# Homework Help: Question regarding the subspace status of a vector in R^2

1. Mar 7, 2013

### MoreDrinks

1. The problem statement, all variables and given/known data
Is the following set a subspace of R2?

2. Relevant equations

(a,b2)

3. The attempt at a solution

I'm exhausted and stumped.

2. Mar 7, 2013

### Zondrina

(a, b2) is a set? Not a vector?

3. Mar 7, 2013

### Staff: Mentor

No, this represents a set of vectors of this form.

4. Mar 7, 2013

### MoreDrinks

Sorry, sorry - I miswrote. The question was "Are the following sets subspaces of R2? The set of all vectors of the form...

and then (a,b2)

I'm trying to think of every reason. There's a zero vector for sure...I don't think scalar multiplication is a problem...but I don't know.

5. Mar 7, 2013

### Staff: Mentor

You still need to show what you have tried. This problem is similar to the other one you posted today - you need to show that for vectors in the set, addition is closed, and scalar multiplication is closed.

6. Mar 7, 2013

### MoreDrinks

I'll be more explicit in the future. I did things, and I didn't know "showing attempt" was such a hard and fast rule.

a and b could both be zero, so there is a zero vector.

2(a,b2) I do not see any problem with.

(a,b2) + (c,d2) = (a+c, b2+d2)

I'm not sure if I'm dealing with a subspace here or not. I suspect not, and that it's not closed under addition.

Last edited: Mar 7, 2013
7. Mar 7, 2013

### Staff: Mentor

Are there any restrictions on a and b? IOW, are they restricted to integers or can they be any real numbers?

Clearly (1, 4) is in the set, since it could be written as (1, 22). What about (2, -9)?

8. Mar 7, 2013

### MoreDrinks

They can be any real numbers.

9. Mar 7, 2013

### MoreDrinks

I had thought that we had to begin by assuming a and b are within R2? I mean, I see where you're going - the square root of negative nine is not within R2, so the vector is not a subspace. There just weren't any ground rules laid down in this class for how we could go about this. Do you, perhaps, have a link to a webpage that really goes over vector spaces and subspaces?

10. Mar 7, 2013

### Staff: Mentor

No, a and b are real numbers. A vector <a, b> with these as its components would be in R2.
More accurately, it's not in R (i.e., it's not real).
So the set of vectors does not form a subspace (of R2).
I don't have a link, but I know there's an article on wikipedia, and no doubt many other sites. The basic idea is that a vector space is a set of things with two operations: addition and scalar multiplication. A vector space must satisfy ~10 vector space axioms, most of which deal with vector addition and scalar multiplication.

A subspace of a vector space is a vector space in its own right. To show that a subset of a vector space is in fact a subspace of that vector space, you need to establish three things:
1. The 0 vector is contained in the set.
2. If u and v are arbitrary vectors in the set, then u + v is also in the set.
3. If u is an arbitrary vector in the set and k is a scalar, then ku is in the set.

That's really about it.

11. Mar 7, 2013

### MoreDrinks

But your suggestion for proving that (a,b2) are not a set of vectors involves making b a non-real number, correct? So we're not actually beginning with a subset of R2 - or do I misunderstand? Thank you for your help thus far.

12. Mar 7, 2013

### vela

Staff Emeritus
You're given the set S=$\{(a,b^2)\in \mathbb{R}^2: a, b \in \mathbb{R}\}$. It's a collection of ordered pairs, right? So any element in S is also in $\mathbb{R}^2$, so it's a subset of $\mathbb{R}^2$. The vector (1,4) = (1,22) is in the set. The vector (2,-9), however, isn't, because you can't express it in the form (a,b2) where a and b are real numbers. S is a set of vectors, but it doesn't contain every vector in $\mathbb{R}^2$.

To prove S is a subspace, you have to show the three things Mark44 listed in his post.

You've already shown that (0,0) is an element of the set because a=0, b=0 works.

Now take two elements of S, (a,b2) and (c,d2). When you add them together, you get (a+c, b2+d2). Can this be written in the form (e, f2) for some real numbers e and f? You need to explain why you know there are such e and f, or you need to find a counterexample to prove you can't.

Now let k be a real number. Is k(a,b2) = (ka, kb2) in S? In other words, can you always find real numbers e and f such that e=ka and f=kb2? Remember k can be any real number.

Last edited: Mar 7, 2013
13. Mar 8, 2013

### MoreDrinks

As far as I can tell, yes, it's closed under addition and scalar multiplication, but the answer is that it's not a subspace. This is why I am stumped.

14. Mar 8, 2013

### Staff: Mentor

It is not closed under scalar multiplication. Try a factor of -1 for any vector with non-zero second component.

15. Mar 8, 2013

### MoreDrinks

Like, -1(a,b2)? So we get (-a,-b2)? Is this a problem - am I viewing the order of operations wrong or something?

16. Mar 8, 2013

### Staff: Mentor

(1,4) is element of your set of vectors (coming from a=1, b=2), but -1(1,4)=(-1,-4) is not an element of your set (as there is no real b where b^2=-4).

17. Mar 8, 2013

### MoreDrinks

It may take me a little longer than the other kids, but I get there. Thank you, sir.

18. Mar 8, 2013

### Staff: Mentor

MoreDrinks, now do you see where I was going with my question?

Last edited: Mar 8, 2013
19. Mar 8, 2013

### MoreDrinks

Surely, Mark. Follow-up question: when you're checking to see if a subset of vectors is closed under scalar multiplication, must a scalar be within the original vector space from which the subset is derived?

Last edited by a moderator: Mar 8, 2013
20. Mar 8, 2013

### vela

Staff Emeritus
It would help if, instead of simply claiming you get it, you demonstrate your understanding by explaining your solution. For example, you said S is closed under vector addition. How do you know? I wonder if you truly understand the concepts because you later claimed S was closed under scalar multiplication.

21. Mar 8, 2013

### MoreDrinks

Well, we start in R2 with the form (a,b2). Anything of the form (a,b2) that we can add to our original (a,b2) is going to leave us with a (a,b2). Any b in R2 that is squared is going to be positive, which doesn't leave us with a second element b2 that is negative and, subsequently, possessing a non-real root outside of R2.

Am I barking up the wrong tree? I'm sorry if the way I'm going about this is totally non-standard - there has never been any sort of "how to prove it" instruction in my educational history, so I'm sort of just jumping into this. It's not even required for my major: I just want to know, and I've enjoyed the matrix algebra in the course hitherto. Thank you for your help thus far.

22. Mar 8, 2013

### vela

Staff Emeritus
It looks like you have the basic idea, but you need to be more careful about how you write it out.
OK.

Here you're using the symbols a and b to represent three different quantities because you're talking about three different vectors. A symbol should represent only one thing at a time.

You're also just restating what it is you're trying to prove, that the sum of two vectors is in the right form.

The phrase in red doesn't make sense because b is just a regular old number while R2 is a set of order pairs, so b can't be in R2. You mean b is in R, not R2.

The rest of the sentence isn't very clear because, again, you're using b to mean different things.

You seem to be on the right track, and there's no need to apologize as long as you're trying, which you are. I'll mention that a lot of students struggle with learning how to write proofs, so you're not alone. It's a skill you can pick up as you go along, though.

23. Mar 8, 2013

### MoreDrinks

Thank you. Cheers.

24. Mar 8, 2013

### Staff: Mentor

There is always some field (this is a precise mathematical term) from which components of a vector come. The field is usually the real numbers or the complex numbers, but finite fields such as {0, 1} can be used. In vector spaces such as R2 and R3, the field is the real numbers. In vector spaces such as C2, the field is the complex numbers, so in Cartesian form, each vector in C2 looks like this: <a1 + b1i, a2 + b2i>.

25. Mar 8, 2013

### Staff: Mentor

As vela said, many students struggle when it comes to doing proofs, as it isn't taught much or at all in courses before you get to linear algebra. IMO, it's a useful skill to have, though, as it fosters logical thinking that's not limited just to mathematics.