Basic matrices

1. Oct 14, 2013

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

2. Oct 14, 2013

tiny-tim

hi converting1!
nooo …

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

3. Oct 14, 2013

converting1

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
4. Oct 14, 2013

tiny-tim

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

5. Oct 14, 2013

converting1

hmm, I'm not so sure I understand, what if t = 20? i.e. $\delta _{1999,20} \delta _{1999,20} = 0.0 = 0 ?$

6. Oct 14, 2013

converting1

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

7. Oct 14, 2013

tiny-tim

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

8. Oct 14, 2013

converting1

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. Oct 14, 2013

tiny-tim

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

10. Oct 14, 2013

converting1

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

11. Oct 14, 2013

tiny-tim

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

12. Oct 15, 2013

converting1

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?

13. Oct 15, 2013

tiny-tim

what is part c) ?

14. Oct 15, 2013

converting1

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

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

15. Oct 15, 2013

tiny-tim

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 )

16. Oct 15, 2013

converting1

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

is everything else correct that I've written?

17. Oct 15, 2013

tiny-tim

nearly

(what happened to the δjk ? )

18. Oct 15, 2013

converting1

is it not 1 if j = k??

19. Oct 15, 2013

tiny-tim

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

20. Oct 15, 2013

converting1

I'm really not sure what to put