Show that $\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}$ Question: Show that $\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}$ where $\epsilon_{ijk}$ is the Levi-civita symbol and $\delta_{ij}$ is the Kronecker delta symbol.
My attempt: $\epsilon_{ijk}$ assumes non-zero values only when $i\not = j\not =k$ i.e.where $i,j,k$ is a permutation of $1,2,3$.  


*

*If $i \not = l$ then $l=j$ or $l=k$, and then $\epsilon_{ijk}\epsilon_{ljk}=0$

*When $i=l$, then $\epsilon_{ijk}\epsilon_{ljk}=\epsilon_{ijk}\epsilon_{ijk}=(\epsilon_{ijk})^2=6$ since $i,j,k$ can have $3!=6$ permutations.


Then we have finally $\epsilon_{ijk}\epsilon_{ljk}=6\delta_{il}$ and not $\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}$.
Is the question itself wrong? Or is there a flaw in my attempt to solve? 
Please help.
 A: The summation convention says that you sum any letter that appears twice.
So $\epsilon_{ijk}\epsilon_{ljk}$, it really means $\sum_j\sum_k\epsilon_{ijk}\epsilon_{ljk}$
You are right that $\epsilon_{ijk}\epsilon_{ijk}=6$, but then $\epsilon_{ijk}\epsilon_{ijk}=\sum_i\sum_j\sum_k\epsilon_{ijk}\epsilon_{ijk}$
For the same reason, $\delta_{ii}=3,\delta_{11}=1$
A: First to know that
$\epsilon_{ijm}\epsilon_{klm}=\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk} \, (*)$,
the proof of this equation is in Proof relation between Levi-Civita symbol and Kronecker deltas in Group Theory.
Then for the $\epsilon_{ijk}\epsilon_{ljk}$, we can make some substitutions in the equation $(*)$ as follows
$$
\begin{aligned}
m&:= k \\ k&:=l \\ l&:=j
\end{aligned}
$$
With facts that $\delta_{ii}=3$ and $\delta_{ik}\delta_{kj}=\delta_{ij}$, then equation $(*)$ becomes
$$
\begin{aligned}
\epsilon_{ijk}\epsilon_{ljk}&=\delta_{il}\delta_{jj}-\delta_{ij}\delta_{jl}\\
&=3\delta_{il}-\delta_{il}\\
&=2\delta_{il}
\end{aligned}
$$
Therefore $\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}$.
