A way of writing the Leibniz formula for determinants

Question

Is it right to write the Leibniz formula for the determinant the following way? \begin{equation} \det{X} = \frac{1}{n!}\sum_{i_1=1}^n\cdots\sum_{i_n=1}^n \sum_{j_1=1}^n\cdots\sum_{j_n=1}^n\epsilon_{i_1\cdots i_n}\epsilon_{j_1\cdots j_n} \prod_{k=1}^{n}X_{i_kj_k}\,, \end{equation}

Answer 1

You are correct. The usual definition of determinant $$ \det X = \sum_{j_1=1}^n \sum_{j_2=1}^n \cdots \sum_{j_n=1}^n \epsilon_{j_1 \cdots j_n} \prod_{i=1}^n X_{ij_i} $$ is the case of your summation where $\ i_k = k \ $ for all $k$. If another choice of permutation $\ \{i_k\}_{k=1}^n \ $ is used then the summation will be the determinant multiplied by the signature of the permutation. If the sequence $\ \{i_k\}_{k=1}^n \ $ has repeated values, then the Levi-Civita Symbol is zero. Thus the total over all sequences has the determinant repeated $\ n! \ $ times and hence the reason that dividing by $\ n! \ $ gives the determinant itself. Now a critical part is played by the two Levi-Civita symbols. An easy way to see this is to examine the special case of the identity matrix where $\ X_{i_kj_k} = \delta_{i_kj_k} \ $ which is $0$ unless $\ i=j \ $ and here the two Levi-Civita symbols are the same with product $1$. Without the second Levi-Civita symbol, the sum would be $\ 0\ $ for $\ n>1 \ $ since half of permutations in $\ S_n \ $are even and the other half are odd. Another easy way to see the need for the two symbols is that the determinant of the transpose of a matrix is the same as the determinant of the original matrix. In the summation formula this just swaps the $i$ and $j$ index lists which implies the two symbols are swapped also.

Answer 2

I just found this paper where equation 9 for Hyperdeterminants can be rewritten as: \begin{equation} \det{{X}} = \frac{1}{n!}\sum_{i_1^{(1)}=1}^n\cdots\sum_{i_n^{(1)}=1}^n\cdots \sum_{i_1^{(q)}=1}^n\cdots\sum_{i_n^{(q)}=1}^n\epsilon_{i_1^{(1)}\cdots i_n^{(1)}}\cdots\epsilon_{i_1^{(q)}\cdots i_n^{(q)}} \prod_{k=1}^{n}{X}_{i_k^{(1)}\cdots i_k^{(q)}} \end{equation} So, my formula is correct for matrices, that is, when order $q$=2.