# Levi-Civita tensor

Show that

$\epsilon_{ijk} A_{il} A_{jm} A_{kn} = \det(A) \epsilon_{lmn}$

where $\epsilon$ epsilon is the standard Levi-Civita symbol and A is a three dimensional matrix.

I found the above property in a book used for my course about turbulence, where it is used to prove that the $\epsilon$ is isotropic. The proof for the property however is not provided, and I'm struggling to prove it myself. Thanks.

• I would argue that this "property" is actually the honest definition of the determinant. The expansion by minors due to Laplace is far more complicated even if it is how you first see the determinant defined. Of course, mathematicians who are found of permutations can differ with my opinion, but, I prefer a sum of indices as opposed to a sum over permutations (the other good way to define the determinant) – James S. Cook May 24 '15 at 19:24

Consider the following matrix $$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ Then the $\det(\mathbf{A})$ can be written as \begin{align} \det(\mathbf{A}) &= (-1)^0a_{11}(a_{22}a_{33}-a_{23}a_{32})+(-1)^1a_{12}(a_{21}a_{33}-a_{23}a_{31})\\ &+(-1)^2a_{13}(a_{21}a_{32}-a_{22}a_{31}) \end{align} Now, the minor of a determinant is denoted as $A_{nm}$ where $n,m$ correspond to the row and column of the deleted entries. Therefore, $a_{22}a_{33}-a_{23}a_{32} = A_{11}$. What are the other minors? Recall that $$\epsilon_{ijk} = \begin{cases} 0, & \text{when } i = j\text{ or } i = k\text{ or } j = k\\ 1, & \text{if we have an even permutation}\\ -1, & \text{if we have an odd permutation} \end{cases}$$ Can you figure the rest out now?
$A=(A_{ij})$, for definition
$$\det(\mathbf{A}) = \varepsilon_{ijk} A_{1i}A_{2j}A_{3k}$$
$$3!\det(\mathbf{A}) = \varepsilon_{lmn}\varepsilon_{ijk} A_{il}A_{jm}A_{kn}$$