# Basic matrices

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

tiny-tim
Homework Helper
hi converting1! 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}## hi converting1! nooo …

##(\epsilon (i,j) \epsilon (k,l))_{r,s} = (\epsilon (i,j))_{r,t} (\epsilon (k,l))_{t,s}## 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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tiny-tim
Homework Helper
hi converting1! 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 ## similary ##\delta _{10t} \delta _{500t} = 0 ##​

once you've convinced yourself of this, you should be able to do the general case hi converting1! ah, you're still not thinking in terms of the summation convention:

##\delta _{1999t} \delta _{1999t} = 1 ## similary ##\delta _{10t} \delta _{500t} = 0 ##​

once you've convinced yourself of this, you should be able to do the general case hmm, I'm not so sure I understand, what if t = 20? i.e. ##\delta _{1999,20} \delta _{1999,20} = 0.0 = 0 ? ##

also take a 2x2 matrix as an example, ## \epsilon (1,2) _{22} = \delta _{12} \delta_{22} = 0 \times 1 = 0 ##

tiny-tim
Homework Helper
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 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 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 ##?

tiny-tim
Homework Helper
has your lecturer not taught you the summation convention (the einstein summation convention)?

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).

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

oooh 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 oooh 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 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?

tiny-tim
Homework Helper
ok using all for part c):

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

" hence generalise for ##\epsilon (i,j) \epsilon (k,l)##"

tiny-tim
Homework Helper
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 … ? (btw, you don't need to write "\displaystyle" on this forum )

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 … ? (btw, you don't need to write "\displaystyle" on this forum )

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

is everything else correct that I've written?

tiny-tim
Homework Helper
## = 1\times \epsilon (i,l) ## ?

nearly (what happened to the δjk ? )

nearly (what happened to the δjk ? )

is it not 1 if j = k??

tiny-tim
Homework Helper
yes, but you want a general formula for ## \epsilon (i,j) \epsilon (k,l) ## yes, but you want a general formula for ## \epsilon (i,j) \epsilon (k,l) ## I'm really not sure what to put

tiny-tim
Homework Helper
say it (partly) in words, starting "ε(i,j)ε(k.l) is … ", and then we'll put it completely into symbols say it (partly) in words, starting "ε(i,j)ε(k.l) is … ", and then we'll put it completely into symbols it's epsilon(i,l) if j = k else it's 0,

tiny-tim
Homework Helper
it's epsilon(i,l) if j = k else it's 0,

ok, so it's epsilon(i,l) times … ? ok, so it's epsilon(i,l) times … ? ## \delta _{jk} ## ?

tiny-tim
ε(i,j)ε(k,l) = δjkε(i,l) work your way through it, and compare it with parts (i) and (ii), until you're convinced how it works 