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I Tensor Algebras - Cooperstein Theorem 10.8

  1. Apr 22, 2016 #1
    I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

    I am focused on Section 10.3 The Tensor Algebra ... ...

    I need help in order to get a basic understanding of Theorem 10.8 which is a Theorem concerning the direct sum of a family of subspaces as a solution to a UMP ... the theorem is preliminary to tensor algebras ...

    I am struggling to understand how the function ##G## as defined in the proof actually gives us ##G \circ \epsilon_i = g_i## ... ... if I see the explicit mechanics of this I may understand the functions involved better ... and hence the whole theorem better ...


    Theorem 10.8 (plus some necessary definitions and explanations) reads as follows:


    ?temp_hash=418f57e2318ea815d5165ddfe81dbe73.png



    In the above we read the following:


    " ... ... Then define


    ##G(f) = \sum_{j = 1}^t g_{i_j} (f(i_j)) ##


    We leave it to the reader to show that this is a linear transformation and if ##G## exists then it must be defined in this way, that is, it is unique. ... ... "


    Can someone please help me to ...

    (1) demonstrate explicitly, clearly and in detail that ##G(f) = \sum_{j = 1}^t g_{i_j} (f(i_j)) ## satisfies ##G \circ \epsilon_i = g_i## (if I understand the detail of this then I may well understand the functions involved better, and in turn, understand the theorem better ...)


    (2) show that ##G## is a linear transformation and, further, that if ##G## exists then it must be defined in this way, that is, it is unique.



    Hope that someone can help ...

    Peter
     

    Attached Files:

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  3. Apr 22, 2016 #2

    andrewkirk

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    We are given that ##spt(f)=\{i_1,....,i_t\}##. That means that ##\exists v_{i_1}\in V_{i_1},...,v_{i_t}\in V_{i_t}## such that
    $$f=\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$

    Then for ##k\in \{1,...,t\}## we have, by replacing ##f## by ##\epsilon_{i}(v_{i})## in the definition of ##G(f)##:

    $$G\circ\epsilon_{i}(v_{i})\equiv G(\epsilon_{i}(v_{i}))\equiv\sum_{j=1}^t g_{i_j}(\epsilon_{i}(v_{i})(i_j))$$

    In your earlier thread ##\epsilon_k## was defined as the function from ##V_k## to the direct product ##V## that maps ##v_k## to
    ##(0,0,.....0,v_k,0,....,0)## where ##v_k## is in the ##k##th position. That only covers finite direct sums. However it looks from the above as though Cooperstein was - between the two sections you quoted - moved on to defining and allowing infinite direct sums (because of his use of an index set ##I## of unspecified cardinality, rather than just labelling the component spaces as ##V_1## to ##V_n##). That means that ##\epsilon_k## needs an appropriately modified definition. I'm guessing the definition he's using is something like that ##\epsilon_i:V_i\to V## such that ##\epsilon_i(v_i)(j)## is zero for all ##j\in I## except ##i##, for which it gives ##v_i##.

    Applying that to the above equation, we have

    $$G\circ\epsilon_{i}(v_{i})\equiv \sum_{j=1}^t g_{i_j}(\epsilon_{i}(v_{i})(i_j))
    =g_{i}(v_i)$$
    as required.
     
    Last edited: Apr 22, 2016
  4. Apr 22, 2016 #3
    Thanks so much, Andrew ...

    Just working through your post and reflecting on what you have said ...

    Thanks again for your help ...

    Peter
     
  5. Apr 22, 2016 #4

    andrewkirk

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    And the linearity of [itex]G[/itex] follows from the linearity of the [itex]g_k[/itex]. Say we have [itex]f_1=\sum_{k=1}^{t_1}\epsilon_{1k}(v_{1k}),f_2=\sum_{k=1}^{t_2}\epsilon_{2k}(v_{2k})[/itex] where [itex]spt(f_1)=a_{11},...,a_{1t_1},spt(f_2)=a_{21},...,a_{2t_2}[/itex]
    so that [itex]spt(f_1+f_2)\subseteq spt(f_1)\cup spt(f_2)=i_1...i_t[/itex] ([itex]t\leq t_1+t_2[/itex]) so that we can write
    $$f_1=\sum_{k=1}^{t}\epsilon_{3k}(u_{1k}),f_2=\sum_{k=1}^{t}\epsilon_{3k}(u_{2k})$$
    where, for [itex]r\in\{1,2\}[/itex], [itex]u_{rk}=v_{rj}[/itex] for the [itex]j[/itex] such that [itex]a_{rj}=i_k[/itex] if [itex]i_k\in spt(f_r)[/itex] and otherwise [itex]u_{rk}=0[/itex].

    Then we have
    $$G(f_1+f_2)=
    G\left(\sum_{k=1}^{t}\epsilon_{3k}(u_{1k})+\sum_{k=1}^{t}\epsilon_{3k}(u_{2k})\right)
    =\sum_{j=1}^t g_{i_j}\left(\sum_{k=1}^{t}\epsilon_{3k}(u_{1k})(i_j)+\sum_{k=1}^{t}\epsilon_{3k}(u_{2k})(i_j)\right)$$

    $$=\sum_{j=1}^t g_{i_j}\left(\sum_{k=1}^{t}\epsilon_{3k}(u_{1k})(i_j)\right)+
    \sum_{j=1}^t g_{i_j}\left(\sum_{k=1}^{t}\epsilon_{3k}(u_{2k})(i_j)\right)
    $$

    $$=\sum_{j=1}^t g_{i_j}\left(f_1(i_j)\right)+
    \sum_{j=1}^t g_{i_j}\left(f_2(i_j)\right)
    =G(f_1)+G(f_2)$$

    Proving that [itex]G(\lambda f)=\lambda G(f)[/itex] follows the general pattern of this proof but is much easier.
     
  6. Apr 22, 2016 #5

    andrewkirk

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    Finally, uniqueness. We use the fact that ##\{\epsilon_i(v_i)\ :\ i\in I\wedge v_i\in V_i\}## is a spanning set for ##V##. I used that in the first line of the previous post but omitted to point that out. The proof is straightforward, based on the fact that all elements of ##V## have finite support.

    Say we have another linear map ##G':V\to W## such that ##\forall i\in I:\ G'\circ \epsilon_i =g_i##. Then, for any ##f\in V##, writing ##f=\sum_{k=1}^t \epsilon_k(v_k)##, we have

    $$G'(f)=G'\left(\sum_{k=1}^t \epsilon_k(v_k)\right)
    =\sum_{k=1}^t G'\circ \epsilon_k(v_k)
    =\sum_{k=1}^t g_k(v_k)
    =\sum_{k=1}^t G\circ \epsilon_k(v_k)
    =G\left(\sum_{k=1}^t \epsilon_k(v_k)\right)=G(f)
    $$

    So ##G'=G##.
     
  7. Apr 22, 2016 #6

    Hi Andrew,

    Thanks again for your help ...

    Just a couple of clarifications, though ...

    1. I know that ##f## has finite support which means ##f## is non-zero at only finite ##i \in I## ... but ... I cannot follow how/why we have

    $$f=\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$


    Could you please explain ( perhaps, if you would, slowly and in detail :frown: ,,, ) why/how this is true ... and maybe what it means ..



    2. You write:

    " ... ... something like that ##\epsilon_i:V_i\to V## such that ##\epsilon_i(v_i)(j)## is zero for all ##j\in I## except ##i##, for which it gives ##v_i##. ... ... "


    ##\epsilon_i## has domain ##V_i## and so I understand an expression like ##\epsilon_i (v_i)## ... but your expression ##\epsilon_i(v_i)(j)## has two arguments, namely ##v_i## and ##j## ... ??? ... can you explain what is going on ...


    Sorry to be slow and perhaps over-careful ... but I am trying to ensure that I fully understand the material ...

    Peter
     
  8. Apr 22, 2016 #7

    andrewkirk

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    It comes from the adaptation of the projection functions ##\pi_i:V\to V_i##, defined in your earlier thread, to the infinite-dimensional case. A little reflection shows that the natural adaptation (which may perhaps be in the intervening passages of Cooperstein) is to define ##\pi_i## by ##\pi_i(f)\equiv f(i)##.

    There is a little work to be done to re-prove (a) and (b) from your External Direct Sum thread for the infinite-dimensional case (although I note that Cooperstein didn't even bother proving them in the finite-dimensional case. I think he's a bit slack.), but it should be pretty straightforward.

    Taking that as read, we proceed as follows:

    ##f:I\to \bigcup_{i\in I}V_i## has finite support, so let the support be ##i_1,...,i_t## and let ##v_{i_k}\equiv f(i_k)##.

    Next, we use (b)

    $$\sum_{i\in I}\epsilon_i\circ \pi_i=I_V$$

    to get

    $$f=I_Vf=\sum_{i\in I}\epsilon_i\circ \pi_i(f)
    =\sum_{i\in I}\epsilon_i\left(\pi_i(f)\right)
    =\sum_{i\in I}\epsilon_i\left(f(i)\right)$$
    Note that, by the linearity of ##\epsilon_i##, the elements of this last sum are all zero except when ##i\in\{i_1,...,i_t\}##, so we have

    $$f=\sum_{k=1}^t\epsilon_{i_k}\left(f(i_k)\right)
    =\sum_{k=1}^t\epsilon_{i_k}\left(v_{i_k}\right)$$

    as required.
    ##\epsilon_i(v_i)## is an element of the direct sum ##V##. Recall that the elements of the direct sum are functions from the index set ##I## to ##\bigcup_{i\in I}V_i##. So ##\epsilon_i(v_i)## is such a function, and can thus be applied to an element ##j## of ##I##. When we do this, we write it as ##\epsilon_i(v_i)(j)##. It can help avoid confusion to write this as ##\left(\epsilon_i(v_i)\right)(j)##
     
  9. Apr 23, 2016 #8

    Hi Andrew ... thanks again for the help ...

    But ... just a clarification ... ...

    You write:

    "... ...
    We are given that ##spt(f)=\{i_1,....,i_t\}##. That means that ##\exists v_{i_1}\in V_{i_1},...,v_{i_t}\in V_{i_t}## such that
    $$f=\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$ ... ... "

    and you follow this by writing ... :

    " ... ... Then for ##k\in \{1,...,t\}## we have, by replacing ##f## by ##\epsilon_{i}(v_{i})## in the definition of ##G(f)##:"


    I do not follow this ... shouldn't you be replacing f by $$\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$ ... ... ... ???

    Can you explain ...

    Peter
     
  10. Apr 24, 2016 #9

    andrewkirk

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    In the definition of ##G##, the symbol ##f## stands for any arbitrary element of ##V##. Now ##\epsilon_i(v_i)## is such an element and thus can be slotted in as the argument to ##G## in that definition. Perhaps I should have added that ##v_i## is an arbitrary element of ##V_i##. Note that the support of ##\epsilon_i(v_i):I\to \bigcup_{i\in I}V_i## is the singleton ##\{i\}##, so the sum in the RHS of the definition of ##G## only has one element when the argument is ##\epsilon_i(v_i)##, so we can discard the summation symbol.

    I think Cooperstein has confused the issue by defining ##f## before he defines ##G##, and thereby implying that ##G## somehow depends on ##f##, which it doesn't! It would be better if he had written the following instead:

    Define function ##G:V\to W## such that, ##\forall f\in V##, ##G(f)\equiv \sum_{j\in spt(f)} g_j(f(j))##.

    Applying that to ##\epsilon_i(v_i)## then gives

    $$G(\epsilon_i(v_i))\equiv \sum_{j\in spt(\epsilon_i(v_i))} g_j((\epsilon_i(v_i))(j))
    =\sum_{j\in \{i\}} g_j((\epsilon_i(v_i))(j))
    =g_i((\epsilon_i(v_i))(i))
    =g_i(v_i)
    $$
     
    Last edited: Apr 24, 2016
  11. Apr 24, 2016 #10

    Hi Andrew,

    Thanks to your posts I now have a much better understanding of what is going on ...

    But just one further (minor) issue ...

    You write:

    " ... ... In the definition of ##G##, the symbol ##f## stands for any arbitrary element of ##V##. Now ##\epsilon_i(v_i)## is such an element and thus can be slotted in as the argument to ##G## in that definition. ... ... "

    But ... is ##\epsilon_i(v_i)## really an arbitrary element? ... ... it has the special form of being an element with support equal to a one element set ... shouldn't we be taking a general element - that is an element with support equal to an n-element set where n is any integer ... ...

    Can you help clarify this issue?

    Peter
     
  12. Apr 24, 2016 #11

    andrewkirk

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    ##f## stands for the arbitrary element of ##V##, not ##\epsilon_i(v_i)##. The latter is a specific element.

    What we are doing is substituting a specific element of ##V## for the arbitrary element ##f##, in the first-order logical formula:

    $$\forall f\in V:\ G(f)= \sum_{j\in spt(f)} g_j(f(j))$$

    This is the type of operation enabled by the axiom schema of substitution (aka instantiation) which is Q5 in this axiomatisation of first order logic. The universal quantifier ##\forall## is what enables this substitution.

    It's the same as if we take the formula ##\forall x\in\mathbb{R}:\ x^2\geq 0## and substitute the specific element -2 for the arbitrary element ##x##. That gives us the valid formula ##(-2)^2\geq 0##.
     
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