The Canonical Forms of Permutation Matrices

Abstract

We address classification of permutation matrices, in terms of permutation similarity relations, which play an important role in investigating the reducible solutions of some symmetric matrix equations. We solve the three problems. First, what is the canonical form of a permutation similarity class? Second, how to obtain the standard form of arbitrary permutation matrix? Third, for any permutation matrix A, how to find the permutation matrix T, such that $T^{-1}AT$ is in canonical form? Besides, the decomposition theorem of permutation matrices and the factorization theorem of both permutation matrices and monomial matrices are demonstrated.

1. Introduction

The incidence matrix of a projective plane of order n is a 0-1 matrix of order $n^2+n+1$. Two projective planes are isomorphic if the incidence matrix of one projective plane can be transformed into the incidence matrix of the other one by permutation of rows and/or columns. After sorting the rows and columns, the incidence matrix of a projective plane can be reduced to (not unique) a standard form. In the reduced form, the incidence matrix can be split into blocks. Most blocks are permutation matrices (see [1]). If we keep the position of every block of the reduced form and perform permutations of the rows and columns, every permutation matrix is transformed into another matrix that is permutationally similar to the original one.

The members in the symmetry group ${\mathrm{S}}_n$ of order n are called permutations. They are tightly connected with permutation matrices of order n. Permutation matrices are powerful tools in the representation theory of groups, discrete mathematics, applied mathematics, and some engineering technology (see [2,3,4,5]). They play an important role in the study of the reducible solutions of matrix equations (see [6]). Since the elementary row (or column) transformations are inevitable in solving matrix equations, which are equivalent to the multiplication by permutation matrices or diagonal matrices. The tricks of matrix transformations (especially the row or column permutations) are applicable.

This paper is devoted to the permutational similarity relation and to the classification of the permutation matrices. In particular, we focus on the standard structure of a general permutation matrix, on the canonical form of a permutation similarity class, and on how to generate the canonical form. Furthermore, a theorem is presented about the decomposition of a permutation matrix into a diagonal matrix and some generalized cycle matrices of type II. A factorization theorem shows that an arbitrary non-identity permutation matrix is the product of some generalized cycle matrices of type I. These contents are represented in Section 3 which is the main part of this paper.

The number of permutational similarity classes of a permutation matrices of order n is discussed in Section 4. A similar factorization for monomial matrices is discussed at the end of the paper.

2. Preliminary

Let n be a positive integer, P be a square matrix of order n. If P is a binary matrix (i.e. elements are either 0 or 1, also referred to as 0-1 matrix or (0, 1) matrix) and there is a unique “1” in every row and every column, then P is called a permutation matrix. If we substitute the “1”s in a permutation matrix by other non-zero elements, we obtain a monomial matrix, also referred to as a generalized permutation matrix.

As a matter of fact, there is a reason for the name “permutation matrix”. If a matrix T of size n × r is multiplied by a permutation matrix P of order n (from the left side of T), we obtain a permutation of the rows of T. If U is a matrix of size t by n, and P acts on U on the right, we have a permutation of the columns of U. The inverse of a permutation matrix $P^{-1}$ coincides with the transpose $P^{\mathrm{T}}$ while $P^{-1}$ itself is a permutation matrix.

Let k be a positive integer greater than 1, C be an invertible (0, 1) matrix of order k, if $C^k$ = $I_k$ ($I_k$ is the identity matrix of order k) and $C^i$≠$I_k$ for any i ($1\le i<k$), then C will be referred to as a cycle matrix of order k. A cycle matrix of order k of the form

$$\left[\begin{array}{ccccc}0& & & & 1\\ 1& 0\\ & 1& \ddots \\ & & \ddots & \ddots \\ & & & 1& 0\end{array}\right]$$

is a standard cycle matrix. The identity matrix of order 1 represents a cycle matrix of order 1.

If $C_1$ is a permutation matrix of order n, and there are exactly k zero diagonal elements (here $2⩽k⩽n$), if $C^k$ = $I_n$ and $C^i$≠$I_n$ for any i ($1⩽i<k$), then $C_1$ is termed a generalized cycle matrix of Type I with cycle order k.

If $C_2$ is a (0, 1) matrix of order n, rank $C_2$ = k, with k non-zero entries, ($2⩽k⩽n$), if $C^k$ is a diagonal of rank k, and $C^i$ is non-diagonal ($1⩽i<k$), then $C_2$ will be called a generalized cycle matrix of type II with cycle order k. Obviously, a generalized cycle matrix of type II plus some suitable diagonal (0, 1) matrix gives a generalized cycle matrix of type I with the same cycle order.

Let A and B be two monomial matrices of order n, if there is a permutation matrix T such that $B=T^{-1}AT$, then A and B are permutationally similar. The permutation similarity relation is an equivalence relation. Hence the set of the permutation matrices (or monomial matrices) or order n may be naturally split into equivalence classes.

3. Main Results

In this section, we attend to give 3 main theorems about the canonical form, the decomposition and the factorization of a permutation matrix, respectively.

Theorem 1 solves the following three problems (which arise naturally from the definitions),

(a)

What is the canonical form of a permutation similarity class?

(b)

How to generate the canonical form of a given permutation matrix?

(c)

If B is the canonical form of the permutation matrix A, how to find the permutation matrix T, such that $B=T^{-1}AT$?

Now we give some theorems that would solve these problems.

Theorem 1. (Similarity Theorem)

For any permutation matrix A of order n, there is a permutation matrix T, such that, $T^{-1}AT$=$\mathrm{diag}$$\{I_t$, $N_{k_1}$, ⋯, $N_{k_r}\}$, where

$$N_{k_i}=\left[\begin{array}{ccccc}0& & & & 1\\ 1& 0\\ & 1& \ddots \\ & & \ddots & \ddots \\ & & & 1& 0\end{array}\right]$$

is a cycle matrix of order$k_i$in standard form, (i=1, 2, ⋯, r), 2 ⩽ $k_1$⩽$k_2$⋯⩽$k_r$, $0⩽r⩽⌊{\displaystyle \frac{n}{2}}⌋$, $0⩽t⩽n$, and ${\displaystyle \sum _{i=1}^r}{k}_i$+t=n. T, t, r, $k_r$ are determined by A.

If A is an identity matrix, then t = n, r = 0. When A is a cycle matrix, $t=0$, $r=1$, $k_1$ = n. In this theorem, the quasi-diagonal matrix (or block-diagonal matrices) $\mathrm{diag}$ $\{I_t$, $N_{k_1}$, ⋯, $N_{k_r}\}$ will be called the canonical form of a permutation matrix in permutational similarity relation.

The main idea of this proof is similar to that concerning the decomposition of a root subspace into cyclic subspaces.

In a root subspace $V_{\lambda }$ associated with a linear transformation $\mathcal{B}$ and the eigenvalue $\lambda$ of a matrix B, if v is a root vector of height n belonging to $\mathcal{B}$, then the subspace spanned by $\{{\left(\mathcal{B}-\lambda I\right)}^{n-1}v$, ${\left(\mathcal{B}-\lambda I\right)}^{n-2}v$, ⋯, $\left(\mathcal{B}-\lambda I\right)v$, $v\}$ is a cyclic subspace, and $V_{\lambda }$ is the direct sum of some cyclic subspaces.

Proof. 

For any permutation matrix A of order n, let $\mathcal{A}$ be a linear transformation defined on the vector space ${\mathbb{R}}^n$ with bases

$$\mathcal{B}=\left\{e_1,e_2,\cdots ,e_n\right\},$$

(1)

where

$$e_i={(\underset{i-1}{\underbrace{0,\cdots ,0}},1,\underset{n-i}{\underbrace{0,\cdots ,0}})}^{\mathrm{T}},(i=1,2,\cdots ,n).$$

(2)

Here the regular letter “T” in the upper index means transposition. Suppose A is the matrix of the transformation $\mathcal{A}$ in the basis $\mathcal{B}$, and for any vector $\alpha \in {\mathbb{R}}^n$ with coordinates x (in the basis $\mathcal{B}$), the coordinates of $\mathcal{A}\alpha$ is $Ax$, i.e., $\mathcal{A}\alpha =\mathcal{B}Ax$. Here the coordinates are written as a column vector.

It is clear that the coordinates of $e_i$ in the basis $\mathcal{B}$ is ${(\underset{i-1}{\underbrace{0,\cdots ,0}},1,\underset{n-i}{\underbrace{0,\cdots ,0}})}^{\mathrm{T}}$. Since A is a permutation matrix, $A{e}_i$ is the i’th column of A.

We decompose ${\mathbb{R}}^n$ into some subspaces. In each subspace $V_i$, there is a basis $\{e_i$, $A{e}_i$, $A^2{e}_i$, ⋯, $A^{k_i-1}{e}_i\}$, where $A^{k_i}{e}_i=e_i$. The positive $k_i$ is the minimal integer satisfying this condition, i.e. the dimension of the cyclic subspace. Using this basis, the matrix of the transformation $\mathcal{A}$ restricted in $V_i$, can be written by

$${\left[\begin{array}{ccccc}0& & & & 1\\ 1& 0\\ & 1& \ddots \\ & & \ddots & \ddots \\ & & & 1& 0\end{array}\right]}_{k_i\times k_i}.$$

Let us now find all these cyclic subspaces. In order to describe the procedure precisely and concisely, we will use some auxiliary variables.

Step 1: Let S = {1, 2, ⋯, n}, $\mathcal{C}$ = $\left\{e_i|i\in S\right\}$, $a_{11}$ = min S, $F_1$ = $\left[a_{11}\right]$, $G_1=\left[e_{a_{11}}\right]$. (Here $F_1$ and $G_1$ are sequences, or sets equipped with precedence).

For the first cyclic subspace, of course $A{e}_{a_{11}}\in \mathcal{C}$. If $A{e}_{a_{11}}\ne e_{a_{11}}$, assume $e_{a_{12}}=A{e}_{a_{11}}$, then put $a_{12}$ and $e_{a_{12}}$ at the end of the sequences $F_1$ and $G_1$, respectively. If $A{e}_{a_{1,j}}\ne e_{a_{11}}$, assume $e_{a_{1,(j+1)}}=A{e}_{a_{1,j}}$, (i.e., $A^j{e}_{a_{11}}=e_{a_{1,(j+1)}}$), then add $a_{1,(j+1)}$ and $e_{a_{1,(j+1)}}$ at the end of sequences $F_1$ and $G_1$, respectively (j = 1, 2, ⋯ ). Since $A{e}_i\in \mathcal{C}$ ($\forall i\in S$), there is an integer $h_1$ such that $A{e}_{a_{1,h_1}}=e_{a_{11}}$ (otherwise the sequence $e_{a_{11}}$, $A{e}_{a_{11}}$, $A^2{e}_{a_{11}}$, $A^3{e}_{a_{11}}$, ⋯ is infinite). Suppose that $h_1$ is the minimal integer satisfying this condition ($1⩽h_1⩽n$). It is clear that $A^{h_1}{e}_{a_{11}}=e_{a_{11}}$, $A^{h_1}{e}_{a_{1j}}=e_{a_{1j}}$, (1 ⩽ j ⩽ $h_1$). It is possible that $h_1=1$ or $h_1=n$. At last $\left|F_1\right|$ = $\left|G_1\right|$ = $h_1$. Finally, remove the elements of $G_1$ from $\mathcal{C}$, and the elements of $F_1$ from S.

The first cyclic subspace is thus spanned by the basis $G_1$ = $[e_{a_{11}}$, $e_{a_{12}}$, ⋯, $e_{a_{1h_1}}]$. Usually, a basis of a linear space is denoted by braces, not brackets. However, braces denote sets, and this disregards the precedence. In order to avoid ambiguities, here we use brackets, which stand for sequences, where precedence is relevant. The dimension of this subspace is $\left|F_1\right|=h_1$. The matrix of the transformation $\mathcal{A}$, restricted to this cyclic subspace, is

$$N_{h_1}={\left[\begin{array}{ccccc}0& & & & 1\\ 1& 0\\ & 1& \ddots \\ & & \ddots & \ddots \\ & & & 1& 0\end{array}\right]}_{h_1\times h_1}.$$

Let us now search for the next cyclic subspace, if it exists.

Step 2: If S≠ Ø, let $a_{21}$ = min S, $F_2$ = $\left[a_{21}\right]$, $G_2=\left[e_{a_{21}}\right]$. It is clear that $A{e}_{a_{21}}\in \mathcal{C}$. (Otherwise we would have $A{e}_{a_{21}}\in G_1$. However, since all the elements in $G_1$ are removed from $\mathcal{C}$, then it exists $k_0$, s.t. $A^{k_0}{e}_{a_{11}}=A{e}_{a_{21}}$ with $k_0\ne 0$, so, $A^{k_0-1}{e}_{a_{11}}=e_{a_{21}}$ as A is invertible, which means that $e_{a_{21}}$ = $A^{k_0-1}{e}_{a_{11}}$ is in the set $G_1$, which is a contradiction.) If $A^{i-1}{e}_{a_{21}}\ne e_{a_{21}}$, suppose $A^{i-1}{e}_{a_{21}}=e_{a_{2i}}$ (i = 2, 3, ⋯), then add $a_{2i}$ and $e_{a_{2i}}$ at the end of the sequences $F_2$ and $G_2$, respectively. There will be a $h_2$, such that $A^{h_2}{e}_{a_{21}}=e_{a_{21}}$ (let $h_2$ be the minimal integer satisfying this condition. It is possible that $h_2$ = 1 or $h_2=n-h_1$). Obviously, $A^{h_2}{e}_{a_{2i}}=e_{a_{2i}}$, (1 ⩽ i ⩽ $h_2$). Then remove the elements of $G_2$ from $\mathcal{C}$, and remove the elements of $F_2$ from S.

Now another cyclic subspace is spanned by the basis $G_2$ = $[e_{a_{21}}$, $e_{a_{22}}$, ⋯, $e_{a_{2h_2}}]$. The dimension of this subspace is $\left|F_2\right|=h_2$. The matrix of the transformation $\mathcal{A}$ restricted to this cyclic subspace is $N_{h_2}$.

Step 3: If S≠Ø, goto step 2 and construct $F_3$, $F_4$, ⋯ and $G_3$, $G_4$, ⋯. This leads to other cyclic subspaces, their basis, and the matrices of the transformation $\mathcal{A}$ restricted to these cyclic subspaces. The procedure stops after a finite number of steps since n is finite.

Assume that we have $F_1$, $F_2$, ⋯, $F_u$ and $G_1$, $G_2$, ⋯, $G_u$, such that ${\displaystyle \bigcup _{i=1}^u}{F}_i$ = { 1, 2, ⋯, n}, ${\displaystyle \bigcup _{i=1}^u}{G}_i$ = { $e_1$, $e_2$, ⋯, $e_n$ }, $F_i\cap F_j$ = $G_i\cap G_j$ = Ø, ($1⩽i\ne j⩽u$).

There is a possibility that u = 1 (when A is a cycle matrix of order n) or n (when A is an identity matrix).

Step 4: Sort $F_1$, $F_2$, ⋯, $F_u$ by candinality, s.t. $\left|F_1^{\prime }\right|$⩽$\left|F_2^{\prime }\right|$ ⩽⋯⩽$\left|F_u^{\prime }\right|$. Then sort $G_i$ correspondingly, i.e., $G_i^{\prime }$ = $\left\{e_x\left|x\in F_i^{\prime }\right.\right\}$ (i = 1, 2, ⋯, u).

Suppose $\left|F_1^{\prime }\right|$ = $\left|F_2^{\prime }\right|$ = ⋯ = $\left|F_t^{\prime }\right|$ = 1. If t = n then A is an identity matrix. It is possible that $t=0$.

Let $r=u-t$. Denote the unique element in $G_i^{\prime }$ by $e_i^{\prime }$ (i = 1, 2, ⋯, t). Let $k_j$ = $\left|G_{t+j}^{\prime }\right|$, and denote the elements in $G_{t+j}^{\prime }$ by $e_{j,v}^{\prime }$ (j = 1, 2, ⋯, r; v = 1, 2, ⋯, $k_j$). Then, the matrix of $\mathcal{A}$ restricted to the subspace spanned by the bases ${\mathcal{D}}_0$ = $\left\{e_1^{\prime }\right.$, $e_2^{\prime }$, ⋯, $\left.e_t^{\prime }\right\}$ } is $I_t$ since $\mathcal{A}{e}_i^{\prime }=e_i^{\prime }$ (i = 1, 2, ⋯, t), or $\mathcal{A}{\mathcal{D}}_0$ = ${\mathcal{D}}_0{I}_t$; and the matrix of $\mathcal{A}$ restricted in the subspace spanned by the bases ${\mathcal{D}}_j$ = $G_{t+j}^{\prime }$ = $\{e_{j,1}^{\prime }$, $e_{j,2}^{\prime }$, ⋯, $e_{j,k_j}^{\prime }\}$ (j = 1, 2, ⋯, r) is

$$N_{k_j}=\left[\begin{array}{ccccc}0& & & & 1\\ 1& 0\\ & 1& \ddots \\ & & \ddots & \ddots \\ & & & 1& 0\end{array}\right],$$

which is a cycle matrix of order $k_j$, as $\mathcal{A}{e}_{j,v}^{\prime }=e_{j,v+1}^{\prime }$, (v = 1, 2, ⋯, $k_j-1$), and $\mathcal{A}{e}_{j,k_j}^{\prime }=e_{j,1}^{\prime }$, i.e., $\mathcal{A}{\mathcal{D}}_j$ = ${\mathcal{D}}_j{N}_{k_j}$ (j = 1, 2, ⋯, r). So, the matrix of $\mathcal{A}$ with bases

$$\mathcal{D}=\left\{e_1^{\prime },e_2^{\prime },\cdots ,e_t^{\prime };e_{1,1}^{\prime },e_{1,2}^{\prime },\cdots ,e_{1,k_1}^{\prime };\cdots \cdots ;e_{r,1}^{\prime },e_{r,2}^{\prime },\cdots ,e_{r,k_r}^{\prime }\right\}$$

(3)

is

$$B=I_t\oplus N_{k_1}\oplus \cdots \oplus N_{k_r}=\mathrm{diag}\{I_t,N_{k_1},\cdots ,N_{k_r}\}.$$

Since $\mathcal{D}$ is a reordering of $\mathcal{B}$, there is a permutation matrix T, such that $\mathcal{D}$ = $\mathcal{B}T$. Then $B=T^{-1}AT$ and Theorem 1 is proved.

Take the matrix

$$P_2=\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 1& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0\\ 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1\end{array}\right]$$

as an example, and assume it is the matrix of a transformation ${\mathcal{P}}_2$ in the basis $\{e_1$, $e_2$, $e_3$, ⋯, $e_7\}$ in ${\mathbb{R}}^7$.

Let us now search for the first cyclic subspace.

Since $P_2{e}_1$ = $e_6$, $P_2{e}_6$ = $e_1$, we have $F_1$ = [1, 6], and the first cyclic subspace is spanned by the basis $\{e_1$, $e_6\}$. Its dimension is $\left|F_1\right|=2$. The matrix of the transformation ${\mathcal{P}}_2$ restricted to this cyclic subspace is $N_2$ = $\left[\begin{array}{cc}& 1\\ 1\end{array}\right]$.

Since $P_2{e}_2$ = $e_3$, $P_2{e}_3$ = $e_4$, $P_2{e}_4$ = $e_2$, so $F_2$ = [2, 3, 4], $\left|F_2\right|=3$. The second cyclic subspace is spanned by the basis $\{e_2$, $e_3$, $e_4\}$, and its dimension is $\left|F_2\right|=3$. The matrix of the transformation ${\mathcal{P}}_2$ restricted to this cyclic subspace is $N_3$ = $\left[\begin{array}{ccc}& & 1\\ 1\\ & 1\end{array}\right]$.

Since $P_2{e}_5$ = $e_5$, $F_3$ = [5], $\left|F_3\right|=1$. The third cyclic subspace is spanned by the basis $\left\{e_5\right\}$, and the dimension is $\left|F_3\right|=1$. The matrix of ${\mathcal{P}}_2$ restricted to this cyclic subspace is $N_1$ = $\left[1\right]$. Finally, since $P_2{e}_7$ = $e_7$, $F_4$ = [7], $\left|F_4\right|=1$. The fourth cyclic subspace is spanned by the basis $\left\{e_7\right\}$, the dimension is $\left|F_4\right|=1$. The matrix of ${\mathcal{P}}_2$ restricted to this cyclic subspace is $N_1$ = $\left[1\right]$.

Overall, we have that $P_2$ is permutationally similar to the canonical form

$$B_2=\mathrm{diag}\{I_2,N_2,N_3\}=\left[\begin{array}{c}1\\ & 1\\ & & 0& 1\\ & & 1& 0\\ & & & & 0& 0& 1\\ & & & & 1& 0& 0\\ & & & & 0& 1& 0\end{array}\right],$$

or

$$\begin{array}{cc}& P_2\left\{e_5;e_7;e_1,e_6;e_2,e_3,e_4\right\}\hfill \\ \hfill =& \left\{e_5;e_7;e_6,e_1;e_3,e_4,e_2\right\}\hfill \\ \hfill =& \left\{e_5;e_7;e_1,e_6;e_2,e_3,e_4\right\}\left[\begin{array}{c}1\\ & 1\\ & & 0& 1\\ & & 1& 0\\ & & & & 0& 0& 1\\ & & & & 1& 0& 0\\ & & & & 0& 1& 0\end{array}\right].\hfill \end{array}$$

Now we find $T_2$, such that $B_2=T_2^{-1}{P}_2{T}_2$.

It follows from

$$(e_5,e_7,e_1,e_6,e_2,e_3,e_4)=(e_1,e_2,e_3,e_4,e_5,e_6,e_7)\left[\begin{array}{ccccccc}& & 1& 0& 0& 0& 0\\ & & & 0& 1& 0& 0\\ & & & 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 1\\ 1& & & 0\\ 0& & & 1\\ 0& 1& 0& 0\end{array}\right],$$

denote

$$T_2=\left[\begin{array}{ccccccc}& & 1& 0& 0& 0& 0\\ & & & 0& 1& 0& 0\\ & & & 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 1\\ 1& & & 0\\ 0& & & 1\\ 0& 1& 0& 0\end{array}\right],$$

which implying

$$T_2^{-1}=T_2^{\mathrm{T}}=\left[\begin{array}{ccccccc}& & & & 1& 0& 0\\ & & & & & 0& 1\\ 1& 0& 0& 0& 0& 0& 0\\ 0& & & & & 1& 0\\ 0& 1\\ 0& 0& 1\\ 0& 0& 0& 1\end{array}\right],$$

so $P_2$ = $T_2$$B_2$$T_2^{-1}$, that is,

$\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 1& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0\\ 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccccccc}& & 1& 0& 0& 0& 0\\ & & & & 1& 0& 0\\ & & & & & 1& 0\\ 0& 0& 0& 0& 0& 0& 1\\ 1\\ 0& & & 1\\ 0& 1& 0& 0\end{array}\right]\left[\begin{array}{c}1\\ & 1\\ & & 0& 1\\ & & 1& 0\\ & & & & 0& 0& 1\\ & & & & 1& 0& 0\\ & & & & 0& 1& 0\end{array}\right]\left[\begin{array}{ccccccc}& & & & 1& 0& 0\\ & & & & & 0& 1\\ 1& 0& 0& 0& 0& 0& 0\\ 0& & & & & 1& 0\\ 0& 1\\ 0& 0& 1\\ 0& 0& 0& 1\end{array}\right].$

Then, $B_2=T_2^{-1}{P}_2{T}_2$, i.e.,

$$\begin{array}{cc}& \left[\begin{array}{c}1\\ & 1\\ & & 0& 1\\ & & 1& 0\\ & & & & 0& 0& 1\\ & & & & 1& 0& 0\\ & & & & 0& 1& 0\end{array}\right]\hfill \\ \hfill =& \left[\begin{array}{ccccccc}& & & & 1& 0& 0\\ & & & & & 0& 1\\ 1& 0& 0& 0& 0& 0& 0\\ 0& & & & & 1& 0\\ 0& 1\\ 0& 0& 1\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 1& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0\\ 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{ccccccc}& & 1& 0& 0& 0& 0\\ & & & & 1& 0& 0\\ & & & & & 1& 0\\ 0& 0& 0& 0& 0& 0& 1\\ 1\\ 0& & & 1\\ 0& 1& 0& 0\end{array}\right].\hfill \end{array}$$

□

Theorem 2. (Decomposition Theorem)

For any permutation matrix A of order n, if A is not the identity, then there are some generalized cycle matrices $Q_1$, $Q_2$, ⋯, $Q_r$ of type II and a diagonal matrix $D_t$ of rank t, such that, A=$Q_1+Q_2+\cdots +Q_r+D_t$, where the non-zero elements in $D_t$ are all ones, ${\displaystyle \sum _{i=1}^r}\mathrm{rank}{Q}_i$+t= n; $1⩽r⩽⌊{\displaystyle \frac{n}{2}}⌋$, r, $Q_i$ (i=1, 2, ⋯, r) and $D_t$ are determined by A.

If the cycle order of $Q_i$ is $k_i$, (i = 1, 2, ⋯, r), then $2⩽{\displaystyle \sum _{i=1}^r}{k}_i⩽n$. If A is a cycle matrix, then $t=0$, $r=1$, $k_1$ = n.

The main idea of the proof may be summarized as follow.

Denote by $O_m$ a zero square matrix of order $m.$ By Theorem 1, there is a permutation matrix T, such that, $T^{-1}AT$=$\mathrm{diag}$$\{I_t$, $N_{k_1}$, ⋯, $N_{k_r}\}$. Then consider the matrices

$M_0$ = $\mathrm{diag}$$\{I_t$, $O_{k_1}$, ⋯, $O_{k_r}\}$,

$M_1$ = $\mathrm{diag}$$\{O_t$, $N_{k_1}$, $O_{k_2}$, ⋯, $O_{k_r}\}$,

$M_2$ = $\mathrm{diag}$$\{O_t$, $O_{k_1}$, $N_{k_2}$, $O_{k_3}$, ⋯, $O_{k_r}\}$,

$\cdots \cdots$,

$M_r$ = $\mathrm{diag}$$\{O_t$, $O_{k_1}$, $O_{k_2}$, ⋯, $O_{k_{r-1}},N_{k_r}\}$,

Clearly,

$$\mathrm{diag}\{I_t,N_{k_1},\cdots ,N_{k_r}\}=M_0+M_1+M_2+\cdots +M_r,$$

and

$$\begin{array}{cc}\hfill A& =T\mathrm{diag}\{I_t,N_{k_1},\cdots ,N_{k_r}\}{T}^{-1}\hfill \\ & =T{M}_0{T}^{-1}+T{M}_1{T}^{-1}+T{M}_2{T}^{-1}+\cdots +T{M}_r{T}^{-1}.\hfill \end{array}$$

It is clear that $\mathrm{rank}{M}_i$ = $k_i$(i = 1, 2, ⋯, r), and $\mathrm{rank}{M}_0$ = t. The matrix $M_i$ is a generalized cycle matrix of type II with cycle order $k_i$. Since T is invertible, $\mathrm{rank}$ $T{M}_i{T}^{-1}$ = $\mathrm{rank}{M}_i$ = $k_i$. T and $T^{-1}$ are permutation matrices, so $Q_i$ = $T{M}_i{T}^{-1}$ is also a 0-1 matrix with the same rank. Since $N_{k_i}^{k_i}=I_{k_i}$,

$$M_i^{k_i}=\mathrm{diag}\{O_t,O_{k_1},\cdots ,I_{k_i},\cdots ,O_{k_r}\},$$

is a diagonal matrix of rank $k_i$, and ${\left(T{M}_i{T}^{-1}\right)}^{k_i}$ = $T{M}_i^{k_i}{T}^{-1}$ is a diagonal matrix of rank $k_i$, too. If the exponent is less than $k_i$, the conclusion does not hold (but it will cost us some more words to prove this proposition). Then,

$$Q_i=T{M}_i{T}^{-1}$$

(4)

is a generalized cycle matrix of type II with cycle order $k_i$. Analogously,

$$D_t=T{M}_0{T}^{-1}=T\mathrm{diag}\{I_t,O_{k_1},\cdots ,O_{k_r}\}{T}^{-1}$$

(5)

is a diagonal matrix of rank $t.$

Following this idea, we may prove Theorem 2 in a different way. However, this requires to obtain $Q_i$ and $D_t$ directly, which may be challenging. We prefer to move on with another proof following the idea of the proof of Theorem 1. In this way, we construct $Q_i$ and $D_t$ more conveniently.

Proof. 

Starting from the $F_i^{\prime }$ generated above, one may construct a 0-1 matrix $D_t$ of order n, such that the j’th column of $D_t$ is the j’th column of A ($\forall j\in {\displaystyle \bigcup _{i=1}^t}{F}_i^{\prime }$) and the other columns of $D_t$ are 0 vectors. Of course, $D_t$ is a diagonal matrix of rank t, as the j’th column of A is $e_j$ (by definition, $A{e}_j=e_j$).

Then, construct a 0-1 matrix $Q_i$(i = 1, 2, ⋯, r) of order n, such that the j’th column of $Q_i$ is the j’th column of A ($j\in F_{t+i}^{\prime }$) and the other columns of $Q_i$ are 0 vectors. As

$$\left(\bigcup _{i=1}^t{F}_i^{\prime }\right)\bigcup \left(\bigcup _{i=1}^r{F}_{t+i}^{\prime }\right)=\bigcup _{i=1}^u{F}_i=\{1,2,\cdots ,n\},$$

(6)

and

$$F_{i_1}\cap F_{i_2}=⌀(1⩽i_1\ne i_2⩽u),$$

every column of A appears exact once in a matrix ($D_t$ or $Q_i$, denoted by M) in the expression ${\displaystyle \sum _{i=1}^r}{Q}_i+D_t$, in the same position as it appears in A. Besides, the columns in the same position in the matrices other than M appeared in the sum ${\displaystyle \sum _{i=1}^r}{Q}_i+D_t$ are all 0 vectors. Overall, we have

$$\sum _{i=1}^r{Q}_i+D_t=A.$$

Let us now prove that $Q_i$ is a generalized cycle matrix of type II with cycle order $k_i$.

Assume that the members in $F_{t+i}^{\prime }$ (i = 1, 2, ⋯, r) are $a_{i,1}^{\prime }$, $a_{i,2}^{\prime }$, ⋯, $a_{i,k_i}^{\prime }$, and that $F_{t+i}^{\prime }=F_s$ for some s ($1⩽s⩽u$). Then we have a relation about the members in $G_{t+i}^{\prime }$ and the members in a certain $G_s$, i.e., $e_{i,v}^{\prime }=e_{a_{i,v}^{\prime }}\in G_s,v=1,2,\cdots ,k_i$. By the definition of $F_s$, we know that $A{e}_{a_{i,v}^{\prime }}$ = $e_{a_{i,v+1}^{\prime }}$, (v = 1, 2, ⋯, $k_i-1$), $A{e}_{a_{i,k_i}^{\prime }}$ = $e_{a_{i,1}^{\prime }}$, so $e_{a_{i,v+1}^{\prime }}$ is the $a_{i,v}^{\prime }$’th column of A.

As $Q_i$ is made of some 0 vectors and $k_i$ columns of A, and the columns of A are linearly independent, the rank of $Q_i$ is $k_i$. The $a_{i,v}^{\prime }$’th column of $Q_i$ is the $a_{i,v}^{\prime }$’th column of A, so $Q_i{e}_{a_{i,v}^{\prime }}$ = $e_{a_{i,v+1}^{\prime }}$, (v = 1, 2, ⋯, $k_i-1$), $Q_i{e}_{a_{i,k_i}^{\prime }}$ = $e_{a_{i,1}^{\prime }}$, $Q_i{e}_l$ = 0 ($\forall e_l\in \mathcal{D}∖G_{t+i}^{\prime }$). Therefore $Q_i^v{e}_{a_{i,1}^{\prime }}$ = $e_{a_{i,v+1}^{\prime }}$ (v = 1, 2, ⋯, $k_i-1$), $Q_i^{k_i}{e}_{a_{i,1}^{\prime }}$ = $e_{a_{i,1}^{\prime }}$, (so $Q_i^v$ is not diagonal as $Q_i^v{e}_{a_{i,1}^{\prime }}$ = $e_{a_{i,v+1}^{\prime }}$≠$e_{a_{i,1}^{\prime }}$). Then

$$Q_i^{k_i}{e}_{a_{i,v}^{\prime }}=Q_i^{v-1}\left(Q_i^{k_i-v+1}{e}_{a_{i,v}^{\prime }}\right)=Q_i^{v-1}\left(e_{a_{i,1}^{\prime }}\right)=e_{a_{i,v}^{\prime }},(v=1,2,\cdots ,k_i).$$

Hence $Q_i^{k_i}$ is a diagonal matrix of rank $k_i$. Therefore $Q_i$ is a generalized cycle matrix of type II with cycle order $k_i$. □

Theorem 3. (Factorization Theorem)

For any permutation matrix A of order n, if A is not the identity, then there are some generalized cycle matrices $P_1$, $P_2$, ⋯, $P_r$ of type I, such that, A = $P_1{P}_2\cdots P_r$, where $1⩽r⩽⌊{\displaystyle \frac{n}{2}}⌋$; r, $P_i$(i=1, 2, ⋯, r) are determined by A. $P_{i_1}$ and $P_{i_2}$ commute ($1⩽i_1\ne i_2⩽r$).

If the cycle order of $P_i$ is $k_i$, (i = 1, 2, ⋯, r), then $2⩽\sum _{i=1}^r{k}_i⩽n$.

Since $T^{-1}AT$=$\mathrm{diag}$$\{I_t$, $N_{k_1}$, ⋯, $N_{k_r}\}$, for convenience, we denote

$Y_1$ = $\mathrm{diag}$$\{I_t$, $N_{k_1}$, $I_{k_2}$, ⋯, $I_{k_r}\}$,

$Y_2$ = $\mathrm{diag}$$\{I_t$, $I_{k_1}$, $N_{k_2}$, $I_{k_3}$, ⋯, $I_{k_r}\}$,

$\cdots \cdots$,

$Y_r$ = $\mathrm{diag}$$\{I_t$, $I_{k_1}$, $I_{k_2}$, ⋯, $I_{k_{r-1}},N_{k_r}\}$.

Obviously,

$$\mathrm{diag}\{I_t,N_{k_1},\cdots ,N_{k_r}\}=Y_1{Y}_2\cdots Y_r$$

and

$$\begin{array}{cc}\hfill A& =T\mathrm{diag}\{I_t,N_{k_1},\cdots ,N_{k_r}\}{T}^{-1}=T{Y}_1{Y}_2\cdots Y_r{T}^{-1}\hfill \\ & =\left(T{Y}_1{T}^{-1}\right)\left(T{Y}_2{T}^{-1}\right)\cdots \left(T{Y}_r{T}^{-1}\right).\hfill \end{array}$$

Obviously, ${\left(N_{k_i}\right)}^{k_i}=I_{k_i}$, ${\left(N_{k_i}\right)}^{k_i-j}\ne I_{k_i}$, $0<j<k_i$.

So ${\left(Y_i\right)}^{k_i}=I_n$, ${\left(Y_i\right)}^{k_i-j}\ne I_n$, $0<j<k_i$.

It is clear that $Y_i$ is a generalized cycle matrix of type I with cycle order $k_i$ (i = 1, 2, ⋯, r), and it is thus sufficient to prove that

$$P_i=T{Y}_i{T}^{-1}$$

(7)

is a generalized cycle matrix of type I with cycle order $k_i$. Rather obviously, $P_i^{k_i}$ = $T{Y}_i^{k_i}{T}^{-1}$ = $T{I}_n{T}^{-1}$ = $I_n$, however, it is not easy to prove that there are exact $k_i$ vanishing entries in the diagonal of $P_i$, and that $k_i$ is the minimal positive integer satisfying the condition $P_i^{k_i}$ = $I_n$.

Proof. 

Let $D_t^{\left(b\right)}$ = $I_n-D_t$, where $D_t$ is determined by Equation (5). Then

$$\mathrm{rank} D_t=t,\mathrm{rank}{D}_t^{\left(b\right)}=n-t.$$

Now build a 0-1 matrix $J_i^a$ (i = 1, 2, ⋯, r) of order n, such that the j’th column of $J_i^a$ is the j’th column of $I_n$ ($j\in F_{t+i}^{\prime }$), and the other columns of $J_i^a$ are 0 vectors. So

$$\sum _{i=1}^r{J}_i^{\left(a\right)}+D_t=I_n.$$

Let $J_i^{\left(b\right)}$ = $I_n-J_i^{\left(a\right)}$, then

$$\mathrm{rank}{J}_i^{\left(a\right)}=k_i,\mathrm{rank}{J}_i^{\left(b\right)}=n-k_i.$$

We have

$$\begin{array}{ccc}{D}_t^{\left(b\right)}{D}_t=D_t{D}_t^{\left(b\right)}=0,& J_i^{\left(a\right)}{J}_i^{\left(b\right)}=J_i^{\left(b\right)}{J}_i^{\left(a\right)}=0,& Q_i{J}_i^{\left(b\right)}=J_i^{\left(b\right)}{Q}_i=0,\\ Q_{i_1}{J}_{i_2}^{\left(a\right)}=J_{i_2}^{\left(a\right)}{Q}_{i_1}=0,& J_{i_1}^{\left(a\right)}{J}_{i_2}^{\left(a\right)}=J_{i_2}^{\left(a\right)}{J}_{i_1}^{\left(a\right)}=0,& Q_{i_1}{Q}_{i_2}=Q_{i_2}{Q}_{i_1}=0,\end{array}$$

and

$$\begin{array}{cc}{Q}_{i_1}{J}_{i_2}^{\left(b\right)}=J_{i_2}^{\left(b\right)}{Q}_{i_1}=Q_{i_1}\ne 0,& J_{i_1}^{\left(a\right)}{J}_{i_2}^{\left(b\right)}=J_{i_2}^{\left(b\right)}{J}_{i_1}^{\left(a\right)}=J_{i_1}^{\left(a\right)}\ne 0,\end{array}$$

where $1⩽i_1\ne i_2⩽r$, $Q_i$ is defined above Equation (6) on page 6.

It is not difficult to prove that

$$J_{i_1}^{\left(b\right)}{J}_{i_2}^{\left(b\right)}=J_{i_2}^{\left(b\right)}{J}_{i_1}^{\left(b\right)}=I_n-J_{i_2}^{\left(a\right)}-J_{i_1}^{\left(a\right)}.$$

If we denote $P_i=Q_i+J_i^{\left(b\right)}$, we have

$$\mathrm{rank}{P}_i=n,I_n+Q_i=P_i+J_i^{\left(a\right)}.$$

Clearly,

$$P_{i_1}{P}_{i_2}=\left(Q_{i_1}+I_n-J_{i_1}^{\left(a\right)}\right)\left(Q_{i_2}+I_n-J_{i_2}^{\left(a\right)}\right)=Q_{i_1}+Q_{i_2}+I_n-J_{i_1}^{\left(a\right)}-J_{i_2}^{\left(a\right)},$$

and

$$P_{i_2}{P}_{i_1}=\left(Q_{i_2}+I_n-J_{i_2}^{\left(a\right)}\right)\left(Q_{i_1}+I_n-J_{i_1}^{\left(a\right)}\right)=Q_{i_2}+Q_{i_1}+I_n-J_{i_2}^{\left(a\right)}-J_{i_1}^{\left(a\right)}.$$

So, $P_{i_1}{P}_{i_2}=P_{i_2}{P}_{i_1}$, i.e., $P_{i_1}$ and $P_{i_2}$ commute.

Hence

$$\prod _{i=1}^r{P}_i=\prod _{i=1}^r\left(Q_i+I_n-J_i^{\left(a\right)}\right)=\sum _{i=1}^r{Q}_i+I_n-\sum _{i=1}^r{J}_i^{\left(a\right)}=\sum _{i=1}^r{Q}_i+D_t=A.$$

We can also prove the equality above in a different way.

Because $Q_{i_1}{Q}_{i_2}=0$, $Q_{i_1}{D}_t$ = $D_t{Q}_{i_1}$ = 0, ($1⩽i_1\ne i_2⩽r$), then

$$\left(I_n+D_t\right)\prod _{i=1}^r\left(I_n+Q_i\right)=I_n+\sum _{i=1}^r{Q}_i+D_t=I_n+A.$$

(8)

Since $D_t{Q}_i$ = $Q_i{D}_t$ = 0, we have $D_t{J}_i^{\left(a\right)}$ = $J_i^{\left(a\right)}{D}_t$ = 0.

By construction, when $1⩽i⩽r$, $v\in F_{t+i}^{\prime }\bigcup \left({\displaystyle \bigcup _{i=1}^u}{F}_i^{\prime }\right)$, the v’th column (or the v’th row) of $P_j$ ($1⩽j⩽r$, $j\ne i$) is equal to the v’th column (or the v’th row) of $I_n$, so the v’th column (or the v’th row) of ${\displaystyle \prod _{\begin{array}{c}{}_{1⩽j⩽r}\\ {}^{j\ne i}\end{array}}^r}{P}_j$ is equal to the v’th column (or the v’th row) of $I_n$.

Therefore, when $J_i^{\left(a\right)}$ is multiplied by ${\displaystyle \prod _{\begin{array}{c}{}_{1⩽j⩽r}\\ {}^{j\ne i}\end{array}}^r}{P}_j$, the v’th column does not change, while the other columns of $J_i^{\left(a\right)}$ are 0 vectors, such that

$$J_i^{\left(a\right)}\prod _{\begin{array}{c}{}_{1⩽j⩽r}\\ {}^{j\ne i}\end{array}}^r{P}_j=J_i^{\left(a\right)}.$$

For the same reason, $D_t{\displaystyle \prod _{i=1}^r}{P}_i$ = $D_t$.

Noting that

$$\begin{array}{cc}& \left(I_n+D_t\right)\prod _{i=1}^r\left(I_n+Q_i\right)\hfill \\ \hfill =& \left(I_n+D_t\right)\prod _{i=1}^r\left(P_i+J_i^{\left(a\right)}\right)\hfill \\ \hfill =& \left(I_n+D_t\right)\left(\prod _{i=1}^r{P}_i+\sum _{i=1}^r\left(J_i^{\left(a\right)}\prod _{\begin{array}{c}{}_{1⩽j⩽r}\\ {}^{j\ne i}\end{array}}^r{P}_j\right)\right)(J_{i_1}^{\left(a\right)}{J}_{i_2}^{\left(a\right)}=J_{i_2}^{\left(a\right)}{J}_{i_1}^{\left(a\right)}=0,i_1\ne i_2)\hfill \\ \hfill =& \left(I_n+D_t\right)\left(\prod _{i=1}^r{P}_i+\sum _{i=1}^r{J}_i^{\left(a\right)}\right)\hfill \\ \hfill =& \prod _{i=1}^r{P}_i+\sum _{i=1}^r{J}_i^{\left(a\right)}+D_t\prod _{i=1}^r{P}_i+D_t\sum _{i=1}^r{J}_i^{\left(a\right)}\hfill \\ \hfill =& \prod _{i=1}^r{P}_i+\sum _{i=1}^r{J}_i^{\left(a\right)}+D_t+0=\prod _{i=1}^r{P}_i+I_n,\hfill \end{array}$$

we have that

$$\left(I_n+D_t\right)\prod _{i=1}^r\left(I_n+Q_i\right)=\prod _{i=1}^r{P}_i+I_n.$$

(9)

It follows from Equations (8) and (9), that

$$\prod _{i=1}^r{P}_i+I_n=I_n+A,$$

thus ${\displaystyle \prod _{i=1}^r}{P}_i=A$.

Now, we prove that $P_i$ is a generalized cycle matrix of type II with cycle order $k_i$.

Since $P_i=Q_i+J_i^{\left(b\right)}$, $Q_i{J}_i^{\left(b\right)}$ = $J_i^{\left(b\right)}{Q}_i$ = 0, then $P_i^m$ = $Q_i^m+{\left(J_i^{\left(b\right)}\right)}^m$ = $Q_i^m+J_i^{\left(b\right)}$ ($\forall m\in {\mathbb{Z}}^{+}$), and $P_i^{k_i}=Q_i^{k_i}+J_i^{\left(b\right)}$.

It follows from $Q_i^{k_i}{e}_{a_{i,v}^{\prime }}$ = $A^{k_i}{e}_{a_{i,v}^{\prime }}$ = $e_{a_{i,v}^{\prime }}$ (v = 1, 2, ⋯, $k_i$) that

$$\forall e_l\in \mathcal{D}\setminus G_{t+i}^{\prime },Q_i{e}_l=0⟹Q_i^{k_i}{e}_l=0.$$

Here $\mathcal{D}$ is defined in Equation (3).

On the other hand, $J_i^{\left(b\right)}{e}_{a_{i,v}^{\prime }}$ = 0 (v = 1, 2, ⋯, $k_i$), $J_i^{\left(b\right)}{e}_l$ = $e_l$, ($\forall e_l\in \mathcal{D}∖G_{t+i}^{\prime }$).

So, for any $e_l$ in $\mathcal{B}$, if $e_l\in G_i^{\prime }$, then $\left(Q_i^{k_i}+J_i^{\left(b\right)}\right)e_l$ = $Q_i^{k_i}{e}_l$ = $e_l$; otherwise, $e_l\notin G_i^{\prime }$, then $\left(Q_i^{k_i}+J_i^{\left(b\right)}\right)e_l$ = $J_i^{\left(b\right)}{e}_l$ = $e_l$.

This means that $P_i^{k_i}{e}_l$ = ($Q_i^{k_i}+J_i^{\left(b\right)})e_l$ = $e_l$ ($\forall e_l\in \mathcal{B}$), i.e. $P_i^{k_i}$$(e_1,e_2,\cdots ,e_n)$ = $(e_1,e_2,$$\cdots ,e_n$), or $P_i^{k_i}$$I_n$ = $I_n$. (Actually, by $Q_i^{k_i}$ = $J_i^{\left(a\right)}$, we have that $P_i^{k_i}=Q_i^{k_i}+J_i^{\left(b\right)}=J_i^{\left(a\right)}+J_i^{\left(b\right)}=I_n.$)

When $1⩽m<k_i$, $Q_i^m$ is not diagonal, and neither is $Q_i^m+J_i^{\left(b\right)}$ = $P_i^m$. So, $P_i$ is a generalized cycle matrix of type II with cycle order $k_i$. □

4. On the Number of Permutation Similarity Classes

The number of permutation similarity classes of permutation matrices of order n is the partition number $p\left(n\right)$. There is a recursion formula for $p\left(n\right)$,

$$\begin{array}{cc}\hfill p\left(n\right)& =p(n-1)+p(n-2)-p(n-5)-p(n-7)+\cdots +\hfill \\ \hfill & {(-1)}^{k-1}p\left(n-{\displaystyle \frac{3k^2\pm k}{2}}\right)+\cdots \cdots \hfill \\ \hfill & =\sum _{k=1}^{k_1}{(-1)}^{k-1}p\left(n-{\displaystyle \frac{3k^2+k}{2}}\right)+\sum _{k=1}^{k_2}{(-1)}^{k-1}p\left(n-{\displaystyle \frac{3k^2-k}{2}}\right),\hfill \end{array}$$

(10)

(see [7], p. 55), where

$$k_1=⌊{\displaystyle \frac{\sqrt{24n+1}-1}{6}}⌋,k_2=⌊{\displaystyle \frac{\sqrt{24n+1}+1}{6}}⌋,$$

(11)

and $p\left(0\right)=1$. In the above formula, $⌊x⌋$ denotes the floor function, i.e. the maximum integer that is less than or equal to the real number x.

Asymptotically, we have (see e.g., [8,9])

$$p\left(n\right)\sim {\displaystyle \frac{1}{4n\sqrt{3}}}exp\left(\sqrt{\frac{2}{3}}\pi n^{1/2}\right).$$

(12)

This formula has been obtained by Godfrey H. Hardy and Srinivasa Ramanujan in 1918 [10] (In [11,12], one may find two different proofs. The evaluation of the constants can be found in [13]).

Formula (12) is relevant for theoretical analysis and very convenient to estimate the value of $p\left(n\right)$ by simple means. However, the accuracy of the asymptotic Formula (12) is limited when n is small. Another celebrated formula, given in term of a convergent series, has been found by Rademacher in 1937, based on the work of Hardy and Srinivasa Ramanujan, see [7,14].

In [15], several other formulae modified from Formula (12) have been obtained, showing high accuracy and yet expressed in terms of elementary functions, e.g.

$$p\left(n\right)\approx ⌊{\displaystyle \frac{exp\left(\sqrt{\frac{2}{3}}\pi \sqrt{n}\right)}{4\sqrt{3}\left(n+C_2^{\prime }\left(n\right)\right)}}+{\displaystyle \frac{1}{2}}⌋,1⩽n⩽80$$

(13)

with a relative error less than 0.004%, where

$$C_2^{\prime }\left(n\right)=\left\{\begin{array}{cc}0.4527092482\times \sqrt{n+4.35278}-0.05498719946,\hfill & n=3,5,7,\cdots ,79;\hfill \\ 0.4412187317\times \sqrt{n-2.01699}+0.2102618735,\hfill & n=4,6,8,\cdots ,80.\hfill \end{array}\right.$$

and

$$p\left(n\right)\approx ⌊{\displaystyle \frac{exp\left(\sqrt{\frac{2}{3}}\pi \sqrt{n}\right)}{4\sqrt{3}\left(n+a_2\sqrt{n+c_2}+b_2\right)}}+{\displaystyle \frac{1}{2}}⌋,n⩾80$$

(14)

with a relative error less than $5\times {10}^{-8}$ when $n⩾180$, where $a_2=0.4432884566$, $b_2=0.1325096085$ and $c_2=0.274078$.

5. Results for Monomial Matrices

Any monomial matrix M can be written as a product of a permutation matrix P and an invertible diagonal matrix D. Turn all the non-zero elements of M into 1, then we have a permutation matrix P. Suppose that the unique non-zero elements in the i’th row of M is $c_i$, and the unique non-zero element in the i’th column of M is $d_i$, i = 1, 2, ⋯, n. Let $D_1$ = $\mathrm{diag}$ { $c_1$, $c_2$, ⋯, $c_n$ }, $D_2$ = $\mathrm{diag}$ { $d_1$, $d_2$, ⋯, $d_n$ }, then we have M = $P{D}_2$ = $D_{1}P$.

For the permutation matrix P, there is a permutation matrix T such that $T^{-1}PT$ = Y has the canonical form diag $\{I_t$, $N_{k_1}$, ⋯, $N_{k_r}\}$ as proved in Theorem 1. In the expression $T^{-1}PT$, the permutation matrix $T^{-1}$ changes only the position of the rows, and T just changes the position of the columns of P. Since the non-zero elements of M and P share the same locations in the matrices, so do $T^{-1}MT$ and $T^{-1}PT$. Denote the unique non-zero element in the i’th row of $T^{-1}MT$ by $a_i$, and the unique non-zero element in the i’th column of $T^{-1}MT$ by $b_i$, i = 1, 2, ⋯, n. Let $D_3$ = $\mathrm{diag}$ { $a_1$, $a_2$, ⋯, $a_n$ }, $D_4$ = $\mathrm{diag}$ { $b_1$, $b_2$, ⋯, $b_n$ }, then $T^{-1}MT$ = $D_{3}Y$ = $Y{D}_4$.

Finally, we have that

$$\begin{array}{cc}\hfill M& =D_{1}T\left[\begin{array}{c}{I}_t\\ & N_1\\ & & \ddots \\ & & & N_r\end{array}\right]T^{-1}=T\left[\begin{array}{c}{I}_t\\ & N_1\\ & & \ddots \\ & & & N_r\end{array}\right]T^{-1}{D}_2\hfill \\ & =T{D}_3\left[\begin{array}{c}{I}_t\\ & N_1\\ & & \ddots \\ & & & N_r\end{array}\right]T^{-1}=T\left[\begin{array}{c}{I}_t\\ & N_1\\ & & \ddots \\ & & & N_r\end{array}\right]D_4{T}^{-1}.\hfill \end{array}$$

$D_1$, $D_2$, $D_3$ and $D_4$ could be easily obtained from M directly. Their relations can be stated as below.

$D_2=P^{-1}{D}_{1}P$, $D_3=T^{-1}{D}_{1}T$, $D_4=Y^{-1}{D}_{3}Y$.

6. Conclusions

For any permutation matrix A of order n, we can obtain its canonical form

$B=\mathrm{diag}\{I_t,N_{k_1},\cdots ,N_{k_r}\}$ and a permutation matrix T by the algorithm described in the proof of Theorem 1, such that, $B=T^{-1}AT$, where t, r, $k_1$, ⋯, $k_r$ and T are uniquely determined from A. Any matrix permutationally similar to A has the same canonical form.

The permutation matrix A can be written as the sum of some generalized cycle matrices $Q_1$, $Q_2$, ⋯, $Q_r$ of type II and a diagonal matrix $D_t$ of rank t, where t and r are the same as that mentioned above, $Q_1$, $Q_2$, ⋯, $Q_r$ and $D_t$ are determined from A by Equations (4) and (5) in the proof of Theorem 2.

We can also denote A as the product of some generalized cycle matrices $P_1$, $P_2$, ⋯, $P_r$ of type I, where t is the same as that mentioned above, $P_1$, $P_2$, ⋯, $P_r$ can be constructed from the Equation (7) in the proof of Theorem 3.

7. Concluding Remark

We can also prove Theorem 1 by the combinatorial method, which may seem easier. But the other two theorems could not be easily proved in the same way. Theorem 1 could be written in the form of permutation transformations (which are the members of the symmetry group ${\mathrm{S}}_n$). If L is a Latin square, every row (or column) of L could be considered as a permutation transformation. When searching for the invariant isotopism group of L, we will encounter the canonical form of the permutational similarity relations (of permutation matrices or of permutation transformations in ${\mathrm{S}}_n$). So the conclusions obtained here could be applied in Latin squares or projective planes.