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Basic matrices

  1. Oct 14, 2013 #1
    for fixed ## m \geq 2 ## let ## \epsilon (i,j) ## denote the mxm matrix ## \epsilon (i,j)_{rs} = \delta _{ir} \delta _{js} ##

    when m = 2000 show by formal calculations that

    i) ## \epsilon (500,199) \epsilon (1999,10) = \epsilon (500,10) ##

    ii) ## \epsilon (1999,10) \epsilon (500,1999) = 0##

    hence generalise for ##\epsilon (i,j) \epsilon (k,l)## and find a single equation using the kroneeker delta symbol for the generalisation

    my attempt thus far:

    i) rewriting this as ##\delta _{500r} \delta _{1999s} \delta _{1999r} \delta _{10s} ## then we note that ## \delta _{1999s} \delta _{1999r} = \epsilon (1999,1999) ## which gives a 0 everywhere except the 1999th ith position and 1999th jth position but I'm not sure how I can conclude that this gives ##\epsilon (500,10)##

    that's all Ihave so far unfortunately
     
  2. jcsd
  3. Oct 14, 2013 #2

    tiny-tim

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    hi converting1! :smile:
    nooo …

    ##(\epsilon (i,j) \epsilon (k,l))_{r,s} = (\epsilon (i,j))_{r,t} (\epsilon (k,l))_{t,s}## :wink:
     
  4. Oct 14, 2013 #3
    oh damn im so stupid, so here is what I got so far let me know if my logic is flawed:

    i) ## (\epsilon (500,1999))_{r,t} (\epsilon (1999,10))_{t,s} = \delta _{500r} \delta _{1999t} \delta _{1999t} \delta _{10s} ##

    we know that ##\delta _{1999t} \delta _{1999t} = 1 ## if t = 1999, else it will be 0, so we get the required result

    for ii) ## (\epsilon (1999,10))_{r,t} (\epsilon (500,1999))_{t,s} = \delta _{1999r} \delta _{10t} \delta _{500t} \delta _{1999s} ## we now see that ## \delta _{10t} \delta _{500t} = 0 ## as if t = 10, then ## \delta _{500t} = 0 ## and if t = 500 then ## \delta _{10t} = 0 ## hence we will get 0 always, the desired result

    now generalising,

    consider ## (\epsilon (i,j) \epsilon (k,l))_{rs} = (\epsilon (i,j))_{r,t} (\epsilon (k,l))_{t,s} = \delta _{ir} \delta _{jt} \delta _{kt} \delta _{ls} ##

    ## \delta _{jt} \delta _{kt} = 1 ## if j = k, else it's 0, hence we get ## \delta _{ir} \delta _{ls}## if j = k, i.e. we get ## \epsilon (i,l) ## if j = k, else we get 0.

    now for the final part I am not sure how to express it as a single equation, could you give me a little hint?
     
    Last edited: Oct 14, 2013
  5. Oct 14, 2013 #4

    tiny-tim

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    hi converting1! :smile:
    ah, you're still not thinking in terms of the summation convention:

    ##\delta _{1999t} \delta _{1999t} = 1 ## :wink:

    similary ##\delta _{10t} \delta _{500t} = 0 ## ​

    once you've convinced yourself of this, you should be able to do the general case :smile:
     
  6. Oct 14, 2013 #5
    hmm, I'm not so sure I understand, what if t = 20? i.e. ##\delta _{1999,20} \delta _{1999,20} = 0.0 = 0 ? ##
     
  7. Oct 14, 2013 #6
    also take a 2x2 matrix as an example, ## \epsilon (1,2) _{22} = \delta _{12} \delta_{22} = 0 \times 1 = 0 ##
     
  8. Oct 14, 2013 #7

    tiny-tim

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    ah, that's right, you don't understand …

    you can't substitute one value for t into δ1999,tδ1999,t

    the einstein summation convention means that that's a ∑ …

    (over all values of t) δ1999,tδ1999,t :wink:
     
  9. Oct 14, 2013 #8
    really? our lecturer defined ## \epsilon (i,j) _{r,s} = 1 ## if r = i and j = s , else it is 0, so I am just using that definition

    so how do you know ## \delta_{1999,t} \delta_{1999,t} = 1 ##?
     
  10. Oct 14, 2013 #9

    tiny-tim

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    has your lecturer not taught you the summation convention (the einstein summation convention)?
     
  11. Oct 14, 2013 #10
    No, not yet, however in my applied mathematics lecturer we just started it today - so I know very little about it (almost nothing).
     
  12. Oct 14, 2013 #11

    tiny-tim

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    oooh :redface:

    ok: this sort of question is very difficult to do without either the summation convention or a ∑
    whenever you see a δr,s, you know it's going to be multiplying something else with either an r or an s (or both)

    and since all these things are parts of matrices, you know it's actually going to be a matrix multiplication:

    A = BC means Ar,s = ∑(all values of t) Br,tCt,s

    you can either write it with a ∑ (like that),

    or you can use the summation convention and just write Ar,s = Br,tCt,s (with a "∑" being understood but not written)

    that enables you to say "δr,s replaces any s in the thing next to it by r", eg δr,sAs,u = Ar,u

    alternatively, use a ∑ : ∑ δr,sAs,u = Ar,u

    in particular, δr,sδs,u = δr,u

    (or ∑s δr,sδs,u = δr,u)


    ok, try all that in the original question :smile:
     
  13. Oct 15, 2013 #12
    ok using all for part c):

    ## (\epsilon (i,j) \epsilon (k,l) )_{r,s} = \displaystyle \sum_{t=1} ^{n} \epsilon (i,j) _{r,t} \epsilon (k,l) _{t,s} = \displaystyle \sum_{t=1}^{n} \delta _{ir} \delta _{jt} \delta _{kt} \delta _{ls} = \displaystyle \sum_{t=1}^n \delta _{ir} \delta _{jt} \delta _{jt} \delta _{ls}## if j = k, which then is ## = \delta _{ir} \delta _{jj} \delta _{jj} \delta _{ls} + \displaystyle \sum_{t \not= j } \delta_ {ir} \delta _{jt} \delta _{kt} \delta _{ls} = \delta _{ir} \delta _{ls} + 0 = \epsilon (i,l) ## if ## j \not= k ## then we get ## \displaystyle \sum_{t=1}^n \delta _{ir} \delta _{jt} \delta _{kt} \delta _{ls} = 0 ## as t cannot equal k and j simultaneously

    is that ok?
     
  14. Oct 15, 2013 #13

    tiny-tim

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    what is part c) ? :confused:
     
  15. Oct 15, 2013 #14
    sorry I didn't label in the original post part c is this:

    " hence generalise for ##\epsilon (i,j) \epsilon (k,l)##"
     
  16. Oct 15, 2013 #15

    tiny-tim

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    no, that last item should be ## = \displaystyle \delta _{ir} \delta _{jk} \delta _{ls}##

    and then rewrite that as ##= \delta _{jk} (\delta _{ir} \delta _{ls})##

    so now you have ##(\epsilon (i,j) \epsilon (k,l) )_{r,s}= \delta _{jk} (\delta _{ir} \delta _{ls})##

    sooo … ? :smile:

    (btw, you don't need to write "\displaystyle" on this forum :wink:)
     
  17. Oct 15, 2013 #16
    ## = 1\times \epsilon (i,l) ## ?

    is everything else correct that I've written?
     
  18. Oct 15, 2013 #17

    tiny-tim

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    nearly :smile:

    (what happened to the δjk ? :wink:)
     
  19. Oct 15, 2013 #18
    is it not 1 if j = k??
     
  20. Oct 15, 2013 #19

    tiny-tim

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    yes, but you want a general formula for ## \epsilon (i,j) \epsilon (k,l) ## :wink:
     
  21. Oct 15, 2013 #20
    I'm really not sure what to put
     
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