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Prove vector space postulate 1.X = X is independent of others

  1. Jul 5, 2006 #1
    Hi everyone,

    I would like to seek help in proving that the vector space postulate 1.X = X cannot be derived from the other postulates, e.g. X + 0 = X, X + (Y + Z) = (X + Y) + Z.

    The only hint I am given is to construct the "pseudo-scalar product"
    c # X = the projection of c.X on a fixed line.

    This problem is from Birkhoff and Maclane's "A Survey of Modern Algebra".
     
  2. jcsd
  3. Jul 5, 2006 #2

    StatusX

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    The vector space axioms are:

    1. V is an abelian group under addition, which means:
    a) (X+Y)+Z=X+(Y+Z)
    b) there is a vector 0 with X+0=X
    c) For each X, there is a vector -X with X+(-X)=0
    d) X+Y=Y+X

    2. Let F be the underlying field (eg, the real numbers). Then there is a map from F x V into V (ie, for any r in F and X in V, there is a unique vector rX in V), which satisfies:
    a) r(X+Y) = rX+rY
    b) (r+s)X = rX+sX
    c) (rs)X = r(sX)
    d) 1X = X

    So what you want to do is show that 2 d) is independent from the other axioms, right? What you need to do is come up with a structure that satisfies all other axioms, but not 2d). That is, find a field F (probably the real numbers) and an abelian group V (probably Rn under vector addition), and find a new definition scalar product such that 2 a-c are satisfied, but not 2). It sounds like they already did the hard part by giving you such a scalar product. All you need to do is verify it has these properties.
     
  4. Jul 5, 2006 #3
    Hi StatusX,

    Thanks a many. While reading your reply, I was thinking: is it possible to define this pseudo-scalar product

    r # X = 0, where X is a vector in R^N, and so forms an abelian group under addition ?

    In this case, we have
    2a) r # (X + Y) = r # (Z) = 0 = 0 + 0 = r # X + r # Y, where Z = X + Y

    b) (r + s) # X = k # X = 0 = 0 + 0 = r # X + s # X, k = r + s

    c) (rs) # X = k # X = 0 = r # 0 = r # (s # X)

    d) 1 # X = 0, which is not necessarily equal to X.

    Thus 2a) - c) are satisfied, but not d).

    really appreciate the help, thanks.
     
  5. Jul 5, 2006 #4

    StatusX

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    Yes, that works nicely. It looks as if the scalar products that satisfy axioms 2a-c are exactly those that are projections onto some subspace of the vector space. If this is the whole space, the last axiom is satisfied, and if it is any smaller, it is not. The example they gave as a hint used a 1 dimensional subspace (a line), where as your example uses a zero dimensional subspace, and so is even simpler.
     
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