I Vector Subtraction and Topology

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Why is the operation of subtracting vectors not defined?
I learned in a vector calculus class that the operation of vectors is not defined. The professor mentioned it had to do with topology. How does the operation of vector subtraction relate to topology and how does topological properties prevent vector subtraction from being defined?
 

fresh_42

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Could you be more specific? What do you mean by operation of vectors?

Vectors are usually defined as an additive group over a field.

This means we have addition, subtraction, associativity, zero element, and distributive multiplication by e.g. reals, or rationals. There is no topology at prior, although there are topological vector spaces. They play an important role in physics. But whatever vector space we consider, vector addition and subtraction is fundamental for all of them. It is why we consider it in the first place.
 

scottdave

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Vectors can be added and subtracted. I'm not sure of the context. Perhaps if you are expressing vectors in Polar form, then addition and subtraction is not just straight add/subtract of the elements, anymore
 
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Could you be more specific? What do you mean by operation of vectors?

Vectors are usually defined as an additive group over a field.

This means we have addition, subtraction, associativity, zero element, and distributive multiplication by e.g. reals, or rationals. There is no topology at prior, although there are topological vector spaces. They play an important role in physics. But whatever vector space we consider, vector addition and subtraction is fundamental for all of them. It is why we consider it in the first place.
I could be a misunderstanding on my part.

From what I understood, we call vector subtraction the addition of the inverse, but it is not defined the same way as it is for real numbers. But this just confused me, as I could not see why that would be the case.
 

fresh_42

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I could be a misunderstanding on my part.

From what I understood, we call vector subtraction the addition of the inverse, but it is not defined the same way as it is for real numbers. But this just confused me, as I could not see why that would be the case.
This is true. The additive inverse element of a vector $$v$$ is $$-v$$ and $$w-v=w+(-v)$$ is the vector addition. We cannot divide vectors!
 

Math_QED

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By definition, a vector space ##V## is in particular an additive abelian group. Thus every vector has an additive inverse:
$$v+(-v)=0=(-v)+v$$
 

WWGD

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Summary: Why is the operation of subtracting vectors not defined?

I learned in a vector calculus class that the operation of vectors is not defined. The professor mentioned it had to do with topology. How does the operation of vector subtraction relate to topology and how does topological properties prevent vector subtraction from being defined?
What class is this , for context? Maybe you mean vectors based at different points?
 
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What class is this , for context? Maybe you mean vectors based at different points?
I'm taking vector calculus, but the professor just mentioned that and went on. So I was curious as to why that statement was true.
 

mathwonk

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what do you suppose would happen if you raised your hand and asked?
 

WWGD

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Strange thing is that if v is a vector in the sense of a vector space, then the set of vectors is an Abelian group.
 

fresh_42

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Strange thing is that if v is a vector in the sense of a vector space, then the set of vectors is an Abelian group.
I assume he spoke about another operation. Of course we can only guess, so a few possibilities are:
- no multiplication instead of addition: ##\vec{u}\cdot \vec{v}##
- no addition in GL(V): ##A+B##
- no addition in a group representation: ##(g+h).\vec{v}##
- no addition by scalars: ##c+\vec{v}##
All not allowed (at prior).
 
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WWGD

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I assume he spoke about another operation. Of course we can only guess, so a few possibilities are:
- no multiplication instead of addition: ##\vec{u}\cdot \vec{v}##
- no addition in GL(V): ##A+B##
- no addition in a group representation: ##(g+h).\vec{v}##
- no addition by scalars: ##c+\vec{v}##
All not allowed (at prior).
Yes, but the OP is about subtraction of vectors not being defined.
 

fresh_42

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Yes, but the OP is about subtraction of vectors not being defined.
I meant that the teacher might have meant something else than the OP means he meant. Don't you mean?

(I love this word. It has so many mean meanings that you can almost use it alone to say something.)
 

pasmith

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How do you do differential calculus without being able to assign a meaning to [itex]f(t + h) - f(t)[/itex]?
 

WWGD

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How do you do differential calculus without being able to assign a meaning to [itex]f(t + h) - f(t)[/itex]?
They may have been referring to vectors based at different points. Not too clear from the OP.
 

Erland

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Or maybe the professor said "division" and not "subtraction"....?

If the professor really said "subtraction", the OP should insist that the professor explains what (s)he means, and if necessary refer to the posts in this thread.
 

FactChecker

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I'm taking vector calculus, but the professor just mentioned that and went on. So I was curious as to why that statement was true.
As others have said, that statement is not true. So you need to ask him what he meant or we are left to guess.
 

mathwonk

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I refer you to my post in the thread:


but the point is not to let anyone get away easily with not answering a question, they are taking your money to teach you.

remember that students do matter. you are not just fodder. you are at that age where you are becoming a person, i.e. someone who has weight, i urge you to seize the opportunity.

of course the more diplomatic and less confrontational you can make it the better, but your version of it makes it sound as if this guy is a hard case. but it could be you who is being too meek. be polite but courageous, and good luck.
 
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HallsofIvy

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It's clear that you misunderstood. Perhaps what you professor was saying was that (in any algebraic system, not just vector spaces) we don't consider "subtraction" a separate operation since, unlike addition, subtraction is neither "commutative" (a- b is not, in general, equal to b- a) nor "associative" (a+ (b+ c) is not, in general, equal to (a+ b)+ c.
 

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