I've been asked to prove that:
$\epsilon_{ijk}$$\epsilon_{kqr}$ = $\delta_{iq}$$\delta_{jr}$ - $\delta_{ir}$$\delta_{jq}$
We start with:
$\epsilon_{ijk}$$\epsilon_{kqr}$ = $\delta_{ip}$($\delta_{jq}$$\delta_{kr}$ - $\delta_{jr}$$\delta_{kq}$) + $\delta_{iq}$($\delta_{jr}$$\delta_{kp}$ - $\delta_{jp}$$\delta_{kr}$) + $\delta_{ir}$($\delta_{jp}$$\delta_{kq}$ - $\delta_{jq}$$\delta_{kp}$)
We start by setting p = k, as this is essential for the proof:
We thus have: $\delta_{ik}$($\delta_{jq}$$\delta_{kr}$ - $\delta_{jr}$$\delta_{kq}$) + $\delta_{iq}$($\delta_{jr}$$\delta_{kk}$ - $\delta_{jk}$$\delta_{kr}$) + $\delta_{ir}$($\delta_{jk}$$\delta_{kq}$ - $\delta_{jq}$$\delta_{kk}$)
This is defined to be in R^3, so we know that $\delta_{ik}$ = 3, and we also know that $\delta_{ij}$ $\delta_{jk}$ = $\delta_{ik}$, we thus implement both of these and simplify to get:
$\delta_{ik}$ $\delta_{jq}$ $\delta_{kr}$ - $\delta_{ik}$ $\delta_{jr}$ $\delta_{kq}$ + 2 $\delta_{iq}$ $\delta_{jr}$ - 2 $\delta_{ir}$ $\delta_{jq}$
I know that for this to work, that $\delta_{ik}$ $\delta_{jq}$ $\delta_{kr}$ must equal $\delta_{ir}$ $\delta_{jq}$
and $\delta_{ik}$ $\delta_{jr}$ $\delta_{kq}$ must equal $\delta_{iq}$ $\delta_{jr}$
However, I can't see or understand why this would be the case. And honestly, I don't really understand the explanation for why $\delta_{ij}$ $\delta_{ij}$ is $\delta_{ik}$. Any help appreciated.