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Linear Algebra

  1. Dec 13, 2006 #1
    Could someone please double-check the accuracy of my solutions to two linear algebra problems. Thank you.
    Last edited: Jan 22, 2008
  2. jcsd
  3. Dec 13, 2006 #2


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    There are definite problems with the first solution. First, you can't just pick I0 contained in G, it must also contain I. Also, you say V is partially ordered by set inclusion, but V is the vector space, so this isn't right. In general, however, the proof is a little confusing and overly long. Let B be the collection of linearly independent sets between I and G. This is nonempty (since it contains I) and is partially ordered by inclusion. If C is a totally ordered subset of B, we show that the union of elements in C, denoted W, is an upper bound of C in B. As the union of elements in C, W clearly includes every element of C, so it is an upper bound. Is it an element of B? Well since each element of C is between I and G, W is between I and G. And it's linearly independent because if we could pick some finite collection of vectors in W with a non-trivial linear combination equal to 0, then by total orderedness of C, some element of C would contain each of these vectors, making that element of C linearly dependent, contradiction.

    By Zorn's Lemma, B has a maximal element, call it B. If B spans V, then we're done. Else, there is some v in G \ span(B). But then B U {v} is between I and G, linearly independent, (i.e. it's an element of B) and properly containing B, contradicting the maximality of B. So G \ span(B) is empty, hence span(B) contains span(G) = V.

    Your proof seems to do something like the above in some parts, but seems to be missing some things, and at the same time doing some extra unnecessary things. One definite error is that you don't show that EVERY totally ordered subset has an upper bound, you simply construct one totally ordered subset, and then attempt to prove that it has an upper bound. Moreoever, you don't show that it has an upper bound, you just say that because it's contained in G, it does.

    Also, you don't explicitly say what collection you're applying Zorn's Lemma to. In my proof, I defined a collection B, and so I get a maximal element in B. What is your maximal element a maximal element of?


    Could you define L(V1, ..., Vk; W)?
    Last edited: Dec 13, 2006
  4. Dec 13, 2006 #3
    Thank you so much! L(V1, ..., Vk; W) is the set of all linear mappings from V1 X V2 X...X Vk to W.
    Last edited: Dec 13, 2006
  5. Dec 13, 2006 #4


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    First problem:

    You're actually mixing paradigms here. You need to invoke transfinite induction to construct the chain

    [tex]I_0 \subset I_1 \subset \cdots \subset I_\alpha \subset \cdots G.[/tex]

    (There's no reason to think G-I is countable) But once you have it, you switch over to Zorn's lemma. I think your proof would flow much better if you stuck with one or the other -- either use transfinite induction to directly construct your basis, or gear your proof towards the use of Zorn's lemma right from the beginning. (either way, your proof will be very similar to the proof that any vector space has a basis)

    Incidentally, the above chain is not a chain of linearly independent sets: it contains G!

    For the second problem, you can make it notationally similar by noticing that the general case is essentially an immediate consequence of the case that p = 2, s = 1, i1 = 1, and i2 = 2.

    What is your definition of natural in natural isomorphism? Is it meant informally, or is there actually a technical condition that it implies? In the context of category theory, the isomorphism

    L(A, B ; C) ---> L(A ; L(B, C))

    is natural in A iff, for any map A --> A', the following diagram is commutative.

    Code (Text):

    L(A, B ; C)  ------> L(A ; L(B, C))
         ^                     ^
         |                     |
         |                     |
         |                     |
         |                     |
    L(A', B ; C) ------> L(A' ; L(B, C))
    Where the horizontal maps are the natural isomorphisms, and the vertical maps are the linear maps induced by the map A --> A'.

    And, we have similar diagrams for naturality in B and C. (The vertical arrows point downwards for naturality in C)
    Last edited: Dec 13, 2006
  6. Dec 13, 2006 #5
    Natural isomorphism means an isomorphism independent of the choice of a basis. So I take it that I screwed up the first question but got the second question right (or almost right)?
    Last edited: Dec 13, 2006
  7. Dec 13, 2006 #6


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    Your solution to 2.14 looks good.
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