Can the Levi-Civita Kronecker Delta relation be proven using a matrix approach?

[LoopQG]

This isn't a HW question just something I am curious about. I was looking on wikipedia and found a way to prive the Levi-Citiva Kronecker Delta relation that I hadn't seen before.

The site claims
$$\epsilon_{ijk}\epsilon_{lmn} = \det \begin{bmatrix} \delta_{il} \delta_{im} \delta_{in}\\ \delta_{jl} \delta_{jm} \delta_{jn}\\ \delta_{kl} \delta_{km} \delta_{kn}\\ \end{bmatrix}$$

= $$\delta_{il}\left( \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}\right) - \delta_{im}\left( \delta_{jl}\delta_{kn} - \delta_{jn}\delta_{kl} \right) + \delta_{in} \left( \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl} \right) \$$

the website is:
http://en.wikipedia.org/wiki/Levi-Civita_symbol

The way I have proved the relation before is showing that all 81 components of the tensor are zero accept the $$\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}$$ terms.

Taking the determinate of the matrix I do get the correct answer,just not sure why you can write that matrix down.

Can anybody offer a proof of the matrix ? If so it is much easier than the way I have previously done it.

 

[Tedjn]

I took a look at the Wikipedia page, and I can offer only what I came up with as an explanation:

On the Wikipedia page, below the definition of $\varepsilon_{ijk}$, there is the claim

$$\det A = \sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3 \varepsilon_{ijk}a_{1i}a_{2j}a_{3k}.$$​

This is a well-known equivalent form of the determinant that generalizes easily to higher dimensions (in the form of even and odd permutations). We will take it for granted.

Two other facts about the determinant we take for granted are that det(AB) = det(A)det(B) and that det(A^T) = det(A). This all actually can be proved rigorously, beginning with the determinant function defined by properties, but it is a fair amount of work.

Consider the matrix A to be the matrix with a_1i, a_2j, a_3k = 1 and everything else 0. From the above, $\det A = \varepsilon_{ijk}$. We may define a similar matrix B whose determinant is $\varepsilon_{lmn}$.

The product $\varepsilon_{ijk}\varepsilon_{lmn}$ equals $\det(AB^T)$. What are the entries of this matrix? Take the top left entry for illustration. It equals

a₁₁b₁₁ + a₁₂b₂₁ + a₁₃b₃₁.
​

Only when i=l do we have a_1i and b_l1 matching up to give 1. Otherwise, the sum is 0. Therefore, the top left entry is equal to $\delta_{il}$. Likewise reasoning gives us the other 8 matrix entries.

This generalizes immediately to the higher dimensional case.