# Determinant with Levi-Civita Symbol?

From Schaum's Outline in Tensor Calculus

If $A = [a_{ij}]_{nn}$ is any square matrix, then define $\text{ det } A = \epsilon_{i_1i_2i_3...i_{n-1}i_n}a_{1 \, \cdot \, i_1}a_{2 \, \cdot \, i_2}...a_{(n - 1) \, \cdot \, i_{n - 1}}a_{n \, \cdot \, i_n}$.

I can check this by expanding the product and sum in full, but what's the derivation or motivation behind this formula? I tried to find something on the Internet about this.

• I bet this is explained (and/or used as a definition, in fact) in every single textbook which treats determinants! – Mariano Suárez-Álvarez May 7 '15 at 5:19
• @MarianoSuárez-Álvarez I don't think the levi cevita is as common as you're saying to define determinants. It's more common to see it in physics, in particular in differential geometry, as physicists like to use index notation. Otherwise, I have seen it in courses defined with permutations, defined using the Laplace expansion, etc. In fact Strang's linear algebra book, which is relatively common, mentions both of these but not the levi cevita definition. – snulty Sep 30 '16 at 9:24
• @snulty, the Levi-Civita symbol is simply the signature of the permutation of its indices (or zero, if two indices coincides) so the definition the OP has in mind is exactly the same thing as the one with permiutations. – Mariano Suárez-Álvarez Sep 30 '16 at 9:40
• @MarianoSuárez-Álvarez I do understand that, I mean the answer below shows that explicitly also. While something like $\sin(\pi/2)$ is exactly $1$, it would be unhelpful to claim that every single book teaching arithmetic tells you that $\sin(\pi/2)=1$. – snulty Sep 30 '16 at 10:21
• @MarianoSuárez-Álvarez Also apologies for commenting on something over a year ago :) – snulty Sep 30 '16 at 10:22

This is the Leibniz formula $$\det(A)=\sum_{\sigma\in\mathcal S_n}\text{sign}(σ)a_{1σ(1)}a_{2σ(2)}...a_{nσ(n)},$$ the eps-tensor is the sign of the permutation.
It follows from repeated application of the Laplace formula, essentially the multilinear nature of the determinant, so that in the end it is represented as a linear combination of determinants of permutation matrices. $$\det(A)=\det(a_1,...,a_n)=\sum_{i_1,...,i_n\in\{1,...,n\}}a_{1i_1}...a_{ni_n}\det(e_{i_1},...,e_{i_n})$$ where $\det(e_{i_1},e_{i_1}...,e_{i_n})=ϵ_{i_1i_2...i_n}$ is only different from zero if $(i_1,i_2,...,i_n)$ is a permutation of $(1,2,...,n)$, so that $(e_{i_1},e_{i_1}...,e_{i_n})$ is a permutation matrix, and the determinant of a permutation matrix is the sign of the permutation is the value of the eps-tensor for this permutation.