Is This a Vector Space?

[sarah22]

He gets only the positive vectors. But I don't get which is not a vector space. What I understand is vector space maybe a R^2, R^3 or R^n.

Can anyone here explain it more clearly? I don't get what he said.

http://www.youtube.com/watch?v=JibVXBElKL0" @ 29:55

 

[D H]

Can you explain the context? I, for one, don't comment on videos -- particularly ones where the problem is 30 minutes in (and given the lack of response, I'm not alone).

 

[Landau]

"positive vectors" makes no sense. He takes a subset S of \mathbb{R}^2, namely S=\{(x,y)\in\mathbb{R}^2|x,y\geq 0\}.
If S were a vector space, it would be closed under scalar multiplication. That is, whenever (x,y) is in S, it would follow that r*(x,y), i.e. (rx,ry), is in S, for every real number r. But this is not true, e.g. take r=-1: while (1,1) is in S, -1*(1,1)=(-1,-1) is not in S.

 

[sarah22]

@D H

Landau got it.

@Landau

So it is not a vector space because when you multiply it to a negative scalar, it will go down to the 3rd quadrant? Isn't it adding negative vector would make it go outside the 1st quadrant? So it should follow that also?

If I understand it correctly, for it to be called a vector space it must follow the rule that when you multiply and adding it to all real scalar, the resulting vector should be inside the vector space?

Sorry if I'm wrong. I'm studying it on my own and my brain can't pick up what he said about the vector space. :(

 

[Lord Crc]

If I understand it correctly, for it to be called a vector space it must follow the rule that when you multiply and adding it to all real scalar, the resulting vector should be inside the vector space?

Yes, almost. When you multiply an element by a real scalar or add two elements of the vector space, the result must also be in the (same) vector space. Adding a real scalar to an element is not defined.

 

[Rasalhague]

So it is not a vector space because when you multiply it to a negative scalar, it will go down to the 3rd quadrant?

That's right.

Isn't it adding negative vector would make it go outside the 1st quadrant? So it should follow that also?

Yes, it's another of the rules for a vector space that every vector in the set must have an additive inverse in the set. That means, for every vector u in the set, there must be some vector v, also in the set, such that u + v = 0 (the zero vector); that's to say, v = -u. In other words for every vector u, there is some vector -u. But this isn't the case for the set of vectors belonging only to the first quadrant of R², so it can't be a vector space.

On the other hand, the set of vectors he's asking about all have a positive x component and a positive y component. If you add anyone of these vectors to any other, the resulting vector will still be in the first quadrant. It's true that adding either of the vectors (-1,0) or (0,-1), or any linear combination of these, to any vector (x,y) in this set would result in a vector not in the set, but this isn't a problem because neither (-1,0) nor (0,-1) belong to the set he's asking about (the first quadrant of R²).

If I understand it correctly, for it to be called a vector space it must follow the rule that when you multiply and adding it to all real scalar, the resulting vector should be inside the vector space?

When you multiply a vector in the set by a scalar, the resulting vector has to be in the set. Also when you add any vector in the set to any other vector in the set, the resulting vector has to be in the set. (But adding a scalar to a vector isn't defined.) A general name for this property is "closure"; the set is said to be "closed under" some operation (in this case closed under scalar multiplication and vector addition) if the result belongs to the set.

Sorry if I'm wrong. I'm studying it on my own and my brain can't pick up what he said about the vector space. :(

Don't be sorry! Asking questions and testing out whether you've got them right is a great way to learn. I'm studying on my own too. A few months ago I didn't understand much. I remember when I watched lecture one of that series for the first time it made no sense to me at all! I think it's starting to get clearer, although I still have a lot to learn (I got up to lecture 13, but I took a break to do some background reading...), so hopefully more knowledgeable people here can correct me if I've made any mistakes.

 

[Rasalhague]

Rules for a Vector Space. (Gilbert Strang's example R² fails to meet conditions 4 and 6, although I think it satisfies all of the other conditions.)

A vector space V must form a "group" under the operation of vector addition. This means:

1) Closure under vector addition. Given that u + v = w, if u is in V and v is in V, then w must also be in V.

2) Associativity under vector addition: u + (v + w) = (u + v) + w.

3) There exists some identity element, written 0, in V such that for any vector u in V, u + 0 = u.

4) For every vector u in V there is an additive inverse, which is to say some vector v such that u + v = 0. (The inverse of u is denoted -u.)

The group is commutative:

5) u + v = v + u.

(Another name for a commutative group is "Abelian group".)

There is operation called scalar multiplication, relating elements of F with elements of V, where F is a field (see definition below)--called the base field of the vector space; elements of the field are called scalars--such that for any scalars a and b in F and any vectors u and v in V,

6) V is closed under scalar multiplication: for all a in F and all u in V, if a u = v then v is in V.

7) a (b u) = (a b) u.

8) (a + b) u = a u + b u.

9) a(u + v) = a u + a v.

10) 1 u = u.

Note, a field in this sense is an algebraic structure (not the same thing as a "scalar field" or "vector field"; I think the similar name is a coincidence) consisting of a set which forms a commutative group with some operation (which we call addition). The same set, excluding its additive identity element, 0, must also form a commutative group with another operation (which we call multiplication). And, to complete the definition, multiplication is distributive over addition, thus: a(b+c) = ab+ac.

Examples of fields: the real numbers, the complex numbers (in each case with addition and multiplication as they're usually defined). Either of these fields can be used as the scalars for a vector field.