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What makes a zero vector?

  1. Apr 28, 2012 #1
    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 :)
     
  2. jcsd
  3. Apr 28, 2012 #2


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

    DonAntonio
     
  4. Apr 28, 2012 #3

    HallsofIvy

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    The "zero vector" in any vector space, V, with vector addition [itex]\oplus[/itex] is the vector, a, such that, for any v in V, [itex]a\oplus v= v[/itex]. 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 [itex](a, 1)\oplus (0, 1)= (a+0, 1)= (a, 1)=v[/itex].
     
    Last edited: Apr 28, 2012
  5. Apr 29, 2012 #4
    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 :)
     
  6. Apr 29, 2012 #5
    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.
     
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