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# Super matrix of minors and calculation of the determinant by expanding along multiple rows or columns

David Pigeon

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David Pigeon. Super matrix of minors and calculation of the determinant by expanding along multiple rows or columns. 2024. hal-0459454v3

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Preprint submitted on 7 Jun 2024

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HAL
open science

# Super matrix of minors and calculation of the determinant by expanding along multiple rows or columns

David Pigeon

<figure><img src="image_2.png" /></figure>

## To cite this version:

David Pigeon. Super matrix of minors and calculation of the determinant by expanding along multiple rows or columns. 2025. hal-0459454v7

HAL Id: hal-04594544

https://hal.science/hal-04594544v7

Preprint submitted on 30 Jan 2025

HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

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Distributed under a Creative Commons Attribution 4.0 International License
# Super matrix of minors and calculation of the determinant by expanding along multiple rows or columns

David Pigeon

7 juin 2024

**Keywords** : determinant, inverse of a matrix, cofactor matrix, cotranspose, minors of a matrix, super matrix of minors.

## Résumé

We provide a detailed proof of the calculation of the determinant by expanding along multiple rows or columns. Thus, we define the super matrix of minors associated with a square matrix, which allows us to generalize the usual formula relating the determinant, the matrix, and the cofactor matrix.

## Table des matières

<table><tr><td>1 Introduction</td><td>2</td></tr><tr><td>2 Minors of a Matrix</td><td>7</td></tr><tr><td>2.1 Order on the Set of Subsets of a Set</td><td>7</td></tr><tr><td>2.2 Minor of a Matrix</td><td>10</td></tr><tr><td>3 Super matrices</td><td>17</td></tr><tr><td>3.1 Notations</td><td>17</td></tr><tr><td>3.2 Kronecker Matrices</td><td>17</td></tr><tr><td>3.3 Super matrices of minors and cofactors</td><td>19</td></tr><tr><td>3.4 Further Column Expansions</td><td>25</td></tr><tr><td>3.5 Case of dimension 3</td><td>27</td></tr><tr><td>3.6 An Example in Dimension 4</td><td>29</td></tr><tr><td>A Permutation Matrices</td><td>31</td></tr><tr><td>A.1 Symmetric Groups</td><td>31</td></tr><tr><td>A.2 Determinant of a Square Matrix</td><td>33</td></tr><tr><td>A.3 Permutation Matrices</td><td>35</td></tr><tr><td>A.4 Examples</td><td>37</td></tr><tr><td>B The Transpose and the Cotranspose</td><td>39</td></tr><tr><td>B.1 Notations</td><td>39</td></tr><tr><td>B.2 Properties</td><td>41</td></tr><tr><td>Références</td><td>44</td></tr></table>
# Super matrix of minors and calculation of the determinant by expanding along multiple rows or columns

David Pigeon*

30 janvier 2025

**Keywords** : determinant, inverse of a matrix, cofactor matrix, cotranspose, minors of a matrix, super matrix of minors, super cofactor matrix.

## Résumé

We provide a detailed proof of the calculation of the determinant by expanding along several rows or columns. We thus define the supermatrix of minors associated with a square matrix. This allows us to generalize the usual formula relating the determinant, the matrix, and the cofactor matrix.

## Table des matières

<table><tr><td>1 Introduction</td><td>2</td></tr><tr><td>2 Minors of a Matrix</td><td>6</td></tr><tr><td>2.1 Order on the Set of Subsets of a Set</td><td>6</td></tr><tr><td>2.2 Minor of a Matrix</td><td>9</td></tr><tr><td>3 Super matrices</td><td>17</td></tr><tr><td>3.1 Notations</td><td>17</td></tr><tr><td>3.2 Kronecker Matrices</td><td>17</td></tr><tr><td>3.3 Super matrices of minors and cofactors</td><td>19</td></tr><tr><td>3.4 Further Column Expansions</td><td>25</td></tr><tr><td>3.5 Case of dimension 3</td><td>27</td></tr><tr><td>3.6 An Example in Dimension 4</td><td>29</td></tr><tr><td>A Permutation Matrices</td><td>31</td></tr><tr><td>A.1 Symmetric Groups</td><td>31</td></tr><tr><td>A.2 Determinant of a Square Matrix</td><td>33</td></tr><tr><td>A.3 Permutation Matrices</td><td>34</td></tr><tr><td>A.4 Examples</td><td>37</td></tr><tr><td>B The Transpose and the Cotranspose</td><td>39</td></tr><tr><td>B.1 Notations</td><td>39</td></tr><tr><td>B.2 Properties</td><td>41</td></tr><tr><td>Références</td><td>44</td></tr></table>

<footnote>*https://sites.google.com/view/davidpigeon/</footnote>
# 1 Introduction

Let's start with an example. Let's calculate the determinant of the 4 × 4 square matrix :

$$A := (a_{ij})_{ij} := \begin{pmatrix} 1 & 2 & 1 & -2 \\ 1 & 3 & 2 & -1 \\ -2 & 1 & 3 & 4 \\ 1 & -1 & 3 & 1 \end{pmatrix}$$

A common technique for calculating the determinant of A is to expand along a row (or a column). The expansion along the i-th row is given by :

$$\det(A) = \sum_{j=1}^{4} (-1)^{i+j} a_{ij} \Delta_{\{i\},\{j\}}(A)$$

where $\Delta_{\{i\},\{j\}}(A)$ is the 3 × 3 determinant of the matrix A with the i-th row and the j-th column removed. We calculate the determinant of A by expanding along the third row of A, then expanding each 3 × 3 determinant along their first rows, and using the formula for the 2 × 2 determinant :

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.$$

We have :

$$
\begin{aligned}
\det(A) &= (-1)^{3+1} \times (-2) \times \begin{vmatrix} 2 & 1 & -2 \\ 3 & 2 & -1 \\ -1 & 3 & 1 \end{vmatrix} + (-1)^{3+2} \times 1 \times \begin{vmatrix} 1 & 1 & -2 \\ 1 & 2 & -1 \\ 1 & 3 & 1 \end{vmatrix} \\
&\quad + (-1)^{3+3} \times 3 \times \begin{vmatrix} 1 & 2 & -2 \\ 1 & 3 & -1 \\ 1 & -1 & 1 \end{vmatrix} + (-1)^{3+4} \times 4 \times \begin{vmatrix} 1 & 2 & 1 \\ 1 & 3 & 2 \\ 1 & -1 & 3 \end{vmatrix} \\
&= -2 \left( (-1)^{1+1} \times 2 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \right) \\
&\quad - \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \right) \\
&\quad + 3 \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \right) \\
&\quad - 4 \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + (-1)^{1+3} \times 1 \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \right) \\
&= -2(2(2+3) - (3-1) - 2(9+2)) - ((2+3) - (1+1) - 2(3-2)) \\
&\quad + 3((3-1) - 2(1+1) - 2(-1-3)) - 4((9+2) - 2(3-2) + (-1-3)) \\
&= 28 - 1 + 18 - 20 \\
&= 25
\end{aligned}
$$

Looking more closely at this calculation, we realize that we perform the same calculation multiple times. Each 2 × 2 determinant appears twice. Let's highlight identical 2 × 2 determinants with the same color and keep the $(-1)^{\bullet}$ with all the details in the exponent. We have :

$$
\det(A) = (-1)^{3+1} \times (-2) \times \left( (-1)^{1+1} \times 2 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \right) \\
+ (-1)^{3+2} \times 1 \times \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \right)
$$
# 1 Introduction

Let's start with an example. Let's calculate the determinant of the 4 × 4 square matrix :

$$A := (a_{ij})_{ij} := \begin{pmatrix} 1 & 2 & 1 & -2 \\ 1 & 3 & 2 & -1 \\ -2 & 1 & 3 & 4 \\ 1 & -1 & 3 & 1 \end{pmatrix}$$

A common technique for calculating the determinant of A is to expand along a row (or a column). The expansion along the i-th row is given by :

$$\det(A) = \sum_{j=1}^{4} (-1)^{i+j} a_{ij} \Delta_{\{i\},\{j\}}(A)$$

where $\Delta_{\{i\},\{j\}}(A)$ is the 3 × 3 determinant of the matrix A with the i-th row and the j-th column removed. We calculate the determinant of A by expanding along the third row of A, then expanding each 3 × 3 determinant along their first rows, and using the formula for the 2 × 2 determinant :

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.$$

We have :

$$
\begin{aligned}
\det(A) &= (-1)^{3+1} \times (-2) \times \begin{vmatrix} 2 & 1 & -2 \\ 3 & 2 & -1 \\ -1 & 3 & 1 \end{vmatrix} + (-1)^{3+2} \times 1 \times \begin{vmatrix} 1 & 1 & -2 \\ 1 & 2 & -1 \\ 1 & 3 & 1 \end{vmatrix} \\
&\quad + (-1)^{3+3} \times 3 \times \begin{vmatrix} 1 & 2 & -2 \\ 1 & 3 & -1 \\ 1 & -1 & 1 \end{vmatrix} + (-1)^{3+4} \times 4 \times \begin{vmatrix} 1 & 2 & 1 \\ 1 & 3 & 2 \\ 1 & -1 & 3 \end{vmatrix} \\
&= -2 \left( (-1)^{1+1} \times 2 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \right) \\
&\quad - \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \right) \\
&\quad + 3 \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \right) \\
&\quad - 4 \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + (-1)^{1+3} \times 1 \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \right) \\
&= -2(2(2+3) - (3-1) - 2(9+2)) - ((2+3) - (1+1) - 2(3-2)) \\
&\quad + 3((3-1) - 2(1+1) - 2(-1-3)) - 4((9+2) - 2(3-2) + (-1-3)) \\
&= 28 - 1 + 18 - 20 \\
&= 25
\end{aligned}
$$

By looking more closely at this calculation, we realize that we perform the same calculation multiple times. Each 2 × 2 determinant appears twice. Let's highlight identical 2 × 2 determinants with the same color and keep the $(-1)^{\bullet}$ with all the details in the exponent. We have :

$$
\begin{aligned}
\det(A) &= (-1)^{3+1} \times (-2) \times \left( (-1)^{1+1} \times 2 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \right) \\
&\quad + (-1)^{3+2} \times 1 \times \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \right)
\end{aligned}
$$
$$ + (-1)^{3+3} \times 3 \times \left( \begin{aligned} & (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \\ + (-1)^{3+4} \times 4 \times \left( \begin{aligned} & (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + (-1)^{1+3} \times 1 \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \end{aligned} \right) $$

Grouping identical 2 × 2 determinants, we have :

$$ \begin{aligned} \det(A) = & ((-1)^{3+2+1+1} \times (-1) \times 1 + (-1)^{3+1+1+1} \times (-2) \times 2) \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \\ & + ((-1)^{3+3+1+1} \times 3 \times 1 + (-1)^{3+1+1+2} \times (-2) \times 1) \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} \\ & + ((-1)^{3+3+1+3} \times 3 \times (-2) + (-1)^{3+4+1+1} \times 4 \times 1) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \\ & + ((-1)^{3+2+1+3} \times 1 \times (-2) + (-1)^{3+4+1+2} \times 4 \times 2) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \\ & + ((-1)^{3+4+1+3} \times 4 \times 1 + (-1)^{3+3+1+3} \times 3 \times (-2)) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \\ & + ((-1)^{3+1+1+2} \times (-1) \times 1 + (-1)^{3+3+1+2} \times 3 \times 2) \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \end{aligned} $$

Thus, we can compute the 2 × 2 determinants only once, multiplying the result by their multiplicity and sign.

$$ \begin{aligned} \det(A) = & ((-1)^{3+2+1+1} \times (-1) \times 1 + (-1)^{3+1+1+1} \times (-2) \times 2) \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \\ & + ((-1)^{3+3+1+1} \times 3 \times 1 + (-1)^{3+1+1+2} \times (-2) \times 1) \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} \\ & + ((-1)^{3+3+1+3} \times 3 \times (-2) + (-1)^{3+4+1+1} \times 4 \times 1) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \\ & + ((-1)^{3+2+1+3} \times 1 \times (-2) + (-1)^{3+4+1+2} \times 4 \times 2) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \\ & + ((-1)^{3+4+1+3} \times 4 \times 1 + (-1)^{3+3+1+3} \times 3 \times (-2)) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \\ & + ((-1)^{3+1+1+2} \times (-1) \times 1 + (-1)^{3+3+1+2} \times 3 \times 2) \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \\ = & ((-1) \times (-1) \times (-1) + 1 \times (-2) \times 2) \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \\ & + (1 \times 3 \times 1 + (-1) \times (-2) \times 1) \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} \\ & + ((-1) \times 3 \times (-2) + (-1) \times 4 \times 1) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \\ & + ((-1) \times 1 \times (-2) + 1 \times 4 \times 2) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \\ & + ((-1) \times 4 \times 1 + (-1) \times 3 \times (-2)) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \\ & + (1 \times (-1) \times 1 + (-1) \times 3 \times 2) \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \end{aligned} $$
$$ + (-1)^{3+3} \times 3 \times \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \right) \\ + (-1)^{3+4} \times 4 \times \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + (-1)^{1+3} \times 1 \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \right) $$

Grouping identical $2 \times 2$ determinants, we have:

$$ \begin{aligned} \det(A) = & ((-1)^{3+2+1+1} \times (-1) \times 1 + (-1)^{3+1+1+1} \times (-2) \times 2) \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \\ & + ((-1)^{3+3+1+1} \times 3 \times 1 + (-1)^{3+1+1+2} \times (-2) \times 1) \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} \\ & + ((-1)^{3+4+1+1} \times 4 \times 1 + (-1)^{3+1+1+3} \times (-2) \times (-2)) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \\ & + ((-1)^{3+3+1+2} \times 3 \times 2 + (-1)^{3+2+1+2} \times 1 \times 1) \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \\ & + ((-1)^{3+4+1+2} \times 4 \times 2 + (-1)^{3+2+1+3} \times 1 \times (-2)) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \\ & + ((-1)^{3+4+1+3} \times 4 \times 1 + (-1)^{3+3+1+3} \times 3 \times (-2)) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \end{aligned} $$

The terms in front of each identical $2 \times 2$ determinant are also $2 \times 2$ determinants. We have:

$$ \begin{aligned} \det(A) = & (-1)^{3+2+1+1} \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{3+3+1+1} \begin{vmatrix} 1 & 1 \\ -2 & 3 \end{vmatrix} \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} \\ & + (-1)^{3+4+1+1} \begin{vmatrix} 1 & -2 \\ -2 & 4 \end{vmatrix} \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} + (-1)^{3+3+1+2} \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \\ & + (-1)^{3+4+1+2} \begin{vmatrix} 2 & -2 \\ 1 & 4 \end{vmatrix} \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + (-1)^{3+4+1+3} \begin{vmatrix} 1 & -2 \\ 3 & 4 \end{vmatrix} \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \end{aligned} $$

For all $i, j, k, l \in \{1, 2, 3, 4\}$ such that $i \neq j$ and $k \neq l$, let:

$$ \Delta_{\{i,j\},\{k,l\}}(A) $$

be the $2 \times 2$ determinant of the matrix $A$ obtained by removing rows $i, j$ and columns $k, l$. By rearranging the elements in the exponents of $-1$, we have:

$$ \begin{aligned} \det(A) = & (-1)^{1+3+1+2} \Delta_{\{2,4\},\{3,4\}}(A) \times \Delta_{\{1,3\},\{1,2\}}(A) + (-1)^{1+3+1+3} \Delta_{\{2,4\},\{2,4\}}(A) \times \Delta_{\{1,3\},\{1,3\}}(A) \\ & + (-1)^{1+3+1+4} \Delta_{\{2,4\},\{2,3\}}(A) \times \Delta_{\{1,3\},\{1,4\}}(A) + (-1)^{1+3+2+3} \Delta_{\{2,4\},\{1,4\}}(A) \times \Delta_{\{1,3\},\{2,3\}}(A) \\ & + (-1)^{1+3+2+4} \Delta_{\{2,4\},\{1,3\}}(A) \times \Delta_{\{1,3\},\{2,4\}}(A) + (-1)^{1+3+3+4} \Delta_{\{2,4\},\{1,2\}}(A) \times \Delta_{\{1,3\},\{3,4\}}(A) \end{aligned} $$

The elements in the exponents of $-1$ are exactly the sums of the elements of the sets appearing in the subscripts of the $\Delta$ on the right, i.e., we have terms of the type:

$$ (-1)^{a+b+c+d} \times \cdots \times \Delta_{\{a,b\},\{c,d\}}(A). $$

For the middle terms, we see in the subscripts the complements of the sets appearing in the $\Delta$ terms on the right. Let's then denote for any subset $I$ of $\{1, 2, 3, 4\}$:

* $\bar{I}$ as the complement of $I$ in $\{1, 2, 3, 4\}$;
$$
\begin{align*}
&= (1 - 4) \times (2 + 3) + (3 + 2) \times (3 - 1) + (6 - 4) \times (9 + 2) \\
&\quad + (2 + 8) \times (3 - 2) + (-4 + 6) \times (-1 - 3) + (-1 - 6) \times (1 + 1) \\
&= -3 \times 5 + 5 \times 2 + 2 \times 11 + 10 \times 1 + 2 \times (-4) + (-7) \times 2 \\
&= -15 + 10 + 22 + 10 - 8 - 14 \\
&= 25.
\end{align*}
$$

By looking more closely at this calculation, we realize that we perform the same calculation multiple times. Each 2 × 2 determinant appears twice. Let's highlight identical 2 × 2 determinants with the same color and keep the (-1)• with all the details in the exponent. We have :

$$
\begin{align*}
\det(A) = (-1)^{3+1} \times (-2) \times \left( (-1)^{1+1} \times 2 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \right. \\
\qquad + (-1)^{3+2} \times 1 \times \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{1+2} \times 1 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \right) \\
\qquad + (-1)^{3+3} \times 3 \times \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + (-1)^{1+3} \times (-2) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \right) \\
\qquad \left. + (-1)^{3+4} \times 4 \times \left( (-1)^{1+1} \times 1 \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} + (-1)^{1+2} \times 2 \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + (-1)^{1+3} \times 1 \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix} \right)
\end{align*}
$$

Grouping identical 2 × 2 determinants, we have :

$$
\begin{align*}
\det(A) = & ((-1)^{3+2+1+1} \times (-1) \times 1 + (-1)^{3+1+1+1} \times (-2) \times 2) \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \\
& + ((-1)^{3+3+1+1} \times 3 \times 1 + (-1)^{3+1+1+2} \times (-2) \times 1) \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} \\
& + ((-1)^{3+4+1+1} \times 4 \times 1 + (-1)^{3+1+1+3} \times (-2) \times (-2)) \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} \\
& + ((-1)^{3+3+1+2} \times 3 \times 2 + (-1)^{3+2+1+2} \times 1 \times 1) \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \\
& + ((-1)^{3+4+1+2} \times 4 \times 2 + (-1)^{3+2+1+3} \times 1 \times (-2)) \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \\
& + ((-1)^{3+4+1+3} \times 4 \times 1 + (-1)^{3+3+1+3} \times 3 \times (-2)) \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix}
\end{align*}
$$

The terms in front of each identical 2 × 2 determinant are also 2 × 2 determinants. We have :

$$
\begin{align*}
\det(A) = (-1)^{3+2+1+1} \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} \times \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} + (-1)^{3+3+1+1} \begin{vmatrix} 1 & 1 \\ -2 & 3 \end{vmatrix} \times \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} \\
& + (-1)^{3+4+1+1} \begin{vmatrix} 1 & -2 \\ -2 & 4 \end{vmatrix} \times \begin{vmatrix} 3 & 2 \\ -1 & 3 \end{vmatrix} + (-1)^{3+3+1+2} \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \\
& + (-1)^{3+4+1+2} \begin{vmatrix} 2 & -2 \\ 1 & 4 \end{vmatrix} \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + (-1)^{3+4+1+3} \begin{vmatrix} 1 & -2 \\ 3 & 4 \end{vmatrix} \times \begin{vmatrix} 1 & 3 \\ 1 & -1 \end{vmatrix}
\end{align*}
$$

For all i,j,k,l ∈ {1,2,3,4} such that i ≠ j and k ≠ l, let :

$$
\Delta_{\{i,j\},\{k,l\}}(A)
$$
• $|I|$ as the sum of the elements of $I$, i.e. :

$$|I| := \sum_{i \in I} i.$$

Each term is thus of the type :

$$(-1)^{|I|+|J|} \Delta_{\overline{I}, \overline{J}}(A) \Delta_{I, J}(A)$$

with $I$ and $J$ being two subsets of $\{1, 2, 3, 4\}$ of cardinality 2.

The set $\{1, 3\}$ and its complement $\{2, 4\}$ appear in each term, so let :

$$K := \{1, 3\}.$$

We have :

$$\det(A) = (-1)^{|K|+|\{1,2\}|} \Delta_{\overline{K}, \overline{\{1,2\}}}(A) \times \Delta_{K, \{1,2\}}(A) + (-1)^{|K|+|\{1,3\}|} \Delta_{\overline{K}, \overline{\{1,3\}}}(A) \times \Delta_{K, \{1,3\}}(A) \\ + (-1)^{|K|+|\{1,4\}|} \Delta_{\overline{K}, \overline{\{1,4\}}}(A) \times \Delta_{K, \{1,4\}}(A) + (-1)^{|K|+|\{2,3\}|} \Delta_{\overline{K}, \overline{\{2,3\}}}(A) \times \Delta_{K, \{2,3\}}(A) \\ + (-1)^{|K|+|\{2,4\}|} \Delta_{\overline{K}, \overline{\{2,4\}}}(A) \times \Delta_{K, \{2,4\}}(A) + (-1)^{|K|+|\{3,4\}|} \Delta_{\overline{K}, \overline{\{3,4\}}}(A) \times \Delta_{K, \{3,4\}}(A)$$

Final simplification, there appear six sets :

$$\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}.$$

These six sets are exactly the subsets with 2 elements of $\{1, 2, 3, 4\}$. Let $\mathcal{P}_{4,2}$ denote the set of subsets with 2 elements of $\{1, 2, 3, 4\}$. We then obtain the formula :

$$\det(A) = \sum_{L \in \mathcal{P}_{4,2}} (-1)^{|K|+|L|} \Delta_{\overline{K}, \overline{L}}(A) \times \Delta_{K,L}(A)$$

called the **expansion along the rows of $\{1, 3\}$ of $A$**.

We will generalize the formula to the expansion along $k$ rows and by transposition to the expansion along $k$ columns.

Let's revisit the formula :

$$ (*) \det(A) = \sum_{j=1}^{4} (-1)^{i+j} a_{ij} \Delta_{\{i\}, \{j\}}(A)$$

We denote :

$$\mathcal{P}_{4,1} := \{\{1\}, \{2\}, \{3\}, \{4\}\}$$

the set of subsets with 1 element of $\{1, 2, 3, 4\}$. Formula (*) thus becomes :

$$\det(A) = \sum_{L \in \mathcal{P}_{4,1}} (-1)^{i+|L|} \Delta_{\overline{\{i\}}, \overline{L}}(A) \times \Delta_{\{i\}, L}(A)$$

called the **expansion along row $i$ of $A$**.

Similarly, we denote :

$$\mathcal{P}_{4,3} := \{\{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}\}$$

the set of subsets with 3 elements of $\{1, 2, 3, 4\}$. We then have the final formula :

$$\det(A) = \sum_{L \in \mathcal{P}_{4,1}} (-1)^{i+|L|} \Delta_{\overline{\{i\}}, \overline{L}}(A) \times \Delta_{\{i\}, L}(A) \\ = \sum_{L \in \mathcal{P}_{4,1}} (-1)^{10-i+10-|L|} \Delta_{\overline{\{i\}}, \overline{L}}(A) \times \Delta_{\{i\}, L}(A)$$
be the 2 × 2 determinant of the matrix A obtained by removing rows $i, j$ and columns $k, l$. By rearranging the elements in the exponents of $-1$, we have:

$$ \det(A) = (-1)^{1+3+1+2} \Delta_{\{2,4\},\{3,4\}}(A) \times \Delta_{\{1,3\},\{1,2\}}(A) + (-1)^{1+3+1+3} \Delta_{\{2,4\},\{2,4\}}(A) \times \Delta_{\{1,3\},\{1,3\}}(A) \\ + (-1)^{1+3+1+4} \Delta_{\{2,4\},\{2,3\}}(A) \times \Delta_{\{1,3\},\{1,4\}}(A) + (-1)^{1+3+2+3} \Delta_{\{2,4\},\{1,4\}}(A) \times \Delta_{\{1,3\},\{2,3\}}(A) \\ + (-1)^{1+3+2+4} \Delta_{\{2,4\},\{1,3\}}(A) \times \Delta_{\{1,3\},\{2,4\}}(A) + (-1)^{1+3+3+4} \Delta_{\{2,4\},\{1,2\}}(A) \times \Delta_{\{1,3\},\{3,4\}}(A) $$

The elements in the exponents of $-1$ are exactly the sums of the elements of the sets appearing in the subscripts of the $\Delta$ on the right, i.e., we have terms of the type:

$$ (-1)^{a+b+c+d} \times \cdots \times \Delta_{\{a,b\},\{c,d\}}(A). $$

For the middle terms, we see in the subscripts the complements of the sets appearing in the $\Delta$ terms on the right. Let's then denote for any subset $I$ of $\{1, 2, 3, 4\}$:

* $\bar{I}$ as the complement of $I$ in $\{1, 2, 3, 4\}$;

* $|I|$ as the sum of the elements of $I$, i.e.:

$$ |I| := \sum_{i \in I} i. $$

Each term is thus of the type:

$$ (-1)^{|I|+|J|} \Delta_{\bar{I}, \bar{J}}(A) \Delta_{I, J}(A) $$

with $I$ and $J$ being two subsets of $\{1, 2, 3, 4\}$ of cardinality 2.

The set $\{1, 3\}$ and its complement $\{2, 4\}$ appear in each term, so let:

$$ K := \{1, 3\}. $$

We have:

$$ \det(A) = (-1)^{|K|+|\{1,2\}|} \Delta_{\overline{K}, \overline{\{1,2\}}}(A) \times \Delta_{K, \{1,2\}}(A) + (-1)^{|K|+|\{1,3\}|} \Delta_{\overline{K}, \overline{\{1,3\}}}(A) \times \Delta_{K, \{1,3\}}(A) \\ + (-1)^{|K|+|\{1,4\}|} \Delta_{\overline{K}, \overline{\{1,4\}}}(A) \times \Delta_{K, \{1,4\}}(A) + (-1)^{|K|+|\{2,3\}|} \Delta_{\overline{K}, \overline{\{2,3\}}}(A) \times \Delta_{K, \{2,3\}}(A) \\ + (-1)^{|K|+|\{2,4\}|} \Delta_{\overline{K}, \overline{\{2,4\}}}(A) \times \Delta_{K, \{2,4\}}(A) + (-1)^{|K|+|\{3,4\}|} \Delta_{\overline{K}, \overline{\{3,4\}}}(A) \times \Delta_{K, \{3,4\}}(A) $$

Final simplification, there appear six sets:

$$ \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}. $$

These six sets are exactly the subsets with 2 elements of $\{1, 2, 3, 4\}$. Let $\mathcal{P}_{4,2}$ denote the set of subsets with 2 elements of $\{1, 2, 3, 4\}$. We then obtain the formula:

$$ \det(A) = \sum_{L \in \mathcal{P}_{4,2}} (-1)^{|K|+|L|} \Delta_{\overline{K}, \overline{L}}(A) \times \Delta_{K, L}(A) $$

called the **expansion along the rows of $\{1, 3\}$ of A**.

We will generalize the formula to the expansion along $k$ rows and by transposition to the expansion along $k$ columns.

Let's revisit the formula:

$$ (*) \det(A) = \sum_{j=1}^{4} (-1)^{i+j} a_{ij} \Delta_{\{i\}, \{j\}}(A) $$

We denote:

$$ \mathcal{P}_{4,1} := \{\{1\}, \{2\}, \{3\}, \{4\}\} $$
$$ 
\begin{aligned}
&= \sum_{L \in \mathcal{P}_{4,1}} (-1)^{|\overline{\{i\}}|+|\overline{L}|} \Delta_{\overline{K},\overline{L}}(A) \times \Delta_{\{i\},L}(A) \\
&= \sum_{L \in \mathcal{P}_{4,3}} (-1)^{|K|+|L|} \Delta_{\overline{K},\overline{L}}(A) \times \Delta_{\{i\},L}(A)
\end{aligned}
$$

with $K := \overline{\{i\}}$ called the **expansion along the rows of K** of A.

We have just seen that with a matrix of size 4, we can expand the determinant of the matrix along one row, two rows, or three rows. In this example, we also observe that the formula for expanding the determinant along two rows can be derived from the formula for expanding it along one row; we will utilize this in the general case (we will provide a proof by induction).

The following section will address the general case.

**Reference.** The idea for this text came to me while searching for an alternative formula for the determinant of 4 × 4 matrices for my students. Reflecting on the possibility of expanding the determinant along two rows, I discovered, through manual calculation, the formula introduced earlier in this document. This discovery led me to wonder whether a more general formula existed. I then conjectured the formula stated in Proposition 2.2.4, which I subsequently programmed in Python, yielding satisfactory results.

During my online research, I found only one source, the Wikipedia page [2], where the proposed formula is presented with different notations and references the book [1]. This book offers a very brief outline of a proof for this formula. The purpose of this text is to provide a complete and detailed demonstration while introducing the notion of a supermatrix to arrive at a simpler formula.
the set of subsets with 1 element of {1, 2, 3, 4}. Formula (*) thus becomes :

$$ \det(A) = \sum_{L \in \mathcal{P}_{4,1}} (-1)^{i+|L|} \Delta_{\overline{\{i\}}, \overline{L}}(A) \times \Delta_{\{i\}, L}(A) $$

called the **expansion along row i of A**.

Similarly, we denote :

$$ \mathcal{P}_{4,3} := \{\{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}\} $$

the set of subsets with 3 elements of {1, 2, 3, 4}. We then have the final formula :

$$ \begin{aligned} \det(A) &= \sum_{L \in \mathcal{P}_{4,1}} (-1)^{i+|L|} \Delta_{\overline{\{i\}}, \overline{L}}(A) \times \Delta_{\{i\}, L}(A) \\ &= \sum_{L \in \mathcal{P}_{4,1}} (-1)^{10-i+10-|L|} \Delta_{\overline{\{i\}}, \overline{L}}(A) \times \Delta_{\{i\}, L}(A) \\ &= \sum_{L \in \mathcal{P}_{4,1}} (-1)^{|\overline{\{i\}}|+|\overline{L}|} \Delta_{\overline{K}, \overline{L}}(A) \times \Delta_{\{i\}, L}(A) \\ &= \sum_{L \in \mathcal{P}_{4,3}} (-1)^{|K|+|L|} \Delta_{\overline{K}, \overline{L}}(A) \times \Delta_{\{i\}, L}(A) \end{aligned} $$

with $K := \overline{\{i\}}$ called the **expansion along the rows of K of A**.

We have seen that with a 4 × 4 matrix, we can expand the determinant of the matrix along one row, two rows, or three rows. The next section will deal with the general case. In the book [1], one can find an idea of proof (see also [2]).
## 2 Minors of a Matrix

We generalize what was seen in the introduction to square matrices of any size. Unless otherwise specified, $n$ and $k$ are natural numbers such that $k \le n$ and $n > 0$.

### 2.1 Order on the Set of Subsets of a Set

We begin by defining notations for the subsets of $\{1, \dots, n\}$.

**Notation 2.1.1**

We denote $\mathcal{P}_n$ the set of subsets of $\{1, \dots, n\}$ and $\mathcal{P}_{n,k}$ the set of subsets of $\{1, \dots, n\}$ with cardinality $k$. Therefore, we have:

$$ \mathrm{Card}(\mathcal{P}_n) = 2^n $$

$$ \mathrm{Card}(\mathcal{P}_{n,k}) = \binom{n}{k} $$

(1) We define two orders.

(a) On $\mathcal{P}_{n,k}$, we define the lexicographical order $\leq$, i.e. for any elements $I := \{i_1, \dots, i_k\}$ and $J := \{j_1, \dots, j_k\}$ of $\mathcal{P}_{n,k}$ such that $i_1 < \dots < i_k$ and $j_1 < \dots < j_k$, we denote $I \leq J$ if:

$$ I = J \lor i_1 < j_1 \lor [\exists k \in \{2, \dots, n-1\}, i_k < j_k \land (\forall l \in \{1, \dots, k-1\}, i_l = j_l)]. $$

(b) On $\mathcal{P}_n$, we define the order $\leq$ which extends the order on $\mathcal{P}_{n,k}$ defined in (a) : for all elements $I$ and $J$ of $\mathcal{P}_n$, we write $I \leq J$ if:

$$ \mathrm{Card}(I) < \mathrm{Card}(J) \lor [\mathrm{Card}(I) = \mathrm{Card}(J) \land I \leq J]. $$

(2) For any $I \in \mathcal{P}_n$, we denote $\bar{I}$ as the complement of $I$ in $\{1, \dots, n\}$;

(3) For any $I \in \mathcal{P}_n$, we denote $|I|$ as the sum of the elements of $I$, i.e.:

$$ |I| := \sum_{i \in I} i. $$

We set:

$$ |\emptyset| := 0. $$

(4) For any ordered set, we will use parentheses to indicate that the set is ordered, i.e. a set $E := \{i_1, \dots, i_k\}$ with:

$$ i_1 < \dots < i_k. $$

Then $E$ will be denoted $(i_1, \dots, i_k)$. For example, we will denote:

$$ (1, 3, 4, 7) := \{3, 7, 4, 1\} $$

and with the total order given on $\mathcal{P}_2$ in (1.b), we have:

$$ \mathcal{P}_2 = (\emptyset, \{1\}, \{2\}, \{1, 2\}). $$
## 2 Minors of a Matrix

We generalize what was seen in the introduction to square matrices of any size. Unless otherwise specified, $n$ and $k$ are natural numbers such that $k \le n$ and $n > 0$.

### 2.1 Order on the Set of Subsets of a Set

We begin by defining notations for the subsets of $\{1, \dots, n\}$.

#### Notation 2.1.1

We denote $\mathcal{P}_n$ the set of subsets of $\{1, \dots, n\}$ and $\mathcal{P}_{n,k}$ the set of subsets of $\{1, \dots, n\}$ with cardinality $k$. Therefore, we have:

$$ \mathrm{Card} (\mathcal{P}_n) = 2^n $$

$$ \mathrm{Card} (\mathcal{P}_{n,k}) = \binom{n}{k} $$

(1) We define two orders.

(a) On $\mathcal{P}_{n,k}$, we define the lexicographic order $\leq$, i.e., for all elements $I := \{i_1, \dots, i_k\}$ and $J := \{j_1, \dots, j_k\}$ of $\mathcal{P}_{n,k}$ such that $i_1 < \dots < i_k$ and $j_1 < \dots < j_k$, we write $I \leq J$ if:

$$ i_1 < j_1 \lor [\exists k \in \{1, \dots, n-1\}, i_k < j_k \land (\forall l \in \{1, \dots, k\}, i_l = j_l)]. $$

(b) On $\mathcal{P}_n$, we define the order $\leq$ which extends the order on $\mathcal{P}_{n,k}$ defined in (a) : for all elements $I$ and $J$ of $\mathcal{P}_n$, we write $I \leq J$ if:

$$ \mathrm{Card}(I) < \mathrm{Card}(J) \lor [\mathrm{Card}(I) = \mathrm{Card}(J) \land I \leq J]. $$

(2) For any $I \in \mathcal{P}_n$, we denote $\bar{I}$ as the complement of $I$ in $\{1, \dots, n\}$;

(3) For any $I \in \mathcal{P}_n$, we denote $|I|$ as the sum of the elements of $I$, i.e.:

$$ |I| := \sum_{i \in I} i. $$

We set:

$$ |\emptyset| := 0. $$

(4) For any ordered set, we will use parentheses to indicate that the set is ordered, i.e. $E := \{i_1, \dots, i_k\}$ is a set if we have:

$$ i_1 < \dots < i_k. $$

Then $E$ will be denoted $(i_1, \dots, i_k)$. For example, we will denote:

$$ (1, 3, 4, 7) := \{3, 7, 4, 1\} $$

and with the total order given on $\mathcal{P}_2$ in (1.b), we have:

$$ \mathcal{P}_2 = (\emptyset, \{1\}, \{2\}, \{1, 2\}). $$

#### Example 2.1.2

Let's take the case where $n := 4$. We have:

$$ \mathcal{P}_4 = (\emptyset, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, $$
### Example 2.1.2

Let's take the case where $n := 4$. We have :

$$ \mathcal{P}_4 = (\emptyset, \{1\}, \{2\}, \{3\}, \{4\}, \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, \\ \{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}, \{1,2,3,4\}) $$

and

$$ \mathcal{P}_{4,0} = (\emptyset) $$

$$ \mathcal{P}_{4,1} = (\{1\}, \{2\}, \{3\}, \{4\}) $$

$$ \mathcal{P}_{4,2} = (\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}) $$

$$ \mathcal{P}_{4,3} = (\{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}) $$

$$ \mathcal{P}_{4,4} = (\{1,2,3,4\}) $$

In the proofs, we will use another notation that allows us to identify the elements in an ordered set.

### Notation 2.1.3

Let the ordered set $E := (x_1, \dots, x_k)$, i.e. :

$$ x_1 < \dots < x_k. $$

For any $x \in E$, there exists $l \in \{1, \dots, k\}$ such that $x = x_l$, we denote :

$$ \mathfrak{p}(x, E) := l $$

called the **position of x in E**.

### Example 2.1.4

Let's take the case where $n := 4$. We have :

$$ \mathcal{P}_4 = (\emptyset, \{1\}, \{2\}, \{3\}, \{4\}, \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, \\ \{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}, \{1,2,3,4\}) $$

and

$$ \mathcal{P}_{4,0} = (\emptyset) $$

$$ \mathcal{P}_{4,1} = (\{1\}, \{2\}, \{3\}, \{4\}) $$

$$ \mathcal{P}_{4,2} = (\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}) $$

$$ \mathcal{P}_{4,3} = (\{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}) $$

$$ \mathcal{P}_{4,4} = (\{1,2,3,4\}) $$

In the proofs, we will use another notation that allows us to identify the elements in an ordered set.

### Notation 2.1.5

Let the ordered set $E := (x_1, \dots, x_k)$, i.e. :

$$ x_1 < \dots < x_k. $$
$$ \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\}) $$

and

$$ \mathcal{P}_{4,0} = (\emptyset) $$

$$ \mathcal{P}_{4,1} = (\{1\}, \{2\}, \{3\}, \{4\}) $$

$$ \mathcal{P}_{4,2} = (\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}) $$

$$ \mathcal{P}_{4,3} = (\{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}) $$

$$ \mathcal{P}_{4,4} = (\{1,2,3,4\}) $$

In the proofs, we will use another notation that allows us to identify the elements in an ordered set.

### Notation 2.1.3

Let the ordered set $E := (x_1, \dots, x_k)$, i.e. :

$$ x_1 < \dots < x_k. $$

For any $x \in E$, there exists $l \in \{1, \dots, k\}$ such that $x = x_l$, we denote :

$$ \mathfrak{p}(x, E) := x $$

called the **position of x in E**.

### Example 2.1.4

Let's take the case where $n := 4$. We have :

$$ \mathcal{P}_4 = (\emptyset, \{1\}, \{2\}, \{3\}, \{4\}, \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, \\ \{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}, \{1,2,3,4\}) $$

and

$$ \mathcal{P}_{4,0} = (\emptyset) $$

$$ \mathcal{P}_{4,1} = (\{1\}, \{2\}, \{3\}, \{4\}) $$

$$ \mathcal{P}_{4,2} = (\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}) $$

$$ \mathcal{P}_{4,3} = (\{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}) $$

$$ \mathcal{P}_{4,4} = (\{1,2,3,4\}) $$

In the proofs, we will use another notation that allows us to identify the elements in an ordered set.

### Notation 2.1.5

Let the ordered set $E := (x_1, \dots, x_k)$, i.e. :

$$ x_1 < \dots < x_k. $$

For any $x \in E$, there exists $l \in \{1, \dots, k\}$ such that $x = x_l$, we denote :

$$ \mathfrak{p}(x, E) := x $$

called the **position of x in E**.
For any $x \in E$, there exists $l \in \{1, \dots, k\}$ such that $x = x_l$, we denote :

$$ \mathfrak{p}(x, E) := x $$

called the **position of x in E**.

### Example 2.1.6

(1) Let $I := (i_1, \dots, i_k)$ (ordered) be a subset of $\{1, \dots, n\}$ of cardinality $k \in \{1, \dots, n\}$ and $j \in I$, i.e., there exists $l \in \{1, \dots, k\}$ such that $j = i_l$. Thus, we have :

$$ \mathfrak{p}(j, I) := l $$

Let $I' := I - \{j\}$, thus $j \in \overline{I}$. We have the formula :

$$ \mathfrak{p}(j, I) + \mathfrak{p}(j, \overline{I'}) = j + 1. \qquad (2.1.1) $$

Let's see on an example why there is a +1 (it comes from the fact that $j$ is counted twice). For $I := \{2, 4, 5, 6, 10\}$ a subset of $\{1, \dots, 12\}$ and $j := 6$. The position of 6 in $I$ is $\mathfrak{p}(6, I) = 4$. Since $I' := I - \{6\} = \{2, 4, 5, 10\}$, we have $\overline{I'} = \{1, 3, 6, 7, 8, 9, 11, 12\}$ and $\mathfrak{p}(6, \overline{I'}) = 3$. Thus, we have :

$$ \mathfrak{p}(j, I) + \mathfrak{p}(j, \overline{I'}) = 4 + 3 = 6 + 1 = j + 1. $$

(2) Let $I \in \mathcal{P}_{n,k}$. We have $\overline{I} \in \mathcal{P}_{n,n-k}$ and :

$$ \mathfrak{p}(I, \mathcal{P}_{n,k}) + \mathfrak{p}(\overline{I}, \mathcal{P}_{n,n-k}) = \binom{n}{k} + 1. \qquad (2.1.2) $$

Let's see on two examples why there is a +1. We take the example 2.1.4, we have :

$$ \mathcal{P}_{4,1} = (\{1\}, \{2\}, \{3\}, \{4\}) $$

$$ \mathcal{P}_{4,2} = (\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}) $$

$$ \mathcal{P}_{4,3} = (\{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}) $$

(a) For $I := \{2\} \in \mathcal{P}_{4,1}$, we have $\overline{I} = \{1,3,4\} \in \mathcal{P}_{4,3}$ and :

$$ \mathfrak{p}(I, \mathcal{P}_{4,1}) + \mathfrak{p}(\overline{I}, \mathcal{P}_{4,3}) = 2 + 3 = 4 + 1 = \binom{4}{1} + 1. $$

(b) For $I := \{1,4\} \in \mathcal{P}_{4,2}$, we have $\overline{I} = \{2,3\} \in \mathcal{P}_{4,2}$ and :

$$ \mathfrak{p}(I, \mathcal{P}_{4,2}) + \mathfrak{p}(\overline{I}, \mathcal{P}_{4,2}) = 3 + 4 = 6 + 1 = \binom{4}{2} + 1. $$

(3) Let $I \in \mathcal{P}_n$. We have $\overline{I} \in \mathcal{P}_n$ and :

$$ \mathfrak{p}(I, \mathcal{P}_n) + \mathfrak{p}(\overline{I}, \mathcal{P}_n) = 2^n + 1. \qquad (2.1.3) $$

Look at two examples to understand why there is a +1. Let's take example 2.1.4, we have :

$$ \mathcal{P}_4 = (\emptyset, \{1\}, \{2\}, \{3\}, \{4\}, \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, \\ \{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}, \{1,2,3,4\}) $$
Example 2.1.6

(1) Let $I := (i_1, \dots, i_k)$ (ordered) be a subset of $\{1, \dots, n\}$ of cardinality $k \in \{1, \dots, n\}$ and $j \in I$, i.e., there exists $l \in \{1, \dots, k\}$ such that $j = i_l$. Thus, we have :

$$
\mathfrak{p}(j, I) := l
$$

Let $I' := I - \{j\}$, thus $j \in \overline{I}'$. We have the formula :

$$
\mathfrak{p}(j, I) + \mathfrak{p}(j, \overline{I}') = j + 1. \tag{2.1.1}
$$

Let's see on an example why there is a +1 (it comes from the fact that $j$ is counted twice). For $I := \{2, 4, 5, 6, 10\}$ a subset of $\{1, \dots, 12\}$ and $j := 6$. The position of 6 in $I$ is $\mathfrak{p}(6, I) = 4$. Since $I' := I - \{6\} = \{2, 4, 5, 10\}$, we have $\overline{I}' = \{1, 3, 6, 7, 8, 9, 11, 12\}$ and $\mathfrak{p}(6, \overline{I}') = 3$. Thus, we have :

$$
\mathfrak{p}(j, I) + \mathfrak{p}(j, \overline{I}') = 4 + 3 = 6 + 1 = j + 1.
$$

(2) Let $I \in \mathcal{P}_{n,k}$. We have $\overline{I} \in \mathcal{P}_{n,n-k}$ and :

$$
\mathfrak{p}(I, \mathcal{P}_{n,k}) + \mathfrak{p}(\overline{I}, \mathcal{P}_{n,n-k}) = \binom{n}{k} + 1. \quad (2.1.2)
$$

Let's see on two examples why there is a +1. We take the example 2.1.4, we have :

$$
\mathcal{P}_{4,1} = (\{1\}, \{2\}, \{3\}, \{4\})
$$

$$
\mathcal{P}_{4,2} = (\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\})
$$

$$
\mathcal{P}_{4,3} = (\{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\})
$$

(a) For $I := \{2\} \in \mathcal{P}_{4,1}$, we have $\overline{I} = \{1,3,4\} \in \mathcal{P}_{4,3}$ and :

$$
\mathfrak{p}(I, \mathcal{P}_{4,1}) + \mathfrak{p}(\overline{I}, \mathcal{P}_{4,3}) = 2 + 3 = 4 + 1 = \binom{4}{1} + 1.
$$

(b) For $I := \{1,4\} \in \mathcal{P}_{4,2}$, we have $\overline{I} = \{2,3\} \in \mathcal{P}_{4,2}$ and :

$$
\mathfrak{p}(I, \mathcal{P}_{4,2}) + \mathfrak{p}(\overline{I}, \mathcal{P}_{4,2}) = 3 + 4 = 6 + 1 = \binom{4}{2} + 1.
$$

(3) Let $I \in \mathcal{P}_n$. We have $\overline{I} \in \mathcal{P}_n$ and :

$$
\mathfrak{p}(I, \mathcal{P}_n) + \mathfrak{p}(\bar{I}, \mathcal{P}_n) = 2^n + 1. \tag{2.1.3}
$$

Look at two examples to understand why there is a +1. Let's take example 2.1.4, we have :

$$
\mathcal{P}_4 = (\emptyset, \{1\}, \{2\}, \{3\}, \{4\}, \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, \\
\{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}, \{1,2,3,4\})
$$

(a) For $I := \{2\} \in \mathcal{P}_4$, we have $\overline{I} = \{1,3,4\} \in \mathcal{P}_4$ and :

$$
\mathfrak{p}(I, \mathcal{P}_4) + \mathfrak{p}(\overline{I}, \mathcal{P}_4) = 3 + 14 = 16 + 1 = 2^4 + 1.
$$

(b) For $I := \{1,4\} \in \mathcal{P}_4$, we have $\overline{I} = \{2,3\} \in \mathcal{P}_4$ and :

$$
\mathfrak{p}(I, \mathcal{P}_4) + \mathfrak{p}(\overline{I}, \mathcal{P}_4) = 8 + 9 = 16 + 1 = 2^4 + 1.
$$
(a) For $I := \{2\} \in \mathcal{P}_4$, we have $\bar{I} = \{1, 3, 4\} \in \mathcal{P}_4$ and :

$$ \mathfrak{p}(I, \mathcal{P}_4) + \mathfrak{p}(\bar{I}, \mathcal{P}_4) = 3 + 14 = 16 + 1 = 2^4 + 1. $$

(b) For $I := \{1, 4\} \in \mathcal{P}_4$, we have $\bar{I} = \{2, 3\} \in \mathcal{P}_4$ and :

$$ \mathfrak{p}(I, \mathcal{P}_4) + \mathfrak{p}(\bar{I}, \mathcal{P}_4) = 8 + 9 = 16 + 1 = 2^4 + 1. $$

## 2.2 Minor of a Matrix

In this subsection, we study the submatrices associated with a matrix (see Appendix ?? for the notations used on matrices). We define the **empty matrix** (without row and without column) :

$$ \mathfrak{M}(\emptyset) \in \mathcal{M}_0(\mathbb{C}). $$

and we set :

$$ \det(\mathfrak{M}(\emptyset)) := 1. $$

### Definition 2.2.1

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, J \in \mathcal{P}_{n,k}$.

(i) The **submatrix** $A_{I,J}$ is the matrix of $\mathcal{M}_{n-k}(\mathbb{C})$ where we removed the rows numbered by $I$ and the columns numbered by $J$.

(ii) The **minor** $(I, J)$ **of** $A$ is the determinant of the matrix $A_{I,J}$. We denote it :

$$ \Delta_{I,J}(A) := \det(A_{I,J}). $$

(iii) The **cofactor** $(I, J)$ **of** $A$ is defined by :

$$ \mathrm{Cof}_{I,J}(A) := (-1)^{|I|+|J|} \Delta_{I,J}(A). $$

### Example 2.2.2

Let $A := (a_{ij})_{ij} \in \mathcal{M}_n(\mathbb{C})$.

(1) Case $I := J := \emptyset$. We have :

$$ \begin{align*} A_{\emptyset, \emptyset} &= A \\ \Delta_{\emptyset, \emptyset}(A) &= \det(A_{\emptyset, \emptyset}) \\ &= \det(A) \\ \mathrm{Cof}_{\emptyset, \emptyset}(A) &= (-1)^{|\emptyset|+|\emptyset|} \Delta_{\emptyset, \emptyset}(A) \\ &= \det(A) \end{align*} $$

(2) Case $I := \{1, \dots, n\} - \{i\} = \{\overline{i}\}$ and $J := \{1, \dots, n\} - \{j\} = \{\overline{j}\}$. We have :

$$ \begin{align*} A_{\overline{\{i\}}, \overline{\{i\}}} &= a_{i,j} \\ \Delta_{\overline{\{i\}}, \overline{\{i\}}} (A) &= \det(a_{i,j}) \\ &= a_{i,j} \\ \mathrm{Cof}_{\overline{\{i\}}, \overline{\{i\}}} (A) &= (-1)^{|\overline{\{i\}}|+|\overline{\{i\}}|} \Delta_{\overline{\{i\}}, \overline{\{i\}}} (A) \end{align*} $$
## 2.2 Minor of a Matrix

In this subsection, we study the submatrices associated with a matrix (see Appendix ?? for the notations used on matrices). We define the **empty matrix** (without row and without column) :

$$ \mathfrak{M}(\emptyset) \in \mathcal{M}_0(\mathbb{C}). $$

and we set :

$$ \det(\mathfrak{M}(\emptyset)) := 1. $$

### Definition 2.2.1

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, J \in \mathcal{P}_{n,k}$.

(i) The **submatrix** $A_{I,J}$ is the matrix of $\mathcal{M}_{n-k}(\mathbb{C})$ where we removed the rows numbered by $I$ and the columns numbered by $J$.

(ii) The **minor** $(I, J)$ **of** $A$ is the determinant of the matrix $A_{I,J}$. We denote it :

$$ \Delta_{I,J}(A) := \det(A_{I,J}). $$

(iii) The **cofactor** $(I, J)$ **of** $A$ is defined by :

$$ \mathrm{Cof}_{I,J}(A) := (-1)^{|I|+|J|} \Delta_{I,J}(A). $$

### Example 2.2.2

Let $A := (a_{ij})_{ij} \in \mathcal{M}_n(\mathbb{C})$.

(1) Case $I := J := \emptyset$. We have :

$$ \begin{aligned} A_{\emptyset, \emptyset} &= A \\ \Delta_{\emptyset, \emptyset}(A) &= \det(A_{\emptyset, \emptyset}) \\ &= \det(A) \\ \mathrm{Cof}_{\emptyset, \emptyset}(A) &= (-1)^{|\emptyset|+|\emptyset|} \Delta_{\emptyset, \emptyset}(A) \\ &= \det(A) \end{aligned} $$

(2) Case $I := \{1, \dots, n\} - \{i\} = \overline{\{i\}}$ and $J := \{1, \dots, n\} - \{j\} = \overline{\{j\}}$. We have :

$$ \begin{aligned} A_{\overline{\{i\}}, \overline{\{i\}}} &= a_{i,j} \\ \Delta_{\overline{\{i\}}, \overline{\{i\}}} (A) &= \det (a_{i,j}) \\ &= a_{i,j} \\ \mathrm{Cof}_{\overline{\{i\}}, \overline{\{i\}}} (A) &= (-1)^{|\overline{\{i\}}|+|\overline{\{i\}}|} \Delta_{\overline{\{i\}}, \overline{\{i\}}} (A) \\ &= (-1)^{(n(n+1)/2 - i + (n(n+1)/2 - j))} a_{i,j} \\ &= (-1)^{i+j} a_{i,j} \end{aligned} $$

(3) Case $I := J := \{1, \dots, n\}$. We have :

$$ \begin{aligned} A_{\{1, \dots, n\}, \{1, \dots, n\}} &= \mathfrak{M}(\emptyset) \\ \Delta_{\{1, \dots, n\}, \{1, \dots, n\}} (A) &= \det(A_{\{1, \dots, n\}, \{1, \dots, n\}}) \\ &= \det(\mathfrak{M}(\emptyset)) \end{aligned} $$
$$ 
\begin{aligned}
&= (-1)^{(n(n+1)/2 - i + (n(n+1)/2 - j)a_{i,j}} \\
&= (-1)^{i+j} a_{i,j}
\end{aligned}
$$

(3) Case $I := J := \{1, \dots, n\}$. We have :

$$ 
\begin{aligned}
A_{\{1, \dots, n\}, \{1, \dots, n\}} &= \mathfrak{M}(\emptyset) \\
\Delta_{\{1, \dots, n\}, \{1, \dots, n\}}(A) &= \det(A_{\{1, \dots, n\}, \{1, \dots, n\}}) \\
&= \det(\mathfrak{M}(\emptyset)) \\
&= 1 \\
\mathrm{Cof}_{\{1, \dots, n\}, \{1, \dots, n\}}(A) &= (-1)^{|\{1, \dots, n\}|+|\{1, \dots, n\}|} \Delta_{\emptyset, \emptyset}(A) \\
&= (-1)^{(n(n+1)/2+(n(n+1)/2)\det(A))} \\
&= \det(A)
\end{aligned}
$$

We start by proving the usual lemma for expansion along a row. For any elements $a$ and $b$, we denote $\delta_{a,b}$ the **Kronecker symbol** defined by :

$$ \delta_{a,b} := \begin{cases} 1 & \text{if } a = b \\ 0 & \text{if } a \neq b \end{cases} $$

### Lemma 2.2.3

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $i, l \in \{1, \dots, n\}$. Then we have :

$$ 
\begin{aligned}
\det(A)\delta_{i,l} &= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \Delta_{\{i\},\{j\}}(A) \\
&= \sum_{j=1}^{n} [A]_{l,j} \mathrm{Cof}_{\{i\},\{j\}}(A).
\end{aligned}
$$

**Proof.**

Let $A := (a_{ij})_{ij}$. We prove the result in two parts.

(1) Case $i = l$. We show that :

$$ \det(A) = \sum_{j=1}^{n} (-1)^{i+j} a_{i,j} \Delta_{\{i\},\{j\}}(A). $$

(2) Case $i \neq l$. We show that :

$$ \sum_{j=1}^{n} (-1)^{i+j} a_{i,j} \Delta_{\{i\},\{j\}}(A) = 0. $$

(1) **Case $i = l$.** By the linearity of the determinant with respect to the $i$-th row of $A$ (see subsection A.3 for notations on permutation matrices), we have :

$$ 
\begin{aligned}
\det(A) &= \det(\mathcal{L}_1(A)^T, \dots, \mathcal{L}_n(A)^T) \\
&= \det\left(\mathcal{L}_1(A)^T, \dots, \mathcal{L}_{i-1}(A)^T, \sum_{j=1}^n a_{i,j} e_k, \mathcal{L}_{i-1}(A)^T, \dots, \mathcal{L}_n(A)^T\right) \\
&= \sum_{j=1}^n a_{i,j} \det\left(\mathcal{L}_1(A)^T, \dots, \mathcal{L}_{i-1}(A)^T, e_j, \mathcal{L}_{i-1}(A)^T, \dots, \mathcal{L}_n(A)^T\right)
\end{aligned}
$$
$$
\begin{align*}
&= 1 \\
\operatorname{Cof}_{\{1,...,n\},\{1,...,n\}}(A) &= (-1)^{|\{1,...,n\}|+|\{1,...,n\}|} \Delta_{\emptyset,\emptyset}(A) \\
&= (-1)^{\frac{n(n+1)}{2}+\frac{n(n+1)}{2}} \det(A) \\
&= \det(A)
\end{align*}
$$

We start by proving the usual lemma for expansion along a row. For any elements $a$ and $b$, we denote $\delta_{a,b}$ the
**Kronecker symbol** defined by :

$$
\delta_{a,b} := \begin{cases} 1 & \text{if } a = b \\ 0 & \text{if } a \neq b \end{cases}
$$

**Lemma 2.2.3**

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $i, l \in \{1, \dots, n\}$. Then we have :

$$
\begin{align*}
\det(A)\delta_{i,l} &= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \Delta_{\{i\},\{j\}}(A) \\
&= \sum_{j=1}^{n} [A]_{l,j} \mathrm{Cof}_{\{i\},\{j\}}(A).
\end{align*}
$$

**Proof.**

Let $A := (a_{ij})_{ij}$. We prove the result in two parts.

(1) Case $i = l$. We show that :

$$
\det(A) = \sum_{j=1}^{n} (-1)^{i+j} a_{i,j} \Delta_{\{i\},\{j\}}(A).
$$

(2) Case $i \neq j$. We show that :

$$
\sum_{k=1}^{n} (-1)^{i+k} a_{l,j} \Delta_{\{i\},\{j\}}(A) = 0.
$$

(1) **Case** $i = j$. By linearity of the determinant with respect to the $i$-th row (see subsection A.3 for the notations on permutation matrices), we have :

$$
\begin{align*}
\det(A) &= \det(\mathcal{L}_1(A)^T, \dots, \mathcal{L}_n(A)^T) \\
&= \det\left(\mathcal{L}_1(A)^T, \dots, \mathcal{L}_{i-1}(A)^T, \sum_{j=1}^n a_{i,j} e_k, \mathcal{L}_{i-1}(A)^T, \dots, \mathcal{L}_n(A)^T\right) \\
&= \sum_{j=1}^n a_{i,j} \det\left(\mathcal{L}_1(A)^T, \dots, \mathcal{L}_{i-1}(A)^T, e_j, \mathcal{L}_{i-1}(A)^T, \dots, \mathcal{L}_n(A)^T\right) \\
&= \sum_{j=1}^n a_{i,j} \det\left(\begin{array}{c}
\mathcal{L}_1(A) \\
\vdots \\
\mathcal{L}_{i-1}(A) \\
e_j^T \\
\mathcal{L}_{i+1}(A) \\
\vdots \\
\mathcal{L}_n(A)
\end{array}\right)
\end{align*}
$$
$$= \sum_{j=1}^{n} a_{i,j} \det \begin{pmatrix} \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ e_j^T \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix}$$

$$= \sum_{j=1}^{n} a_{i,j} \det \left( \mathbf{P}_{(1,2,...,i)} \mathbf{P}_{(1,2,...,i)}^{-1} \right) \det \begin{pmatrix} \left( \begin{array}{c} \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ e_j^T \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{array} \right) \end{pmatrix}$$

$$= \sum_{j=1}^{n} a_{i,j} \det \left( \mathbf{P}_{(1,2,...,i)} \right) \det \begin{pmatrix} \mathbf{P}_{(1,2,...,i)} \begin{pmatrix} \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ e_j^T \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \end{pmatrix}$$

because $\mathbf{P}_{(1,2,...,i)}^{-1} = \mathbf{P}_{(1,2,...,i)}^{T}$

$$= \sum_{j=1}^{n} a_{i,j} \varepsilon((1, 2, \dots, i)) \det \begin{pmatrix} \mathbf{P}_{(1,2,...,i)} \begin{pmatrix} \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ e_j^T \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \end{pmatrix}$$

by A.3.1

$$= \sum_{j=1}^{n} a_{i,j} (-1)^{i-1} \det \begin{pmatrix} e_j^T \\ \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix}$$

by proposition A.4.1

$$= \sum_{j=1}^{n} a_{i,j} (-1)^{i-1} \det \begin{pmatrix} e_j^T \\ \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \det \left( \mathbf{P}_{(1,2,...,j)}^{-1} \mathbf{P}_{(1,2,...,j)} \right)$$
$$
\begin{align*}
&= \sum_{j=1}^{n} a_{i,j} \det(\mathbf{P}_{(1,2,...,i)}) \det \left( \mathbf{P}_{(1,2,...,i)} \begin{pmatrix} \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ e_j^T \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \right) \\
&= \sum_{j=1}^{n} a_{i,j} \varepsilon((1,2,...,i)) \det \left( \mathbf{P}_{(1,2,...,i)} \begin{pmatrix} \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ e_j^T \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \right) \tag*{by A.3.1}
\end{align*}
$$

$$
= \sum_{j=1}^{n} a_{i,j} (-1)^{i-1} \det \begin{pmatrix} e_j^T \\ \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \quad \text{by proposition A.4.1}
$$

$$
= \sum_{j=1}^{n} a_{i,j} (-1)^{i-1} \det \left( \begin{pmatrix} e_j^T \\ \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \mathbf{P}_{(1,2,...,j)} \right) \det(\mathbf{P}_{(1,2,...,j)})
$$
$$
= \sum_{j=1}^{n} a_{i,j} (-1)^{i-1} \varepsilon((1,2,...,j)) \det \left( \begin{pmatrix} e_j^T \\ \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \mathbf{P}_{(1,2,...,j)} \right) \quad \text{by A.3.1}
$$

$$
\begin{align*}
&= \sum_{j=1}^{n} a_{i,j} (-1)^{i-1} (-1)^{j-1} \det \begin{pmatrix} 1 & 0 \\ * & A_{\{i\},\{j\}} \end{pmatrix} && \text{by proposition A.4.1} \\
&= \sum_{j=1}^{n} (-1)^{i+j} a_{i,j} \det \begin{pmatrix} 1 & 0 \\ * & A_{\{i\},\{j\}} \end{pmatrix} &&
\end{align*}
$$

Let's show that for all $j \in \{1, \dots, n\}$ :

$$
\det \begin{pmatrix} 1 & 0 \\ * & A_{\{i\},\{j\}} \end{pmatrix} = \Delta_{\{i\},\{j\}}(A).
$$

Let's define :

$$
B := (b_{k,l})_{k,l} := \begin{pmatrix} 1 & 0 \\ * & A_{\{i\},\{j\}} \end{pmatrix}
$$
$$ = \sum_{j=1}^{n} a_{i,j} (-1)^{i-1} \det \left( \begin{pmatrix} e_j^T \\ \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \mathbf{P}_{(1,2,...,j)} \right) \det (\mathbf{P}_{(1,2,...,j)}) \quad \text{because } \mathbf{P}_{(1,2,...,j)}^{-1} = \mathbf{P}_{(1,2,...,j)}^T $$

$$ = \sum_{j=1}^{n} a_{i,j} (-1)^{i-1} \varepsilon((1,2,...,j)) \det \left( \begin{pmatrix} e_j^T \\ \mathcal{L}_1(A) \\ \vdots \\ \mathcal{L}_{i-1}(A) \\ \mathcal{L}_{i+1}(A) \\ \vdots \\ \mathcal{L}_n(A) \end{pmatrix} \mathbf{P}_{(1,2,...,j)} \right) \quad \text{by A.3.1} $$

$$ = \sum_{j=1}^{n} a_{i,j} (-1)^{i-1} (-1)^{j-1} \det \begin{pmatrix} 1 & 0 \\ * & A_{\{i\},\{j\}} \end{pmatrix} \quad \text{by proposition A.4.1} $$

$$ = \sum_{j=1}^{n} (-1)^{i+j} a_{i,j} \det \begin{pmatrix} 1 & 0 \\ * & A_{\{i\},\{j\}} \end{pmatrix} $$

Let's show that for all $j \in \{1, \dots, n\}$:

$$ \det \begin{pmatrix} 1 & 0 \\ * & A_{\{i\},\{j\}} \end{pmatrix} = \Delta_{\{i\},\{j\}}(A). $$

Let's define:

$$ B := (b_{k,l})_{k,l} := \begin{pmatrix} 1 & 0 \\ * & A_{\{i\},\{j\}} \end{pmatrix} $$

Let $j \in \{1, \dots, n\}$. We have:

$$ \begin{aligned} \det(B) &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} b_{i,\sigma(i)} \\ &= \sum_{\substack{\sigma \in \mathfrak{S}_n \\ \sigma(1)=1}} \varepsilon(\sigma) b_{1,1} \prod_{i=2}^{n} b_{i,\sigma(i)} \\ &= \sum_{\substack{\sigma \in \mathfrak{S}_n \\ \sigma(1)=1}} \varepsilon(\sigma) \prod_{i=2}^{n} [A_{\{i\},\{j\}}]_{i-1,\sigma(i)-1} \\ &= \sum_{\substack{\sigma' \in \mathfrak{S}_{n-1} \\ \sigma'(1)=1}} \varepsilon(\sigma') \prod_{i=1}^{n-1} [A_{\{i\},\{j\}}]_{i,\sigma'(i)} \\ &= \Delta_{\{i\},\{j\}}(A) \end{aligned} $$

So we have:

$$ \begin{aligned} \det(A) &= \sum_{j=1}^{n} (-1)^{i+j} a_{i,j} \det \begin{pmatrix} 1 & 0 \\ * & A_{\{i\},\{j\}} \end{pmatrix} \\ &= \sum_{j=1}^{n} (-1)^{i+j} a_{i,j} \Delta_{\{i\},\{j\}}(A) \end{aligned} $$
Let $j \in \{1, \dots, n\}$. We have :

$$
\begin{align*}
\det(B) &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} b_{i, \sigma(i)} \\
&= \sum_{\substack{\sigma \in \mathfrak{S}_n \\ \sigma(1)=1}} \varepsilon(\sigma) b_{1,1} \prod_{i=2}^{n} b_{i, \sigma(i)} \\
&= \sum_{\substack{\sigma \in \mathfrak{S}_n \\ \sigma(1)=1}} \varepsilon(\sigma) \prod_{i=2}^{n} [A_{\{i\}, \{j\}}]_{i-1, \sigma(i)-1} \\
&= \sum_{\substack{\sigma' \in \mathfrak{S}_{n-1} \\ \sigma'(1)=1}} \varepsilon(\sigma') \prod_{i=1}^{n-1} [A_{\{i\}, \{j\}}]_{i, \sigma'(i)} \\
&= \Delta_{\{i\}, \{j\}}(A)
\end{align*}
$$

So we have :

$$
\begin{align*}
\det(A) &= \sum_{j=1}^{n} (-1)^{i+j} a_{i,j} \det \begin{pmatrix} 1 & 0 \\ * & A_{\{i\}, \{j\}} \end{pmatrix} \\
&= \sum_{j=1}^{n} (-1)^{i+j} a_{i,j} \Delta_{\{i\}, \{j\}}(A)
\end{align*}
$$

(2) Case $i \neq l$. Let $B$ be the matrix defined by the rows :

$$
\forall j \in \{1, \dots, n\}, \mathcal{L}_j(B) := \begin{cases} \mathcal{L}_j(A) & \text{if } j \neq i \\ \mathcal{L}_l(A) & \text{if } j = i \end{cases}
$$

So the rows $i$ and $l$ of $B$ are equal (hence $\det(B) = 0$). Using (1) with the matrix $B$ and as $B_{\{i\},\{j\}} = A_{\{i\},\{j\}}$,
we have :

$$
\begin{align*}
0 &= \det(B) \\
&= \sum_{j=1}^{n} (-1)^{i+j} [B]_{i,j} \Delta_{\{i\},\{j\}}(B) \\
&= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \det(B_{\{i\},\{j\}}) \\
&= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \det(A_{\{i\},\{j\}}) \\
&= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \Delta_{\{i\},\{j\}}(A)
\end{align*}
$$

<figure><img src="image_2.png" /></figure>

We can also write the formula as :

$$
\begin{align*}
\det(A)\delta_{i,l} &= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \Delta_{\{i\},\{j\}}(A) \\
&= \sum_{J \in \mathcal{P}_{n,1}} (-1)^{i+|J|} \Delta_{\overline{\{l\}},\overline{J}}(A) \Delta_{\{i\},J}(A) \\
&= \sum_{J \in \mathcal{P}_{n,1}} \Delta_{\overline{\{l\}},\overline{J}}(A) \mathrm{Cof}_{\{i\},J}(A).
\end{align*}
$$

We generalize this last equality.
(2) Case $i \neq l$. Let $B$ be the matrix defined by the rows :

$$ \forall j \in \{1, \dots, n\}, \mathcal{L}_j(B) := \begin{cases} \mathcal{L}_j(A) & \text{if } j \neq i \\ \mathcal{L}_l(A) & \text{if } j = i \end{cases} $$

So the rows $i$ and $l$ of $B$ are equal (hence $\det(B) = 0$). Using (1) with the matrix $B$ and as $B_{\{i\},\{j\}} = A_{\{i\},\{j\}}$, we have :

$$ \begin{aligned} 0 &= \det(B) \\ &= \sum_{j=1}^{n} (-1)^{i+j} [B]_{i,j} \Delta_{\{i\},\{j\}}(B) \\ &= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \det(B_{\{i\},\{j\}}) \\ &= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \det(A_{\{i\},\{j\}}) \\ &= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \Delta_{\{i\},\{j\}}(A) \end{aligned} $$

□

We can also write the formula as :

$$ \begin{aligned} \det(A)\delta_{i,l} &= \sum_{j=1}^{n} (-1)^{i+j} [A]_{l,j} \Delta_{\{i\},\{j\}}(A) \\ &= \sum_{J \in \mathcal{P}_{n,1}} (-1)^{i+|J|} \Delta_{\overline{\{l\}},\overline{J}}(A) \Delta_{\{i\},J}(A) \\ &= \sum_{J \in \mathcal{P}_{n,1}} \Delta_{\overline{\{l\}},\overline{J}}(A) \mathrm{Cof}_{\{i\},J}(A). \end{aligned} $$

We generalize this last equality.

### Proposition 2.2.4

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, L \in \mathcal{P}_{n,k}$. Then we have :

$$ \begin{aligned} \det(A)\delta_{I,L} &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\overline{L},\overline{J}}(A)\Delta_{I,J}(A) \\ &= \sum_{J \in \mathcal{P}_{n,k}} \Delta_{\overline{L},\overline{J}}(A)\mathrm{Cof}_{I,J}(A) \end{aligned} $$

**Proof.** Let's prove the result in two parts.

(1) Case $I = L$. We show that :

$$ \det(A) = \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\overline{I},\overline{J}}(A) \Delta_{I,J}(A). $$

(2) Case $I \neq L$. We show that :

$$ \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\overline{L},\overline{J}}(A) \Delta_{I,J}(A) = 0. $$
### Proposition 2.2.4

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, L \in \mathcal{P}_{n,k}$. Then we have :

$$ \det(A)\delta_{I,L} = \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{L},\bar{J}}(A)\Delta_{I,J}(A) \\ = \sum_{J \in \mathcal{P}_{n,k}} \Delta_{\bar{L},\bar{J}}(A)\mathrm{Cof}_{I,J}(A) $$

**Proof.** Let's prove the result in two parts.

(1) Case $I = L$. We show that :

$$ \det(A) = \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{I},\bar{J}}(A)\Delta_{I,J}(A). $$

(2) Case $I \neq L$. We show that :

$$ \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{L},\bar{J}}(A)\Delta_{I,J}(A) = 0. $$

(1) Case $I = L$. We have :

Let's prove by strong double induction on $(n, k)$ such that $n \in \mathbb{N}_*$ and $k \in \{0, \dots, n\}$ the property :

$$ \mathcal{Q}(n,k): \forall A \in \mathcal{M}_n(\mathbb{C}), \forall I \in \mathcal{P}_{n,k}, \det(A) = \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{I},\bar{J}}(A)\Delta_{I,J}(A). $$

Here is the proof outline.

(i) **Base case for $n := 1$.** We show that $\mathcal{Q}(1,0)$ and $\mathcal{Q}(1,1)$ hold.

(ii) **Induction step for $n \in \mathbb{N}_*$.** Let $n \in \mathbb{N}_*$. We assume that $\mathcal{Q}(n', k')$ holds for all $(n', k') \in \mathbb{N}_* \times \mathbb{N}$ such that $0 \le k' \le n' \le n$ (strong double induction).

(I) **Base case for $k := 0$ and $k := 1$.** We show that $\mathcal{Q}(n+1,0)$ holds and by Lemma 2.2.3, we have $\mathcal{Q}(n+1,1)$.

(II) **Induction step for $k \in \{1, \dots, n\}$.** Let $k \in \{1, \dots, n\}$. We assume that $\mathcal{Q}(n+1,l)$ holds for all $l \in \{0, \dots, k\}$. We show that $\mathcal{Q}(n+1,k+1)$ holds.

(III) **Conclusion for $k \in \{0, \dots, n+1\}$.** By the principle of strong induction, $\mathcal{Q}(n+1,k)$ holds for all $k \in \{0, \dots, n+1\}$.

(iii) **Conclusion for $n \in \mathbb{N}_*$.** By the principle of strong double induction, $\mathcal{Q}(n,k)$ holds for all $(n,k)$ such that $n \in \mathbb{N}_*$ and $k \in \{0, \dots, n\}$.

(i) Let $A \in \mathcal{M}_1(\mathbb{C})$. Since $\mathcal{P}_{1,0} = \{\emptyset\}$ and $\mathcal{P}_{1,1} = \{\{1\}\}$, we have :

$$ \begin{aligned} (-1)^{|\emptyset|+|\emptyset|} \Delta_{\bar{\emptyset},\bar{\emptyset}}(A)\Delta_{\emptyset,\emptyset}(A) &= (-1)^0 \Delta_{\{1\},\{1\}}(A)\Delta_{\emptyset,\emptyset}(A) \\ &= \det(A) \end{aligned} $$

$$ \begin{aligned} (-1)^{|\{1\}|+|\{1\}|} \Delta_{\overline{\{1\}},\overline{\{1\}}} (A) \Delta_{\{1\},\{1\}} (A) &= (-1)^2 \Delta_{\emptyset,\emptyset}(A) \Delta_{\{1\},\{1\}}(A) \\ &= \det(A) \end{aligned} $$

So $\mathcal{Q}(1,0)$ and $\mathcal{Q}(1,1)$ hold.

(ii) Let $n \in \mathbb{N}_*$ and $A \in \mathcal{M}_{n+1}(\mathbb{C})$. Suppose $\mathcal{Q}(n', k')$ holds for all $(n', k') \in \mathbb{N}_* \times \mathbb{N}$ such that $0 \le k' \le n' \le n$.
(1) **Case I = L.** We have:

Let's prove by strong double induction on $(n, k)$ such that $n \in \mathbb{N}_*$ and $k \in \{0, \dots, n\}$ the property:

$$ \mathcal{Q}(n,k): \quad \forall A \in \mathcal{M}_n(\mathbb{C}), \quad \forall I \in \mathcal{P}_{n,k}, \quad \det(A) = \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{I},\bar{J}}(A) \Delta_{I,J}(A). $$

Here is the proof outline.

(i) **Base case for $n := 1$.** We show that $\mathcal{Q}(1,0)$ and $\mathcal{Q}(1,1)$ hold.

(ii) **Induction step for $n \in \mathbb{N}_*$.** Let $n \in \mathbb{N}_*$. We assume that $\mathcal{Q}(n', k')$ holds for all $(n', k') \in \mathbb{N}_* \times \mathbb{N}$ such that $0 \le k' \le n' \le n$ (strong double induction).

(I) **Base case for $k := 0$ and $k := 1$.** We show that $\mathcal{Q}(n+1,0)$ holds and by Lemma 2.2.3, we have $\mathcal{Q}(n+1,1)$.

(II) **Induction step for $k \in \{1, \dots, n\}$.** Let $k \in \{1, \dots, n\}$. We assume that $\mathcal{Q}(n+1,l)$ holds for all $l \in \{0, \dots, k\}$. We show that $\mathcal{Q}(n+1,k+1)$ holds.

(III) **Conclusion for $k \in \{0, \dots, n+1\}$.** By the principle of strong induction, $\mathcal{Q}(n+1,k)$ holds for all $k \in \{0, \dots, n+1\}$.

(iii) **Conclusion for $n \in \mathbb{N}_*$.** By the principle of strong double induction, $\mathcal{Q}(n,k)$ holds for all $(n,k)$ such that $n \in \mathbb{N}_*$ and $k \in \{0, \dots, n\}$.

(i) Let $A \in \mathcal{M}_1(\mathbb{C})$. Since $\mathcal{P}_{1,0} = \{\emptyset\}$ and $\mathcal{P}_{1,1} = \{\{1\}\}$, we have:

$$ \begin{aligned} (-1)^{|\emptyset|+|\emptyset|} \Delta_{\bar{\emptyset},\bar{\emptyset}}(A) \Delta_{\emptyset,\emptyset}(A) &= (-1)^0 \Delta_{\{1\},\{1\}}(A) \Delta_{\emptyset,\emptyset}(A) \\ &= \det(A) \end{aligned} $$

$$ \begin{aligned} (-1)^{|\{1\}|+|\{1\}|} \Delta_{\overline{\{1\}},\overline{\{1\}}} (A) \Delta_{\{1\},\{1\}} (A) &= (-1)^2 \Delta_{\emptyset,\emptyset} (A) \Delta_{\{1\},\{1\}} (A) \\ &= \det(A) \end{aligned} $$

So $\mathcal{Q}(1,0)$ and $\mathcal{Q}(1,1)$ hold.

(ii) Let $n \in \mathbb{N}_*$ and $A \in \mathcal{M}_{n+1}(\mathbb{C})$. Suppose $\mathcal{Q}(n', k')$ holds for all $(n', k') \in \mathbb{N}_* \times \mathbb{N}$ such that $0 \le k' \le n' \le n$.

(I) Since $\mathcal{P}_{n+1,0} = \{\emptyset\}$, we have:

$$ \begin{aligned} (-1)^{|\emptyset|+|\emptyset|} \Delta_{\bar{\emptyset},\bar{\emptyset}}(A) \Delta_{\emptyset,\emptyset}(A) &= (-1)^0 \Delta_{\{1,\dots,n+1\},\{1,\dots,n+1\}}(A) \Delta_{\emptyset,\emptyset}(A) \\ &= \det(A) \end{aligned} $$

So $\mathcal{Q}(n+1,0)$ holds.

By Lemma 2.2.3, $\mathcal{Q}(n+1,1)$ holds.

(II) Let $k \in \{1, \dots, n\}$ and $I \in \mathcal{P}_{n+1,k+1}$. Suppose $\mathcal{Q}(n+1,l)$ holds for all $l \in \{0, \dots, k\}$. Since $n+1 \ge k+1 \ge 1$, there exists $i_0 \in I$ and $I' := I - \{i_0\} \in \mathcal{P}_{n+1,k}$. For all $J' \in \mathcal{P}_{n+1,k}$, we have by property $\mathcal{Q}(n+1-k,1)$ applied to $A_{\overline{I'},\overline{J'}}$ with $\{i_0\} \in \overline{I'}$:

$$ \begin{aligned} (**) : & \Delta_{\overline{I'},\overline{J'}}(A) \\ &= \det(A_{\overline{I'},\overline{J'}}) \\ &= \sum_{K \in \mathcal{P}_{n+1-k,1}} (-1)^{|\mathfrak{p}(i_0,\overline{I'})|+|K|} \Delta_{\overline{\{\mathfrak{p}(i_0,\overline{I'})\}},K} (A_{\overline{I'},\overline{J'}}) \Delta_{\{\mathfrak{p}(i_0,\overline{I'})\},K} (A_{\overline{I'},\overline{J'}}) \quad \text{by } \mathcal{Q}(n+1-k,1) \\ &= \sum_{j \in \overline{J'}} (-1)^{\mathfrak{p}(i_0,\overline{I'})+\mathfrak{p}(j,\overline{J'})} a_{i_0,j} \Delta_{\mathfrak{p}(i_0,\overline{I'}),\mathfrak{p}(j,\overline{J'})} (A_{\overline{I'},\overline{J'}}) \\ &= \sum_{j \in \overline{J'}} (-1)^{\mathfrak{p}(i_0,\overline{I'})+\mathfrak{p}(j,\overline{J'})} a_{i_0,j} \Delta_{\overline{I},\overline{J'} \sqcup \{j\}} (A) \end{aligned} $$
(I) Since $\mathcal{P}_{n+1,0} = \{\emptyset\}$, we have :

$$(-1)^{|\emptyset|+|\bar{\emptyset}|} \Delta_{\bar{\emptyset}, \bar{\emptyset}}(A) \Delta_{\emptyset, \emptyset}(A) = (-1)^0 \Delta_{\{1,...,n+1\}, \{1,...,n+1\}}(A) \Delta_{\emptyset, \emptyset}(A) \\ = \det(A)$$

So $\mathcal{Q}(n+1,0)$ holds.

By Lemma 2.2.3, $\mathcal{Q}(n+1,1)$ holds.

(II) Let $k \in \{1, ..., n\}$ and $I \in \mathcal{P}_{n+1,k+1}$. Suppose $\mathcal{Q}(n+1, l)$ holds for all $l \in \{0, ..., k\}$. Since $n+1 \ge k+1 \ge 1$, there exists $i_0 \in I$ and $I' := I - \{i_0\} \in \mathcal{P}_{n+1,k}$. For all $J' \in \mathcal{P}_{n+1,k}$, we have by property $\mathcal{Q}(n+1-k, 1)$ applied to $A_{\bar{I}', \bar{J}'}$ with $\{i_0\} \in \bar{I}'$:

$$(**): \begin{aligned} & \Delta_{\bar{I}', \bar{J}'}(A) \\ &= \det(A_{\bar{I}', \bar{J}'}) \\ &= \sum_{K \in \mathcal{P}_{n+1-k, 1}} (-1)^{|\mathfrak{P}(i_0, \bar{I}')|+|K|} \Delta_{\overline{\mathfrak{P}(i_0, \bar{I}')}, K}(A_{\bar{I}', \bar{J}'}) \Delta_{\overline{\mathfrak{P}(i_0, \bar{I}')}, K}(A_{\bar{I}', \bar{J}'}) \quad \text{by } \mathcal{Q}(n+1-k, 1) \\ &= \sum_{j \in \bar{J}'} (-1)^{|\mathfrak{P}(i_0, \bar{I}') + \mathfrak{P}(j, \bar{J}')|} a_{i_0, j} \Delta_{\overline{\mathfrak{P}(i_0, \bar{I}')}, \mathfrak{P}(j, \bar{J}')}(A_{\bar{I}', \bar{J}'}) \\ &= \sum_{j \in \bar{J}'} (-1)^{|\mathfrak{P}(i_0, \bar{I}') + \mathfrak{P}(j, \bar{J}')|} a_{i_0, j} \Delta_{\overline{\bar{I}, \bar{J}' \cup \{j\}}}(A) \end{aligned}$$

**Remark :** we could have $n+1-k=n+1$ for $k:=0$ if we had not done a two-term initialization with $\mathcal{Q}(n+1,1)$ true. So it was useful to prove Lemma 2.2.3 first.

We have :

$$\det(A) = \sum_{J' \in \mathcal{P}_{n+1,k}} (-1)^{|I'|+|J'|} \Delta_{\bar{I}', \bar{J}'}(A) \Delta_{I', J'}(A) \quad \text{by } \mathcal{Q}(n+1, k)$$

$$= \sum_{J' \in \mathcal{P}_{n+1,k}} (-1)^{|I'|+|J'|} \left( \sum_{j \in \bar{J}'} (-1)^{|\mathfrak{P}(i_0, \bar{I}') + \mathfrak{P}(j, \bar{J}')|} a_{i_0, j} \Delta_{\overline{\bar{I}, J' \cup \{j\}}} (A) \right) \Delta_{I', J'} (A) \quad \text{by } (**)$$

$$= \sum_{J' \in \mathcal{P}_{n+1,k}} \sum_{j \in \bar{J}'} (-1)^{|I'|+|\mathfrak{P}(i_0, \bar{I}')|+|J'|+|\mathfrak{P}(j, \bar{J}')|} a_{i_0, j} \Delta_{\overline{\bar{I}, J' \cup \{j\}}} (A) \Delta_{I', J'} (A)$$

$$= \sum_{J' \in \mathcal{P}_{n+1,k}} \sum_{j \in \bar{J}'} (-1)^{|I'|+|i_0|+|J'|+|j|} \Delta_{\overline{\bar{I}, J' \cup \{j\}}} \times ((-1)^{-|\mathfrak{P}(i_0, I)-\mathfrak{P}(j, J' \cup \{j\})|} a_{i_0, j} (A) \Delta_{I', J'} (A)) \quad \text{by (2.1.1)}$$

$$= \sum_{J' \in \mathcal{P}_{n+1,k}} \sum_{j \in \bar{J}'} (-1)^{|I'|+|J'|+|j|} \Delta_{\overline{\bar{I}, J' \cup \{j\}}} \times ((-1)^{|\mathfrak{P}(i_0, I)+\mathfrak{P}(j, J' \cup \{j\})|} [A_{I,J}]_{\mathfrak{P}(i_0, I)+\mathfrak{P}(j, J' \cup \{j\})} \Delta_{\{\mathfrak{P}(i_0, I)\}, \{\mathfrak{P}(j, J' \cup \{j\})\}} (A_{I,J}))$$

$$= \sum_{J' \in \mathcal{P}_{n+1,k}} \sum_{j \in \bar{J}'} (-1)^{|I'|+|J'|+|j|} \Delta_{\overline{\bar{I}, J' \cup \{j\}}} (A) \det(A_{I, \overline{J' \cup \{j\}}}) \quad \text{by } \mathcal{Q}(n-k, 1)$$

$$= \sum_{J \in \mathcal{P}_{n+1,k+1}} (-1)^{|I'|+|J'|} \Delta_{\bar{I}, \bar{J}}(A) \det(A_{I, \bar{J}})$$

$$= \sum_{J \in \mathcal{P}_{n+1,k+1}} (-1)^{|I'|+|J'|} \Delta_{\bar{I}, \bar{J}}(A) \Delta_{I, \bar{J}}(A)$$

So $\mathcal{Q}(n+1, k+1)$ holds.
**Remark :** we could have $n + 1 - k = n + 1$ for $k := 0$ if we had not done a two-term initialization with $\mathcal{Q}(n+1, 1)$ true. So it was useful to prove Lemma 2.2.3 first.

We have :

$$ \det(A) = \sum_{J' \in \mathcal{P}_{n+1,k}} (-1)^{|I'|+|J'|} \Delta_{\overline{I}, \overline{J'}}(A) \Delta_{I', J'}(A) \quad \text{by } \mathcal{Q}(n+1, k) $$

$$ = \sum_{J' \in \mathcal{P}_{n+1,k}} (-1)^{|I'|+|J'|} \left( \sum_{j \in J'} (-1)^{\mathfrak{P}(i_0, \overline{I}') + \mathfrak{P}(j, \overline{J'})} a_{i_0, j} \Delta_{\overline{I}, \overline{J'} \cup \{j\}}(A) \right) \Delta_{I', J'}(A) \quad \text{by } (\ast) $$

$$ = \sum_{J' \in \mathcal{P}_{n+1,k}} \sum_{j \in J'} (-1)^{|I'|+|\mathfrak{P}(i_0, \overline{I}')|+|J'|+|\mathfrak{P}(j, \overline{J'})|} a_{i_0, j} \Delta_{\overline{I}, \overline{J'} \cup \{j\}}(A) \Delta_{I', J'}(A) $$

$$ = \sum_{J' \in \mathcal{P}_{n+1,k}} \sum_{j \in J'} (-1)^{|I'|+|i_0|+|J'|+|j|} \Delta_{\overline{I}, \overline{J'} \cup \{j\}} \times ((-1)^{-|\mathfrak{P}(i_0, I)|-|\mathfrak{P}(j, J' \cup \{j\})|} a_{i_0, j}(A) \Delta_{I', J'}(A)) \quad \text{by (2.1.1)} $$

$$ = \sum_{J' \in \mathcal{P}_{n+1,k}} \sum_{j \in J'} (-1)^{|I'|+|J'|+|j|} \Delta_{\overline{I}, \overline{J'} \cup \{j\}} \times ((-1)^{\mathfrak{P}(i_0, I)+\mathfrak{P}(j, J' \cup \{j\})} [A_{I,J}]_{\mathfrak{P}(i_0, I)+\mathfrak{P}(j, J' \cup \{j\})} \Delta_{\{\mathfrak{P}(i_0, I)\}, \{\mathfrak{P}(j, J' \cup \{j\})\}} (A_{I,J})) $$

$$ = \sum_{J' \in \mathcal{P}_{n+1,k}} \sum_{j \in J'} (-1)^{|I'|+|J'|+|j|} \Delta_{\overline{I}, \overline{J'} \cup \{j\}}(A) \det(A_{I, \overline{J'} \cup \{j\}}) \quad \text{by } \mathcal{Q}(n-k, 1) $$

$$ = \sum_{J \in \mathcal{P}_{n+1,k+1}} (-1)^{|I'|+|J|} \Delta_{\overline{I}, \overline{J}}(A) \det(A_{I,J}) $$

$$ = \sum_{J \in \mathcal{P}_{n+1,k+1}} (-1)^{|I'|+|J|} \Delta_{\overline{I}, \overline{J}}(A) \Delta_{I,J}(A) $$

So $\mathcal{Q}(n+1, k+1)$ holds.

(III) By the principle of strong induction, $\mathcal{Q}(n+1, k)$ holds for all $k$ such that $k \in \{0, \dots, n+1\}$.

(iii) By the principle of strong double induction, $\mathcal{Q}(n, k)$ holds for all $(n, k)$ such that $n \in \mathbb{N}_*$ and $k \in \{0, \dots, n\}$.

(2) Case $I \neq L$. Let $I := (i_1, \dots, i_k)$, $L := (l_1, \dots, l_k)$, and $B$ the matrix defined by the rows :

$$ \forall k \in \{1, \dots, n\}, \quad \mathcal{L}_k(B) := \begin{cases} \mathcal{L}_k(A) & \text{if } k \in \overline{I} \\ \mathcal{L}_{l_j}(A) & \text{if } k := i_j \in I \end{cases} $$

Since $I$ and $L$ have the same cardinality and are distinct, there exists an element in $L$ that is not in $I$ (and vice versa), i.e., there exists $l' := l_j \in L - I = L \cap \overline{I}$. Since $l' \neq i_j$ (because $l' \in \overline{I}$ and $i_j \in I$) and since :

$$ \mathcal{L}_{l'}(B) = \mathcal{L}_{l'}(A) $$

$$ \mathcal{L}_{i_j}(B) = \mathcal{L}_{l_j}(A) = \mathcal{L}_{l'}(A) $$

the rows $l'$ and $i_j$ of $B$ are equal (thus $\det(B) = 0$). Using point (1) with matrix $B$ and since :

$$ B_{I,J} = A_{I,J} \quad , \quad B_{\overline{I},\overline{J}} = A_{\overline{I},\overline{J}} $$

we have :

$$ \begin{aligned} 0 &= \det(B) \\ &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I'|+|J'|} \Delta_{\overline{I}, \overline{J}}(B) \Delta_{I,J}(B) \end{aligned} $$
(III) By the principle of strong induction, $\mathcal{Q}(n+1, k)$ holds for all $k$ such that $k \in \{0, \dots, n+1\}$.

(iii) By the principle of strong double induction, $\mathcal{Q}(n, k)$ holds for all $(n, k)$ such that $n \in \mathbb{N}_*$ and $k \in \{0, \dots, n\}$.

(2) Case $I \neq L$. Let $I := (i_1, \dots, i_k)$, $L := (l_1, \dots, l_k)$, and $B$ the matrix defined by the rows :

$$ \forall k \in \{1, \dots, n\}, \quad \mathcal{L}_k(B) := \begin{cases} \mathcal{L}_k(A) & \text{if } k \in \bar{I} \\ \mathcal{L}_{l_j}(A) & \text{if } k := i_j \in I \end{cases} $$

Since $I$ and $L$ have the same cardinality and are distinct, there exists an element in $L$ that is not in $I$ (and vice versa), i.e., there exists $l' := l_j \in L - I = L \cap \bar{I}$. Since $l' \neq i_j$ (because $l' \in \bar{I}$ and $i_j \in I$) and since :

$$ \begin{aligned} \mathcal{L}_{l'}(B) &= \mathcal{L}_{l'}(A) \\ \mathcal{L}_{i_j}(B) &= \mathcal{L}_{l_j}(A) = \mathcal{L}_{l'}(A) \end{aligned} $$

the rows $l'$ and $i_j$ of $B$ are equal (thus $\det(B) = 0$). Using point (1) with matrix $B$ and since :

$$ B_{I,J} = A_{I,J} \quad , \quad B_{\bar{I},\bar{J}} = A_{\bar{I},\bar{J}} $$

we have :

$$ \begin{aligned} 0 &= \det(B) \\ &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{I},\bar{J}}(B) \Delta_{I,J}(B) \\ &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \det(B_{\bar{I},\bar{J}}) \det(B_{I,J}) \\ &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \det(A_{\bar{I},\bar{J}}) \det(A_{I,J}) \\ &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{I},\bar{J}}(A) \Delta_{I,J}(A) \end{aligned} $$

□

In the case where $I = L$, we have the determinant on the left-hand side, hence the following definition.

### Definition 2.2.5

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I \in \mathcal{P}_{n,k}$. The **determinant calculation of A** by expanding along the rows of $I$ is :

$$ \begin{aligned} \det(A) &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{I},\bar{J}}(A) \Delta_{I,J}(A) \\ &= \sum_{J \in \mathcal{P}_{n,k}} \Delta_{\bar{I},\bar{J}}(A) \mathrm{Cof}_{I,J}(A) \\ &= \sum_{J \in \mathcal{P}_{n,k}} \mathrm{Cof}_{\bar{I},\bar{J}}(A) \Delta_{I,J}(A). \end{aligned} $$

For the third equality, we use the fact that :

$$ (-1)^{|\bar{I}|+|\bar{J}|} = (-1)^{n(n+1)/2 - |I| + n(n+1)/2 - |J|} = (-1)^{|I|+|J|}. $$
$$
\begin{align*}
&= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I^+|+|J|} \det(B_{\bar{I}, \bar{J}}) \det(B_{I, J}) \\
&= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I^+|+|J|} \det(A_{\bar{L}, \bar{J}}) \det(A_{I, J}) \\
&= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I^+|+|J|} \Delta_{\bar{L}, \bar{J}}(A) \Delta_{I, J}(A)
\end{align*}
$$

<figure><img src="image_4.png" /></figure>

In the case where $I = L$, we have the determinant on the left-hand side, hence the following definition.

**Definition 2.2.5**

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I \in \mathcal{P}_{n,k}$. The **determinant calculation of A** by expanding along the rows of $I$ is :

$$
\begin{align*}
\det(A) &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I^+|+|J|} \Delta_{\bar{I}, \bar{J}}(A) \Delta_{I,J}(A) \\
&= \sum_{J \in \mathcal{P}_{n,k}} \Delta_{\bar{I}, \bar{J}}(A) \mathrm{Cof}_{I,J}(A) \\
&= \sum_{J \in \mathcal{P}_{n,k}} \mathrm{Cof}_{\bar{I}, \bar{J}}(A) \Delta_{I,J}(A).
\end{align*}
$$

For the third equality, we use the fact that :

$$
(-1)^{|\bar{I}|+|\bar{I}|} = (-1)^{n(n+1)/2 - |I^+|n(n+1)/2 - |J|} = (-1)^{|I^+|+|J|}.
$$
# 3 Super matrices

## 3.1 Notations

We now define the concept of minor matrices associated with a square matrix. We begin by defining the concept of square matrices indexed by an ordered set.

**Notation 3.1.1**

Let $E := (x_1, \dots, x_N)$ be an ordered set and let $U : E^2 \to \mathbb{C}$ be a function. We can represent $\Gamma$ in matrix form. For all $i, j \in \{1, \dots, N\}$, we denote :

$$U_{x_i, x_j} := U(x_i, x_j).$$

We then denote $U$ by :

$$U = \begin{pmatrix} U_{x_1, x_1} & \cdots & U_{x_1, x_N} \\ \vdots & \ddots & \vdots \\ U_{x_N, x_1} & \cdots & U_{x_N, x_N} \end{pmatrix}.$$

We will say informally that $U$ is an element of $\mathcal{M}_N(\mathbb{C})$.

Using the orders defined on $\mathcal{P}_{n,k}$ and $\mathcal{P}_n$ (see notations 2.1.1), we informally denote :

$$(V_{I,J})_{I,J \in \mathcal{P}_{n,k}} := (V_{I,J})_{\mathfrak{p}(I, \mathcal{P}_{n,k}), \mathfrak{p}(J, \mathcal{P}_{n,k})} \in \mathcal{M}_{(n)k}(\mathbb{C})$$

$$(W_{I,J})_{I,J \in \mathcal{P}_n} := (W_{I,J})_{\mathfrak{p}(I, \mathcal{P}_n), \mathfrak{p}(J, \mathcal{P}_n)} \in \mathcal{M}_{2n}(\mathbb{C})$$

**Example 3.1.2: Case $n := 2$**

As :

$$\mathcal{P}_{2,1} = (\{1\}, \{2\})$$

$$\mathcal{P}_2 = (\emptyset, \{1\}, \{2\}, \{1, 2\})$$

we have :

$$(V_{I,J})_{I,J \in \mathcal{P}_{2,1}} = \begin{pmatrix} V_{\{1\}, \{1\}} & V_{\{1\}, \{2\}} \\ V_{\{2\}, \{1\}} & V_{\{2\}, \{2\}} \end{pmatrix}$$

$$(W_{I,J})_{I,J \in \mathcal{P}_2} = \begin{pmatrix} W_{\emptyset, \emptyset} & W_{\emptyset, \{1\}} & W_{\emptyset, \{2\}} & W_{\emptyset, \{1,2\}} \\ W_{\{1\}, \emptyset} & W_{\{1\}, \{1\}} & W_{\{1\}, \{2\}} & W_{\{1\}, \{1,2\}} \\ W_{\{2\}, \emptyset} & W_{\{2\}, \{1\}} & W_{\{2\}, \{2\}} & W_{\{1\}, \{1,2\}} \\ W_{\{1,2\}, \emptyset} & W_{\{1,2\}, \{1\}} & W_{\{1,2\}, \{2\}} & W_{\{1,2\}, \{1,2\}} \end{pmatrix}$$

## 3.2 Kronecker Matrices

In this subsection, we define the concept of Kronecker matrix.

**Notation 3.2.1**

Let $n \in \mathbb{N}_*$ and $k \in \{0, \dots, n\}$.

(1) The $(n, k)$ Kronecker matrix is the $\binom{n}{k}$-sized matrix defined by :

$$\mathbf{K}_{n,k} := ((-1)^{|I|} \delta_{I,J})_{I,J \in \mathcal{P}_{n,k}}$$
# 3 Super matrices

## 3.1 Notations

We now define the concept of minor matrices associated with a square matrix. We begin by defining the concept of square matrices indexed by an ordered set.

**Notation 3.1.1**

Let $E := (x_1, \dots, x_N)$ be an ordered set and $U : E^2 \to \mathbb{C}$ a function. We can represent $U$ in matrix form. For all $i, j \in \{1, \dots, N\}$, let :

$$U_{x_i, x_j} := U(x_i, x_j).$$

We then denote $U$ by :

$$U = \begin{pmatrix} U_{x_1, x_1} & \cdots & U_{x_1, x_N} \\ \vdots & \ddots & \vdots \\ U_{x_N, x_1} & \cdots & U_{x_N, x_N} \end{pmatrix}.$$

We will say informally that $U$ is an element of $\mathcal{M}_N(\mathbb{C})$.

Using the orders defined on $\mathcal{P}_{n,k}$ and $\mathcal{P}_n$ (see notations 2.1.1), we informally denote :

$$(V_{I,J})_{I,J \in \mathcal{P}_{n,k}} := (V_{I,J})_{\mathfrak{p}(I, \mathcal{P}_{n,k}), \mathfrak{p}(J, \mathcal{P}_{n,k})} \in \mathcal{M}_{(k)}^{(n)}(\mathbb{C})$$

$$(W_{I,J})_{I,J \in \mathcal{P}_n} := (W_{I,J})_{\mathfrak{p}(I, \mathcal{P}_n), \mathfrak{p}(J, \mathcal{P}_n)} \in \mathcal{M}_{2^n}^{(\mathbb{C})}$$

**Example 3.1.2: Case $n := 2$**

As :

$$\mathcal{P}_{2,1} = (\{1\}, \{2\})$$

$$\mathcal{P}_2 = (\emptyset, \{1\}, \{2\}, \{1, 2\})$$

we have :

$$(V_{I,J})_{I,J \in \mathcal{P}_{2,1}} = \begin{pmatrix} V_{\{1\}, \{1\}} & V_{\{1\}, \{2\}} \\ V_{\{2\}, \{1\}} & V_{\{2\}, \{2\}} \end{pmatrix}$$

$$(W_{I,J})_{I,J \in \mathcal{P}_2} = \begin{pmatrix} W_{\emptyset, \emptyset} & W_{\emptyset, \{1\}} & W_{\emptyset, \{2\}} & W_{\emptyset, \{1,2\}} \\ W_{\{1\}, \emptyset} & W_{\{1\}, \{1\}} & W_{\{1\}, \{2\}} & W_{\{1\}, \{1,2\}} \\ W_{\{2\}, \emptyset} & W_{\{2\}, \{1\}} & W_{\{2\}, \{2\}} & W_{\{1\}, \{1,2\}} \\ W_{\{1,2\}, \emptyset} & W_{\{1,2\}, \{1\}} & W_{\{1,2\}, \{2\}} & W_{\{1,2\}, \{1,2\}} \end{pmatrix}$$

## 3.2 Kronecker Matrices

In this subsection, we define the concept of Kronecker matrix.

**Notation 3.2.1**

Let $n \in \mathbb{N}_*$ and $k \in \{0, \dots, n\}$.

(1) The $(n, k)$ Kronecker matrix is the $\binom{n}{k}$-sized matrix defined by :

$$\mathbf{K}_{n,k} := ((-1)^{|I|} \delta_{I,J})_{I,J \in \mathcal{P}_{n,k}}$$
with the (lexicographical) order defined on $\mathcal{P}_{n,k}$ (see 2.1.1).

(2) The $n$ Kronecker matrix is the $2^n$-sized matrix defined by :

$$ K_n := ((-1)^{|I|} \delta_{I,J})_{I,J \in \mathcal{P}_n} := \begin{pmatrix} K_{n,0} & & & (0) \\ & \ddots & & \\ & & K_{n,n} & \end{pmatrix} $$

with the order defined on $\mathcal{P}_n$ (see 2.1.1).

### Example 3.2.2: Case $n := 2$

(1) As :

$$ \mathcal{P}_{2,0} = (\emptyset) $$

$$ \mathcal{P}_{2,1} = (\{1\}, \{2\}) $$

$$ \mathcal{P}_{2,2} = (\{1, 2\}) $$

we have :

$$ K_{2,0} = ((-1)^{|\emptyset|}) = ((-1)^0) = (1) $$

$$ K_{2,1} = \begin{pmatrix} (-1)^{|\{1\}|} & 0 \\ 0 & (-1)^{|\{2\}|} \end{pmatrix} = \begin{pmatrix} (-1)^1 & 0 \\ 0 & (-1)^2 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} $$

$$ K_{2,2} = ((-1)^{|\{1,2\}|}) = ((-1)^{1+2}) = (-1) $$

(2) As :

$$ \mathcal{P}_2 = (\emptyset, \{1\}, \{2\}, \{1, 2\}). $$

we have :

$$ K_2 = \begin{pmatrix} K_{2,0} & 0 & 0 \\ 0 & K_{2,1} & 0 \\ 0 & 0 & K_{2,2} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $$

We have the simple properties of $K_n$.

### Lemma 3.2.3

The matrix $K_n$ is invertible and we have :

$$ K_n^{-1} = K_n = K_n^T = (-1)^{n(n+1)/2} K_n^C $$

where $\bullet^C$ denotes the cotranspose (see appendix B).

**Proof.** The first two equalities are due to the fact that $K_n$ is diagonal with $\pm 1$ on its diagonal, so it is symmetric and involutive (i.e., it equals its inverse).

For the last equality, we only need to examine the coefficients on the diagonal because the cotranspose of a
with the (lexicographical) order defined on $\mathcal{P}_{n,k}$ (see 2.1.1).

(2) The $n$ Kronecker matrix is the $2^n$-sized matrix defined by :

$$ \mathbf{K}_n := ((-1)^{|I|} \delta_{I,J})_{I,J \in \mathcal{P}_n} := \begin{pmatrix} \mathbf{K}_{n,0} & & (0) \\ & \ddots & \\ (0) & & \mathbf{K}_{n,n} \end{pmatrix} $$

with the order defined on $\mathcal{P}_n$ (see 2.1.1).

### Example 3.2.2: Case $n := 2$

(1) As :

$$ \mathcal{P}_{2,0} = (\emptyset) $$

$$ \mathcal{P}_{2,1} = (\{1\}, \{2\}) $$

$$ \mathcal{P}_{2,2} = (\{1, 2\}) $$

we have :

$$ \mathbf{K}_{2,0} = ((-1)^{|\emptyset|}) = ((-1)^0) = (1) $$

$$ \mathbf{K}_{2,1} = \begin{pmatrix} (-1)^{|\{1\}|} & 0 \\ 0 & (-1)^{|\{2\}|} \end{pmatrix} = \begin{pmatrix} (-1)^1 & 0 \\ 0 & (-1)^2 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} $$

$$ \mathbf{K}_{2,2} = ((-1)^{|\{1,2\}|}) = ((-1)^{1+2}) = (-1) $$

(2) As :

$$ \mathcal{P}_2 = (\emptyset, \{1\}, \{2\}, \{1, 2\}). $$

we have :

$$ \mathbf{K}_2 = \begin{pmatrix} \mathbf{K}_{2,0} & 0 & 0 \\ 0 & \mathbf{K}_{2,1} & 0 \\ 0 & 0 & \mathbf{K}_{2,2} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $$

We have the simple properties of $\mathbf{K}_n$.

### Lemma 3.2.3

The matrix $\mathbf{K}_n$ is invertible and we have :

$$ \mathbf{K}_n^{-1} = \mathbf{K}_n = \mathbf{K}_n^T = (-1)^{n(n+1)/2} \mathbf{K}_n^C $$

where $\bullet^C$ denotes the cotranspose (see appendix B).

**Proof.** The first two equalities are due to the fact that $\mathbf{K}_n$ is diagonal with $\pm 1$ on its diagonal, so it is symmetric and involutive (i.e., it equals its inverse).

For the last equality, we only need to examine the coefficients on the diagonal because the cotranspose of a
diagonal matrix is diagonal. Let $I \in \mathcal{P}_n$. Using equation (2.1.3) and notation (3.1.1), we have :

$$
\begin{align*}
[\mathbf{K}_n^C]_{I,I} &= [\mathbf{K}_n^C]_{\mathfrak{P}(I,\mathcal{P}_n),\mathfrak{P}(I,\mathcal{P}_n)} \\
&= [\mathbf{K}_n]_{2^{n+1}-\mathfrak{P}(I,\mathcal{P}_n),2^{n+1}-\mathfrak{P}(I,\mathcal{P}_n)} \\
&= [\mathbf{K}_n]_{\mathfrak{P}(\bar{I},\mathcal{P}_n),\mathfrak{P}(\bar{I},\mathcal{P}_n)} \\
&= [\mathbf{K}_n]_{\bar{I},\bar{I}} \\
&= (-1)^{|\bar{I}|} \\
&= (-1)^{n(n+1)/2-|I|} \\
&= (-1)^{n(n+1)/2} [\mathbf{K}_n]_{I,I}
\end{align*}
$$

Hence the result.

3.3 Super matrices of minors and cofactors

We can then define the matrices of minors and cofactors associated with a matrix.

Definition 3.3.1

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $k \in \{0, \dots, n\}$.

(i) (a) The $k$-matrix of minors of $A$ is the $\binom{n}{k}$-sized matrix defined by :

$$
\mathrm{Spam}_k(A) := (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_{n,k}}
$$

with the lexicographic order defined on $\mathcal{P}_{n,k}$ (see 2.1.1).

(b) The **super matrix of minors of** A is the 2<sup>n</sup>-sized matrix defined by :

$$
\text{Spam}(A) := (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_n} := \begin{pmatrix} \text{Spam}_0(A) & & (0) \\ & \ddots & \\ (0) & & \text{Spam}_n(A) \end{pmatrix}
$$

with the order defined on $\mathcal{P}_n$ (see 2.1.1) and considering the determinant of a non-square matrix
as zero (thus, $\Delta_{I,J}(A) = 0$ if $\text{Card}(I) \neq \text{Card}(J)$).

(ii) (a) The $k$-matrix of cofactors of $A$ is the $\binom{n}{k}$-sized matrix defined by :

$$
\mathrm{Spof}_k(A) := (\mathrm{Cof}_{I,J}(A))_{I,J \in \mathcal{P}_{n,k}}
$$

with the lexicographic order defined on $\mathcal{P}_{n,k}$ (see 2.1.1).

(b) The **super matrix of cofactors of** A is the 2<sup>n</sup>-sized matrix defined by :

$$
\mathrm{Spof}(A) := (\mathrm{Cof}_{I,J}(A))_{I,J \in \mathcal{P}_n} := \begin{pmatrix} \mathrm{Spof}_0(A) & & (0) \\ & \ddots & \\ (0) & & \mathrm{Spof}_n(A) \end{pmatrix}
$$

with the order defined on $\mathcal{P}_n$ (see 2.1.1) and considering the determinant of a non-square matrix
as zero (thus, $\Delta_{I,J}(A) = 0$ if $\mathrm{Card}(I) \neq \mathrm{Card}(J)$).

Therefore, we have :

$$
\mathrm{Spof}(A) = \mathbf{K}_n \mathrm{Spam}(A) \mathbf{K}_n \tag{3.3.1}
$$
diagonal matrix is diagonal. Let $I \in \mathcal{P}_n$. Using equation (2.1.3) and notation (3.1.1), we have :

$$
\begin{align*}
[\mathbf{K}_n^C]_{I,I} &= [\mathbf{K}_n^C]_{\mathfrak{P}(I,\mathcal{P}_n),\mathfrak{P}(I,\mathcal{P}_n)} \\
&= [\mathbf{K}_n]_{2^{n+1}-\mathfrak{P}(I,\mathcal{P}_n),2^{n+1}-\mathfrak{P}(I,\mathcal{P}_n)} \\
&= [\mathbf{K}_n]_{\mathfrak{P}(\bar{I},\mathcal{P}_n),\mathfrak{P}(\bar{I},\mathcal{P}_n)} \\
&= [\mathbf{K}_n]_{\bar{I},\bar{I}} \\
&= (-1)^{|\bar{I}|} \\
&= (-1)^{n(n+1)/2-|I|} \\
&= (-1)^{n(n+1)/2} [\mathbf{K}_n]_{I,I}
\end{align*}
$$

Hence the result.

3.3 Super matrices of minors and cofactors

We can then define the matrices of minors and cofactors associated with a matrix.

Definition 3.3.1

Let $A \in \mathcal{M}_n(C)$ and $k \in \{0, \dots, n\}$.

(i) (a) The $k$-matrix of minors of $A$ is the $\binom{n}{k}$-sized matrix defined by :

$$
\mathrm{Spam}_k(A) := (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_{n,k}}
$$

with the lexicographic order defined on $\mathcal{P}_{n,k}$ (see 2.1.1).

(b) The **super matrix of minors of** A is the 2ⁿ-sized matrix defined by :

$$
\text{Spam}(A) := (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_n} := \begin{pmatrix} \text{Spam}_0(A) & & (0) \\ & \ddots & \\ (0) & & \text{Spam}_n(A) \end{pmatrix}
$$

with the order defined on $\mathcal{P}_n$ (see 2.1.1) and considering the determinant of a non-square matrix
as zero (thus, $\Delta_{I,J}(A) = 0$ if $\text{Card}(I) \neq \text{Card}(J)$).

(ii) (a) The $k$-matrix of cofactors of $A$ is the $\binom{n}{k}$-sized matrix defined by :

$$
\mathrm{Spof}_k(A) := (\mathrm{Cof}_{I,J}(A))_{I,J \in \mathcal{P}_{n,k}}
$$

with the lexicographic order defined on $\mathcal{P}_{n,k}$ (see 2.1.1).

(b) The **super matrix of cofactors of** A is the 2ⁿ-sized matrix defined by :

$$
\mathrm{Spof}(A) := (\mathrm{Cof}_{I,J}(A))_{I,J \in \mathcal{P}_n} := \begin{pmatrix} \mathrm{Spof}_0(A) & & (0) \\ & \ddots & \\ (0) & & \mathrm{Spof}_n(A) \end{pmatrix}
$$

with the order defined on $\mathcal{P}_n$ (see 2.1.1) and considering the determinant of a non-square matrix
as zero (thus, $\Delta_{I,J}(A) = 0$ if $\text{Card}(I) \neq \text{Card}(J)$).

Therefore, we have :

$$
\mathrm{Spof}(A) = \mathbf{K}_n \mathrm{Spam}(A) \mathbf{K}_n \tag{3.3.1}
$$
i.e., for all $k \in \{0, \dots, n\}$:

$$ \mathrm{Spof}_k(A) = \mathbf{K}_{n,k} \ \mathrm{Spam}_k(A) \ \mathbf{K}_{n,k} \qquad (3.3.2) $$

**Lemma 3.3.2**

We have :

$$ \mathrm{Spam}(A) = (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_n} = \begin{pmatrix} \det(A) & & & (0) \\ & \ddots & A^{CT} & \\ (0) & & & 1 \end{pmatrix} $$

where $\bullet^C$ denotes the cotranspose (see appendix B).

**Proof.** Using example 2.2.2 and since $\mathcal{P}_{n,n-1} = (\overline{\{n\}}, \dots, \overline{\{1\}})$, we have :

$$ \begin{aligned} \mathrm{Spam}_0(A) &= (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_{n,0}} = \Delta_{\emptyset,\emptyset}(A) \\ &= \det(A) \end{aligned} $$

$$ \begin{aligned} \mathrm{Spam}_{n-1}(A) &= (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_{n,n-1}} \\ &= \begin{pmatrix} \Delta_{\overline{\{n\}},\overline{\{n\}}}(A) & \cdots & \Delta_{\overline{\{n\}},\overline{\{1\}}}(A) \\ \vdots & \ddots & \vdots \\ \Delta_{\overline{\{1\}},\overline{\{n\}}}(A) & \cdots & \Delta_{\overline{\{1\}},\overline{\{1\}}}(A) \end{pmatrix} \\ &= \begin{pmatrix} \det(A_{\overline{\{n\}},\overline{\{n\}}}) & \cdots & \det(A_{\overline{\{n\}},\overline{\{1\}}}) \\ \vdots & \ddots & \vdots \\ \det(A_{\overline{\{1\}},\overline{\{n\}}}) & \cdots & \det(A_{\overline{\{1\}},\overline{\{1\}}}) \end{pmatrix} \\ &= \begin{pmatrix} [A]_{nn} & \cdots & [A]_{n1} \\ \vdots & \ddots & \vdots \\ [A]_{1n} & \cdots & [A]_{11} \end{pmatrix} \\ &= A^{CT} \end{aligned} $$

$$ \begin{aligned} \mathrm{Spam}_n(A) &= (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_{n,n}} \\ &= \Delta_{\{1,\dots,n\},\{1,\dots,n\}}(A) = 1 \end{aligned} $$

□

Thus, we have :

$$ \begin{aligned} \mathrm{Spam}(A)^C &= \begin{pmatrix} 1 & & & (0) \\ & A^T & & \\ & & \ddots & \\ (0) & & & \det(A) \end{pmatrix} \\ \mathrm{Spof}(A) &= \begin{pmatrix} \mathbf{K}_0 & \det(A) & \mathbf{K}_0 & & & (0) \\ & & \ddots & & & \\ & & & \mathbf{K}_{n-1} & A^{CT} & \mathbf{K}_{n-1} \\ & & & & \mathbf{K}_n & \mathbf{K}_n \end{pmatrix} \\ &= \begin{pmatrix} \det(A) & & & & (0) \\ & \ddots & & & \\ & & \mathbf{K}_{n-1} & A^{CT} & \mathbf{K}_{n-1} \\ & & & 1 & \end{pmatrix} \end{aligned} $$
i.e., for all $k \in \{0, \dots, n\}$:

$$ \mathrm{Spof}_k(A) = \mathbf{K}_{n,k} \mathrm{Spam}_k(A) \mathbf{K}_{n,k} \qquad (3.3.2) $$

### Lemma 3.3.2

We have:

$$ \mathrm{Spam}(A) = (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_n} = \begin{pmatrix} \det(A) & & & (0) \\ & \ddots & & \\ & & A^{CT} & \\ (0) & & & 1 \end{pmatrix} $$

where $\bullet^C$ denotes the cotranspose (see appendix B).

**Proof.** Using example 2.2.2 and since $\mathcal{P}_{n,n-1} = (\overline{\{n\}}, \dots, \overline{\{1\}})$, we have:

$$ \begin{aligned} \mathrm{Spam}_0(A) &= (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_{n,0}} = \Delta_{\emptyset,\emptyset}(A) \\ &= \det(A) \end{aligned} $$

$$ \begin{aligned} \mathrm{Spam}_{n-1}(A) &= (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_{n,n-1}} \\ &= \begin{pmatrix} \Delta_{\overline{\{n\}},\overline{\{n\}}}(A) & \cdots & \Delta_{\overline{\{n\}},\{1\}}(A) \\ \vdots & \ddots & \vdots \\ \Delta_{\overline{\{1\}},\overline{\{n\}}}(A) & \cdots & \Delta_{\overline{\{1\}},\{1\}}(A) \end{pmatrix} \\ &= \begin{pmatrix} \det(A_{\overline{\{n\}},\{n\}}) & \cdots & \det(A_{\overline{\{n\}},\{1\}}) \\ \vdots & \ddots & \vdots \\ \det(A_{\overline{\{1\}},\{n\}}) & \cdots & \det(A_{\overline{\{1\}},\{1\}}) \end{pmatrix} \\ &= \begin{pmatrix} [A]_{nn} & \cdots & [A]_{n1} \\ \vdots & \ddots & \vdots \\ [A]_{1n} & \cdots & [A]_{11} \end{pmatrix} \\ &= A^{CT} \end{aligned} $$

$$ \begin{aligned} \mathrm{Spam}_n(A) &= (\Delta_{I,J}(A))_{I,J \in \mathcal{P}_{n,n}} \\ &= \Delta_{\{1,\dots,n\},\{1,\dots,n\}}(A) = 1 \end{aligned} $$

Thus, we have:

$$ \begin{aligned} \mathrm{Spam}(A)^C &= \begin{pmatrix} 1 & & & (0) \\ & A^T & & \\ & & \ddots & \\ (0) & & & \det(A) \end{pmatrix} \\ \mathrm{Spof}(A) &= \begin{pmatrix} \mathbf{K}_0 & \det(A) & \mathbf{K}_0 & & & (0) \\ & \ddots & & & & \\ & & \mathbf{K}_{n-1} & A^{CT} & \mathbf{K}_{n-1} & \\ (0) & & & & \mathbf{K}_n & \mathbf{K}_n \\ & & & & & \end{pmatrix} \\ &= \begin{pmatrix} \det(A) & & & & (0) \\ & \ddots & & & \\ & & \mathbf{K}_{n-1} & A^{CT} & \mathbf{K}_{n-1} & \\ (0) & & & & 1 & \end{pmatrix} \end{aligned} $$
### Example 3.3.3: Case $n := 2$

Let :

$$A := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathcal{M}_2(\mathbb{C}).$$

We have :

$$ \text{Spam}(A) = \begin{pmatrix} \det(A) & 0 & 0 \\ 0 & A^{CT} & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \det(A) & 0 & 0 \\ 0 & d & c \\ 0 & b & a \\ 0 & 0 & 1 \end{pmatrix} $$

$$ \text{Spam}(A)^C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & A^T & 0 \\ 0 & 0 & \det(A) \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & a & c \\ 0 & b & d \\ 0 & 0 & \det(A) \end{pmatrix} $$

with :

$$ \det(A) = ad - bc. $$

Before proving the general result, we prove the following simple lemma.

### Lemma 3.3.4

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, J \in \mathcal{P}_n$. We have :

$$ [\text{Spam}(A)]_{\bar{L}, \bar{J}} = [\text{Spam}(A)^C]_{J, L} $$

**Proof.** Using equation (2.1.3) and notation (3.1.1), we have :

$$ \begin{aligned} [\text{Spam}(A)^C]_{J,L} &= [\text{Spam}(A)^C]_{\mathfrak{p}(J, \mathcal{P}_n), \mathfrak{p}(L, \mathcal{P}_n)} \\ &= [\text{Spam}(A)]_{2^{n+1}-\mathfrak{p}(L, \mathcal{P}_n), 2^{n+1}-\mathfrak{p}(J, \mathcal{P}_n)} \\ &= [\text{Spam}(A)]_{\mathfrak{p}(\bar{L}, \mathcal{P}_n), \mathfrak{p}(\bar{J}, \mathcal{P}_n)} \\ &= [\text{Spam}(A)]_{\bar{L}, \bar{J}} \end{aligned} $$

<figure><img src="image_3.png" /></figure>

### Theorem 3.3.5

Let $A \in \mathcal{M}_n(\mathbb{C})$. We have :

$$ \begin{aligned} \det(A) I_{2^n} &= \mathbf{K}_n \text{ Spam}(A) \mathbf{K}_n \text{ Spam}(A)^C \\ &= \text{Spof}(A) \text{ Spam}(A)^C \\ &= \text{Spam}(A) \text{ Spof}(A)^C \end{aligned} $$

**Proof.** Let $I, J \in \mathcal{P}_n$. Using proposition 2.2.4 and lemma 3.3.4, we have :

$$ \begin{aligned} [\mathbf{K}_n \text{ Spam}(A) \mathbf{K}_n \text{ Spam}(A)^C]_{I,L} &= \sum_{J,J',J'' \in \mathcal{P}_n} [\mathbf{K}_n]_{I,J''} [\text{Spam}(A)]_{J'',J'} [\mathbf{K}_n]_{J',J} [\text{Spam}(A)^C]_{J,L} \\ &= \sum_{J,J',J'' \in \mathcal{P}_n} (-1)^{|I|} \delta_{I,J''} [\text{Spam}(A)]_{J'',J'} (-1)^{|J'|} \delta_{J',J} [\text{Spam}(A)^C]_{J,L} \end{aligned} $$
### Example 3.3.3: Case $n := 2$

Let :

$$A := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathcal{M}_2(\mathbb{C}).$$

We have :

$$\text{Spam}(A) = \begin{pmatrix} \det(A) & 0 & 0 \\ 0 & A^{CT} & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \det(A) & 0 & 0 \\ 0 & d & c \\ 0 & b & a \\ 0 & 0 & 1 \end{pmatrix}$$

$$\text{Spam}(A)^C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & A^T & 0 \\ 0 & 0 & \det(A) \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & a & c \\ 0 & b & d \\ 0 & 0 & \det(A) \end{pmatrix}$$

with :

$$\det(A) = ad - bc.$$

Before proving the general result, we prove the following simple lemma.

### Lemma 3.3.4

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, J \in \mathcal{P}_n$. We have :

$$[\text{Spam}(A)]_{\bar{L}, \bar{J}} = [\text{Spam}(A)^C]_{J,L}$$

**Proof.** Using equation (2.1.3) and notation (3.1.1), we have :

$$\begin{align*}
[\text{Spam}(A)^C]_{J,L} &= [\text{Spam}(A)^C]_{\mathfrak{p}(J, \mathcal{P}_n), \mathfrak{p}(L, \mathcal{P}_n)} \\
&= [\text{Spam}(A)]_{2^{n+1}-\mathfrak{p}(L, \mathcal{P}_n), 2^{n+1}-\mathfrak{p}(J, \mathcal{P}_n)} \\
&= [\text{Spam}(A)]_{\mathfrak{p}(\bar{L}, \mathcal{P}_n), \mathfrak{p}(\bar{J}, \mathcal{P}_n)} \\
&= [\text{Spam}(A)]_{\bar{L}, \bar{J}}
\end{align*}$$

□

### Theorem 3.3.5

Let $A \in \mathcal{M}_n(\mathbb{C})$. We have :

$$\begin{align*}
\det(A) I_{2^n} &= \mathbf{K}_n \text{ Spam}(A) \mathbf{K}_n \text{ Spam}(A)^C \\
&= \text{Spof}(A) \text{ Spam}(A)^C \\
&= \text{Spam}(A) \text{ Spof}(A)^C
\end{align*}$$

**Proof.** Let $I, J \in \mathcal{P}_n$. Using proposition 2.2.4 and lemma 3.3.4, we have :

$$\begin{align*}
[\mathbf{K}_n \text{ Spam}(A) \mathbf{K}_n \text{ Spam}(A)^C]_{I,L} &= \sum_{J,J',J'' \in \mathcal{P}_n} [\mathbf{K}_n]_{I,J''} [\text{Spam}(A)]_{J'',J'} [\mathbf{K}_n]_{J',J} [\text{Spam}(A)^C]_{J,L} \\
&= \sum_{J,J',J'' \in \mathcal{P}_n} (-1)^{|I|} \delta_{I,J''} [\text{Spam}(A)]_{J'',J'} (-1)^{|J'|} \delta_{J',J} [\text{Spam}(A)^C]_{J,L}
\end{align*}$$
$$
\begin{align*}
&= \sum_{J \in \mathcal{P}_n} (-1)^{|I|+|J|} [\text{Spam}(A)]_{I,J} [\text{Spam}(A)]_{\bar{L},\bar{J}} \\
&= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{I,J}(A) \Delta_{\bar{L},\bar{J}}(A) \\
&= \det(A) \delta_{I,L}.
\end{align*}
$$

For the last equality, we use lemma 3.2.3. We have :

$$
\begin{align*}
\text{Spam}(A) \text{Spof}(A)^C &= \text{Spam}(A) (\mathbf{K}_n \text{Spam}(A)\mathbf{K}_n)^C \\
&= \text{Spam}(A) \mathbf{K}_n^C \text{Spam}(A)^C \mathbf{K}_n^C \\
&= \text{Spam}(A) (-1)^{n(n+1)/2} \mathbf{K}_n \text{Spam}(A)^C (-1)^{n(n+1)/2} \mathbf{K}_n \\
&= \mathbf{K}_n (\mathbf{K}_n \text{Spam}(A)^C \mathbf{K}_n \text{Spam}(A)) \mathbf{K}_n \\
&= \mathbf{K}_n \det(A) I_{2^n} \mathbf{K}_n \\
&= \det(A) I_{2^n} \\
&= \text{Spof}(A) \text{Spam}(A)^C
\end{align*}
$$

Example 3.3.6

Let's revisit example (1) from 3.3.3 :

$$
A := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathcal{M}_2(\mathbb{C}).
$$

We have :

$$
\begin{align*}
& \mathbf{K}_2 \text{Spam}(A) \mathbf{K}_2 \text{Spam}(A)^C \\
&= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \det(A) & 0 & 0 & 0 \\ 0 & d & c & 0 \\ 0 & b & a & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & a & c & 0 \\ 0 & b & d & 0 \\ 0 & 0 & 0 & \det(A) \end{pmatrix} \\
&= \begin{pmatrix} \det(A) & 0 & 0 & 0 \\ 0 & \det(A) & 0 & 0 \\ 0 & 0 & \det(A) & 0 \\ 0 & 0 & 0 & \det(A) \end{pmatrix} \\
&= \det(A) I_4
\end{align*}
$$

This implies the invertibility of the super matrix of an invertible matrix.

Corollary 3.3.7

Let $A \in \text{GL}_n(\mathbb{C})$. We have $\text{Spam}(A) \in \text{GL}_{2^n}(\mathbb{C})$ and :

$$
\begin{align*}
\text{Spam}(A)^{-1} &= \frac{1}{\det(A)} \mathbf{K}_n \text{Spam}(A)^C \mathbf{K}_n \\
&= \frac{1}{\det(A)} \text{Spof}(A)^C
\end{align*}
$$
$$
\begin{align*}
&= \sum_{J \in \mathcal{P}_n} (-1)^{|I|+|J|} [\text{Spam}(A)]_{I,J} [\text{Spam}(A)]_{\bar{L},\bar{J}} \\
&= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{I,J}(A) \Delta_{\bar{L},\bar{J}}(A) \\
&= \det(A) \delta_{I,L}.
\end{align*}
$$

For the last equality, we use lemma 3.2.3. We have :

$$
\begin{align*}
\text{Spam}(A) \text{Spof}(A)^C &= \text{Spam}(A) (\mathbf{K}_n \text{Spam}(A)\mathbf{K}_n)^C \\
&= \text{Spam}(A) \mathbf{K}_n^C \text{Spam}(A)^C \mathbf{K}_n^C \\
&= \text{Spam}(A) (-1)^{n(n+1)/2} \mathbf{K}_n \text{Spam}(A)^C (-1)^{n(n+1)/2} \mathbf{K}_n \\
&= \mathbf{K}_n (\mathbf{K}_n \text{Spam}(A)^C \mathbf{K}_n \text{Spam}(A)) \mathbf{K}_n \\
&= \mathbf{K}_n \det(A) I_{2^n} \mathbf{K}_n \\
&= \det(A) I_{2^n} \\
&= \text{Spof}(A) \text{Spam}(A)^C
\end{align*}
$$

Example 3.3.6

Let's revisit example (1) from 3.3.3 :

$$
A := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathcal{M}_2(\mathbb{C}).
$$

We have :

$$
\begin{align*}
& \mathbf{K}_2 \text{Spam}(A) \mathbf{K}_2 \text{Spam}(A)^C \\
&= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \det(A) & 0 & 0 & 0 \\ 0 & d & c & 0 \\ 0 & b & a & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & a & c & 0 \\ 0 & b & d & 0 \\ 0 & 0 & 0 & \det(A) \end{pmatrix} \\
&= \begin{pmatrix} \det(A) & 0 & 0 & 0 \\ 0 & \det(A) & 0 & 0 \\ 0 & 0 & \det(A) & 0 \\ 0 & 0 & 0 & \det(A) \end{pmatrix} \\
&= \det(A) I_4
\end{align*}
$$

This implies the invertibility of the super matrix of minors of an invertible matrix.

Corollary 3.3.7

Let $A \in \text{GL}_n(\mathbb{C})$. We have $\text{Spam}(A) \in \text{GL}_{2^n}(\mathbb{C})$ and :

$$
\begin{align*}
\text{Spam}(A)^{-1} &= \frac{1}{\det(A)} \mathbf{K}_n \text{Spam}(A)^C \mathbf{K}_n \\
&= \frac{1}{\det(A)} \text{Spof}(A)^C
\end{align*}
$$
i.e. for all $k \in \{0, \dots, n\}$:

$$ \text{Spam}_k(A)^{-1} = \frac{1}{\det(A)} \mathbf{K}_{n-k} \text{Spam}_{n-k}(A)^C \mathbf{K}_{n-k}. $$

For $k := n - 1$, we have:

$$ \begin{aligned} A^{-1} &= \frac{1}{\det(A)} \mathbf{K}_1 \text{Spam}_1(A)^C \mathbf{K}_1 \\ &= \frac{1}{\det(A)} \text{Cof}_1(A)^T \end{aligned} $$

**Proof.** Since $\det(A) \neq 0$ and $\mathbf{K}_n^{-1} = \mathbf{K}_n$, we have:

$$ I_{2^n} = \frac{1}{\det(A)} \text{Spam}(A) \mathbf{K}_n \text{Spam}(A)^C \mathbf{K}_n $$

i.e. by Lemma 3.2.3:

$$ \begin{aligned} \text{Spam}(A)^{-1} &= \frac{1}{\det(A)} \mathbf{K}_n \text{Spam}(A)^C \mathbf{K}_n \\ &= \frac{1}{\det(A)} (\mathbf{K}_n^C \text{Spam}(A) \mathbf{K}_n^C)^C \\ &= \frac{1}{\det(A)} ((-1)^{n(n+1)/2} \mathbf{K}_n \text{Spam}(A) (-1)^{n(n+1)/2} \mathbf{K}_n)^C \\ &= \frac{1}{\det(A)} (\mathbf{K}_n \text{Spam}(A) \mathbf{K}_n)^C \\ &= \frac{1}{\det(A)} \text{Spof}(A)^C \end{aligned} $$

Since:

$$ \begin{aligned} \text{Spam}(A)^{-1} &= \begin{pmatrix} \text{Spam}_0(A) & & & (0) \\ & \ddots & & \\ (0) & & \text{Spam}_n(A) & \\ & & & \text{Spam}_n(A)^{-1} \end{pmatrix}^{-1} \\ &= \begin{pmatrix} \text{Spam}_0(A)^{-1} & & & (0) \\ & \ddots & & \\ (0) & & \text{Spam}_n(A)^{-1} & \\ & & & \text{Spam}_n(A)^{-1} \end{pmatrix} \\ &= \begin{pmatrix} 1/\det(A) & & & (0) \\ & \ddots & & \\ & & (A^{CT})^{-1} & \\ (0) & & & 1 \end{pmatrix} \end{aligned} $$

$$ \begin{aligned} \text{Spof}(A)^C &= \begin{pmatrix} \mathbf{K}_0 \text{Spam}_0(A) & \mathbf{K}_0 & & (0) \\ & (0) & \mathbf{K}_n \text{Spam}_n(A) & \mathbf{K}_n \\ & & & \text{Spam}_n(A)^C \end{pmatrix}^C \\ &= \begin{pmatrix} \mathbf{K}_n \text{Spam}_n(A)^C & \mathbf{K}_n & & (0) \\ & (0) & \mathbf{K}_0 \text{Spam}_0(A)^C & \mathbf{K}_0 \end{pmatrix} \end{aligned} $$
i.e. for all $k \in \{0, \dots, n\}$:

$$ \text{Spam}_k(A)^{-1} = \frac{1}{\det(A)} \mathbf{K}_{n-k} \text{Spam}_{n-k}(A)^C \mathbf{K}_{n-k}. $$

For $k := n - 1$, we have:

$$ \begin{aligned} A^{-1} &= \frac{1}{\det(A)} \mathbf{K}_1 \text{Spam}_1(A)^C \mathbf{K}_1 \\ &= \frac{1}{\det(A)} \text{Cof}_1(A)^T \end{aligned} $$

**Proof.** Since $\det(A) \neq 0$ and $\mathbf{K}_n^{-1} = \mathbf{K}_n$, we have:

$$ I_{2^n} = \frac{1}{\det(A)} \text{Spam}(A) \mathbf{K}_n \text{Spam}(A)^C \mathbf{K}_n $$

i.e. by Lemma 3.2.3:

$$ \begin{aligned} \text{Spam}(A)^{-1} &= \frac{1}{\det(A)} \mathbf{K}_n \text{Spam}(A)^C \mathbf{K}_n \\ &= \frac{1}{\det(A)} (\mathbf{K}_n^C \text{Spam}(A) \mathbf{K}_n^C)^C \\ &= \frac{1}{\det(A)} ((-1)^{n(n+1)/2} \mathbf{K}_n \text{Spam}(A) (-1)^{n(n+1)/2} \mathbf{K}_n)^C \\ &= \frac{1}{\det(A)} (\mathbf{K}_n \text{Spam}(A) \mathbf{K}_n)^C \\ &= \frac{1}{\det(A)} \text{Spof}(A)^C \end{aligned} $$

Since:

$$ \begin{aligned} \text{Spam}(A)^{-1} &= \begin{pmatrix} \text{Spam}_0(A) & & & (0) \\ & \ddots & & \\ (0) & & \text{Spam}_n(A) & \\ & & & \text{Spam}_n(A)^C \end{pmatrix}^{-1} \\ &= \begin{pmatrix} \text{Spam}_0(A)^{-1} & & & (0) \\ & \ddots & & \\ (0) & & \text{Spam}_n(A)^{-1} & \\ & & & \text{Spam}_n(A)^C \end{pmatrix} \\ &= \begin{pmatrix} 1/\det(A) & & & (0) \\ & \ddots & & \\ & & (A^{CT})^{-1} & \\ (0) & & & 1 \end{pmatrix} \end{aligned} $$

$$ \begin{aligned} \text{Spof}(A)^C &= \begin{pmatrix} \mathbf{K}_0 \text{Spam}_0(A) \mathbf{K}_0 & & & (0) \\ & \ddots & & \\ (0) & & \mathbf{K}_n \text{Spam}_n(A) \mathbf{K}_n & \\ & & & \text{Spam}_n(A)^C \end{pmatrix}^C \\ &= \begin{pmatrix} \mathbf{K}_n \text{Spam}_n(A)^C \mathbf{K}_n & & & (0) \\ & \ddots & & \\ (0) & & \mathbf{K}_0 \text{Spam}_0(A)^C \mathbf{K}_0 & \\ & & & \text{Spam}_0(A)^C \end{pmatrix} \end{aligned} $$
$$ = \begin{pmatrix} 1 & & & & (0) \\ & \mathbf{K}_{n-1} & A^T & \mathbf{K}_{n-1} & \\ & (0) & & \det(A) & \end{pmatrix} $$

We have for all $k \in \{0, \dots, n\}$:

$$ \text{Spam}_k(A)^{-1} = \frac{1}{\det(A)} \mathbf{K}_{n-k} \text{Spam}_{n-k}(A)^C \mathbf{K}_{n-k} $$

For $k := n - 1$, we have:

$$ (A^{-1})^{CT} = (A^{CT})^{-1} = \frac{1}{\det(A)} \mathbf{K}_1 \text{Spam}_1(A)^C \mathbf{K}_1 $$

i.e. we have:

$$ \begin{aligned} A^{-1} &= \left( \frac{1}{\det(A)} \mathbf{K}_1 \text{Spam}_1(A)^C \mathbf{K}_1 \right)^{CT} \\ &= \frac{1}{\det(A)} \mathbf{K}_1 \text{Spam}_1(A)^T \mathbf{K}_1 \\ &= \frac{1}{\det(A)} (\mathbf{K}_1 \text{Spam}_1(A) \mathbf{K}_1)^T \\ &= \frac{1}{\det(A)} \text{Cof}_1(A)^T \end{aligned} $$

### Example 3.3.8

Let's revisit example 3.3.6 with $\det(A) \neq 0$ i.e.:

$$ A := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \text{GL}_2(\mathbb{C}) $$

We have:

$$ \begin{aligned} \begin{pmatrix} \det(A) & 0 & 0 \\ 0 & A^{CT} & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1} &= \text{Spam}(A)^{-1} \\ &= \frac{1}{\det(A)} \mathbf{K}_2 \text{Spam}(A)^C \mathbf{K}_2 \\ &= \frac{1}{\det(A)} \begin{pmatrix} \mathbf{K}_{2,0} & 0 & 0 \\ 0 & \mathbf{K}_{2,1} & 0 \\ 0 & 0 & \mathbf{K}_{2,2} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & A^T & 0 \\ 0 & 0 & \det(A) \end{pmatrix} \begin{pmatrix} \mathbf{K}_{2,0} & 0 & 0 \\ 0 & \mathbf{K}_{2,1} & 0 \\ 0 & 0 & \mathbf{K}_{2,2} \end{pmatrix} \\ &= \frac{1}{\det(A)} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \mathbf{K}_{2,1} A^T \mathbf{K}_{2,1} & 0 \\ 0 & 0 & \det(A) \end{pmatrix} \end{aligned} $$

So we have:

$$ \begin{aligned} (A^{-1})^{CT} &= (A^{CT})^{-1} \\ &= \mathbf{K}_{2,1} A^T \mathbf{K}_{2,1} \end{aligned} $$
$$ = \begin{pmatrix} 1 & & & & (0) \\ & \mathbf{K}_{n-1} & A^T & \mathbf{K}_{n-1} & \\ & (0) & & \det(A) & \end{pmatrix} $$

We have for all $k \in \{0, \dots, n\}$:

$$ \text{Spam}_k(A)^{-1} = \frac{1}{\det(A)} \mathbf{K}_{n-k} \text{Spam}_{n-k}(A)^C \mathbf{K}_{n-k} $$

For $k := n - 1$, we have:

$$ (A^{-1})^{CT} = (A^{CT})^{-1} = \frac{1}{\det(A)} \mathbf{K}_1 \text{Spam}_1(A)^C \mathbf{K}_1 $$

i.e. we have:

$$ \begin{aligned} A^{-1} &= \left( \frac{1}{\det(A)} \mathbf{K}_1 \text{Spam}_1(A)^C \mathbf{K}_1 \right)^{CT} \\ &= \frac{1}{\det(A)} \mathbf{K}_1 \text{Spam}_1(A)^T \mathbf{K}_1 \\ &= \frac{1}{\det(A)} (\mathbf{K}_1 \text{Spam}_1(A) \mathbf{K}_1)^T \\ &= \frac{1}{\det(A)} \text{Cof}_1(A)^T \end{aligned} $$

<figure><img src="image_5.png" /></figure>

### Example 3.3.8

Let's revisit example 3.3.6 with $\det(A) \neq 0$ i.e.:

$$ A := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \text{GL}_2(\mathbb{C}) $$

We have:

$$ \begin{aligned} \begin{pmatrix} \det(A) & 0 & 0 \\ 0 & A^{CT} & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1} &= \text{Spam}(A)^{-1} \\ &= \frac{1}{\det(A)} \mathbf{K}_2 \text{Spam}(A)^C \mathbf{K}_2 \\ &= \frac{1}{\det(A)} \begin{pmatrix} \mathbf{K}_{2,0} & 0 & 0 \\ 0 & \mathbf{K}_{2,1} & 0 \\ 0 & 0 & \mathbf{K}_{2,2} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & A^T & 0 \\ 0 & 0 & \det(A) \end{pmatrix} \begin{pmatrix} \mathbf{K}_{2,0} & 0 & 0 \\ 0 & \mathbf{K}_{2,1} & 0 \\ 0 & 0 & \mathbf{K}_{2,2} \end{pmatrix} \\ &= \frac{1}{\det(A)} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \mathbf{K}_{2,1} A^T \mathbf{K}_{2,1} & 0 \\ 0 & 0 & \det(A) \end{pmatrix} \end{aligned} $$

So we have:

$$ \begin{aligned} (A^{-1})^{CT} &= (A^{CT})^{-1} \\ &= \mathbf{K}_{2,1} A^T \mathbf{K}_{2,1} \end{aligned} $$
$$= \frac{1}{\det(A)} \begin{pmatrix} a & -c \\ -b & d \end{pmatrix}$$

i.e. we have :

$$A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} a & -c \\ -b & d \end{pmatrix}^{CT} \\ = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$

We retrieve the usual formula for the inverse of a 2 × 2 matrix.

## 3.4 Further Column Expansions

By taking the transpose, we expand along multiple columns. We start with a simple lemma due to the stability of the determinant under transpose.

### Lemma 3.4.1

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, J \in \mathcal{P}_n$.

(i) We have :

$$(A^T)_{I,J} = (A_{J,I})^T.$$

(ii) We have :

$$\Delta_{I,J}(A^T) = \Delta_{J,I}(A).$$

(iii) We have :

$$\text{Spam}(A^T) = \text{Spam}(A)^T.$$

### Proof.

(i) Removing the rows of $J$ and the columns of $I$ from $A$ is equivalent to removing the rows of $I$ and the columns of $J$ from $A^T$. Hence the result.

(ii) By point (i) and the stability of the determinant under transpose (see proposition B.2.1), we have :

$$\begin{align*} \Delta_{I,J}(A^T) &= \det((A^T)_{I,J}) \\ &= \det((A_{J,I})^T) \\ &= \det(A_{J,I}) \\ &= \Delta_{J,I}(A). \end{align*}$$

(iii) Thus, by point (ii), we have :

$$\begin{align*} \text{Spam}(A^T) &= (\Delta_{I,J}(A^T))_{I,J \in \mathcal{P}_n} \\ &= (\Delta_{J,I}(A))_{I,J \in \mathcal{P}_n} \\ &= \text{Spam}(A)^T \end{align*}$$

We then deduce the expansion of the determinant of a square matrix along multiple columns.
$$ = \frac{1}{\det(A)} \begin{pmatrix} a & -c \\ -b & d \end{pmatrix} $$

i.e. we have :

$$ \begin{aligned} A^{-1} &= \frac{1}{\det(A)} \begin{pmatrix} a & -c \\ -b & d \end{pmatrix}^{CT} \\ &= \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \end{aligned} $$

We retrieve the usual formula for the inverse of a 2 × 2 matrix.

## 3.4 Further Column Expansions

By taking the transpose, we expand along multiple columns. We start with a simple lemma due to the stability of the determinant under transpose.

### Lemma 3.4.1

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, J \in \mathcal{P}_n$.

(i) We have :

$$ (A^T)_{I,J} = (A_{J,I})^T. $$

(ii) We have :

$$ \Delta_{I,J}(A^T) = \Delta_{J,I}(A). $$

(iii) We have :

$$ \text{Spam}(A^T) = \text{Spam}(A)^T. $$

### Proof.

(i) Removing the rows of $J$ and the columns of $I$ from $A$ is equivalent to removing the rows of $I$ and the columns of $J$ from $A^T$. Hence the result.

(ii) By point (i) and the stability of the determinant under transpose (see proposition B.2.1), we have :

$$ \begin{aligned} \Delta_{I,J}(A^T) &= \det((A^T)_{I,J}) \\ &= \det((A_{J,I})^T) \\ &= \det(A_{J,I}) \\ &= \Delta_{J,I}(A). \end{aligned} $$

(iii) Thus, by point (ii), we have :

$$ \begin{aligned} \text{Spam}(A^T) &= (\Delta_{I,J}(A^T))_{I,J \in \mathcal{P}_n} \\ &= (\Delta_{J,I}(A))_{I,J \in \mathcal{P}_n} \\ &= \text{Spam}(A)^T \end{aligned} $$

We then deduce the expansion of the determinant of a square matrix along multiple columns. □
### Theorem 3.4.2

Let $A \in \mathcal{M}_n(\mathbb{C})$. We have :

$$ 
\begin{aligned}
\det(A)I_{2^n} &= \mathbf{K}_n \text{ Spam}(A)^C \mathbf{K}_n \text{ Spam}(A) \\
&= \text{Spof}(A)^C \text{ Spam}(A).
\end{aligned}
$$

**Proof.** As $\mathbf{K}_n$ is symmetric, we have by theorem 3.3.5, by the stability of the determinant under transpose (see proposition B.2.1), and by proposition B.2.1 :

$$ 
\begin{aligned}
\det(A)I_{2^n} &= \det(A^T)I_{2^n} \\
&= \mathbf{K}_n \text{ Spam}(A^T) \mathbf{K}_n \text{ Spam}(A^T)^C \\
&= \mathbf{K}_n \text{ Spam}(A)^T \mathbf{K}_n (\text{Spam}(A)^C)^T \\
&= (\text{Spam}(A)^C \mathbf{K}_n \text{ Spam}(A) \mathbf{K}_n)^T
\end{aligned}
$$

And thus, by taking the transpose and since $\mathbf{K}_n^{-1} = \mathbf{K}_n$ :

$$ 
\begin{aligned}
\det(A)I_{2^n} &= \text{Spam}(A)^C \mathbf{K}_n \text{ Spam}(A) \mathbf{K}_n \\
&= \mathbf{K}_n \text{ Spam}(A)^C \mathbf{K}_n \text{ Spam}(A)
\end{aligned}
$$

<figure><img src="image_4.png" /></figure>

Looking at the coefficients of this matrix equality, we have the following result.

### Corollary 3.4.3

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, L \in \mathcal{P}_{n,k}$. Then we have :

$$ 
\begin{aligned}
\det(A)\delta_{I,L} &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{J},\bar{I}}(A)\Delta_{J,L}(A) \\
&= \sum_{J \in \mathcal{P}_{n,k}} \mathrm{Cof}_{\bar{J},\bar{I}}(A)\Delta_{J,L}(A).
\end{aligned}
$$

**Proof.** Let $I, J \in \mathcal{P}_n$. We have by lemma 3.3.4 :

$$ 
\begin{aligned}
\det(A)\delta_{I,L} &= [\mathbf{K}_n \text{ Spam}(A)^C \mathbf{K}_n \text{ Spam}(A)]_{I,L} \\
&= \sum_{J,J',J'' \in \mathcal{P}_n} [\mathbf{K}_n]_{I,J''}[\text{Spam}(A)^C]_{J'',J'}[\mathbf{K}_n]_{J',J}[\text{Spam}(A)]_{J,L} \\
&= \sum_{J,J',J'' \in \mathcal{P}_n} (-1)^{|I|} \delta_{I,J''}[\text{Spam}(A)^C]_{J'',J'} (-1)^{|J'|} \delta_{J',J}[\text{Spam}(A)]_{J,L} \\
&= \sum_{J \in \mathcal{P}_n} (-1)^{|I|+|J|} [\text{Spam}(A)]_{\bar{J},\bar{I}} [\text{Spam}(A)]_{J,L} \\
&= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{J},\bar{I}}(A) \Delta_{J,L}(A)
\end{aligned}
$$

For the second equality, we use the fact that :

$$ (-1)^{|\bar{I}|+|\bar{I}|} = (-1)^{n(n+1)/2 - |I| + n(n+1)/2 - |J|} = (-1)^{|I|+|J|}. $$

<figure><img src="image_5.png" /></figure>
### Theorem 3.4.2

Let $A \in \mathcal{M}_n(\mathbb{C})$. We have :

$$ 
\begin{aligned}
\det(A)I_{2^n} &= \mathbf{K}_n \text{ Spam}(A)^C \mathbf{K}_n \text{ Spam}(A) \\
&= \text{Spof}(A)^C \text{ Spam}(A).
\end{aligned}
$$

**Proof.** As $\mathbf{K}_n$ is symmetric, we have by theorem 3.3.5, by the stability of the determinant under transpose (see proposition B.2.1), and by proposition B.2.1 :

$$ 
\begin{aligned}
\det(A)I_{2^n} &= \det(A^T) I_{2^n} \\
&= \mathbf{K}_n \text{ Spam}(A^T) \mathbf{K}_n \text{ Spam}(A^T)^C \\
&= \mathbf{K}_n \text{ Spam}(A)^T \mathbf{K}_n (\text{Spam}(A)^C)^T \\
&= (\text{Spam}(A)^C \mathbf{K}_n \text{ Spam}(A) \mathbf{K}_n)^T
\end{aligned}
$$

And thus, by taking the transpose and since $\mathbf{K}_n^{-1} = \mathbf{K}_n$ :

$$ 
\begin{aligned}
\det(A)I_{2^n} &= \text{Spam}(A)^C \mathbf{K}_n \text{ Spam}(A) \mathbf{K}_n \\
&= \mathbf{K}_n \text{ Spam}(A)^C \mathbf{K}_n \text{ Spam}(A)
\end{aligned}
$$

Looking at the coefficients of this matrix equality, we have the following result.

### Corollary 3.4.3

Let $A \in \mathcal{M}_n(\mathbb{C})$ and $I, L \in \mathcal{P}_{n,k}$. Then we have :

$$ 
\begin{aligned}
\det(A)\delta_{I,L} &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{J},\bar{I}}(A)\Delta_{J,L}(A) \\
&= \sum_{J \in \mathcal{P}_{n,k}} \text{Cof}_{\bar{J},\bar{I}}(A)\Delta_{J,L}(A).
\end{aligned}
$$

**Proof.** Let $I, J \in \mathcal{P}_n$. We have by lemma 3.3.4 :

$$ 
\begin{aligned}
\det(A)\delta_{I,L} &= [\mathbf{K}_n \text{ Spam}(A)^C \mathbf{K}_n \text{ Spam}(A)]_{I,L} \\
&= \sum_{J,J',J'' \in \mathcal{P}_n} [\mathbf{K}_n]_{I,J''} [\text{Spam}(A)^C]_{J'',J'} [\mathbf{K}_n]_{J',J} [\text{Spam}(A)]_{J,L} \\
&= \sum_{J,J',J'' \in \mathcal{P}_n} (-1)^{|I|} \delta_{I,J''} [\text{Spam}(A)^C]_{J'',J'} (-1)^{|J'|} \delta_{J',J} [\text{Spam}(A)]_{J,L} \\
&= \sum_{J \in \mathcal{P}_n} (-1)^{|I|+|J|} [\text{Spam}(A)]_{\bar{J},\bar{I}} [\text{Spam}(A)]_{J,L} \\
&= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\bar{J},\bar{I}}(A)\Delta_{J,L}(A)
\end{aligned}
$$

For the second equality, we use the fact that :

$$ (-1)^{|\bar{I}|+|\bar{I}|} = (-1)^{n(n+1)/2 - |I| + n(n+1)/2 - |J|} = (-1)^{|I|+|J|}. $$
In the case where $k := 1$, $I := \{i\}$ and $L := \{l\}$, we retrieve the usual formula for expanding along a column :

$$
\begin{align*}
\det(A)\delta_{i,l} &= \sum_{J \in \mathcal{P}_{n,1}} (-1)^{i+|J|} \Delta_{\overline{J},\{i\}}(A)\Delta_{J,\{l\}}(A) \\
&= \sum_{j=1}^{n} (-1)^{i+j} \Delta_{\overline{\{j\}},\{i\}}(A)\Delta_{\{j\},\{l\}}(A) \\
&= \sum_{j=1}^{n} (-1)^{i+j} [A]_{\overline{\{j\}},\{i\}}\Delta_{\{j\},\{l\}}(A)
\end{align*}
$$

Dans le cas où $I = L$, on a à gauche le déterminant, d'où la définition suivante.

Definition 3.4.4

Soit $A \in \mathcal{M}_n(\mathbb{C})$ et $I \in \mathcal{P}_{n,k}$. Le calcul du déterminant de $A$ en développant suivant les colonnes de $I$ est :

$$
\begin{align*}
\det(A) &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I|+|J|} \Delta_{\overline{J},I}(A) \Delta_{J,I}(A) \\
&= \sum_{J \in \mathcal{P}_{n,k}} \Delta_{\overline{J},I}(A) \mathrm{Cof}_{J,I}(A) \\
&= \sum_{J \in \mathcal{P}_{n,k}} \mathrm{Cof}_{\overline{J},I}(A) \Delta_{J,I}(A)
\end{align*}
$$

3.5 Case of dimension 3

Let :

$$
A := \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \in \mathcal{M}_3(\mathbb{C}).
$$

We have :

$$
\text{Spam}(A) =
\begin{pmatrix}
\det(A)g & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & ei - fh & di - fg & dh - eg & 0 & 0 & 0 & 0 \\
0 & bi - ch & ai - cg & ah - bg & 0 & 0 & 0 & 0 \\
0 & bf - ce & af - cd & ae - bd & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & i & h & g & 0 \\
0 & 0 & 0 & 0 & f & e & d & 0 \\
0 & 0 & 0 & 0 & c & b & a & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}
$$

and

$$
\text{Spam}(A)^C =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & a & d & g & 0 & 0 & 0 & 0 \\
0 & b & e & h & 0 & 0 & 0 & 0 \\
0 & c & f & i & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & ae - bd & ah - bg & dh - eg & 0 \\
0 & 0 & 0 & 0 & af - cd & ai - cg & di - fg & 0 \\
0 & 0 & 0 & 0 & bf - ce & bi - ch & ei - fh & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \det(A)
\end{pmatrix}
$$
<figure><img src="image_6.png" /></figure>

In the case where $k := 1$, $I := \{i\}$ and $L := \{l\}$, we retrieve the usual formula for expanding along a column :

$$
\begin{align*}
\det(A)\delta_{i,l} &= \sum_{J \in \mathcal{P}_{n,1}} (-1)^{i+|J|} \Delta_{\overline{J},\{i\}}(A) \Delta_{J,\{l\}}(A) \\
&= \sum_{j=1}^{n} (-1)^{i+j} \Delta_{\overline{\{j\}},\{i\}}(A) \Delta_{\{j\},\{l\}}(A) \\
&= \sum_{j=1}^{n} (-1)^{i+j} [A]_{\overline{\{j\}},\{i\}} \Delta_{\{j\},\{l\}}(A)
\end{align*}
$$

Dans le cas où $I = L$, on a à gauche le déterminant, d'où la définition suivante.

**Definition 3.4.4**

Soit $A \in \mathcal{M}_n(\mathbb{C})$ et $I \in \mathcal{P}_{n,k}$. Le calcul du déterminant de $A$ en développant suivant les colonnes de $I$ est :

$$
\begin{align*}
\det(A) &= \sum_{J \in \mathcal{P}_{n,k}} (-1)^{|I^+|+|J|} \Delta_{\overline{J},I}(A) \Delta_{J,I}(A) \\
&= \sum_{J \in \mathcal{P}_{n,k}} \Delta_{\overline{J},I}(A) \mathrm{Cof}_{J,I}(A) \\
&= \sum_{J \in \mathcal{P}_{n,k}} \mathrm{Cof}_{\overline{J},I}(A) \Delta_{J,I}(A)
\end{align*}
$$

**3.5 Case of dimension 3**

Let :

$$
A := \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \in \mathcal{M}_3(\mathbb{C}).
$$

We have :

$$
\text{Spam}(A) =
\begin{pmatrix}
\det(A)g & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & ei - fh & di - fg & dh - eg & 0 & 0 & 0 \\
0 & bi - ch & ai - cg & ah - bg & 0 & 0 & 0 \\
0 & bf - ce & af - cd & ae - bd & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & i & h & g \\
0 & 0 & 0 & 0 & f & e & d \\
0 & 0 & 0 & 0 & c & b & a \\
0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}
$$

and

$$
\text{Spam}(A)^C =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & a & d & g & 0 & 0 & 0 & 0 \\
0 & b & e & h & 0 & 0 & 0 & 0 \\
0 & c & f & i & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & ae - bd & ah - bg & dh - eg & 0 \\
0 & 0 & 0 & 0 & af - cd & ai - cg & di - fg & 0 \\
0 & 0 & 0 & 0 & bf - ce & bi - ch & ei - fh & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \det(A)
\end{pmatrix}
$$
with :

$$ \det(A) = (ch - bi)d - (fh - ei)a - (ce - bf)g. $$

As :

$$ \mathcal{P}_{3,0} = (\emptyset) $$

$$ \mathcal{P}_{3,1} = (\{1\}, \{2\}, \{3\}) $$

$$ \mathcal{P}_{3,2} = (\{1, 2\}, \{1, 3\}, \{2, 3\}) $$

$$ \mathcal{P}_{3,3} = (\{1, 2, 3\}) $$

we have :

$$ \mathbf{K}_{3,0} = ((-1)^{|\emptyset|}) = ((-1)^0) = (1) $$

$$ \mathbf{K}_{3,1} = \begin{pmatrix} (-1)^{|\{1\}|} & 0 & 0 \\ 0 & (-1)^{|\{2\}|} & 0 \\ 0 & 0 & (-1)^{|\{3\}|} \end{pmatrix} = \begin{pmatrix} (-1)^1 & 0 & 0 \\ 0 & (-1)^2 & 0 \\ 0 & 0 & (-1)^3 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$

$$ \mathbf{K}_{3,2} = \begin{pmatrix} (-1)^{|\{1,2\}|} & 0 & 0 \\ 0 & (-1)^{|\{1,3\}|} & 0 \\ 0 & 0 & (-1)^{|\{2,3\}|} \end{pmatrix} = \begin{pmatrix} (-1)^{1+2} & 0 & 0 \\ 0 & (-1)^{1+3} & 0 \\ 0 & 0 & (-1)^{2+3} \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$

$$ \mathbf{K}_{3,3} = ((-1)^{|\{1,2,3\}|}) = ((-1)^{1+2+3}) = (1) $$

And as :

$$ \mathcal{P}_3 = (\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}). $$

we have :

$$ \mathbf{K}_3 = \begin{pmatrix} \mathbf{K}_{3,0} & 0 & 0 & 0 \\ 0 & \mathbf{K}_{3,1} & 0 & 0 \\ 0 & 0 & \mathbf{K}_{3,2} & 0 \\ 0 & 0 & 0 & \mathbf{K}_{3,3} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$

Then we have :

$$ \mathbf{K}_{3\text{Spam}}(A)\mathbf{K}_{3\text{Spam}}(A)^C = \begin{pmatrix} \det(A) & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \det(A) & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \det(A) & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \det(A) & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \det(A) & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \det(A) & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \det(A) & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \det(A) \end{pmatrix} \\ = \det(A)I_8 $$
with :

$$ \det(A) = (ch - bi)d - (fh - ei)a - (ce - bf)g. $$

As :

$$ \mathcal{P}_{3,0} = (\emptyset) $$

$$ \mathcal{P}_{3,1} = (\{1\}, \{2\}, \{3\}) $$

$$ \mathcal{P}_{3,2} = (\{1,2\}, \{1,3\}, \{2,3\}) $$

$$ \mathcal{P}_{3,3} = (\{1,2,3\}) $$

we have :

$$ \mathbf{K}_{3,0} = ((-1)^{|\emptyset|}) = ((-1)^0) = (1) $$

$$ \mathbf{K}_{3,1} = \begin{pmatrix} (-1)^{|\{1\}|} & 0 & 0 \\ 0 & (-1)^{|\{2\}|} & 0 \\ 0 & 0 & (-1)^{|\{3\}|} \end{pmatrix} = \begin{pmatrix} (-1)^1 & 0 & 0 \\ 0 & (-1)^2 & 0 \\ 0 & 0 & (-1)^3 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$

$$ \mathbf{K}_{3,2} = \begin{pmatrix} (-1)^{|\{1,2\}|} & 0 & 0 \\ 0 & (-1)^{|\{1,3\}|} & 0 \\ 0 & 0 & (-1)^{|\{2,3\}|} \end{pmatrix} = \begin{pmatrix} (-1)^{1+2} & 0 & 0 \\ 0 & (-1)^{1+3} & 0 \\ 0 & 0 & (-1)^{2+3} \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$

$$ \mathbf{K}_{3,3} = ((-1)^{|\{1,2,3\}|}) = ((-1)^{1+2+3}) = (1) $$

And as :

$$ \mathcal{P}_3 = (\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}). $$

we have :

$$ \mathbf{K}_3 = \begin{pmatrix} \mathbf{K}_{3,0} & 0 & 0 & 0 \\ 0 & \mathbf{K}_{3,1} & 0 & 0 \\ 0 & 0 & \mathbf{K}_{3,2} & 0 \\ 0 & 0 & 0 & \mathbf{K}_{3,3} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$

Then we have :

$$ \mathbf{K}_{3\text{Spam}}(A)\mathbf{K}_{3\text{Spam}}(A)^C = \begin{pmatrix} \det(A) & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \det(A) & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \det(A) & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \det(A) & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \det(A) & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \det(A) & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \det(A) & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \det(A) \end{pmatrix} \\ = \det(A)I_8 $$
3.6 An Example in Dimension 4

Let's reconsider the matrix from the introduction :

$$
A := \begin{pmatrix}
1 & 2 & 1 & -2 \\
1 & 3 & 2 & -1 \\
-2 & 1 & 3 & 4 \\
1 & -1 & 3 & 1
\end{pmatrix}.
$$

We have :

$$
\text{Spam}(A) =
\begin{pmatrix}
25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -43 & 12 & 22 & 35 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -35 & 15 & 15 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -14 & 1 & 6 & 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -5 & -5 & -5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -9 & 5 & 6 & -6 & -9 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 5 & 2 & 11 & 2 & 1 & -4 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 11 & 13 & 7 & 2 & 7 & 7 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 7 & 0 & 7 & 3 & 2 & -3 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 10 & 10 & 5 & 0 & 5 & 5 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 3 & 4 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 3 & -1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 4 & 3 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 2 & 3 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 & 1 & 2 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}
$$

and

$$
\text{Spam}(A)^C =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 2 & 3 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 2 & 3 & 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -2 & -1 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 5 & -3 & 7 & -4 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 5 & 2 & 7 & 1 & -9 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 3 & 2 & 2 & -6 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 5 & 7 & 7 & 11 & 6 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 4 & 10 & 0 & 13 & 2 & 5 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 3 & 10 & 7 & 11 & 5 & -9 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 5 & 25 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -5 & 6 & 15 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -5 & 1 & 15 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -5 & -14 & -35 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25
\end{pmatrix}
$$
3.6 An Example in Dimension 4

Let's reconsider the matrix from the introduction :

$$
A := \begin{pmatrix}
1 & 2 & 1 & -2 \\
1 & 3 & 2 & -1 \\
-2 & 1 & 3 & 4 \\
1 & -1 & 3 & 1
\end{pmatrix}.
$$

We have :

$$
\text{Spam}(A) =
\begin{pmatrix}
25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -43 & 12 & 22 & 35 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -35 & 15 & 15 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -14 & 1 & 6 & 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -5 & -5 & -5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -9 & 5 & 6 & -6 & -9 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 5 & 2 & 11 & 2 & 1 & -4 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 11 & 13 & 7 & 2 & 7 & 7 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 7 & 0 & 7 & 3 & 2 & -3 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 10 & 10 & 5 & 0 & 5 & 5 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 3 & 4 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 3 & -1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 4 & 3 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 2 & 3 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 & 1 & 2 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}
$$

and

$$
\text{Spam}(A)^C =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 2 & 3 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 2 & 3 & 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -2 & -1 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 5 & -3 & 7 & -4 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 5 & 2 & 7 & 1 & -9 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 3 & 2 & 2 & -6 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 5 & 7 & 7 & 11 & 6 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 4 & 10 & 0 & 13 & 2 & 5 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 3 & 10 & 7 & 11 & 5 & -9 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 5 & 25 & 35 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -5 & 6 & 15 & 22 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -5 & 1 & 15 & 12 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -5 & -14 & -35 & -43 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25
\end{pmatrix}
$$
As :

$$
K_4 = \begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}
$$

we have :

$$
\begin{align*}
K_4\text{Spam}(A)K_4\text{Spam}(A)^C &= \begin{pmatrix}
25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25
\end{pmatrix} \\
&= 25I_{16}
\end{align*}
$$
As :

$$
\mathbf{K}_4 =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}
$$

we have :

$$
\begin{align*}
\mathbf{K}_4\text{Spam}(A)\mathbf{K}_4\text{Spam}(A)^C &=
\begin{pmatrix}
25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 25
\end{pmatrix} \\
&= 25I_{16}
\end{align*}
$$
# A Permutation Matrices

The following notations are used for matrices.

* For any matrix $M \in \mathcal{M}_{n,p}(\mathbb{C})$, any $k \in \{1, \dots, n\}$, and any $l \in \{1, \dots, p\}$, $[M]_{k,l}$ denotes the coefficient of $M$ in row $k$ and column $l$.

* The comma in the subscript will be omitted if there is no possible confusion. For example, $a_{ij} := a_{i,j}$, but $a_{i+2,j}$ will remain.

* Let $M := (m_{i,j})_{i,j} \in \mathcal{M}_n(\mathbb{C})$ and $l \in \{1, \dots, n\}$.

(a) $\mathcal{L}_l(M)$ denotes the $l$-th row of $M$, i.e. :

$$ \mathcal{L}_l(M) := (m_{l1} \quad \dots \quad m_{ln}). $$

(b) $\mathfrak{C}_l(M)$ denotes the $l$-th column of $M$, i.e. :

$$ \mathfrak{C}_l(M) := \begin{pmatrix} m_{l1} \\ \vdots \\ m_{ln} \end{pmatrix}. $$

## A.1 Symmetric Groups

### Definition A.1.1

(i) A **permutation of** $\{1, \dots, n\}$ is a bijection $\sigma$ from $\{1, \dots, n\}$ to $\{1, \dots, n\}$. Its support is defined as :

$$ \text{Supp}(\sigma) := \{i \in \{1, \dots, n\}, \sigma(i) \neq i\}. $$

$\mathfrak{S}_n$ denotes the set of permutations of the set $\{1, \dots, n\}$.

(ii) Let $l \in \{2, \dots, n\}$ and $i_1, \dots, i_l$ be distinct elements of $\{1, \dots, n\}$. The **cycle** $\sigma := (i_1, \dots, i_l)$ **of length** $l$ is the permutation of $\{1, \dots, n\}$ such that :

$$ \forall j \in \{1, \dots, l-1\}, \sigma(i_j) = i_{j+1} $$

$$ \sigma(i_l) = i_1 $$

$$ \text{Supp}(\sigma) = \{i_1, \dots, i_l\} $$

(iii) A **transposition** $\tau$ **of** $\{1, \dots, n\}$ is a cycle of length 2, i.e., there exist $i, j \in \{1, \dots, n\}$ distinct such that $\text{Supp}(\tau) = \{i, j\}$. It is denoted as :

$$ \tau_{i,j} := \tau = (i,j). $$

(iv) For any permutation $\sigma$ of $\{1, \dots, n\}$, the **signature of** $\sigma$ is defined as :

$$ \varepsilon(\sigma) := \prod_{1 \le i < j \le n} \frac{\sigma(j) - \sigma(i)}{j - i}. $$

Composition is an internal composition law in $\mathfrak{S}_n$, making $\mathfrak{S}_n$ a group, and :

$$ \text{Card}(\mathfrak{S}_n) = n!. $$

A permutation $\sigma$ can be represented in two ways :
# A Permutation Matrices

The following notations are used for matrices.

* For any matrix $M \in \mathcal{M}_{n,p}(\mathbb{C})$, any $k \in \{1, \dots, n\}$, and any $l \in \{1, \dots, p\}$, $[M]_{k,l}$ denotes the coefficient of $M$ in row $k$ and column $l$.

* The comma in the subscript will be omitted if there is no possible confusion. For example, $a_{ij} := a_{i,j}$, but $a_{i+2,j}$ will remain.

* Let $M := (m_{i,j})_{i,j} \in \mathcal{M}_n(\mathbb{C})$ and $l \in \{1, \dots, n\}$.

(a) $\mathcal{L}_l(M)$ denotes the $l$-th row of $M$, i.e. :

$$ \mathcal{L}_l(M) := (m_{l1} \quad \dots \quad m_{ln}). $$

(b) $\mathfrak{C}_l(M)$ denotes the $l$-th column of $M$, i.e. :

$$ \mathfrak{C}_l(M) := \begin{pmatrix} m_{l1} \\ \vdots \\ m_{ln} \end{pmatrix}. $$

## A.1 Symmetric Groups

### Definition A.1.1

(i) A **permutation of** $\{1, \dots, n\}$ is a bijection $\sigma$ from $\{1, \dots, n\}$ to $\{1, \dots, n\}$. Its support is defined as :

$$ \text{Supp}(\sigma) := \{i \in \{1, \dots, n\}, \sigma(i) \neq i\}. $$

$\mathfrak{S}_n$ denotes the set of permutations of the set $\{1, \dots, n\}$.

(ii) Let $l \in \{2, \dots, n\}$ and $i_1, \dots, i_l$ be distinct elements of $\{1, \dots, n\}$. The **cycle** $\sigma := (i_1, \dots, i_l)$ **of length** $l$ is the permutation of $\{1, \dots, n\}$ such that :

$$ \begin{gathered} \forall j \in \{1, \dots, l-1\}, \quad \sigma(i_j) = i_{j+1} \\ \sigma(i_l) = i_1 \\ \text{Supp}(\sigma) = \{i_1, \dots, i_l\} \end{gathered} $$

(iii) A **transposition** $\tau$ **of** $\{1, \dots, n\}$ is a cycle of length 2, i.e., there exist $i, j \in \{1, \dots, n\}$ distinct such that $\text{Supp}(\tau) = \{i, j\}$. It is denoted as :

$$ \tau_{i,j} := \tau = (i,j). $$

(iv) For any permutation $\sigma$ of $\{1, \dots, n\}$, the **signature of** $\sigma$ is defined as :

$$ \varepsilon(\sigma) := \prod_{1 \le i < j \le n} \frac{\sigma(j) - \sigma(i)}{j - i}. $$

Composition is an internal composition law in $\mathfrak{S}_n$, making $\mathfrak{S}_n$ a group, and :

$$ \text{Card}(\mathfrak{S}_n) = n!. $$

A permutation $\sigma$ can be represented in two ways :
• as a matrix :

$$
\sigma = \begin{pmatrix} 1 & 2 & \cdots & n \\ \sigma(1) & \sigma(2) & \cdots & \sigma(n) \end{pmatrix}
$$

• as a table :

<table>
  <tr>
    <td><i>i</i></td>
    <td>1</td>
    <td>2</td>
    <td>...</td>
    <td><i>n</i></td>
  </tr>
  <tr>
    <td><i>σ</i>(<i>i</i>)</td>
    <td><i>σ</i>(1)</td>
    <td><i>σ</i>(2)</td>
    <td>...</td>
    <td><i>σ</i>(<i>n</i>)</td>
  </tr>
</table>

The following proposition states useful results for understanding the text.

Proposition A.1.2

(i) Any permutation σ can be written as the product of transpositions.

(ii) The signature defines a group morphism :

$$
\varepsilon : \sigma \in (\mathfrak{S}_n, \circ) \mapsto \varepsilon(\sigma) \in (\{\pm 1\}, \times)
$$

and for any transposition τ in $\mathfrak{S}_n$, $\varepsilon(\sigma) = -1$.

It can also be shown that the parity of the number of transpositions appearing in the decomposition does not
depend on the decomposition (i.e., if there are two decompositions of σ into products of transpositions with N and
N' transpositions respectively, then N' ≡ N[2]). We then have :

$$
\varepsilon(\sigma) = (-1)^N
$$

if $\sigma := \tau_1 \circ \cdots \circ \tau_N$. Every cycle $c := (i_1, \dots, i_l)$ can be decomposed into the product of $l-1$ transpositions :

$$
c = (i_1, i_2) \circ \dots \circ (i_{l-1}, i_l)
$$

thus we have :

$$
\varepsilon(c) = (-1)^l.
$$

Two simple examples are given below to understand the defined objects and notions.

Example A.1.3

(1) The symmetric group $\mathfrak{S}_3$ has order 6, and we have :

$$
\begin{align*}
\mathfrak{S}_3 &= \left\{ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \\ 2 & 3 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{pmatrix} \right\} \\
&= \{\text{Id}, (12), (13), (23), (123) = (12)(23), (132) = (13)(32)\} \\
&=: \{\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5, \sigma_6\}
\end{align*}
$$

Like the signature of a transposition is -1, the signatures of the six permutations are :

$$
\varepsilon(\sigma_1) = 1
$$

$$
\varepsilon(\sigma_2) = -1
$$

$$
\varepsilon(\sigma_3) = -1
$$

$$
\varepsilon(\sigma_4) = -1
$$

$$
\varepsilon(\sigma_5) = \varepsilon(\sigma_2) \varepsilon(\sigma_4) = 1
$$

$$
\varepsilon(\sigma_6) = \varepsilon(\sigma_3) \varepsilon(\sigma_4) = 1
$$

Let's demonstrate with the other formula that $\varepsilon(\sigma_6) = 1$. As :

$$
\sigma_6 = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix}
$$
• as a matrix :

$$
\sigma = \begin{pmatrix} 1 & 2 & \cdots & n \\ \sigma(1) & \sigma(2) & \cdots & \sigma(n) \end{pmatrix}
$$

• as a table :

<table>
  <tr>
    <td><i>i</i></td>
    <td>1</td>
    <td>2</td>
    <td>...</td>
    <td><i>n</i></td>
  </tr>
  <tr>
    <td><i>σ</i>(<i>i</i>)</td>
    <td><i>σ</i>(1)</td>
    <td><i>σ</i>(2)</td>
    <td>...</td>
    <td><i>σ</i>(<i>n</i>)</td>
  </tr>
</table>

The following proposition states useful results for understanding the text.

Proposition A.1.2

(i) Any permutation σ can be written as the product of transpositions.

(ii) The signature defines a group morphism :

$$
\varepsilon : \sigma \in (\mathfrak{S}_{n, \circ}) \mapsto \varepsilon(\sigma) \in (\{\pm 1\}, \times)
$$

and for any transposition τ in $\mathfrak{S}_n$, $\varepsilon(\sigma) = -1$.

It can also be shown that the parity of the number of transpositions appearing in the decomposition does not
depend on the decomposition (i.e., if there are two decompositions of σ into products of transpositions with N and
N' transpositions respectively, then N' ≡ N[2]). We then have in this case :

$$
\varepsilon(\sigma) = (-1)^N.
$$

Every cycle $c := (i_1, \dots, i_l)$ can be decomposed into the product of $l-1$ transpositions :

$$
c = (i_1, i_2) \circ \cdots (i_{l-1}, i_l)
$$

thus we have :

$$
\varepsilon(c) = (-1)^l.
$$

Two simple examples are given below to understand the defined objects and notions.

Example A.1.3

(1) The symmetric group $\mathfrak{S}_3$ has order 6, and we have :

$$
\begin{align*}
\mathfrak{S}_3 &= \left\{ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{pmatrix} \right\} \\
&= \{\text{Id}, (12), (13), (23), (123) = (12)(23), (132) = (13)(32)\} \\
&= \{\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5, \sigma_6\}
\end{align*}
$$

Like the signature of a transposition is -1, the signatures of the six permutations are :

$$
\varepsilon(\sigma_1) = 1
$$

$$
\varepsilon(\sigma_2) = -1
$$

$$
\varepsilon(\sigma_3) = -1
$$

$$
\varepsilon(\sigma_4) = -1
$$

$$
\varepsilon(\sigma_5) = \varepsilon(\sigma_2) \varepsilon(\sigma_4) = 1
$$

$$
\varepsilon(\sigma_6) = \varepsilon(\sigma_3) \varepsilon(\sigma_4) = 1
$$

Let's demonstrate with the other formula that $\varepsilon(\sigma_6) = 1$. As :

$$
\sigma_6 = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix}
$$
we have :

$$
\begin{align*}
\varepsilon(\sigma_6) &= \prod_{1 \le i < j \le 3} \frac{\sigma_6(j) - \sigma_6(i)}{j - i} \\
&= \frac{\sigma_6(2) - \sigma_6(1)}{2 - 1} \cdot \frac{\sigma_6(3) - \sigma_6(1)}{3 - 1} \cdot \frac{\sigma_6(3) - \sigma_6(2)}{3 - 2} \\
&= \frac{1 - 3}{2 - 1} \cdot \frac{2 - 3}{3 - 1} \cdot \frac{2 - 1}{3 - 2} \\
&= -1 \times -1 \times 1 \\
&= 1
\end{align*}
$$

(2) Consider the permutation of {1, ..., 12} :

$$
\sigma := \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 2 & 4 & 3 & 10 & 8 & 9 & 7 & 12 & 6 & 1 & 11 & 5 \end{pmatrix}
$$

We have :

$$
\mathrm{Supp}(\sigma) = \{1, 2, 4, 5, 6, 8, 9, 10, 12\}
$$

and

$$
\sigma = c_1 \circ c_2 \circ c_3 = (1, 2, 4, 10) \circ (5, 8, 12) \circ (6, 9)
$$

with

$$
c_1 := (1, 2, 4, 10) = (1, 2) \circ (2, 4) \circ (4, 10)
$$

$$
c_2 := (5, 8, 12) = (5, 8) \circ (8, 12)
$$

$$
c_3 := (6, 9)
$$

So we have :

$$
\sigma = (1, 2) \circ (2, 4) \circ (4, 10) \circ (5, 8) \circ (8, 12) \circ (6, 9)
$$

and

$$
\varepsilon(\sigma) = (-1)^6 = 1
$$

A.2 Determinant of a Square Matrix

In this subsection, we provide important concepts about determinants to understand the text. Let (e₁, ..., eₙ) be the canonical basis of ℂⁿ := 𝒰ₙ,₁(𝒂) over 𝒂 defined by :

$$
e_1 := \begin{pmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \quad e_2 := \begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \dots, e_n := \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix}.
$$

**Definition A.2.1**

Let $f$ be a function from $(\mathbb{C}^n)^n$ to $\mathbb{C}$. We say $f$ is an $n$-linear alternate form if :

(i) (Linearity in each variable) for any $i \in \{1, \dots, n\}$, all $x_1, \dots, x_n, y_i \in \mathbb{C}^n$, and any $\alpha, \beta \in \mathbb{C}$, we have :

$$
f(x_1, \dots, x_{i-1}, \alpha x_i + \beta y_i, x_{i+1}, \dots, x_n) = \alpha f(x_1, \dots, x_n) + \beta f(x_1, \dots, x'_i, \dots, x_n);
$$
we have :

$$
\begin{align*}
\varepsilon(\sigma_6) &= \prod_{1 \le i < j \le 3} \frac{\sigma_6(j) - \sigma_6(i)}{j - i} \\
&= \frac{\sigma_6(2) - \sigma_6(1)}{2 - 1} \cdot \frac{\sigma_6(3) - \sigma_6(1)}{3 - 1} \cdot \frac{\sigma_6(3) - \sigma_6(2)}{3 - 2} \\
&= \frac{1 - 3}{2 - 1} \cdot \frac{2 - 3}{3 - 1} \cdot \frac{2 - 1}{3 - 2} \\
&= -1 \times -1 \times 1 \\
&= 1
\end{align*}
$$

(2) Consider the permutation of {1, ..., 12} :

$$
\sigma := \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 2 & 4 & 3 & 10 & 8 & 9 & 7 & 12 & 6 & 1 & 11 & 5 \end{pmatrix}
$$

We have :

$$
\mathrm{Supp}(\sigma) = \{1, 2, 4, 5, 6, 8, 9, 10, 12\}
$$

and

$$
\sigma = c_1 \circ c_2 \circ c_3 = (1, 2, 4, 10) \circ (5, 8, 12) \circ (6, 9)
$$

with

$$
c_1 := (1, 2, 4, 10) = (1, 2) \circ (2, 4) \circ (4, 10)
$$

$$
c_2 := (5, 8, 12) = (5, 8) \circ (8, 12)
$$

$$
c_3 := (6, 9)
$$

So we have :

$$
\sigma = (1, 2) \circ (2, 4) \circ (4, 10) \circ (5, 8) \circ (8, 12) \circ (6, 9)
$$

and

$$
\varepsilon(\sigma) = (-1)^6 = 1
$$

A.2 Determinant of a Square Matrix

In this subsection, we provide important concepts about determinants to understand the text. Let (e₁, ..., eₙ) be the canonical basis of **C**ⁿ := 𝒰ₙ,₁(**C**) over **C** defined by :

$$
e_1 := \begin{pmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \quad e_2 := \begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \dots, e_n := \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix}.
$$

**Definition A.2.1**

Let $f$ be a function from $(\mathbb{C}^n)^n$ to $\mathbb{C}$. We say $f$ is an $n$-linear alternate form if :

(i) (Linearity in each variable) for any $i \in \{1, \dots, n\}$, all $x_1, \dots, x_n, y_i \in \mathbb{C}^n$, and any $\alpha, \beta \in \mathbb{C}$, we have :

$$
f(x_1, \dots, x_{i-1}, \alpha x_i + \beta y_i, x_{i+1}, \dots, x_n) = \alpha f(x_1, \dots, x_n) + \beta f(x_1, \dots, y_i, \dots, x_n);
$$
(ii) (Alternate) for all $i, j \in \{1, \dots, n\}$ distinct and all $x_1, \dots, x_n \in \mathbb{C}^n$ such that $x_i = x_j$, we have :

$$f(x_1, \dots, x_n) = 0.$$

In $\mathbb{C}$, an $n$-linear form $f$ is alternating if and only if it is **antisymmetric**, that is :

$$\forall x_1, \dots, x_n \in \mathbb{C}^n, \forall i, j \in \{1, \dots, n\}, i < j \implies f(x_1, \dots, x_{i-1}, x_j, x_{i-1}, \dots, x_{j-1}, x_i, x_{j-1}, \dots, x_n) = -f(x_1, \dots, x_n).$$

For any alternating $n$-linear form $f$, we have :

$$f(x_1, \dots, x_n) = \left( \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{j=1}^{n} x_{\sigma(j),j} \right) f(e_1, \dots, e_n).$$

Thus, the set of alternating $n$-linear forms is a $\mathbb{C}$-vector space of dimension 1. It is generated by any nonzero alternating $n$-linear form. Hence the following definitions.

### Definition A.2.2

(i) The **determinant** $\det$ is the unique alternating $n$-linear form such that :

$$\det(e_1, \dots, e_n) = 1.$$

(ii) Let :

$$M := (\mathfrak{C}_1(M) | \dots | \mathfrak{C}_n(M)) \in \mathcal{M}_n(\mathbb{C}).$$

The **determinant** of $M$ is defined by :

$$\det(M) := \det(\mathfrak{C}_1(M), \dots, \mathfrak{C}_n(M)).$$

For $M := (m_{i,j})_{ij}$, we have :

$$\det(M) = \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} m_{i,\sigma(i)}.$$

We have the useful properties of the determinant.

### Proposition A.2.3

Let $M, N \in \mathcal{M}_n(\mathbb{C})$.

(i) We have :

$$\det(MN) = \det(M)\det(N).$$

(ii) We have :

$$\det(M) \neq 0 \iff M \in \mathrm{GL}_n(\mathbb{C}).$$

Taking the transpose, we have that the determinant of a matrix is the determinant of its rows, i.e., for :

$$M := (m_{i,j})_{ij} := \begin{pmatrix} \mathfrak{L}_1(M) \\ \vdots \\ \mathfrak{L}_n(M) \end{pmatrix} \in \mathcal{M}_n(\mathbb{C}).$$

we have :

$$\det(M) = \det(\mathfrak{L}_1(M)^T, \dots, \mathfrak{L}_n(M)^T)$$
(ii) (Alternate) for all $i, j \in \{1, \dots, n\}$ distinct and all $x_1, \dots, x_n \in \mathbb{C}^n$ such that $x_i = x_j$, we have :

$$f(x_1, \dots, x_n) = 0.$$

In $\mathbb{C}$, an $n$-linear form $f$ is alternating if and only if it is **antisymmetric**, that is :

$$\forall x_1, \dots, x_n \in \mathbb{C}^n, \forall i, j \in \{1, \dots, n\}, i < j \implies f(x_1, \dots, x_{i-1}, x_j, x_{i-1}, \dots, x_{j-1}, x_i, x_{j-1}, \dots, x_n) = -f(x_1, \dots, x_n).$$

For any alternating $n$-linear form $f$, we have :

$$f(x_1, \dots, x_n) = \left( \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{j=1}^{n} x_{\sigma(j),j} \right) f(e_1, \dots, e_n).$$

Thus, the set of alternating $n$-linear forms is a $\mathbb{C}$-vector space of dimension 1. It is generated by any nonzero alternating $n$-linear form. Hence the following definitions.

### Definition A.2.2

(i) The **determinant** $\det$ is the unique alternating $n$-linear form such that :

$$\det(e_1, \dots, e_n) = 1.$$

(ii) Let :

$$M := (\mathfrak{C}_1(M) | \dots | \mathfrak{C}_n(M)) \in \mathcal{M}_n(\mathbb{C}).$$

The **determinant** of $M$ is defined by :

$$\det(M) := \det(\mathfrak{C}_1(M), \dots, \mathfrak{C}_n(M)).$$

For $M := (m_{i,j})_{ij}$, we have :

$$\det(M) = \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} m_{i,\sigma(i)}.$$

We have the useful properties of the determinant.

### Proposition A.2.3

Let $M, N \in \mathcal{M}_n(\mathbb{C})$.

(i) We have :

$$\det(MN) = \det(M)\det(N).$$

(ii) We have :

$$\det(M) \neq 0 \iff M \in \mathrm{GL}_n(\mathbb{C}).$$

## A.3 Permutation Matrices

### Definition A.3.1

(i) Let $\sigma \in \mathfrak{S}_n$. The **permutation matrix associated with $\sigma$** is the matrix $P_{\sigma} \in \mathcal{M}_n(\mathbb{C})$ such that for all $k, l \in \{1, \dots, n\}$ :

$$[P_{\sigma}]_{k,l} := \delta_{i,\sigma(j)}.$$
$$
\begin{align*}
&= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [M^T]_{i, \sigma(i)} \\
&= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} m_{\sigma(i), i}.
\end{align*}
$$

A.3 Permutation Matrices

Definition A.3.1

(i) Let $\sigma \in \mathfrak{S}_n$. The permutation matrix associated with $\sigma$ is the matrix $P_\sigma \in \mathcal{M}_n(\mathbb{C})$ such that for all $k,l \in \{1, \dots, n\}$:

$$
[\mathbf{P}_{\sigma}]_{k,l} := \delta_{i,\sigma(j)}.
$$

(ii) Let $i, j \in \{1, \dots, n\}$ such that $i \neq j$. The **permutation matrix** $(i, j)$ **of size** $n$ is the matrix :

$$
\mathbf{P}_{i,j} := \mathbf{P}_{\tau_{i,j}}
$$

with $\tau_{i,j}$ the transposition of $i$ and $j$. It is also called an **elementary permutation matrix** or a **transposition matrix**.

Thus we have :

$$
\mathbf{P}_{i,j} = (e_1 | \cdots | e_{i-1} | e_j | e_{i+1} | \cdots | e_{j-1} | e_i | e_{j+1} | \cdots | e_n).
$$

We revisit Example (1) from A.1.3.

Example A.3.2

We have :

$$
\begin{align*}
\mathfrak{S}_3 &= \{\mathrm{Id}, (12), (13), (23), (123) = (12)(23), (132) = (13)(32)\} \\
&=: \{\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5, \sigma_6\}
\end{align*}
$$

The six associated permutation matrices are :

$$
\begin{align*}
\mathbf{P}_{\sigma_1} &= \mathbf{P}_{\mathrm{Id}} = \begin{pmatrix} \delta_{1,\sigma_1(1)} & \delta_{1,\sigma_1(2)} & \delta_{1,\sigma_1(3)} \\ \delta_{2,\sigma_1(1)} & \delta_{2,\sigma_1(2)} & \delta_{2,\sigma_1(3)} \\ \delta_{3,\sigma_1(1)} & \delta_{3,\sigma_1(2)} & \delta_{3,\sigma_1(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,1} & \delta_{1,2} & \delta_{1,3} \\ \delta_{2,1} & \delta_{2,2} & \delta_{2,3} \\ \delta_{3,1} & \delta_{3,2} & \delta_{3,3} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \\
\mathbf{P}_{\sigma_2} &= \mathbf{P}_{(1,2)} = \begin{pmatrix} \delta_{1,\sigma_2(1)} & \delta_{1,\sigma_2(2)} & \delta_{1,\sigma_2(3)} \\ \delta_{2,\sigma_2(1)} & \delta_{2,\sigma_2(2)} & \delta_{2,\sigma_2(3)} \\ \delta_{3,\sigma_2(1)} & \delta_{3,\sigma_2(2)} & \delta_{3,\sigma_2(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,2} & \delta_{1,1} & \delta_{1,3} \\ \delta_{2,2} & \delta_{2,1} & \delta_{2,3} \\ \delta_{3,2} & \delta_{3,1} & \delta_{3,3} \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \\
\mathbf{P}_{\sigma_3} &= \mathbf{P}_{(1,3)} = \begin{pmatrix} \delta_{1,\sigma_3(1)} & \delta_{1,\sigma_3(2)} & \delta_{1,\sigma_3(3)} \\ \delta_{2,\sigma_3(1)} & \delta_{2,\sigma_3(2)} & \delta_{2,\sigma_3(3)} \\ \delta_{3,\sigma_3(1)} & \delta_{3,\sigma_3(2)} & \delta_{3,\sigma_3(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,3} & \delta_{1,2} & \delta_{1,1} \\ \delta_{2,3} & \delta_{2,2} & \delta_{2,1} \\ \delta_{3,3} & \delta_{3,2} & \delta_{3,1} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \\
\mathbf{P}_{\sigma_4} &= \mathbf{P}_{(2,3)} = \begin{pmatrix} \delta_{1,\sigma_4(1)} & \delta_{1,\sigma_4(2)} & \delta_{1,\sigma_4(3)} \\ \delta_{2,\sigma_4(1)} & \delta_{2,\sigma_4(2)} & \delta_{2,\sigma_4(3)} \\ \delta_{3,\sigma_4(1)} & \delta_{3,\sigma_4(2)} & \delta_{3,\sigma_4(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,1} & \delta_{1,3} & \delta_{1,2} \\ \delta_{2,1} & \delta_{2,3} & \delta_{2,2} \\ \delta_{3,1} & \delta_{3,3} & \delta_{3,2} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \\
\mathbf{P}_{\sigma_5} &= \mathbf{P}_{(1,2,3)} = \begin{pmatrix} \delta_{1,\sigma_5(1)} & \delta_{1,\sigma_5(2)} & \delta_{1,\sigma_5(3)} \\ \delta_{2,\sigma_5(1)} & \delta_{2,\sigma_5(2)} & \delta_{2,\sigma_5(3)} \\ \delta_{3,\sigma_5(1)} & \delta_{3,\sigma_5(2)} & \delta_{3,\sigma_5(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,2} & \delta_{1,3} & \delta_{1,1} \\ \delta_{2,2} & \delta_{2,3} & \delta_{2,1} \\ \delta_{3,2} & \delta_{3,3} & \delta_{3,1} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}
\end{align*}
$$
(ii) Let $i, j \in \{1, \dots, n\}$ such that $i \neq j$. The permutation matrix $(i, j)$ of size $n$ is the matrix :

$$ \mathbf{P}_{i,j} := \mathbf{P}_{\tau_{i,j}} $$

with $\tau_{i,j}$ the transposition of $i$ and $j$. It is also called an **elementary permutation matrix** or a **transposition matrix**.

Thus we have :

$$ \mathbf{P}_{i,j} = (e_1 | \cdots | e_{i-1} | e_j | e_{i+1} | \cdots | e_{j-1} | e_i | e_{j+1} | \cdots | e_n). $$

We revisit Example (1) from A.1.3.

### Example A.3.2

We have :

$$ \begin{aligned} \mathfrak{S}_3 &= \{\mathrm{Id}, (12), (13), (23), (123) = (12)(23), (132) = (13)(32)\} \\ &=: \{\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5, \sigma_6\} \end{aligned} $$

The six associated permutation matrices are :

$$ \mathbf{P}_{\sigma_1} = \mathbf{P}_{\mathrm{Id}} = \begin{pmatrix} \delta_{1,\sigma_1(1)} & \delta_{1,\sigma_1(2)} & \delta_{1,\sigma_1(3)} \\ \delta_{2,\sigma_1(1)} & \delta_{2,\sigma_1(2)} & \delta_{2,\sigma_1(3)} \\ \delta_{3,\sigma_1(1)} & \delta_{3,\sigma_1(2)} & \delta_{3,\sigma_1(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,1} & \delta_{1,2} & \delta_{1,3} \\ \delta_{2,1} & \delta_{2,2} & \delta_{2,3} \\ \delta_{3,1} & \delta_{3,2} & \delta_{3,3} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$

$$ \mathbf{P}_{\sigma_2} = \mathbf{P}_{(1,2)} = \begin{pmatrix} \delta_{1,\sigma_2(1)} & \delta_{1,\sigma_2(2)} & \delta_{1,\sigma_2(3)} \\ \delta_{2,\sigma_2(1)} & \delta_{2,\sigma_2(2)} & \delta_{2,\sigma_2(3)} \\ \delta_{3,\sigma_2(1)} & \delta_{3,\sigma_2(2)} & \delta_{3,\sigma_2(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,2} & \delta_{1,1} & \delta_{1,3} \\ \delta_{2,2} & \delta_{2,1} & \delta_{2,3} \\ \delta_{3,2} & \delta_{3,1} & \delta_{3,3} \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$

$$ \mathbf{P}_{\sigma_3} = \mathbf{P}_{(1,3)} = \begin{pmatrix} \delta_{1,\sigma_3(1)} & \delta_{1,\sigma_3(2)} & \delta_{1,\sigma_3(3)} \\ \delta_{2,\sigma_3(1)} & \delta_{2,\sigma_3(2)} & \delta_{2,\sigma_3(3)} \\ \delta_{3,\sigma_3(1)} & \delta_{3,\sigma_3(2)} & \delta_{3,\sigma_3(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,3} & \delta_{1,2} & \delta_{1,1} \\ \delta_{2,3} & \delta_{2,2} & \delta_{2,1} \\ \delta_{3,3} & \delta_{3,2} & \delta_{3,1} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} $$

$$ \mathbf{P}_{\sigma_4} = \mathbf{P}_{(2,3)} = \begin{pmatrix} \delta_{1,\sigma_4(1)} & \delta_{1,\sigma_4(2)} & \delta_{1,\sigma_4(3)} \\ \delta_{2,\sigma_4(1)} & \delta_{2,\sigma_4(2)} & \delta_{2,\sigma_4(3)} \\ \delta_{3,\sigma_4(1)} & \delta_{3,\sigma_4(2)} & \delta_{3,\sigma_4(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,1} & \delta_{1,3} & \delta_{1,2} \\ \delta_{2,1} & \delta_{2,3} & \delta_{2,2} \\ \delta_{3,1} & \delta_{3,3} & \delta_{3,2} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} $$

$$ \mathbf{P}_{\sigma_5} = \mathbf{P}_{(1,2,3)} = \begin{pmatrix} \delta_{1,\sigma_1(1)} & \delta_{1,\sigma_1(2)} & \delta_{1,\sigma_1(3)} \\ \delta_{2,\sigma_1(1)} & \delta_{2,\sigma_1(2)} & \delta_{2,\sigma_1(3)} \\ \delta_{3,\sigma_1(1)} & \delta_{3,\sigma_1(2)} & \delta_{3,\sigma_1(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,2} & \delta_{1,3} & \delta_{1,1} \\ \delta_{2,2} & \delta_{2,3} & \delta_{2,1} \\ \delta_{3,2} & \delta_{3,3} & \delta_{3,1} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} $$

$$ \mathbf{P}_{\sigma_6} = \mathbf{P}_{(1,3,2)} = \begin{pmatrix} \delta_{1,\sigma_1(1)} & \delta_{1,\sigma_1(2)} & \delta_{1,\sigma_1(3)} \\ \delta_{2,\sigma_1(1)} & \delta_{2,\sigma_1(2)} & \delta_{2,\sigma_1(3)} \\ \delta_{3,\sigma_1(1)} & \delta_{3,\sigma_1(2)} & \delta_{3,\sigma_1(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,3} & \delta_{1,1} & \delta_{1,2} \\ \delta_{2,3} & \delta_{2,1} & \delta_{2,2} \\ \delta_{3,3} & \delta_{3,1} & \delta_{3,2} \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} $$

Here are the fundamental results :

### Lemma A.3.3

(i) Let $\sigma, \tau \in \mathfrak{S}_n$. We have :

$$ \mathbf{P}_{\sigma \circ \tau} = \mathbf{P}_{\sigma} \mathbf{P}_{\tau}. $$

(ii) Let $\sigma \in \mathfrak{S}_n$. We have $\mathbf{P}_{\sigma} \in \mathrm{GL}_n(\mathbb{C})$ and :

$$ \mathbf{P}_{\sigma}^{-1} = \mathbf{P}_{\sigma^{-1}} = \mathbf{P}_{\sigma}^{T}. $$
$$ \mathbf{P}_{\sigma_6} = \mathbf{P}_{(1,3,2)} = \begin{pmatrix} \delta_{1,\sigma_1(1)} & \delta_{1,\sigma_1(2)} & \delta_{1,\sigma_1(3)} \\ \delta_{2,\sigma_1(1)} & \delta_{2,\sigma_1(2)} & \delta_{2,\sigma_1(3)} \\ \delta_{3,\sigma_1(1)} & \delta_{3,\sigma_1(2)} & \delta_{3,\sigma_1(3)} \end{pmatrix} = \begin{pmatrix} \delta_{1,3} & \delta_{1,1} & \delta_{1,2} \\ \delta_{2,3} & \delta_{2,1} & \delta_{2,2} \\ \delta_{3,3} & \delta_{3,1} & \delta_{3,2} \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} $$

Here are the fundamental results :

### Lemma A.3.3

(i) Let $\sigma, \tau \in \mathfrak{S}_n$. We have :

$$ \mathbf{P}_{\sigma \circ \tau} = \mathbf{P}_{\sigma} \mathbf{P}_{\tau}. $$

(ii) Let $\sigma \in \mathfrak{S}_n$. We have $\mathbf{P}_{\sigma} \in \mathrm{GL}_n(\mathbb{C})$ and :

$$ \mathbf{P}_{\sigma}^{-1} = \mathbf{P}_{\sigma^{-1}} = \mathbf{P}_{\sigma}^{T}. $$

(iii) Every permutation matrix is the product of elementary permutation matrices.

(iv) Let $\sigma \in \mathfrak{S}_n$. We have :

$$ \det(\mathbf{P}_{\sigma}) = \varepsilon(\sigma). $$

### Proof.

(i) Since $\tau$ is a bijection, for all $p, q \in \{1, \dots, n\}$, we have :

$$ p = \tau(q) \iff q = \tau^{-1}(p). $$

Thus for all $k, l \in \{1, \dots, n\}$, we have :

$$ \begin{aligned} [\mathbf{P}_{\sigma}\mathbf{P}_{\tau}]_{kl} &= \sum_{m=1}^{n} [\mathbf{P}_{\sigma}]_{km} [\mathbf{P}_{\tau}]_{ml} \\ &= \sum_{m=1}^{n} \delta_{k,\sigma(m)} \delta_{m,\tau(l)} \\ &= \delta_{k,\sigma(\tau(l))} \\ &= [\mathbf{P}_{\sigma\circ\tau}]_{kl} \end{aligned} $$

(ii) For all $k, l \in \{1, \dots, n\}$, we have :

$$ \begin{aligned} [\mathbf{P}_{\sigma}\mathbf{P}_{\sigma}^{T}]_{kl} &= \sum_{m=1}^{n} [\mathbf{P}_{\sigma}]_{km} [\mathbf{P}_{\sigma}^{T}]_{ml} \\ &= \sum_{m=1}^{n} [\mathbf{P}_{\sigma}]_{km} [\mathbf{P}_{\sigma}]_{lm} \\ &= \sum_{m=1}^{n} \delta_{k,\sigma(m)} \delta_{l,\sigma(m)} \\ &= \delta_{k,l} \\ &= [I_{n}]_{kl} \end{aligned} $$

and by bijection of $\sigma$, we have :

$$ \begin{aligned} [\mathbf{P}_{\sigma}\mathbf{P}_{\sigma^{-1}}]_{kl} &= \sum_{m=1}^{n} [\mathbf{P}_{\sigma}]_{km} [\mathbf{P}_{\sigma^{-1}}]_{ml} \\ &= \sum_{m=1}^{n} \delta_{k,\sigma(m)} \delta_{m} \end{aligned} $$
(iii) Every permutation matrix is the product of elementary permutation matrices.

(iv) Let $\sigma \in \mathfrak{S}_n$. We have :

$$ \det(\mathbf{P}_{\sigma}) = \varepsilon(\sigma). $$

### Proof.

(i) Since $\tau$ is a bijection, for all $p, q \in \{1, \dots, n\}$, we have :

$$ p = \tau(q) \iff q = \tau^{-1}(p). $$

Thus for all $k, l \in \{1, \dots, n\}$, we have :

$$ \begin{aligned} [\mathbf{P}_{\sigma}\mathbf{P}_{\tau}]_{kl} &= \sum_{m=1}^{n} [\mathbf{P}_{\sigma}]_{km} [\mathbf{P}_{\tau}]_{ml} \\ &= \sum_{m=1}^{n} \delta_{k,\sigma(m)} \delta_{m,\tau(l)} \\ &= \delta_{k,\sigma(\tau(l))} \\ &= [\mathbf{P}_{\sigma\circ\tau}]_{kl} \end{aligned} $$

(ii) For all $k, l \in \{1, \dots, n\}$, we have :

$$ \begin{aligned} [\mathbf{P}_{\sigma}\mathbf{P}_{\sigma}^{T}]_{kl} &= \sum_{m=1}^{n} [\mathbf{P}_{\sigma}]_{km} [\mathbf{P}_{\sigma}^{T}]_{ml} \\ &= \sum_{m=1}^{n} [\mathbf{P}_{\sigma}]_{km} [\mathbf{P}_{\sigma}]_{lm} \\ &= \sum_{m=1}^{n} \delta_{k,\sigma(m)} \delta_{l,\sigma(m)} \\ &= \delta_{k,l} \\ &= [I_{n}]_{kl} \end{aligned} $$

and by bijection of $\sigma$, we have :

$$ \begin{aligned} [\mathbf{P}_{\sigma}\mathbf{P}_{\sigma^{-1}}]_{kl} &= \sum_{m=1}^{n} [\mathbf{P}_{\sigma}]_{km} [\mathbf{P}_{\sigma^{-1}}]_{ml} \\ &= \sum_{m=1}^{n} \delta_{k,\sigma(m)} \delta_{m,\sigma} \\ &= \sum_{m=1}^{n} \delta_{k,\sigma(\sigma^{-1}(l))} \\ &= \delta_{k,l} \\ &= [I_n]_{kl} \end{aligned} $$

Thus we have $\mathbf{P}_{\sigma} \in \mathrm{GL}_n(\mathbb{C})$ and :

$$ \mathbf{P}_{\sigma}^{-1} = \mathbf{P}_{\sigma^{-1}} = \mathbf{P}_{\sigma}^{T}. $$

(iii) Let $\sigma \in \mathfrak{S}_n$ be a permutation. Then by proposition A.1.2, $\sigma$ is the product of transpositions $\tau_1, \dots, \tau_k$ :

$$ \sigma = \tau_1 \circ \dots \circ \tau_k. $$
$$
\begin{align*}
&= \sum_{m=1}^{n} \delta_{k,\sigma(\sigma^{-1}(l))} \\
&= \delta_{k,l} \\
&= [I_n]_{kl}
\end{align*}
$$

Thus we have $\mathbf{P}_{\sigma} \in \text{GL}_n(\mathbb{C})$ and :

$$
\mathbf{P}_{\sigma}^{-1} = \mathbf{P}_{\sigma^{-1}} = \mathbf{P}_{\sigma}^{T}.
$$

(iii) Let $\sigma \in \mathfrak{S}_n$ be a permutation. Then by proposition A.1.2, $\sigma$ is the product of transpositions $\tau_1, \dots, \tau_k$ :

$$
\sigma = \tau_1 \circ \dots \circ \tau_k.
$$

Thus by point (i) :

$$
\begin{align*}
\mathbf{P}_{\sigma} &= \mathbf{P}_{\tau_1 \circ \dots \circ \tau_k} \\
&= \mathbf{P}_{\tau_1} \cdots \mathbf{P}_{\tau_k}
\end{align*}
$$

Hence the result.

(iv) Let $\tau_{i,j}$ be a transposition with $i,j \in \{1, \dots, n\}$ and $i < j$. Let's show that :

$$
\det(\mathbf{P}_{i,j}) = -1 = \varepsilon(\tau_{i,j}).
$$

By the antisymmetry of the determinant :

$$
\begin{align*}
\det(\mathbf{P}_{i,j}) &= \det(\mathfrak{C}_1(\mathbf{P}_{i,j}), \dots, \mathfrak{C}_1(\mathbf{P}_{i,j})) \\
&= \det(e_1, \dots, e_{i-1}, e_j, e_{i+1}, \dots, e_{j-1}, e_i, e_{j+1}, \dots, e_n) \\
&= -\det(e_1, \dots, e_n) \\
&= -1 \\
&= \varepsilon(\tau_{i,j})
\end{align*}
$$

We reuse the notations from point (iii). Let a permutation decomposed into a product of transpositions be :

$$
\sigma = \tau_1 \circ \dots \circ \tau_k.
$$

Thus we have :

$$
\begin{align*}
\det(\mathbf{P}_{\sigma}) &= \det(\mathbf{P}_{\tau_1} \cdots \mathbf{P}_{\tau_k}) \\
&= \prod_{i=1}^{k} \det(\mathbf{P}_{\tau_i}) \\
&= (-1)^k \\
&= \varepsilon(\sigma)
\end{align*}
$$

<figure><img src="image_6.png" /></figure>

Every cycle $c := (i_1, \dots, i_l)$ decomposes into the product of $l-1$ transpositions :

$$
c = (i_1, i_2) \circ \dots \circ (i_{l-1}, i_l)
$$

thus we have :

$$
\varepsilon(c) = (-1)^l = \det(\mathbf{P}_c). \tag{A.3.1}
$$

A.4 Examples
Thus by point (i) :

$$
\begin{align*}
\mathbf{P}_{\sigma} &= \mathbf{P}_{\tau_1 \circ \cdots \circ \tau_k} \\
&= \mathbf{P}_{\tau_1} \cdots \mathbf{P}_{\tau_k}
\end{align*}
$$

Hence the result.

(iv) Let $\tau_{i,j}$ be a transposition with $i, j \in \{1, \dots, n\}$ and $i < j$. Let's show that :

$$
\det(\mathbf{P}_{i,j}) = -1 = \varepsilon(\tau_{i,j}).
$$

By the antisymmetry of the determinant :

$$
\begin{align*}
\det(\mathbf{P}_{i,j}) &= \det(\mathfrak{C}_1(\mathbf{P}_{i,j}), \dots, \mathfrak{C}_1(\mathbf{P}_{i,j})) \\
&= \det(e_1, \dots, e_{i-1}, e_j, e_{i+1}, \dots, e_{j-1}, e_i, e_{j+1}, \dots, e_n) \\
&= -\det(e_1, \dots, e_n) \\
&= -1 \\
&= \varepsilon(\tau_{i,j})
\end{align*}
$$

We reuse the notations from point (iii). Let a permutation decomposed into a product of transpositions be :

$\sigma = \tau_1 \circ \cdots \circ \tau_k.$

Thus we have :

$$
\begin{align*}
\det(\mathbf{P}_{\sigma}) &= \det(\mathbf{P}_{\tau_1} \cdots \mathbf{P}_{\tau_k}) \\
&= \prod_{i=1}^{k} \det(\mathbf{P}_{\tau_i}) \\
&= (-1)^k \\
&= \varepsilon(\sigma)
\end{align*}
$$

<figure><img src="image_7.png" /></figure>

Every cycle $c := (i_1, \dots, i_l)$ decomposes into the product of $l-1$ transpositions :

$$
c = (i_1, i_2) \circ \dots \circ (i_{l-1}, i_l)
$$

thus we have :

$$
\varepsilon(c) = (-1)^l = \det(\mathbf{P}_c). \tag{A.3.1}
$$

**A.4 Examples**

**Proposition A.4.1**

Permuting the _rows_ (resp. _columns_) of a matrix is equivalent to multiplying the matrix on the _left_ (resp. on the _right_) by a permutation matrix.

Let's illustrate this with an example. Consider the matrix :

$$
M := \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}.
$$
Proposition A.4.1

Permuting the _rows_ (resp. _columns_) of a matrix is equivalent to multiplying the matrix on the _left_ (resp. on the _right_) by a permutation matrix.

Let's illustrate this with an example. Consider the matrix :

$$
M := \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}.
$$

(1)

$[L_2 \leftrightarrow L_3]$

$$
\mathbf{P}_{2,3} M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 7 & 8 & 9 \\ 4 & 5 & 6 \end{pmatrix}
$$

(2)

$[L_1 \leftrightarrow L_3]$

$$
\mathbf{P}_{1,3} M = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix} 7 & 8 & 9 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{pmatrix}
$$

(3)

$[C_2 \leftrightarrow C_3]$

$$
M \mathbf{P}_{2,3} = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 3 & 2 \\ 4 & 6 & 5 \\ 7 & 9 & 8 \end{pmatrix}
$$

(4)

$[C_1 \leftrightarrow C_3]$

$$
M \mathbf{P}_{1,3} = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & 2 & 1 \\ 6 & 5 & 4 \\ 9 & 8 & 7 \end{pmatrix}
$$
(1)

$[L_2 \leftrightarrow L_3]$

$$
\mathbf{P}_{2,3} M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 7 & 8 & 9 \\ 4 & 5 & 6 \end{pmatrix}
$$

(2)

$[L_1 \leftrightarrow L_3]$

$$
\mathbf{P}_{1,3} M = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix} 7 & 8 & 9 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{pmatrix}
$$

(3)

$[C_2 \leftrightarrow C_3]$

$$
M \mathbf{P}_{2,3} = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 3 & 2 \\ 4 & 6 & 5 \\ 7 & 9 & 8 \end{pmatrix}
$$

(4)

$[C_1 \leftrightarrow C_3]$

$$
M \mathbf{P}_{1,3} = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & 2 & 1 \\ 6 & 5 & 4 \\ 9 & 8 & 7 \end{pmatrix}
$$
# B The Transpose and the Cotranspose

The transpose of a matrix is a symmetry with respect to the main diagonal. The cotranspose is defined as the symmetry with respect to the other diagonal.

## B.1 Notations

### Definition B.1.1

Let $M \in \mathcal{M}_{n,p}(\mathbb{C})$ and $N \in \mathcal{M}_n(\mathbb{C})$.

(i) (a) The **transpose** of $M$ is the matrix $M^T \in \mathcal{M}_{p,n}(\mathbb{C})$ defined for all $i \in \{1, \dots, n\}$ and all $j \in \{1, \dots, p\}$ by :

$$[M^T]_{ij} := [M]_{ji}.$$

(b) We say $N$ is **symmetric** if :

$$N^T = N.$$

We denote their set by $\mathcal{S}_n(\mathbb{C})$.

(c) We say $N$ is **antisymmetric** if :

$$N^T = -N.$$

We denote their set by $\mathcal{A}_n(\mathbb{C})$.

(ii) (a) The **cotranspose** of $M$ is the matrix $M^C \in \mathcal{M}_{p,n}(\mathbb{C})$ defined for all $i \in \{1, \dots, n\}$ and all $j \in \{1, \dots, p\}$ by :

$$[M^C]_{ij} := [M]_{p+1-j,n+1-i}.$$

(b) We say $N$ is **cosymmetric** if :

$$N^C = N.$$

We denote their set by $\mathcal{S}_n^C(\mathbb{C})$.

(c) We say $N$ is **coantisymmetric** if :

$$N^C = -N.$$

We denote their set by $\mathcal{A}_n^C(\mathbb{C})$.

The transpose and cotranspose are linear mappings from $\mathcal{M}_{np}(\mathbb{C})$ to $\mathcal{M}_{pn}(\mathbb{C})$ because for all $M, N \in \mathcal{M}_{n,p}(\mathbb{C})$, all $\alpha, \beta \in \mathbb{C}$, all $i \in \{1, \dots, n\}$, and all $j \in \{1, \dots, p\}$, we have :

$$[(\alpha M + \beta N)^T]_{i,j} = [\alpha M + \beta N]_{j,i}$$

$$= \alpha [M]_{j,i} + \beta [N]_{j,i}$$

$$= \alpha [M^T]_{i,j} + \beta [N^T]_{i,j}$$

$$= [\alpha M^T + \beta N^T]_{i,j}$$

$$[(\alpha M + \beta N)^C]_{i,j} = [\alpha M + \beta N]_{p+1-j,n+1-i}$$

$$= \alpha [M]_{p+1-j,n+1-i} + \beta [N]_{p+1-j,n+1-i}$$

$$= \alpha [M^C]_{i,j} + \beta [N^C]_{i,j}$$

$$= [\alpha M^C + \beta N^C]_{i,j}$$

### Example B.1.2

We have :

$$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}^T = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix}$$
# B The Transpose and the Cotranspose

The transpose of a matrix is a symmetry with respect to the main diagonal. The cotranspose is defined as the symmetry with respect to the other diagonal.

## B.1 Notations

### Definition B.1.1

Let $M \in \mathcal{M}_{n,p}(\mathbb{C})$ and $N \in \mathcal{M}_n(\mathbb{C})$.

(i) (a) The **transpose** of $M$ is the matrix $M^T \in \mathcal{M}_{p,n}(\mathbb{C})$ defined for all $i \in \{1, \dots, n\}$ and all $j \in \{1, \dots, p\}$ by :

$$[M^T]_{ij} := [M]_{ji}.$$

(b) We say $N$ is **symmetric** if :

$$N^T = N.$$

We denote their set by $\mathcal{S}_n(\mathbb{C})$.

(c) We say $N$ is **antisymmetric** if :

$$N^T = -N.$$

We denote their set by $\mathcal{A}_n(\mathbb{C})$.

(ii) (a) The **cotranspose** of $M$ is the matrix $M^C \in \mathcal{M}_{p,n}(\mathbb{C})$ defined for all $i \in \{1, \dots, n\}$ and all $j \in \{1, \dots, p\}$ by :

$$[M^C]_{ij} := [M]_{p+1-j,n+1-i}.$$

(b) We say $N$ is **cosymmetric** if :

$$N^C = N.$$

We denote their set by $\mathcal{S}_n^C(\mathbb{C})$.

(c) We say $N$ is **coantisymmetric** if :

$$N^C = -N.$$

We denote their set by $\mathcal{A}_n^C(\mathbb{C})$.

The transpose and cotranspose are linear mappings from $\mathcal{M}_{np}(\mathbb{C})$ to $\mathcal{M}_{pn}(\mathbb{C})$ because for all $M, N \in \mathcal{M}_{n,p}(\mathbb{C})$, all $\alpha, \beta \in \mathbb{C}$, all $i \in \{1, \dots, n\}$, and all $j \in \{1, \dots, p\}$, we have :

$$[(\alpha M + \beta N)^T]_{i,j} = [\alpha M + \beta N]_{j,i}$$

$$= \alpha [M]_{j,i} + \beta [N]_{j,i}$$

$$= \alpha [M^T]_{i,j} + \beta [N^T]_{i,j}$$

$$= [\alpha M^T + \beta N^T]_{i,j}$$

$$[(\alpha M + \beta N)^C]_{i,j} = [\alpha M + \beta N]_{p+1-j,n+1-i}$$

$$= \alpha [M]_{p+1-j,n+1-i} + \beta [N]_{p+1-j,n+1-i}$$

$$= \alpha [M^C]_{i,j} + \beta [N^C]_{i,j}$$

$$= [\alpha M^C + \beta N^C]_{i,j}$$

### Example B.1.2

We have :

$$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}^T = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix}$$
$$ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}^C = \begin{pmatrix} 6 & 3 \\ 5 & 2 \\ 4 & 1 \end{pmatrix} \\ \left( \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \right)^T = \left( \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \right)^C = \begin{pmatrix} 6 & 5 & 4 \\ 3 & 2 & 1 \end{pmatrix}. $$

The two transposes are connected using the co-identity matrix.

### Definition B.1.3

Let $n \in \mathbb{N}_*$. The co-identity matrix of size $n$ is defined by :

$$ \mathbf{C}_n := (\delta_{i,n+1-j})_{i,j} $$

### Example B.1.4

We have :

$$ \mathbf{C}_1 = (1), \quad \mathbf{C}_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \mathbf{C}_3 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} $$

We then have the following simple properties.

### Lemma B.1.5

Let $n \in \mathbb{N}_*$.

(i) We have $\mathbf{C}_n^T = \mathbf{C}_n$, $\mathbf{C}_n^C = \mathbf{C}_n$ and $\mathbf{C}_n^2 = \mathbf{I}_n$ (hence $\mathbf{C}_n^{-1} = \mathbf{C}_n$).

(ii) We have :

$$ \det(\mathbf{C}_n) = (-1)^{|n/2|}. $$

(iii) Let $M \in \mathcal{M}_{n,p}(\mathbb{C})$. We have :

$$ M^C = \mathbf{C}_n M^T \mathbf{C}_p. $$

### Proof.

(i) We have for all $i, j \in \{1, \dots, n\}$ :

$$ \begin{align*} [\mathbf{C}_n^T]_{i,j} &= [\mathbf{C}_n]_{j,i} & [\mathbf{C}_n^C]_{i,j} &= [\mathbf{C}_n]_{n+1-j,n+1-i} & [\mathbf{C}_n^2]_{i,j} &= \sum_{k=1}^{n} [\mathbf{C}_n]_{i,k} [\mathbf{C}_n^2]_{k,j} \\ &= \delta_{j,n+1-i} & &= \delta_{n+1-j,n+1-(n+1-i)} & &= \sum_{k=1}^{n} \delta_{i,n+1-k} \delta_{k,n+1-j} \\ &= \delta_{i,n+1-j} & &= \delta_{n+1-j,i} & &= \delta_{n+1-i,n+1-j} \\ &= [\mathbf{C}_n]_{i,j} & &= [\mathbf{C}_n]_{i,j} & &= [\mathbf{C}_n]_{i,j} \end{align*} $$

(ii) We have :

$$ \det(\mathbf{C}_n) = \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [\mathbf{C}_n]_{i,\sigma(i)} $$
$$ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}^C = \begin{pmatrix} 6 & 3 \\ 5 & 2 \\ 4 & 1 \end{pmatrix} \\ \left( \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \right)^T = \left( \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \right)^C = \begin{pmatrix} 6 & 5 & 4 \\ 3 & 2 & 1 \end{pmatrix}. $$

The two transposes are connected using the co-identity matrix.

### Definition B.1.3

Let $n \in \mathbb{N}_*$. The co-identity matrix of size $n$ is defined by :

$$ \mathbf{C}_n := (\delta_{i,n+1-j})_{i,j} $$

### Example B.1.4

We have :

$$ \mathbf{C}_1 = (1), \quad \mathbf{C}_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \mathbf{C}_3 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} $$

We then have the following simple properties.

### Lemma B.1.5

Let $n \in \mathbb{N}_*$.

(i) We have $\mathbf{C}_n^T = \mathbf{C}_n$, $\mathbf{C}_n^C = \mathbf{C}_n$ and $\mathbf{C}_n^2 = \mathbf{I}_n$ (hence $\mathbf{C}_n^{-1} = \mathbf{C}_n$).

(ii) We have :

$$ \det(\mathbf{C}_n) = (-1)^{\lfloor n/2 \rfloor}. $$

(iii) Let $M \in \mathcal{M}_{n,p}(\mathbb{C})$. We have :

$$ M^C = \mathbf{C}_n M^T \mathbf{C}_p. $$

### Proof.

(i) We have for all $i,j \in \{1, \dots, n\}$ :

$$ \begin{align*} [\mathbf{C}_n^T]_{i,j} &= [\mathbf{C}_n]_{j,i} & [\mathbf{C}_n^C]_{i,j} &= [\mathbf{C}_n]_{n+1-j,n+1-i} & [\mathbf{C}_n^2]_{i,j} &= \sum_{k=1}^{n} [\mathbf{C}_n]_{i,k} [\mathbf{C}_n^2]_{k,j} \\ &= \delta_{j,n+1-i} & &= \delta_{n+1-j,n+1-(n+1-i)} & &= \sum_{k=1}^{n} \delta_{i,n+1-k} \delta_{k,n+1-j} \\ &= \delta_{i,n+1-j} & &= \delta_{n+1-j,i} & &= \delta_{n+1-i,n+1-j} \\ &= [\mathbf{C}_n]_{i,j} & &= [\mathbf{C}_n]_{i,j} & &= [\mathbf{C}_n]_{i,j} \end{align*} $$

(ii) We have :

$$ \det(\mathbf{C}_n) = \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [\mathbf{C}_n]_{i,\sigma(i)} $$
$$
\begin{align*}
&= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} \delta_{i,n+1-\sigma(i)} \\
&= \varepsilon \left( \begin{pmatrix} 1 & 2 & \cdots & n \\ n & n-1 & \cdots & 1 \end{pmatrix} \right) \\
&= \prod_{i=1}^{[n/2]} \varepsilon((i, n+1-i)) \\
&= (-1)^{\lceil n/2 \rceil}
\end{align*}
$$

(iii) We have for all $i \in \{1, \dots, n\}$ and $j \in \{1, \dots, q\}$ :

$$
\begin{align*}
[\mathbf{C}_n M^T \mathbf{C}_p]_{i,j} &= \sum_{k=1}^{n} \sum_{l=1}^{p} [\mathbf{C}_n]_{i,k} [M^T]_{k,l} [\mathbf{C}_p]_{l,j} \\
&= \sum_{k=1}^{n} \sum_{l=1}^{p} \delta_{i,n+1-k} [M]_{l,k} \delta_{l,n+1-j} \\
&= [M]_{n+1-j,n+1-i} \\
&= [M^C]_{i,j}
\end{align*}
$$

B.2 Properties

We have the following simple properties.

Proposition B.2.1

Let $M \in \mathcal{M}_{n,p}(\mathbb{C})$, $N \in \mathcal{M}_{p,q}(\mathbb{C})$, and $A \in \mathcal{M}_n(\mathbb{C})$.

(i) We have :

$$
(MN)^T = N^T M^T , \quad (MN)^C = N^C M^C.
$$

(ii) We have

$$
(M^T)^C = (M^C)^T.
$$

(iii) Suppose that $A \in \text{GL}_n(\mathbb{C})$. Then $A^T, A^C \in \text{GL}_n(\mathbb{C})$ and :

$$
(A^T)^{-1} = (A^{-1})^T , \quad (A^C)^{-1} = (A^{-1})^C .
$$

(iv) We have :

$$
\det(A^T) = \det(A^C) = \det(A).
$$

Proof.

(i) For all $i \in \{1, \dots, n\}$ and $j \in \{1, \dots, q\}$, we have :

$$
\begin{align*}
\left[ (MN)^T \right]_{i,j} &= [MN]_{j,i} \\
&= \sum_{k=1}^{p} [M]_{j,k} [N]_{k,i} \\
&= \sum_{k=1}^{p} [N^T]_{i,k} [M^T]_{k,j}
\end{align*}
\begin{align*}
\left[ (MN)^C \right]_{i,j} &= [MN]_{q+1-j,n+1-i} \\
&= \sum_{k=1}^{p} [M]_{q+1-j,k} [N]_{k,n+1-i} \\
&= \sum_{k=1}^{p} [M]_{q+1-j,p+1-k} [N]_{p+1-k,n+1-i}
\end{align*}
$$
$$
\begin{align*}
&= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} \delta_{i,n+1-\sigma(i)} \\
&= \varepsilon \left( \begin{pmatrix} 1 & 2 & \cdots & n \\ n & n-1 & \cdots & 1 \end{pmatrix} \right) \\
&= \prod_{i=1}^{[n/2]} \varepsilon((i, n+1-i)) \\
&= (-1)^{\lceil n/2 \rceil}
\end{align*}
$$

(iii) We have for all $i \in \{1, \dots, n\}$ and $j \in \{1, \dots, q\}$ :

$$
\begin{align*}
[\mathbf{C}_n M^T \mathbf{C}_p]_{i,j} &= \sum_{k=1}^{n} \sum_{l=1}^{p} [\mathbf{C}_n]_{i,k} [M^T]_{k,l} [\mathbf{C}_p]_{l,j} \\
&= \sum_{k=1}^{n} \sum_{l=1}^{p} \delta_{i,n+1-k} [M]_{l,k} \delta_{l,n+1-j} \\
&= [M]_{n+1-j,n+1-i} \\
&= [M^C]_{i,j}
\end{align*}
$$

<figure><img src="image_8.png" /></figure>

B.2 Properties

We have the following simple properties.

Proposition B.2.1

Let $M \in \mathcal{M}_{n,p}(\mathbb{C})$, $N \in \mathcal{M}_{p,q}(\mathbb{C})$, and $A \in \mathcal{M}_n(\mathbb{C})$.

(i) We have :

$$
(MN)^T = N^T M^T , \quad (MN)^C = N^C M^C .
$$

(ii) We have

$$
(M^T)^C = (M^C)^T .
$$

(iii) Suppose that $A \in \text{GL}_n(\mathbb{C})$. Then $A^T, A^C \in \text{GL}_n(\mathbb{C})$ and :

$$
(A^T)^{-1} = (A^{-1})^T , \quad (A^C)^{-1} = (A^{-1})^C .
$$

(iv) We have :

$$
\det(A^T) = \det(A^C) = \det(A).
$$

Proof.

(i) For all $i \in \{1, \dots, n\}$ and $j \in \{1, \dots, q\}$, we have :

$$
\begin{align*}
\left[ (MN)^T \right]_{i,j} &= [MN]_{j,i} \\
&= \sum_{k=1}^{p} [M]_{j,k} [N]_{k,i} \\
&= \sum_{k=1}^{p} [N^T]_{i,k} [M^T]_{k,j}
\end{align*}
\[
\left[ (MN)^C \right]_{i,j} &= [MN]_{q+1-j,n+1-i} \\
&= \sum_{k=1}^{p} [M]_{q+1-j,k} [N]_{k,n+1-i} \\
&= \sum_{k=1}^{p} [M]_{q+1-j,p+1-k} [N]_{p+1-k,n+1-i}
\end{align*}
$$
$$ = [N^T M^T]_{i,j} $$ $$ = \sum_{k=1}^{p} [N^C]_{i,k} [M^C]_{k,j} $$ $$ = [N^C M^C]_{i,j} $$

(ii) For all $i \in \{1, \dots, n\}$ and $j \in \{1, \dots, q\}$, we have :

$$ \begin{aligned} [(M^T)^C]_{i,j} &= [M^T]_{p+1-j, n+1-i} \\ &= [M]_{n+1-i, p+1-j} \\ &= [M^C]_{j,i} \\ &= [(M^C)^T]_{i,j} \end{aligned} $$

(iii) We have :

$$ \begin{aligned} I_n &= I_n^T & I_n &= I_n^C \\ &= (A^{-1}A)^T & &= (A^{-1}A)^C \\ &= A^C(A^{-1})^T & &= A^C(A^{-1})^C \end{aligned} $$

Thus $A^T, A^C \in \text{GL}_n(\mathbb{C})$ and :

$$ (A^T)^{-1} = (A^{-1})^T, \quad (A^C)^{-1} = (A^{-1})^C. $$

(iv) Let $A := (a_{ij})_{ij}$. The map :

$$ \sigma \in \mathfrak{S}_n \mapsto \sigma^{-1} \in \mathfrak{S}_n $$

is a bijection (it is involutive), so we have :

$$ \begin{aligned} \det(A^T) &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [A^T]_{i, \sigma(i)} \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [A]_{\sigma(i), i} \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma^{-1}) \prod_{i=1}^{n} [A]_{\sigma^{-1}(i), i} \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [A]_{\sigma^{-1}(i), i} \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [A]_{i, \sigma(i)} \\ &= \det(A) \end{aligned} $$

Therefore, by Lemma B.1.5 :

$$ \begin{aligned} \det(A^C) &= \det(C_n A^T C_n) \\ &= \det(A^T) \det(C_n)^2 \\ &= \det(A^T) \\ &= \det(A^T) \end{aligned} $$
$$ = [N^T M^T]_{i,j} $$ 
$$ = \sum_{k=1}^{p} [N^C]_{i,k} [M^C]_{k,j} $$ 
$$ = [N^C M^C]_{i,j} $$

(ii) For all $i \in \{1, \dots, n\}$ and $j \in \{1, \dots, q\}$, we have :

$$ \begin{aligned} [ (M^T)^C ]_{i,j} &= [M^T]_{p+1-j, n+1-i} \\ &= [M]_{n+1-i, p+1-j} \\ &= [M^C]_{j,i} \\ &= [(M^C)^T]_{i,j} \end{aligned} $$

(iii) We have :

$$ \begin{aligned} I_n &= I_n^T & I_n &= I_n^C \\ &= (A^{-1}A)^T & &= (A^{-1}A)^C \\ &= A^C(A^{-1})^T & &= A^C(A^{-1})^C \end{aligned} $$

Thus $A^T, A^C \in \text{GL}_n(\mathbb{C})$ and :

$$ (A^T)^{-1} = (A^{-1})^T, \quad (A^C)^{-1} = (A^{-1})^C. $$

(iv) Let $A := (a_{ij})_{ij}$. The map :

$$ \sigma \in \mathfrak{S}_n \mapsto \sigma^{-1} \in \mathfrak{S}_n $$

is a bijection (it is involutive), so we have :

$$ \begin{aligned} \det(A^T) &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [A^T]_{i, \sigma(i)} \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [A]_{\sigma(i), i} \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma^{-1}) \prod_{i=1}^{n} [A]_{\sigma^{-1}(i), i} \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [A]_{\sigma^{-1}(i), i} \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} [A]_{i, \sigma(i)} \\ &= \det(A) \end{aligned} $$

Therefore, by Lemma B.1.5 :

$$ \begin{aligned} \det(A^C) &= \det(C_n A^T C_n) \\ &= \det(A^T) \det(C_n)^2 \\ &= \det(A^T) \\ &= \det(A^T) \end{aligned} $$
□

We then note :

$$M^{CT} := (M^T)^C = (M^C)^T. \qquad (B.2.1)$$

Furthermore, we have :

$$\begin{aligned} \mathcal{M}_n(\mathbb{C}) &= \mathcal{S}_n(\mathbb{C}) \oplus \mathcal{A}_n(\mathbb{C}) \\ &= \mathcal{S}_n^c(\mathbb{C}) \oplus \mathcal{A}_n^c(\mathbb{C}). \end{aligned}$$

The decompositions are given respectively for all $M \in \mathcal{M}_n(\mathbb{C})$ by :

$$\begin{aligned} M &= \frac{M + M^T}{2} + \frac{M - M^T}{2} \\ &= \frac{M + M^C}{2} + \frac{M - M^C}{2}. \end{aligned}$$
□

We then note :

$$ M^{CT} := (M^T)^C = (M^C)^T. \qquad (B.2.1) $$

Furthermore, we have :

$$ \begin{aligned} \mathcal{M}_n(\mathbb{C}) &= \mathcal{S}_n(\mathbb{C}) \oplus \mathcal{A}_n(\mathbb{C}) \\ &= \mathcal{S}_n^c(\mathbb{C}) \oplus \mathcal{A}_n^c(\mathbb{C}) \\ &= (\mathcal{S}_n(\mathbb{C}) \cap \mathcal{S}_n^c(\mathbb{C})) \oplus (\mathcal{S}_n(\mathbb{C}) \cap \mathcal{A}_n^c(\mathbb{C})) \oplus (\mathcal{A}_n(\mathbb{C}) \cap \mathcal{S}_n^c(\mathbb{C})) \oplus (\mathcal{A}_n(\mathbb{C}) \cap \mathcal{A}_n^c(\mathbb{C})). \end{aligned} $$

The decompositions are given respectively for all $M \in \mathcal{M}_n(\mathbb{C})$ by :

$$ \begin{aligned} M &= \frac{M + M^T}{2} + \frac{M - M^T}{2} \\ &= \frac{M + M^C}{2} + \frac{M - M^C}{2} \\ &= \frac{M + M^T + M^C + M^{CT}}{4} + \frac{M + M^T - M^C - M^{CT}}{4} + \frac{M - M^T + M^C - M^{CT}}{4} + \frac{M - M^T - M^C + M^{CT}}{4}. \end{aligned} $$

Taking the transpose, we have that the determinant of a matrix is the determinant of its rows, i.e., for :

$$ M := (m_{i,j})_{ij} := \begin{pmatrix} \mathcal{L}_1(M) \\ \vdots \\ \mathcal{L}_n(M) \end{pmatrix} \in \mathcal{M}_n(\mathbb{C}). $$

we have :

$$ \begin{aligned} \det(M) &= \det(\mathcal{L}_1(M)^T, \dots, \mathcal{L}_n(M)^T) \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^n [M^T]_{i, \sigma(i)} \\ &= \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^n m_{\sigma(i), i}. \end{aligned} $$
Références

[1] Viktor Vasil evich Prasolov, *Problems and Theorems in Linear Algebra*, American Mathematical Soc., (1994).

[2] Wikipedia (english), *Minor (linear algebra) # Inverse of a matrix*.
Références

[1] Viktor Vasil evich Prasolov, *Problems and Theorems in Linear Algebra*, American Mathematical Soc., (1994).

[2] Wikipedia (english), *Minor (linear algebra) # Inverse of a matrix*.