On the axioms of vector space

[cosmic dust]

Hello, I'd like to make a, probably stupid, question regarding the axioms that define a vetor space. Among them, there are the axioms:

λ$\cdot$(μ$\cdot$X) = (λμ)$\cdot$Χ (1) and 1$\cdot$Χ=Χ (2)

for all λ,μ in the field and for all X in the vector space, where 1 is the identity of the multiplication on the field and "$\cdot$" indicates the multiplication of a scalar with a vector. So, my question is: can axiom (2) derived as a special case of (1) and, therefore, is not really an axiom?

I ask this because, setting λ = μ = 1 in (1), we get:

1$\cdot$(1$\cdot$X) = 1$\cdot$Χ

Since X is an abitrary vector, 1$\cdot$Χ will also be an arbitrary vector, say Y. Then the above equation reads:

1$\cdot$Y = Y

for all Y in the vector space. But this is "axiom" (2)! I understand that, since both axioms are used in the standard definition of vector space, they should indeed be indepentend, despite the above objection. So what is the flaw in my thought about this?

 

[Office_Shredder]

As far as showing it for all vectors in the vector space, is it obvious that if I give you an arbitrary vector Y that you can find some vector X such that 1⋅X = Y?In fact here's a simple counterexample: Let my "vector space" be R² with regular addition and with scalar multiplication defined by

$$\lambda(a,b) = (\lambda a, 0)$$

Let's check the other axioms. Since addition is untouched all the addition only axioms are still true. For scalar multiplication:
Distributivity:
$$(\lambda+\mu)(a,b) = \left( (\lambda+\mu)a,0 \right) = (\lambda a,0) + (\mu a, 0) = \lambda(a,b) + \mu(a,b)$$
Other distributivity:
$$\lambda\left( (a,b) + (c,d) \right) = \lambda(a+c,b+d) = (\lambda a+\lambda c, 0) = (\lambda a,0) + (\lambda c,0) = \lambda(a,b) + \lambda(c,d)$$
Associativity of multiplication:
$$\mu \left( \lambda(a,b) \right) = \mu (\lambda a, 0) = (\mu \lambda a, 0) = \left( \mu \lambda \right) (a,b)$$

So great! We have ourselves a good ol' fashion vector space

 

[D H]

Since X is an abitrary vector, 1$\cdot$Χ will also be an arbitrary vector, say Y. Then the above equation reads:

1$\cdot$Y = Y

for all Y in the vector space. But this is "axiom" (2)! I understand that, since both axioms are used in the standard definition of vector space, they should indeed be indepentend, despite the above objection. So what is the flaw in my thought about this?

The flaw is that Y is not an arbitrary vector in the space. It is a member of the subset of the vector space that can be expressed as $1 \cdot X$, where X is some member of the vector space. What axiom (2) adds is that it says all members of of the vector space can be expressed in this form, and in particular, $1 \cdot X = X$.

 

[cosmic dust]

So, that would be true if the function:

1$\cdot$ : V→V , 1$\cdot$:X→1$\cdot$X

is surjective? But this is an extra axiom, more complicated than it's substitute (that is (2)). Better keep (2) as it is...

Thank's for your replies!

 

[blue_raver22]

I would like to address your question and clarify any confusion you may have about the axioms of vector space. First of all, it is important to understand that the axioms of vector space are the fundamental building blocks that define a vector space. They are necessary conditions that must be satisfied in order for a set to be considered a vector space. Therefore, all of the axioms, including (1) and (2), are equally important and cannot be derived from each other.

To address your specific question, let's look at axiom (2) which states that the scalar 1 multiplied by any vector X must equal X. This axiom is not a special case of (1) because it does not involve any other scalar or vector. It is simply stating that the identity element of the scalar field must behave in a certain way when multiplied by a vector.

On the other hand, axiom (1) involves two scalars, λ and μ, and a vector X. It states that the result of multiplying λ by the result of multiplying μ by X is the same as multiplying the product of λ and μ by X. This axiom is necessary in order to define the distributive property of vector multiplication.

To summarize, both axioms (1) and (2) are necessary and cannot be derived from each other. They serve different purposes in defining a vector space and work together to create a consistent and useful mathematical framework for studying vectors. I hope this helps clarify any confusion and highlights the importance of all the axioms in defining a vector space.