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Are vectors assumed to be with respect to a standard basis?

  1. Aug 26, 2013 #1
    For example, if were given only a vector <5, 3, 1>, is this assumed to be respect with the standard basis of R^3?

    And would this mean that any nonstandard basis is with respect to a standard basis?
     
  2. jcsd
  3. Aug 26, 2013 #2

    Simon Bridge

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    You are free to expand a vector against any basis set you like.
    For your example, any set of three vectors that form a basis for R^3 is fine.

    There is a standard representation though - from the rules for vector spaces.
     
  4. Aug 26, 2013 #3

    Office_Shredder

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    The vector (5,3,1) is defined without reference to any basis at all. R3 is lists of three real numbers, and (5,3,1) is a list of three real numbers. It turns out that this is a vector space and that it has a basis for which (5,3,1) = 5e1 + 3e2 + 1e3 but that's totally incidental to the definition of (5,3,1)
     
  5. Aug 26, 2013 #4
    So (5,3,1) with no reference to a basis is NOT a vector? And the lists of R^3 happen to be vectors because R^3 satisfy the properties of a vector space?
     
  6. Aug 26, 2013 #5
    There's no definition of what is/isn't a vector. There's a definition of a vector space, and given a vector space we call its elements "vectors".

    It's exactly as Office_Shredder put it:
    - We define the set [itex]\mathbb R^3[/itex] as the set of ordered triples of real numbers.
    - For any real number [itex]c \in \mathbb R[/itex] and any ordered triple [itex]x\in \mathbb R^3[/itex], we define what we mean by [itex]cx \in \mathbb R^3[/itex].
    - For any two ordered triples [itex]x,y\in \mathbb R^3[/itex], we define what we mean by [itex]x+y \in \mathbb R^3[/itex].
    Then it turns out that these things together satisfies what we call a vector space.

    All of this makes sense without needing to define the word "vector".
     
  7. Aug 26, 2013 #6

    WannabeNewton

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    A vector exists independent of any basis you choose to represent it in. If ##V## is a vector space then ##v\in V## is a vector, that's all. You can choose a basis ##(e_i)## for ##V## and represent ##v## as ##v = \sum v_i e_i## and further express ##v## relative to this basis in coordinate form ##[v]_{(e_i)} = (x_1,...,x_n)## by passing over to ##K^n## where ##K## is the field that ##V## is over.
     
  8. Aug 26, 2013 #7

    Simon Bridge

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    I'm wondering if this question is more about notation than basis sets.

    if ##\small \{\vec{u},\vec{v},\vec{w}\}## is an arbitrary basis for ##\small \mathbb{R}^3##, then is ## \small <x,y,z>=x\vec{u}+y\vec{v}+z\vec{w}## ?

    I don't know if I'm putting this well.

    If you see <5,3,1> written down, with no other information but that it is a vector, it is safe to expand it as 5<1,0,0>+3<0,1,0>+<0,0,1>.

    Another basis, written in terms of the standard one, may be {<1,1,0>,<1,-1,0>,<0,0,1>} ... we would not expect <5,3,1> to mean 5<1,1,0>+3<1,-1,0>+<0,0,1> unless we are told that it does.

    However, three numbers written like that could be a list of data, or a spherical-polar position, or just a row of text characters - depending on context. If <5,3,1> is the only thing on the page, you have no information.

    Does that answer the question?
     
  9. Aug 27, 2013 #8
    For a bit of context, this arised from an example in Anton's linear algebra text; in the first part he showed how to find the coordinate vector of a given vector (I'm gonna call it <5, 3, 1>) in terms of a given nonstandard basis S.
    Next part, find a vector in R^3 whose coordinate vector with respect to S is <3, 2, 4>.

    So a list of numbers in R^3, (5, 3, 1) exists independent of a basis but happens to have that same representation with respect to the standard basis of R^3 (and also happens to be a vector?)
     
  10. Aug 27, 2013 #9

    Office_Shredder

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    This is correct.
     
  11. Aug 28, 2013 #10

    Stephen Tashi

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  12. Aug 29, 2013 #11

    Simon Bridge

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    Yep - I wondered if this could have been the motivation of the question.
     
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