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I Tensors - Winitzki - Lemma 3

  1. Mar 8, 2016 #1
    I am reading Segei Winitzki's book: Linear Algebra via Exterior Products ...

    I am currently focused on Section 1.7.3 Dimension of a Tensor Product is the Product of the Dimensions ... ...

    I need help in order to get a clear understanding of an aspect of the proof of Lemma 3 in Section 1.7.3 ...

    The relevant part of Winitzki's text reads as follows:


    ?temp_hash=9370430caeb830ed3fa7ba46922fa5c7.png
    ?temp_hash=9370430caeb830ed3fa7ba46922fa5c7.png




    In the above quotation from Winitzki we read the following:

    " ... ... By the result of Exercise 1 in Sec. 6.3 there exists a covector [itex]f^* \in V^*[/itex] such that

    [itex]f^* ( v_j ) = \delta_{ j_1 j }[/itex] for [itex]j = 1, \ ... \ ... \ , \ n[/itex] ... ... "


    I cannot see how to show that there exists a covector [itex]f^* \in V^*[/itex] such that

    [itex]f^* ( v_j ) = \delta_{ j_1 j }[/itex] for [itex]j = 1, \ ... \ ... \ , \ n[/itex] ... ...


    Can someone help me to show this from first principles ... ?


    It may be irrelevant to my problem ... but I cannot see the relevance of Exercise 1 in Section 6 which reads as follows:


    ?temp_hash=9370430caeb830ed3fa7ba46922fa5c7.png


    Exercise 1 refers to Example 2 which reads as follows:


    ?temp_hash=9370430caeb830ed3fa7ba46922fa5c7.png
    ?temp_hash=9370430caeb830ed3fa7ba46922fa5c7.png




    BUT ... since I wish to show the result:

    ... ... ... "there exists a covector [itex]f^* \in V^*[/itex] such that

    [itex]f^* ( v_j ) = \delta_{ j_1 j }[/itex] for [itex]j = 1, \ ... \ ... \ , \ n[/itex] ... ..."


    ... from first principles the above example is irrelevant ... BUT then ... I cannot see its relevance anyway!!!


    Hope someone can help ... ...

    Peter




    ===========================================================

    *** NOTE ***

    To help readers understand Winitzki's approach and notation for tensors I am providing Winitzki's introduction to Section 1.7 ... ... as follows ... ... :


    ?temp_hash=e67c8c3ac372f69967191c4fbbbc85c6.png
    ?temp_hash=e67c8c3ac372f69967191c4fbbbc85c6.png
    ?temp_hash=e67c8c3ac372f69967191c4fbbbc85c6.png
     

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    Last edited: Mar 8, 2016
  2. jcsd
  3. Mar 9, 2016 #2

    Samy_A

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    Read what lavinia explained here:
     
  4. Mar 9, 2016 #3

    andrewkirk

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    Firstly, the author's use of notation is unhelpful, because they are using ##f^*## for two very different maps, one from ##V## to ##\mathbb{R}## and one from ##V\otimes W## to ##\mathbb{R}##. They should use different symbols for the two different maps.

    Let's assume ##f^*## refers only to the first map. We show there exists such a map (which is a covector) simply by defining its operation on a set of basis vectors, and then extending it to cover the whole of the domain ##V## using the linearity rules. To be very clear, let's use a different symbol. Given a basis ##B\equiv \{v_1,...,v_n\}## for ##V##, we first define a function ##g^{j_1}:B\to\mathbb{R}## by

    $$g^{j_1}(v_k)=\delta_k^{j_1}$$

    for ##k\in\{1,...,n\}##.

    Then we define ##f^*## to be the linear function from ##V## to ##\mathbb{R}## that agrees with ##g^{j_1}## on ##B##. It is a basic result of linear algebra (no tensors required) that such a function exists and is unique. If you are not comfortable just accepting that fact, a proof should be easy to find in any decent linear algebra text, or it's easy to prove from scratch. The existence follows from the fact that ##\{v_1,...,v_n\}## spans ##V##. To prove uniqueness, assume there are two such extension functions. Then their difference is also a linear function from ##V## to ##\mathbb{R}## and, by looking at its operation on the basis vectors, it's easily seen to be identically zero.
     
    Last edited: Mar 9, 2016
  5. Mar 9, 2016 #4
    Thanks Samy and Andrew ... appreciate the help ...

    Still working through the Lemma ... but I must say Andrew, your point on the notation really changed things for me ... I now have a pretty good understanding regarding what is going on in the proof of Lemma 3 ...

    Thanks again,

    Peter
     
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