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    • CommentRowNumber1.
    • CommentAuthorTobyBartels
    • CommentTimeAug 17th 2012

    I got tired of not having linear combination, so now we have it.

    • CommentRowNumber2.
    • CommentAuthorTobyBartels
    • CommentTimeAug 29th 2012

    I added variations and also defined linear span, with a redirect and disambiguation at span.

    • CommentRowNumber3.
    • CommentAuthorColin Tan
    • CommentTimeOct 27th 2015
    I'm not clear about the following: Why is the zero element the linear combination of arity 0? Is this by fiat?
    • CommentRowNumber4.
    • CommentAuthorColin Tan
    • CommentTimeOct 27th 2015
    Added an abstract formulation of linear span in analogy to linearly independent subset.
    • CommentRowNumber5.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 27th 2015

    Re #3: I’d say the best resolution is by following Lawvere in his thesis Functorial Semantics of Algebraic Theories: understand the syntactical term “formal linear combinations” in terms of free modules, rather than the other way around. More generally, proceed with the clear understanding that the essential items in the syntax of algebraic theories, namely terms in the language considered modulo provable equality, are really elements of free objects. This gives a presentation-free of handling algebraic theories, and also helps us understand extreme or vacuous cases.

    Thus the set of formal kk-linear combinations generated by the empty set is the (underlying set of the) free kk-module on the empty set. One easily sees this is the initial kk-module which consists of just the zero element. Then, following the abstract formulation you wrote at linear combination, such a formal linear combination, when interpreted in a general kk-module VV by extending a function f:GVf: G \to V to a kk-module map kGVk\langle G\rangle \to V, yields the element 00 in VV when G=G = \emptyset.

    • CommentRowNumber6.
    • CommentAuthorKarol Szumiło
    • CommentTimeOct 27th 2015
    • (edited Oct 27th 2015)

    I don’t think that you need such a high level discussion to justify that the empty sum should be 00. This is simply the only definition consistent with generalized associativity of addition (which means that I would invoke monads rather than Lawvere theories as a high brow explanation). Let x 1,,x nx_1, \ldots, x_n be any non-empty sequence of elements of a vector space group and let ε\varepsilon stand for the empty sum. Then x 1++x n=x 1++x n+εx_1 + \ldots + x_n = x_1 + \ldots + x_n + \varepsilon (since concatenating x 1,,x nx_1, \ldots, x_n with the empty sequence yields x 1,,x nx_1, \ldots, x_n), so ε\varepsilon acts as identity for addition, so ε=0\varepsilon = 0.

    • CommentRowNumber7.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 27th 2015
    • (edited Oct 27th 2015)

    Yes, Karol, that is the usual explanation. High-level explanations are also good to know.

    I think if you read what I wrote, you’ll see I too was emphasizing monads.

    • CommentRowNumber8.
    • CommentAuthorColin Tan
    • CommentTimeOct 27th 2015
    • (edited Oct 27th 2015)
    Thank you, Todd, for providing the explanation in #5! It helped to clarify my understanding.
    • CommentRowNumber9.
    • CommentAuthorColin Tan
    • CommentTimeOct 27th 2015
    Thank you, Karol, for explaining in #6 why the empty sum, if it exists, has to equal zero. However, is there a reason why the empty sum exists in the first place?
    • CommentRowNumber10.
    • CommentAuthorKarol Szumiło
    • CommentTimeOct 27th 2015

    Todd, monads were of course implicit in your second paragraph. I just thought it was a good idea to mention them by name.

    Colin, now we are going into some finer points of mathematical language. Some things are defined as objects of a certain type satisfying certain properties. Some things are defined as values of operations. In the former case, it is a valid question whether such a thing exists. In the latter case, of course it exists, it is a value of the operation we defined. The definition of the empty sum is of the latter kind so it exists because I chose to define it as a value of an operation. However, there would be nothing wrong with choosing not to define it except that things like Todd’s example with free modules would become tedious and unnatural. I do admit that this adds another layer of arbitarity (is this a word?) to the definition of convex sets that we discussed in another thread.