# Homework Help: Symbolic Linear algebra conept

1. Sep 14, 2009

### sleventh

Symbolic Linear algebra concept

1. The problem statement, all variables and given/known data
Evaluate the sum $$\Sigma$$k ( $$\epsilon$$ ijk $$\epsilon$$ lmk (which contains three terms) by considering the result for all possible combinations of i,j,l,m, that is a) i=j b) i=l c) i=m d) j=l e)j=m f)l=m g) i$$\neq$$l or m h) j$$\neq$$l or m

show that

$$\Sigma$$k ( $$\epsilon$$ ijk $$\epsilon$$ lmk = $$\delta$$ il $$\delta$$ jm - $$\delta$$ im $$\delta$$ jl

and use this result to prove
A X (B X C)=(A $$\bullet$$ C)B-(A $$\bullet$$ B)C

2. Relevant equations

3. The attempt at a solution

i have been working on this problem for over three hours and have also gone to my professor which was of little help, the work I have done would most likely only serve to confuse. I would be EXTREMELY grateful to anyone who would be able to help me understand this problem. I understand for part a) the permittivity will always equal 0 for the first Levi-Civita symbol making the entire sum zero. For question like be im lost as to how you could know which values would be +1 or -1 or 0 out of the 81 possible combinations. As for the proofs im lost on the first and the second I am having a hard time manipulating the concept A X B = $$\Sigma$$ j, k $$\epsilon$$ ijk AjBk . Thank you very much to anyone who can shed the slightest of light.
yours truly
sleventh

Last edited: Sep 15, 2009
2. Sep 15, 2009

### tiny-tim

Hi sleventh!

(have a sigma: ∑ and a delta: δ and an epsilon: ε and try using the X2 tag just above the Reply box )
(what does permittivity have to do with it? )

No, there aren't 81 …

for example, for b) i = l, put i = l = 1, then there are only nine combinations for j and k, and most of them are zero!

Try again.
(you mean (AxB)i = …)

Just look at the "matrix" definition of AxB (where ex ey and ez are the basis):
Code (Text):
ex  ey  ez
Ax  Ay  Az
Bx  By  BZ

3. Sep 15, 2009

### sleventh

thank you very much tiny tim, It was silly not to set the matrix into its components, i was only oriented to the summation equation. i still am lost on LaTeX Code: \\Sigma k ( LaTeX Code: \\epsilon ijk LaTeX Code: \\epsilon lmk = LaTeX Code: \\delta il LaTeX Code: \\delta jm - LaTeX Code: \\delta im LaTeX Code: \\delta jl
thank you very much, you helped clear many things up.

4. Sep 16, 2009

### tiny-tim

hmm … your LaTeX looks a bit weird …

the PF version is very simple, it doesn't use \\, and for subscripts just type _{ijk}

(and use \cdot not \bullet)

For more details, see http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000" [Broken]

(but it's easier on the server if you use standard Unicode symbols and the X2 and X2 tags instead )

Last edited by a moderator: May 4, 2017
5. Sep 16, 2009

### lurflurf

If we have n=3 there are 81 as 3^4=81.
The fancy way to do this would be to observe
sum epsilon(ijk)epsilon(lmk)
is isotropic so can be expressed in the form
A*delta(ij)delta(lm)+B*delta(il)delta(jm)+C*delta(im)delta(jl)
since
{delta(ij)delta(lm),delta(il)delta(jm),delta(im)delta(jl)}
is a basis for the isotropic 4th order 3 spance tensors

The tedious method (though not that tedious) as you mentioned is to group 81 componets into groups based on there possible values -1,0,1.

6. Sep 16, 2009

### sleventh

thank you lurflurf, i have been unaware how to put the epsilon notation into delta form, however in retrospect it doesn't seem to have been to difficult. Because of this i did have to sort the 81 possibilities.
I am though lost on how to actually prove the sum of the two epsilons to equal the Kronecker delta equation. If i figure this out i will post the solution.
thanks to everyone who helped
sleventh

7. Sep 16, 2009

### lurflurf

If you take the equation as given, it suffices to show the two sides agree for each of the 81 components. Do you mean how to derive the expression?
from
sum epsilon(ijk)epsilon(lmk)=A*delta(ij)delta(lm)+B*delta(il)delta(jm)+C*delta( im)delta(jl)
consider for example
i=1,j=1,l=2,m=2
sum epsilon(11k)epsilon(22k)=A*delta(11)delta(22)+B*delta(12)delta(12)+C*delta( 12)delta(12)
0=A
i=1,j=2,l=1,m=2
sum epsilon(12k)epsilon(12k)=A*delta(12)delta(12)+B*delta(11)delta(22)+C*delta( 12)delta(12)
1=B
likewise for C