Using diagonality in Einstein notation












1














Given a diagonal matrix $D$, with diagonal elements given by vector $mathbf{d}$. Representing this in Einstein notation gives



$$
D_{ij} = delta_{ijk} d_k
$$



where



$$
delta_{ijk} =
begin{cases}
1 & text{if } i = j = k \
0 & text{ otherwise}
end{cases}
$$



If I now apply this in a matrix multiplication, e.g.



$$
(AD)_{lj} = A_{li} D_{ij} = A_{li} delta_{ijk} d_k = A_{li} d_i
$$
or
$$
(ADA)_{ml} = A_{mi} D_{ij} A_{jl} = A_{mi} delta_{ijk} d_k A_{jl} = A_{mi} d_i A_{il}
$$



The first example makes sense entry-wise, but only if there is no summation over $i$. Also, the indices on the LHS and RHS do not match anymore. The same is true for the second example, but only if there is summation over all three $i$'s.



This obviously violates Einstein notation, but I don't see in which step a false assumption is made. My questions are therefore:




  1. Where do I go wrong in my reasoning?

  2. Is there some other way to exploit the fact that $D$ is diagonal (in index notation)?






linear-algebra index-notation







share|cite|improve this question



























    1














    Given a diagonal matrix $D$, with diagonal elements given by vector $mathbf{d}$. Representing this in Einstein notation gives



    $$
    D_{ij} = delta_{ijk} d_k
    $$



    where



    $$
    delta_{ijk} =
    begin{cases}
    1 & text{if } i = j = k \
    0 & text{ otherwise}
    end{cases}
    $$



    If I now apply this in a matrix multiplication, e.g.



    $$
    (AD)_{lj} = A_{li} D_{ij} = A_{li} delta_{ijk} d_k = A_{li} d_i
    $$
    or
    $$
    (ADA)_{ml} = A_{mi} D_{ij} A_{jl} = A_{mi} delta_{ijk} d_k A_{jl} = A_{mi} d_i A_{il}
    $$



    The first example makes sense entry-wise, but only if there is no summation over $i$. Also, the indices on the LHS and RHS do not match anymore. The same is true for the second example, but only if there is summation over all three $i$'s.



    This obviously violates Einstein notation, but I don't see in which step a false assumption is made. My questions are therefore:




    1. Where do I go wrong in my reasoning?

    2. Is there some other way to exploit the fact that $D$ is diagonal (in index notation)?






    linear-algebra index-notation







    share|cite|improve this question

























      1












      1








      1







      Given a diagonal matrix $D$, with diagonal elements given by vector $mathbf{d}$. Representing this in Einstein notation gives



      $$
      D_{ij} = delta_{ijk} d_k
      $$



      where



      $$
      delta_{ijk} =
      begin{cases}
      1 & text{if } i = j = k \
      0 & text{ otherwise}
      end{cases}
      $$



      If I now apply this in a matrix multiplication, e.g.



      $$
      (AD)_{lj} = A_{li} D_{ij} = A_{li} delta_{ijk} d_k = A_{li} d_i
      $$
      or
      $$
      (ADA)_{ml} = A_{mi} D_{ij} A_{jl} = A_{mi} delta_{ijk} d_k A_{jl} = A_{mi} d_i A_{il}
      $$



      The first example makes sense entry-wise, but only if there is no summation over $i$. Also, the indices on the LHS and RHS do not match anymore. The same is true for the second example, but only if there is summation over all three $i$'s.



      This obviously violates Einstein notation, but I don't see in which step a false assumption is made. My questions are therefore:




      1. Where do I go wrong in my reasoning?

      2. Is there some other way to exploit the fact that $D$ is diagonal (in index notation)?






      linear-algebra index-notation







      share|cite|improve this question













      Given a diagonal matrix $D$, with diagonal elements given by vector $mathbf{d}$. Representing this in Einstein notation gives



      $$
      D_{ij} = delta_{ijk} d_k
      $$



      where



      $$
      delta_{ijk} =
      begin{cases}
      1 & text{if } i = j = k \
      0 & text{ otherwise}
      end{cases}
      $$



      If I now apply this in a matrix multiplication, e.g.



      $$
      (AD)_{lj} = A_{li} D_{ij} = A_{li} delta_{ijk} d_k = A_{li} d_i
      $$
      or
      $$
      (ADA)_{ml} = A_{mi} D_{ij} A_{jl} = A_{mi} delta_{ijk} d_k A_{jl} = A_{mi} d_i A_{il}
      $$



      The first example makes sense entry-wise, but only if there is no summation over $i$. Also, the indices on the LHS and RHS do not match anymore. The same is true for the second example, but only if there is summation over all three $i$'s.



      This obviously violates Einstein notation, but I don't see in which step a false assumption is made. My questions are therefore:




      1. Where do I go wrong in my reasoning?

      2. Is there some other way to exploit the fact that $D$ is diagonal (in index notation)?






      linear-algebra index-notation




      linear-algebra index-notation






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Nov 21 '18 at 11:24









      user495268

      183




      183






















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          What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this



          $$
          D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
          $$



          which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option



          begin{eqnarray}
          (A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
          end{eqnarray}






          share|cite|improve this answer





















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            1 Answer
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            1






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            0














            What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this



            $$
            D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
            $$



            which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option



            begin{eqnarray}
            (A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
            end{eqnarray}






            share|cite|improve this answer


























              0














              What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this



              $$
              D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
              $$



              which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option



              begin{eqnarray}
              (A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
              end{eqnarray}






              share|cite|improve this answer
























                0












                0








                0






                What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this



                $$
                D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
                $$



                which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option



                begin{eqnarray}
                (A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
                end{eqnarray}






                share|cite|improve this answer












                What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this



                $$
                D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
                $$



                which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option



                begin{eqnarray}
                (A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
                end{eqnarray}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 '18 at 20:40









                caverac

                13.8k21130




                13.8k21130






























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