What is the significance of the zero vector in a vector space?

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The significance of the zero vector in the vector space V5 = {(x, 1) | x ∈ R} is that it serves as the additive identity, represented by the vector (0, 1). In this context, for any vector (a, 1) in V5, the equation (a, 1) ⊕ (0, 1) = (a, 1) holds true, confirming that (0, 1) acts as the zero vector. This understanding emphasizes that the zero vector is defined by its role in maintaining the properties of vector addition rather than its numerical value.

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Ok this is from a tutorial I am redoing again.

V5 = {(x, 1) | x ∈ R}, (x1, 1) ⊕ (x2, 1) := (x1 + x2, 1), c.(x, 1) := (cx, 1).

I understand that there exists a zero vector in this vector space, that comes in the form of (0,1). What I do not understand is why that is considered a zero vector for that vector space?

It is hard for me to 'see past' that 1 in the y-coordinate. Please ease my irritations, thank you :)
 
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74197 said:
Ok this is from a tutorial I am redoing again.

V5 = {(x, 1) | x ∈ R}, (x1, 1) ⊕ (x2, 1) := (x1 + x2, 1), c.(x, 1) := (cx, 1).

I understand that there exists a zero vector in this vector space, that comes in the form of (0,1). What I do not understand is why that is considered a zero vector for that vector space?

It is hard for me to 'see past' that 1 in the y-coordinate. Please ease my irritations, thank you :)



According to what you defined \forall (x,1)\in V5\,\,,\,\,(x,1)\oplus (0,1)=(0,1)\oplus (x,1)=(x,1) and thus that is the zero vector in that set (supposedly, a vector space).

DonAntonio
 
The "zero vector" in any vector space, V, with vector addition \oplus is the vector, a, such that, for any v in V, a\oplus v= v. In other words, it is the "additive identity".

In this particular example, the "zero vector" is (0, 1): if v is any vector in this space, of the form (a, 1), then (a, 1)\oplus (0, 1)= (a+0, 1)= (a, 1)=v.
 
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Ah...I see now. It's not about the 'zero', it's more about the fact that the zero vector makes for an additive identity. Thank you Don and HoI :)
 
74197 said:
Ah...I see now. It's not about the 'zero', it's more about the fact that the zero vector makes for an additive identity. Thank you Don and HoI :)

And if you think about it ... what makes the usual zero, "zero?" It's just that zero satisfies the properties of the zero element in the field of the real numbers. Other than that, it's just a point on the real line exactly like any other point.

That's what abstraction does ... it makes you see familiar things in a new way.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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