15
Feb 19

## Leibniz formula for determinants ## Rule for calculating determinant

This is one of those cases when calculations explain the result.

Let $e_j$ denote the $j$th unit row-vector. Row $A_i,$ obviously, can be decomposed as $A_i=\sum_{j=1}^na_{ij}e_j.$ Recall that by Property V, the determinant of $A$ is a multilinear function of its rows. Using Property V $n$ times, we have

(1) $\det A=\det \left(\begin{array}{c}\sum_{j_1=1}^na_{1j_1}e_{j_1}\\A_2\\...\\ A_n\end{array}\right)=\sum_{j_1=1}^na_{1j_1}\det\left(\begin{array}{c}e_{j_1}\\A_2\\...\\ A_{n}\end{array}\right)$ $=\sum_{j_1=1}^na_{1j_1}\det \left(\begin{array}{c}e_{j_1}\\\sum_{j_2=1}^na_{2j_2}e_{j_2}\\...\\A_n\end{array}\right)=\sum_{j_1,j_2=1}^na_{1j_1}a_{2j_2}\det\left(\begin{array}{c}e_{j_1}\\e_{j_2}\\...\\ A_{n}\end{array}\right)=...$ $=\sum_{j_1,...,j_n=1}^na_{1j_1}...a_{nj_n}\det\left(\begin{array}{c} e_{j_1}\\e_{j_2}\\...\\e_{j_n}\end{array}\right).$

Here by Property III $\det\left(\begin{array}{c}e_{j_1}\\e_{j_2}\\...\\e_{j_n}\end{array}\right)=0$

if among rows there are equal vectors. The remaining matrices with nonzero determinants are permutation matrices, so

(2) $\det A=\sum_{j_1,...,j_n:\det P_{j_1,...,j_n}\neq 0}a_{1j_1}...a_{nj_n}\det P_{j_1,...,j_n}.$

Different-rows-different-columns rule. Take a good look at what the condition $\det P_{j_1,...,j_n}\neq 0$ implies about the location of the factors of the product $a_{1j_1}...a_{nj_n}.$ The rows $1,...,n$ to which the factors belong are obviously different. The columns $j_1,...,j_n$ are also different by the definition of a permutation matrix. Conversely, consider any combination $a_{i_1,j_1},...,a_{i_n,j_n}$ of elements such that no two elements belong to the same row or column. Rearrange the first indices $i_1,...,i_n$ in an ascending order, from $1$ to $n.$ This leads to a renumbering $\tilde{j}_{1},...,\tilde{j}_{n}$ of the second indices. The product $a_{i_1,j_1}...a_{i_n,j_n}$ becomes $a_{1,\tilde{j}_1}...a_{n,\tilde{j}_n}.$ Since the original second indices were all different, the new ones will be too. Hence, $P_{\tilde{j}_1,...,\tilde{j}_n}\neq 0$ and such a term must be in (2).

This rule alternatively can be called a cross-out rule because in practice it is used like this: take, say, element $a_{11}$ and cross out the row and column it belongs to. The next factor is selected from the remaining matrix. In that matrix, again cross out the row and column the second factor sits in. Continue like this until you have $n$ factors. Multiply the resulting product by the determinant of an appropriate permutation matrix, and you have one term of the Leibniz formula.

Remark 1. (2) is the Leibniz formula. The formula at the right of (1) is the Levy-Civita formula. The difference is that the Levy-Civita formula contains many more zero terms, while in (2) a term can be zero only if $a_{1j_1}...a_{nj_n}=0$.

Remark 2. Many textbooks instead of $\det P_{j_1,...,j_n}$ write in (2) signatures of permutations. Using $\det P_{j_1,...,j_n}$ is better because a) you save time by skipping the theory of permutations and b) you need a rule to calculate signatures of permutations and $\det P_{j_1,...,j_n}$ is such a rule (see an example).