear space, and similarly for the $b_j$, or at least it requires that addition and scalar multiplication can be defined.

So, assuming that the $a_i$ are vectors in a real or complex linear space, can you think of some examples of linear combinations? What happens when all $a_i$ are zero? What happens when one of the $a_i$ is non-zero?

5. Oct 31, 2015

### ZenchiT

I know there is a theorem which states that a span is a subspace therefore addition and scalar multiplication are defined. Therefore is the number of elements 2?

6. Oct 31, 2015

### fzero

You haven't given enough information to solve the problem, but I think you are also confused on some things. Can you determine the number of elements in the set A? In the set B?

7. Oct 31, 2015

### ZenchiT

Sorry! Oh ok, so does A and B have infinitely many elements?

8. Oct 31, 2015

### fzero

A and B are the sets. The objects in the sets are called elements. You haven't really said what kind of vectors $a_i$ and $b_i$ are or what sort of number field we are working in, so we can't yet say for certain. If they were vectors in Euclidean space like $\mathbb{R}^n$ and linear combinations have coefficients in $\mathbb{R}$, then generally, yes, these sets would have infinitely many elements.

Are you given further information about the vectors involved in the problem?

9. Oct 31, 2015

### ZenchiT

Yes, we are given that they are in Euclidean space R^3. However when we do the union of the two things, surely we just have the two elements, the spans?

10. Oct 31, 2015

### fzero

Have you been given a definition of a union of sets? Given two sets $A$ and $B$, the union of them, $A \cup B$, is a new set, whose elements consist of

1. all of the elements of the set $A$
2. all of the elements of the set $B$
3. including any elements which are in both $A$ and $B$.

So the elements of the union of your sets are vectors, not spans (which are themselves sets of vectors).

11. Oct 31, 2015

### ZenchiT

Oh, I see so does that mean that they're are infinitely many elements? However don't all the elements follow two standard patterns? You'll have elements from span (A) = x1a1+x2a2+x3a3 and elements of span (B) y1b1+y2b2 where x1,x2,x3,y1,y2 are elements of the field?

12. Oct 31, 2015

### fzero

Yes, again the union would generally have an infinite number of elements and they are precisely the linear combinations that you wrote down. But we can try to get a bit more insight if we have more information about the vectors that generate the sets. For instance, are they linearly independent? Can we write any of the $b_i$ as linear combinations of the $a_i$? And so on.

13. Oct 31, 2015

### ZenchiT

How would the vectors being linearly independent affect anything?

14. Oct 31, 2015

### fzero

Well, suppose that the $a_i$ were linearly independent of each other. Since they are vectors in $\mathbb{R}^3$, then they actually form a basis for all vectors in $\mathbb{R}^3$, so we can write $b_1$ and $b_2$ as linear combinations of the $a_i$. Can you use this to conclude something about $A\cup B$ for this case?

15. Oct 31, 2015

### ZenchiT

Ahh okay that makes sense! Would that mean that A U B = A + B?

16. Oct 31, 2015

### ZenchiT

Would that also imply that it is a subspace of R^3?

17. Oct 31, 2015

### fzero

Not quite. The operation $A+B$ for sets isn't well defined. Sets can be defined for elements which are very different. If I had some fruit on my table, I could consider the set of oranges or the set of apples. But how would we add these sets? We can't really add an apple to an orange to get a new element. However, we can take the union of the two sets as defined above. That's about as close to addition as we can get for sets.

We can put additional structure on the elements of a set. For instance if we have a set of real numbers, or vectors, we can define what it means to add two elements together. But this is not at all the same as "adding sets".

Since we are dealing with vectors in $\mathbb{R}^3$, both of the sets $A$ and $B$ are properly subsets of $\mathbb{R}^3$. In the case that $A$ consists of 3 linearly independent vectors, then we can say that $A$ is isomorphic to $\mathbb{R}^3$, since any element of $\mathbb{R}^3$ is also an element of $A$. Furthermore, we can say that $B$ is a subset of $A$, since every vector in $B$ is also in $A$.

$A\cup B$ is surely a subset of $\mathbb{R}^3$, but can you use the observations above to describe it more precisely? Perhaps it would help to think about which vectors we need to span it.

18. Oct 31, 2015

### ZenchiT

I know that it is a subset but is it a subspace? I.e, would this be closed under addition and scalar multiplication? My intuition says that wouldn't be the case because to be closed under addition a+b must be in AUB, however its not always the case that an element from span A and an element of span B would add together to give an element in span A or B. Am I correct in my thinking?

Thank you so much for your help!

19. Oct 31, 2015

### fzero

You are making an important point about the general case. In order to avoid confusion with the special case we've been discussing where the $a_i$ were linearly independent, let's stop assuming that.

Let us start with $A$ and $B$, whose elements are particular vectors in $\mathbb{R}^3$. $A$ and $B$ are individually also subspaces by definition. (I think this is obvious but will explain more if you want.) However, as you say $A\cup B$ is not necessarily a subspace, since a linear combination of, say $a_1$ and $b_2$ is not necessarily in either $A$ or $B$. The set of all linear combinations of $a_i$ and $b_i$ is really $\text{span}(a_1,a_2,a_3,b_1,b_2)$ with general element
$$\sum_i x_i a_i+ \sum_\alpha y_\alpha b_\alpha.$$
This is a subspace of $\mathbb{R}^3$ and we can say that $A\cup B$ is a subset of $\text{span}(a_1,a_2,a_3,b_1,b_2)$.

Your intuition was correct, I just put a little more detail together with it.

However, we are led back to the questions about linear dependence. If we are in $\mathbb{R}^3$, we know that the 5 vectors $\{ a_1,a_2,a_3,b_1,b_2\}$ cannot be linearly independent, since $5>3$. If we knew that the $\{ a_1,a_2,a_3\}$ were linearly independent and therefore a basis for $\mathbb{R}^3$, then we could claim the following:

1. $A = \mathbb{R}^3$
2. $A\cup B=A$
3. $A\cup B$ is a vector space.

It is probably interesting to also consider some cases where $\{ a_1,a_2,a_3\}$ are not linearly independent. Like the example $a_1 = (1,0,0)$, $a_2=(0,1,0)$, $a_3= (1,1,0)$, $b_1 = (0,0,1)$, $b_2=(1,2,1)$.

In any case, your observation that $A\cup B$ isn't closed is probably the important one, once the basics are established.

20. Oct 31, 2015

### ZenchiT

Thank you for your help again! Yes I understand that A and B are subspaces, however the difficulty that I'm finding, is where does this link with linear independence? If we can claim that AUB is a vector space if (a1,a2,a3) are linear independent then does that mean that for any element, a in A and any element b, in B, a+b is in AUB?

21. Oct 31, 2015

### fzero

Yes. The logic is as follows. Under the complete set of definitions and assumptions, $A$ is generated as the span of 3 linearly-independent vectors. Since our total vector space is $\mathbb{R}^3$, this means the 3 vectors are actually a basis for $\mathbb{R}^3$. So any vector in $\mathbb{R}^3$ is also in $A$. Furthermore, $b_1$ and $b_2$ are also elements of $A$, so $B$ is a subset of $A$ (it is also a subspace in this case). In particular, we can write any linear combination
$$y_1 b_1 + y_2 b_2 = x_1 a_1 + x_2 a_2 + x_3 a_3,~~(*)$$
so any combination $a+b$ is actually an element of $A$. So $A\cup B=A$ and this is a vector space, closed under linear combinations of vectors.

Note that it was actually important that $a_1,a_2,a_3$ be a basis in this example (so that $A=\mathbb{R}^3$). However, even if they did not form a basis, as long as (*) were true, then $A\cup B=A$ would still hold.

22. Oct 31, 2015

### ZenchiT

I see! I think that makes sense. What if we consider a case where we have 3 vectors a1,a2,b1 and we have A = span (a1 +a2) and B = span (b1)
Then because neither of them are a basis would the same hold? For this example if b1 was contained within a1 and a2, i.e. x1a1 + x2a2 = b1, only then would AUB be a vector space, am I correct in thinking that?

However if b1 cannot be written as a linear combination of a1 and a2 then there is no way that AUB can be a vector space right? Because of the same argument as before (it is not closed under addition)

23. Oct 31, 2015

### fzero

Yes, both observations are correct. In order for the union to be a vector space, you need to have this kind of linear dependence between the elements of the individual sets.

24. Oct 31, 2015

### ZenchiT

Thank you so much! One last question! How would you go about proving linear dependence between the elements of different sets?
For example if we take a1 = (3,3,0) and a2 = (2,2,0) and b1 (0,0,1)
Then would we have, (3x1,3x1,0) + (2x2,2x2,0) = (0,0,1)
So you would have the following systems of equations:

3x1 +2x2 = 0
3x1 +2x2 = 0
0 =1, which is a contradiction, therefore would this mean that b1 is not contained in a1 and a2?

What would this imply?

Thank you very much for your help!

25. Oct 31, 2015

### fzero

Yes, in this example $b_1$ is linearly independent of $\{a_1,a_2\}$. Furthermore, $a_1$ and $a_2$ are themselves linearly dependent, being multiples of $(1,1,0)$.