Sunday, June 15, 2014

Linear Algebra: #14 Leibniz Formula

  • 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 Sn. In fact, Sn 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 aij , for i, j between 1 and n. Then we have

Linear Algebra: #14 Leibniz Formula equation pic 1

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


Theorem 38
Let A be a diagonal matrix
Linear Algebra: #14 Leibniz Formula equation pic 2

Then det(A) = λ1 λ2 · · · λn .


Theorem 39
Let A be a triangular matrix
Linear Algebra: #14 Leibniz Formula equation pic 3
Then det(A) = a11a22 · · · ann .

Leibniz formula also gives:

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


Theorem 40
det(At) = det(A).


14.1 Special rules for 2 × 2 and 3 × 3 matrices
Linear Algebra: #14 Leibniz Formula equation pic 4
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 , . . . ,εn. Thus

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

Therefore, given that the i-th row in a matrix is
ξi = (ai1ai2· · · ain),

 then we have
Linear Algebra: #14 Leibniz Formula equation pic 5
So let the matrix A be represented by its rows,
Linear Algebra: #14 Leibniz Formula equation pic 6

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

Linear Algebra: #14 Leibniz Formula equation pic 7
To begin with, observe that if εjk = εjl for some jk ≠ jl, then two rows are identical, and therefore the determinant is zero. Thus we need only the sum over all possible permutations (j1, j2, . . . , jn) of the numbers (1, 2, . . . , n). Then, given such a permutation, we have the matrix
Linear Algebra: #14 Leibniz Formula equation pic 8

This can be transformed back into the identity matrix
Linear Algebra: #14 Leibniz Formula equation pic 9
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
Linear Algebra: #14 Leibniz Formula equation pic 10

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