Vector spaces + proving of properties

[JamesGoh]

[SOLVED] vector spaces + proving of properties

Im aware in vector spaces that there are 3 properties associated with it
Note v is an element in a vector space, 0 is the additive identity in the vector space and c is a field element


1) 0.v = 0

2) c.0 = 0

3) (-c).v = c.(-.v) = -(c.v)

atm I am trying to prove 2), however I can only properly prove 2) if I use 1). As mathematical properties are meant to be unique, is my current solution considered as bad practice ?

 

[ice109]

did you already prove 1? if so it could go something like this:

$$c \cdot \vec{0}=c \cdot (0 \cdot \vec{v})$$ by 1
$$(c \cdot 0)\cdot \vec{v}= 0 \cdot \vec{v}=0$$ by 1 again

i'm not sure if this is rigorous though

 

[HallsofIvy]

1) look at $(c+ 0)\vec{v}= c\vec{v}+ 0\vec{v}$

2) Look at $c(\vec{v}+ \vec{0})= c\vec{v}+ c\vec{0}$
3) Look at $(c- c)\vec{v}= c\vec{v}- c\vec{0}$ showing that $-c\vec{0}$ is the inverse of $c\vec{v}$: i.e. (-c)\vec{v}= -(c\vec{v}). Look at c(\vec{v}- \vec{v})= c\vec{v}+ c(-\vec{v})[/itex] to show that c(-\vec{v})= -(c\vec{v}).

 

[mathwonk]

0v + 0v = (0+0)v = 0v, implies that 0v = 0.

 

[ice109]

mine was no good?

mathwonk why does that imply that? ah nm because you can just subtract one of the 0v from the left

 

[JamesGoh]

thanks guys for your insights. I will look into the given suggestions and get back to you

btw Happy 2008 to everyone ! :)

 

[JamesGoh]

1) look at $(c+ 0)\vec{v}= c\vec{v}+ 0\vec{v}$

2) Look at $c(\vec{v}+ \vec{0})= c\vec{v}+ c\vec{0}$
3) Look at $(c- c)\vec{v}= c\vec{v}- c\vec{0}$ showing that $-c\vec{0}$ is the inverse of $c\vec{v}$: i.e. (-c)\vec{v}= -(c\vec{v}). Look at c(\vec{v}- \vec{v})= c\vec{v}+ c(-\vec{v})[/itex] to show that c(-\vec{v})= -(c\vec{v}).

My question arose as a result of using 1). The part which concerned me was the $0\vec{v}$ because that is associated with the 1st property of vector spaces.

However, my textbook gives me the proof for $0\vec{v}$ = 0 (where 0 is the additive identity of the vector space) :D, so I guess it should be okay to use 1), provided you can prove $0\vec{v}$ = 0.

Any thoughts or comments ??

 

[JamesGoh]

Anyone have anything they like to add in regards to my previous post ??

If not, I will mark this as solved

 

[quadraphonics]

Why is it that you want to avoid using property 1 when proving property 2? I don't see any reason to care about that.

 

[JamesGoh]

I thought properties of anything in mathematics are meant to be independent of each other. This was why I initally wanted to avoid the use of property 1 when proving property 2.

 

[ice109]

how bout me, someone tell me if my proof is rigorous?

 

[JamesGoh]

did you already prove 1? if so it could go something like this:

$$c \cdot \vec{0}=c \cdot (0 \cdot \vec{v})$$ by 1
$$(c \cdot 0)\cdot \vec{v}= 0 \cdot \vec{v}=0$$ by 1 again

i'm not sure if this is rigorous though

In light of our discussion, I think your approach would be ok, provided you include the proof of property 1. This would make it more understandable and convincing on how you
reached your answer.

Also for this line of working

$$(c \cdot 0)\cdot \vec{v}= 0 \cdot \vec{v}=0$$ by 1 again

do you mean

$$(c \cdot 0)\cdot \vec{v}= 0 \cdot \vec{v}=vec{0}$$

(where vec0 is the additive identity in the vector space)
instead ?

 

[morphism]

how bout me, someone tell me if my proof is rigorous?

The idea is fine, but I'd specify a specific vector v (for instance the 0 vector).

I thought properties of anything in mathematics are meant to be independent of each other. This was why I initally wanted to avoid the use of property 1 when proving property 2.

Axioms are meant to be independent of each other -- after all, having redundant axioms serves no purpose -- but once you have a set of axioms for some structure and you start establishing some of its properties, then they definitely don't have to be independent.

 

[ice109]

why would you specify the vector v?

 

[morphism]

Because you haven't qualified what v is at all. I suppose you could also say "where v is a fixed vector in our vector space".

 

[JamesGoh]

Axioms are meant to be independent of each other -- after all, having redundant axioms serves no purpose -- but once you have a set of axioms for some structure and you start establishing some of its properties, then they definitely don't have to be independent.

Axiom = Unverified (unproven) property of a structure ?

 

[Hurkyl]

Axiom = Unverified (unproven) property of a structure ?

Axiom = generator of a theory. The theory of vector spaces contains a lot of statements; but they all have the property they are logical consequences of the vector space axioms. They are 'abstract'; they do not refer to any particular structure, mathematical or otherwise.


They are useful becuase if you have some mathematical structure, and you:

(1) Interpret the vector space language in that structure (for example, this inclides specifying what vector addition means in your structure)

(2) Prove that the vector space axioms are valid in this interpretation


Then it follows that the entire theory of vector spaces is valid in the given interpretation.

 

[JamesGoh]

Axiom = generator of a theory. The theory of vector spaces contains a lot of statements; but they all have the property they are logical consequences of the vector space axioms. They are 'abstract'; they do not refer to any particular structure, mathematical or otherwise.

According to wolfram mathworld

An axiom is a proposition regarded as self-evidently true without proof

So, based on your understanding and the mathworld definition, an axiom is basically the proposition that needs a theory to validate it ?

 

[ice109]

no an axiom is a granted truth from which theorems are derived and proven.

 

[JamesGoh]

I just want to verify that I understand what an axiom is

An axiom is basically a math-based statement which is assumed to be true. (i.e. 3 < 4,
2+9=11, 4.0 = 0 etc...).

A theorem is similar to an axiom, except it uses a proof process to validate it (is this correct as well ??). Since properties of anything in mathematics require a proof process to validate them, are properties the same as a theorem ?

thanks everyone so far for your contributions, I am enjoying what I am learning from this discussion.

 

[HallsofIvy]

No, properties are not the same as theorems. "Properties" are characteristics of some mathematical object while theorems are always statements about those properties. You do, of course, have to prove the theorem that the object does have that property (unless that is asserted as an axiom).

An axiom is a statement that a mathematical object does have some defining property. We "accept" it as true because it defines the mathematical object and we can't prove anything until after we have defined the object. If we have any "meta-mathematical" reason to believe that "axiom" is not true, then we are just saying that mathematical object is not valid in this situation.

By "mathematical object" I mean anything we might work with in mathematics, any topological space, or vector space, or group, or field, etc.

 

[JamesGoh]

Ok to hopefully clear up my misunderstandings, is an axiom a fundamental/defining condition that must be satisfied in order for an element/set of elements to be considered as a valid mathematical object ??

i.e. for a set of numbers to be classified as a group, it must contain an identity element (where the latter is the axiom) ??

Please let me know if I am still wrong

mathematical object = anything that can be worked with in mathematics i.e vector spaces, groups, fields etc

 

[HallsofIvy]

Ok to hopefully clear up my misunderstandings, is an axiom a fundamental/defining condition that must be satisfied in order for an element/set of elements to be considered as a valid mathematical object ??

i.e. for a set of numbers to be classified as a group, it must contain an identity element (where the latter is the axiom) ??

Yes, that's true. If you were trying to prove that a certain set (with a binary operation) were a group, you would need to prove that it contained an identity. However, you don't prove that a group has an identity- that's part of the definition of group.

Similarly, in Euclidean geometry, you don't need to prove that "through two distinct points there exist exactly one line" because that's part of the statements that define "Euclidean geometry". That statement alone is not "self evidently" true because it is not true in spherical geometry.

Please let me know if I am still wrong

mathematical object = anything that can be worked with in mathematics i.e vector spaces, groups, fields etc

 

[JamesGoh]

HallsOfIvy, thanks so much for your help so far. I just have a couple of more questions to harass you with :P

I noticed you quoted my statement in your post to let me know if I have still got something wrong or misunderstood. Is this true or just a typo ??

Also, the user morphism (morphism please feel free to add as well) mentioned in his post that once any mathematical structure (anything bounded by axioms in maths) has been defined and some of the mathematical structure's properties have been established, he said that the axioms of that mathematical structure do not have to be independent of each other. Does this dependence statement also apply to the properties of the structure as well (note this may sound like a stupid question) ??

 

[HallsofIvy]

You are referring to :

The idea is fine, but I'd specify a specific vector v (for instance the 0 vector).


Axioms are meant to be independent of each other -- after all, having redundant axioms serves no purpose -- but once you have a set of axioms for some structure and you start establishing some of its properties, then they definitely don't have to be independent.

I believe he was referring to the "properties", things proven to be true by theorems, that don't have to be independent, not axioms. Obviously, theorems don't have to be independent- they have to be proven using axioms and other theorem!

Strictly speaking, axioms don't have to be independent- its just simpler and more efficient not to start with more requirements than you absolutely need. Of course, axioms must be consisten- one axiom must not contradict another.

 

[quadraphonics]

I thought properties of anything in mathematics are meant to be independent of each other.

No, not at all. On the contrary, it's extremely rare for properties of the type in question to be independent of one another. After all, they're all consequences of the same set of axioms.

Using previously-proved properties to prove yet more properties is very much the recommended approach.

 

[JamesGoh]

I have nothing further to add at this point. Thank you everyone for all your help and insights. I shall now mark this as solved

 

[Singularity]

Hello, didn't read everything, but here is one way of showing that 0v = 0 (Please pardon my lack of LaTeX skill).

0v = 0v + 0
= 0v + (0v + (-(0v)))
= (0v + 0v) + (-(0v))
= (0+0)v + (-(0v))
= 0v + (-(0v))
= 0

 

[JamesGoh]

thanks Singularity. Always interested in seeing different approaches to a problem