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I Matrix Rings - Basic Problem with Meaning of Notation

  1. Oct 8, 2016 #1
    I am reading Louis Rowen's book, "Ring Theory" (Student Edition) ...

    I have a problem interpreting Rowen's notation in Section 1.1 Matrix Rings and Idempotents ...

    The relevant section of Rowen's text reads as follows:


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    ?temp_hash=20c8a6acef6aba85cccfabc4dc6aef01.png




    In the above text from Rowen, we read the following:

    " ... ... We obtain a more explicit notation by defining the ##n \times n## matric unit ##e_{ij}## to be the matrix whose ##i-j## entry is ##1##, with all other entries ##0##.

    Thus ##( r_{ij} ) = \sum_{i,j =1}^n r_{ij} e_{ij}## ; addition is componentwise and multiplication is given according to the rule

    ##( r_1 e_{ij} ) ( r_2 e_{uv} ) = \delta_{ju} (r_1 r_2) e_{iv}##

    ... ... ... "




    I am having trouble understanding the rule ##( r_1 e_{ij} ) ( r_2 e_{uv} ) = \delta_{ju} (r_1 r_2) e_{iv}## ... ...

    What are ##r_1## and ##r_2## ... where exactly do they come from ...

    Can someone please explain the rule to me ...?

    To take a specific example ... suppose we are dealing with ##M_2 ( \mathbb{Z} )## and we have two matrices ...


    ##P = \begin{pmatrix} 1 & 3 \\ 5 & 4 \end{pmatrix}##


    and ##Q = \begin{pmatrix} 2 & 1 \\ 3 & 3 \end{pmatrix}## ...


    In this specific case, what are ##r_1## and ##r_2## ... ... and how would the rule in question work ...?


    Hope someone can help ...

    Peter
     

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    Last edited: Oct 8, 2016
  2. jcsd
  3. Oct 8, 2016 #2

    fresh_42

    Staff: Mentor

    In this case, ##r_1## and ##r_2## are simply two arbitrary scalars, i.e. elements of ##R##.
    I wished you would have chosen an example with more zeros ...
    So here we go:

    ##P = 1 \cdot e_{11}+ 3 \cdot e_{12} + 5 \cdot e_{21} + 4 \cdot e_{22}## and ##Q = 2 \cdot e_{11}+ 1 \cdot e_{12} + 3 \cdot e_{21} + 3 \cdot e_{22}## or

    $$P= 1 \cdot \begin{pmatrix} 1 & 0 \\0 & 0 \end{pmatrix}+3 \cdot \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} +5 \cdot \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + 4 \cdot \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$
    and
    $$Q= 2 \cdot \begin{pmatrix} 1 & 0 \\0 & 0 \end{pmatrix}+1 \cdot \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} +3 \cdot \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + 3 \cdot \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$

    And because matrix multiplication is row times column ##e_{ij} \cdot e_{kl} = \delta_{jk} \cdot e_{il}##.
    The scalars ##r_1,r_2## or ##\{1,3,5,4\}## and ##\{2,1,3,3\}## in your cases may stand anywhere, usually left to the basis, as they are the coefficients of the matrices ##P,Q## regarded as elements of the ##\mathbb{Z}-##module ##\mathbb{M}_2(\mathbb{Z})## with the basis ## \{e_{ij}\}##. If you take a field instead of the integers, you get a vector space, which of course is still a ring and therefore an algebra.
     
  4. Oct 8, 2016 #3

    Stephen Tashi

    User Avatar
    Science Advisor

    My guess for an example of that is:

    ##r_1 e_{22} = r_1 \begin{pmatrix} 0&0&0 \\ 0&1&0\\0&0&0 \end{pmatrix}##
    ##r_2 e_{23} = r_2 \begin{pmatrix} 0&0&0 \\ 0&0&1\\0&0&0 \end{pmatrix}##

    ##(r_1 e_{22})(r_2e_{23}) = \delta_{22} r_1 r_2 e_{23} = r_1 r_2 \begin{pmatrix} 0&0&0\\0&0&1\\0&0&0\end{pmatrix}##
    ## = \begin{pmatrix} ( 0&0&0\\0&0&r_1r_2\\0&0&0 \end{pmatrix}##

    I think they are scalars. The notation is confusing - ##(a_{ij})## for a matrix but also ##(r_1 a_{ij})## for a scalar times a matrix. Why not denote a scalar times a matrix by ##r_1(a_{ij})##?


    I think you are expected to express each matrix as a sum - e.g:
    ## P = (1)\begin{pmatrix} 1&0\\0&0 \end{pmatrix} + (3)\begin{pmatrix} 0&1\\0&0\end{pmatrix} + ... ##

    Then you multiply the matrices together using the distributive law for the sums. The result only involves products of form ##e_{ij} e_{uv}##. This is equivalent to the usual way of defining matrix multiplication.
     
  5. Oct 8, 2016 #4

    fresh_42

    Staff: Mentor

    If you now multiply ##P \cdot Q## you get a summation of terms (distributive law) like
    ##(5 \cdot e_{21}) \cdot (2 \cdot e_{11}) = 5 \cdot 2 \cdot e_{21} \cdot e_{11} = 10 \cdot e_{21} \cdot e_{11} = 10 \cdot \delta_{11} \cdot e_{21} = 10 \cdot e_{21} ##.

    Personally I like to think of ##e_{ij} \cdot e_{kl}=\delta_{jk} \cdot e_{il}## as

    "Match the inner indices and forget it, if they aren't equal. If they are equal, drop them and take the outer indices as the new ones."
     
  6. Oct 9, 2016 #5


    Thanks for the help, fresh_42 ... BUT ...

    You write:

    "... ... In this case, ##r_1## and ##r_2## are simply two arbitrary scalars, i.e. elements of ##R##. ... ...


    But the formula ##( r_1 e_{ij} ) ( r_2 e_{uv} ) = \delta_{ju} (r_1 r_2) e_{iv}##"is supposed to represent the multiplication of two matrices and ##( r_1 e_{ij} )## does not seem to represent an arbitrary matrix ... ... ?

    Note that regarding this problem I received some help from on the Math Help Boards ...

    Evgeny writes:

    " ... ... Evgeny.Makarov ... ... ##r_1## and ##r_2## are numbers, or, more precisely, elements of the ring ##R##. They are also coefficients, or coordinates, of a matrix in the basis consisting of ##e_{ij}##. If one matrix is ##\displaystyle\sum_{i,j=1}^nr^{(1)}_{ij}e_{ij}## and another is ##\displaystyle\sum_{i,j=1}^nr^{(2)}_{ij}e_{ij}##, then when you multiply them, you apply distributivity and get ##\displaystyle\sum_{i,j,u,v=1}^nr^{(1)}_{ij}e_{ij}r^{(2)}_{uv}e_{uv}##, and the formula in your quote says how to compute ##r^{(1)}_{ij}e_{ij}r^{(2)}_{uv}e_{uv}##. ... ... "

    Does Evgeny's post look right to you ... ?

    Peter
     
    Last edited: Oct 9, 2016
  7. Oct 9, 2016 #6

    Thanks for the help, Stephen ... ...

    Still reflecting on this issue ... but see the contribution of Evgeny Makarov from the MHB forums ... I have cut Evgeny's post into my reply to fresh_42 ...

    What do you think of Evgeny's interpretation ... ?

    Peter
     
  8. Oct 9, 2016 #7

    fresh_42

    Staff: Mentor

    Yes, it is right. The different naming of ##r_1## and ##r_2## refers to different views of the same thing.
    ##R## is the set, where our matrix entries come from. You might call them numbers, which I avoided, because it doesn't have to be numbers in the usual sense. It could be any ring, e.g. polynomials which usually are not called numbers. Since ##\mathbb{M}_n(R)## forms an ##R-## algebra, I had chosen the word scalar. In the end, they are simply ring elements of ##R##.
    And because ##R## is not necessarily a field and ##\mathbb{M}_n(R)## therefore not necessarily a vector space, but only a finitely generated ##R-##module, I avoided the word coordinate, too. But the difference is marginal in this context. One might say coefficients as well, as the ##e_{ij}## are the generators of ##M=\mathbb{M}_n(R)## and the ##r^{(1)}_{ij}\; , \;r^{(2)}_{ij} \in R## coefficients at the ##e_{ij}##. In the case of ##R## being a field, we call the ##\{e_{ij}\}## a vector basis, ##\mathbb{M}_n(R)## a vector space (or an algebra) and their coefficients coordinates.

    The difference between a vector space and an algebra is, that an algebra ##\mathcal{A}## doesn't need to have a field as scalar set ##R## and even more important, that it has two multiplications: a scalar multiplication ##R \,\cdot\,\mathcal{A}## like vector spaces and modules plus a inner multiplication ##\mathcal{A}\,\cdot\,\mathcal{A}## like rings.

    Edit: Btw, coefficient literally means "number at".
     
  9. Oct 9, 2016 #8
    Thanks again for all your help, fresh_42 ...

    Appreciate your guidance and help ...

    Peter
     
  10. Oct 10, 2016 #9

    Stephen Tashi

    User Avatar
    Science Advisor

    I agree with it. You text is talking about matrices whose elements belong to a ring so a "scalar" is an element of whatever ring is being discussed.
     
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