• Linear Algebra: #14 Leibniz Formula

Definition
A permutation of the numbers {1, . . . , n} is a bijection

σ : {1, . . . , n} → {1, . . . , n}. 

The set of all permutations of the numbers {1, . . . , n} is denoted S_n. In fact, S_n is a group: the symmetric group of order n. Given a permutation σ ∈ S , we will say that a pair of numbers (i, j), with i, j ∈ {1, . . . , n} is a “reversed pair” if i < j, yet σ(i) > σ(j). Let s(σ) be the total number of reversed pairs in σ. Then the sign of sigma is defined to be the number

sign(σ) = (−1)^s(σ)

Theorem 37 (Leibniz)
Let the elements in the matrix A be a_ij , for i, j between 1 and n. Then we have

As a consequence of this formula, the following theorems can be proved:


Theorem 38
Let A be a diagonal matrix

Then det(A) = λ₁ λ₂ · · · λ_n .


Theorem 39
Let A be a triangular matrix

Then det(A) = a₁₁a₂₂ · · · a_nn .

Leibniz formula also gives:

Definition
Let A ∈ M(n × n, F). The transpose A^t of A is the matrix consisting of elements a_ij^t  such that for all i and j we have a_ij^t = a_ji, where a_ji are the elements of the original matrix A.


Theorem 40
det(A^t) = det(A).


14.1 Special rules for 2 × 2 and 3 × 3 matrices

For the 2 × 2 matrix, the Leibniz formula reduces to the simple formula

det(A) =  a₁₁a₂₂ - a₁₂a₂₁ 

For the 3 × 3 matrix, the formula is a little more complicated.

det(A) =  a₁₁a₂₂a₃₃ + a₁₂a₂₃a₃₃ + a₁₃a₂₁a₃₂ - a₁₁a₂₃a₃₂ - a₁₂a₂₁a₃₃ - a₁₁a₂₃a₃₂

14.2 A proof of Leibniz Formula 
Let the rows of the n × n identity matrix be ε₁ , . . . ,ε_n. Thus

ε₁ = (1 0 0 · · · 0), ε₂ = (0 1 0 · · · 0), . . . , ε_n = (0 0 0 · · · 1). 

Therefore, given that the i-th row in a matrix is

ξ_i = (a_i1a_i2· · · a_in),

 then we have

So let the matrix A be represented by its rows,

It was an exercise to show that the determinant function is additive. That is, if B and C are n × n matrices, then we have det(B + C) = det(B) + det(C). Therefore we can write

To begin with, observe that if ε_jk = ε_jl for some j_k ≠ j_l, then two rows are identical, and therefore the determinant is zero. Thus we need only the sum over all possible permutations (j₁, j₂, . . . , j_n) of the numbers (1, 2, . . . , n). Then, given such a permutation, we have the matrix

This can be transformed back into the identity matrix

by means of successively exchanging pairs of rows.

Each time this is done, the determinant changes sign (from +1 to -1, or from -1 to +1). Finally, of course, we know that the determinant of the identity matrix is 1.

Therefore we obtain the Leibniz formula