Basic Fixed Matrices: Proving Equations with Formal Calculations

[converting1]

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]

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}##

 

[converting1]

hi converting1! nooo …

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

oh damn I am 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?

 

[tiny-tim]

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

 

[converting1]

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 ? ##

 

[converting1]

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

 

[tiny-tim]

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 δ_1999,tδ_1999,t …

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

∑_{(over all values of t)} δ_1999,tδ_1999,t ​

 

[converting1]

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

you can't substitute one value for t into δ_1999,tδ_1999,t …

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

∑_{(over all values of t)} δ_1999,tδ_1999,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]

has your lecturer not taught you the summation convention (the einstein summation convention)?

 

[converting1]

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]

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 δ_r,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 A_r,s = ∑_{(all values of t)} B_r,tC_t,s

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

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

that enables you to say "δ_r,s replaces any s in the thing next to it by r", eg δ_r,sA_s,u = A_r,u

alternatively, use a ∑ : ∑ δ_r,sA_s,u = A_r,u

in particular, δ_r,sδ_s,u = δ_r,u

(or ∑_s δ_r,sδ_s,u = δ_r,u)

ok, try all that in the original question ​

 

[converting1]

oooh

ok: this sort of question is very difficult to do without either the summation convention or a ∑
whenever you see a δ_r,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 A_r,s = ∑_{(all values of t)} B_r,tC_t,s

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

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

that enables you to say "δ_r,s replaces any s in the thing next to it by r", eg δ_r,sA_s,u = A_r,u

alternatively, use a ∑ : ∑ δ_r,sA_s,u = A_r,u

in particular, δ_r,sδ_s,u = δ_r,u

(or ∑_s δ_r,sδ_s,u = δ_r,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]

ok using all for part c):

what is part c) ?

 

[converting1]

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]

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 )

 

[converting1]

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]

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

nearly

(what happened to the δ_jk ? )

 

[converting1]

nearly

(what happened to the δ_jk ? )

is it not 1 if j = k??

 

[tiny-tim]

yes, but you want a general formula for ## \epsilon (i,j) \epsilon (k,l) ##

 

[converting1]

yes, but you want a general formula for ## \epsilon (i,j) \epsilon (k,l) ##

I'm really not sure what to put

 

[tiny-tim]

say it (partly) in words, starting "ε(i,j)ε(k.l) is … ", and then we'll put it completely into symbols

 

[converting1]

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,

thanks for your patience

 

[tiny-tim]

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

ok, so it's epsilon(i,l) times … ?

 

[converting1]

ok, so it's epsilon(i,l) times … ?

## \delta _{jk} ## ?

 

[tiny-tim]

(just got up :zzz:)

yes!

ε(i,j)ε(k,l) = δ_jkε(i,l) ​

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

 

[converting1]

thank you tim, really appreciate it.