Basic matrices

  • #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
 

Answers and Replies

  • #2
tiny-tim
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hi converting1! :smile:
i) rewriting this as ##\delta _{500r} \delta _{1999s} \delta _{1999r} \delta _{10s} ## …

nooo …

##(\epsilon (i,j) \epsilon (k,l))_{r,s} = (\epsilon (i,j))_{r,t} (\epsilon (k,l))_{t,s}## :wink:
 
  • #3
hi converting1! :smile:


nooo …

##(\epsilon (i,j) \epsilon (k,l))_{r,s} = (\epsilon (i,j))_{r,t} (\epsilon (k,l))_{t,s}## :wink:

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?
 
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  • #4
tiny-tim
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hi converting1! :smile:
we know that ##\delta _{1999t} \delta _{1999t} = 1 ## if t = 1999, else it will be 0, so we get the required result

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:
 
  • #5
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:

hmm, I'm not so sure I understand, what if t = 20? i.e. ##\delta _{1999,20} \delta _{1999,20} = 0.0 = 0 ? ##
 
  • #6
also take a 2x2 matrix as an example, ## \epsilon (1,2) _{22} = \delta _{12} \delta_{22} = 0 \times 1 = 0 ##
 
  • #7
tiny-tim
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hmm, I'm not so sure I understand, what if t = 20? i.e. ##\delta _{1999,20} \delta _{1999,20} = 0.0 = 0 ? ##

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:
 
  • #8
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:
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 ##?
 
  • #9
tiny-tim
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has your lecturer not taught you the summation convention (the einstein summation convention)?
 
  • #10
has your lecturer not taught you the summation convention (the einstein summation convention)?

No, not yet, however in my applied mathematics lecturer we just started it today - so I know very little about it (almost nothing).
 
  • #11
tiny-tim
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No, not yet, however in my applied mathematics lecturer we just started it today - so I know very little about it (almost nothing).

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:
 
  • #12
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:
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
what is part c) ? :confused:

sorry I didn't label in the original post part c is this:

" hence generalise for ##\epsilon (i,j) \epsilon (k,l)##"
 
  • #15
tiny-tim
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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,

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:)
 
  • #16
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:)

## = 1\times \epsilon (i,l) ## ?

is everything else correct that I've written?
 
  • #17
tiny-tim
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## = 1\times \epsilon (i,l) ## ?

nearly :smile:

(what happened to the δjk ? :wink:)
 
  • #19
tiny-tim
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yes, but you want a general formula for ## \epsilon (i,j) \epsilon (k,l) ## :wink:
 
  • #20
yes, but you want a general formula for ## \epsilon (i,j) \epsilon (k,l) ## :wink:

I'm really not sure what to put
 
  • #21
tiny-tim
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say it (partly) in words, starting "ε(i,j)ε(k.l) is … ", and then we'll put it completely into symbols :smile:
 
  • #22
say it (partly) in words, starting "ε(i,j)ε(k.l) is … ", and then we'll put it completely into symbols :smile:

it's epsilon(i,l) if j = k else it's 0,

thanks for your patience
 
  • #23
tiny-tim
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it's epsilon(i,l) if j = k else it's 0,

ok, so it's epsilon(i,l) times … ? :smile:
 
  • #25
tiny-tim
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(just got up :zzz:)

yes!!

ε(i,j)ε(k,l) = δjkε(i,l) :smile:

work your way through it, and compare it with parts (i) and (ii), until you're convinced how it works :wink:
 

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