Proof that $\epsilon_{ijk}$ is the same as the determinant of the Kronecker delta I just learned about the Kronecker Delta function and $\epsilon_{ijk}$ for the first time, and I still can't wrap my mind around how to prove that $$\epsilon_{ijk} = det\begin{pmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \end{pmatrix}$$ I understand that the RHS is  $$det\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}=1$$ but how is that related to the fact that the $\epsilon$ tensor renders 1 for even, -1 for odd permutations of $i,j,k$ and $0$ for identical ones? And how can it be proven?
The only "proof" I could think of is$$ \delta_{i1}\delta_{j2}\delta_{k3} + \delta_{i2}\delta_{j3}\delta_{k1} + \delta_{i3}\delta_{j1}\delta_{k2} - \delta_{i3}\delta_{j2}\delta_{k1} - \delta_{i2}\delta_{j1}\delta_{k3} - \delta_{i1}\delta_{j3}\delta_{k2}  \\
    =  \begin{cases} 1 \cdot \delta_{ijk} & |\{ijk\} \in \{123\},\{312\},\{231\} \\ -1 \cdot \delta_{ijk} & |\{ijk\} \in \{213\},\{321\},\{132\} \end{cases} \\
    = \epsilon_{ijk} $$ but that doesn't look sound to me.
I am aware of this question, but it's kind of like two steps ahead when I'd like to be able to explain just the first.
Can somebody explain with an Einstein mindset?
 A: A property of the determinant is:


*

*Exchanging two rows while leaving everything else unchanged changes the sign of the determinant, but not the magnitude.


This can be proven from the formula $\det(AB) = (\det A)(\det B)$. The operation of exchanging rows in a matrix $M$ is the same as taking $EM$, where $E$ is the same row exchange on the identity matrix $I$. One can easily show all such matrices $E$ have determinant $-1$.
Now note that if two rows of a matrix $M$ are identical, then exchanging them makes no change to the matrix. Therefore $\det M = -\det M$, from which it follows that $\det M = 0$:


*

*If a matrix has two identical rows, then its determinant is $0$.


Now define $$d_{ijk} = \det \begin{pmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \end{pmatrix}$$
The matrix for $d_{123}$ is the identity matrix, so $d_{123} = 1$. If $i = j$ or $j = k$ or $i = k$, the matrix will have two identical rows, so $d_{ijk} = 0$ in this case. And by the exchange of row rule for matrix determinants, if you exchange the values of any two of $i, j, k$, then $d_{ijk}$ changes sign.
These three facts completely determine all values of $d_{ijk}$. But $\epsilon_{ijk}$ obeys exactly the same set of rules. So the two of them must be equal.
A: As we know, if $A = (a_{ij})$ is an $n\times n$ matrix, its determinant is given by:
$$\mbox{det}A = \sum_{\sigma \in S_{n}}\epsilon_{\sigma}\prod_{i=1}^{n}a_{i,\sigma(i)}$$
where $S_{n}$ is the set of all permutations $\sigma$ of the set $I_{n}:=\{1,...,n\}$ and $\epsilon_{\sigma}$ is the sign of the permutation $\sigma$. In your case, we have $n=3$. 
Let us consider the matrix:
$$\Delta_{ijk} := \begin{pmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \end{pmatrix} $$
For simplicity, let us consider only the case where $i,j,k$ are all different because, if not, it is easy to see that the result follows because at least two rows of $\Delta_{ijk}$ are equal, so that $\det \Delta_{ijk} = 0$ and also is $\epsilon_{ijk}$. Now, as you pointed out, we have:
$$ \epsilon_{ijk} = \mbox{det} \Delta_{ijk} = \sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{i=1}^{3}\delta_{i,\sigma(i)}$$
Now, let $\eta$ be a fixed permutation of $S_{3}$. Every permutation is a bijection from $I_{3}$ to itself, so every permutation has an inverse $\eta^{-1}$. Also, it is easy to prove that $\epsilon_{\eta^{-1}} = \epsilon_{\eta}$. Now, note that:
$$\epsilon_{\eta(i),\eta(j),\eta(k)} = \sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{i=1}^{3}\delta_{\eta(i),\sigma(i)} $$
If we set $\eta(i) = k$, then $i = \eta^{-1}(k)$ and:
$$\sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{i=1}^{3}\delta_{\eta(i),\sigma(i)} = \sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{k=1}^{3}\delta_{k,(\sigma\circ \eta^{-1})(k)} $$
If we sum over every permutation $\sigma$ in $S_{3}$, the composite $\sigma \circ \eta^{-1}$ also covers every permutation of $S_{3}$ so we can redefine $\tilde{\sigma} = \sigma\circ\eta^{-1}$ and:
$$\sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{k=1}^{3}\delta_{k,(\sigma\circ \eta^{-1})(k)}  = \sum_{\tilde{\sigma} \in S_{3}}\epsilon_{\eta}\epsilon_{\tilde{\sigma}}\prod_{k=1}^{3}\delta_{k,\tilde{\sigma}(k)} =\epsilon_{\eta}\mbox{det}\Delta_{ijk}$$
where I've used:
$$\epsilon_{\sigma} = \epsilon_{\tilde{\sigma}}\frac{1}{\epsilon_{\tilde{\sigma}}}\epsilon_{\sigma} = \epsilon_{\tilde{\sigma}}\frac{1}{\epsilon_{\eta}\epsilon_{\sigma}}\epsilon_{\sigma} = \epsilon_{\eta}\epsilon_{\tilde{\sigma}}.$$
The conclusion is that:
$$\epsilon_{\eta(i),\eta(j),\eta(k)}= \epsilon_{\eta}\overbrace{\mbox{det}\Delta_{ijk}}^{=1} = \epsilon_{\eta}$$
Thus, $\epsilon_{\eta(i),\eta(j),\eta(k)}$ coincides with the sign of the permutation $\eta$, i.e. it is $1$ if $\eta$ is an even permutation and $-1$ otherwise.
A: Both sides are $1$ if $i=1,\,j=2,\,k=3$, for the reason you gave. Both sides also multiply by $-1$ if any two indices are exchanged. (The determinant, in particular, is of a matrix that swaps two of its rows in this process.) In the special case where the indices are equal, both expressions must be originally $0$; in the special case where all three indices are unequal, the sign change on each side preserves equality, but of nonzero values. Since $\epsilon_{ijk}$ is completely specified by full antisymmetry together with $\epsilon_{123}=1$, we're done.
