arXiv:1803.02199v3 [math.GM] 4 Jul 2018 h niec arxo rjciepaeo order of plane projective a of matrix incidence The Introduction 1 arxwl etasomdit nte arxta spermu is that 1.2). matr matrix sec. incidence [5] another reduced (refer into the one pos transformed original the of be keep columns and will per and reduced matrix are matrix rows blocks incidence the Most the permuting keep blocks. we r some the If into In incide [4]). divided unique). the (not be columns, form could and standard matrix in rows reduced the to be sorting columns could After plane and/or rows ma one. the incidence other permuting the the by if transformed isomorphic be be could will planes projective Two oeo h lsicto fPruainMatrix Permutation of Classification the on Note A aoia om yl arxdcmoiin yl factori cycle decomposition, matrix cycle form, canonical emtto arxo ooilmatrix. monomial or decomposition matrix matrix permutation cycle the class, canonic equivalence the similar about mainly relation, similarity permutation e Words: Key M20 ujc Classification: Subject AMS2000 hsppri ocnrtdo h lsicto fpermutat of classification the on concentrated is paper This nvriyo cec n ehooyo China of Technology and Science of University emtto arx ooilmti,pruainsimilar permutation matrix, monomial matrix, permutation colo ahmtclScience Mathematical of School [email protected] uy5 2018 5, July IWenwei LI Abstract 1 52,15A36, 15A21, n sa01mti forder of matrix 0-1 a is lfr fapermutational a of form al zation n h atrzto fa of factorization the and rxo n rjcieplane projective one of trix dcdfr,teincidence the form, educed c arxo projective a of matrix nce to feeybokwhen block every of ition ainlysmlrt the to similar tationally (refer matrices mutation o arxwt the with matrix ion h niec arxof matrix incidence the x vr permutation every ix, n 2 ity, + n 1. + 2 Preliminary 2

This paper will focus on the permutational similarity relation and the classification of the permutation matrices. It will demonstrate the standard structure of a general permutation matrix, the canonical form of a permutation similarity class, how to generate the canonical form and solve some other related issues. The main theorems will be proved by two different methods (linear algebra method and the combinatorial method). The number of permutational similarity classes of permutation matrices of order n will be mentioned on Section 5. A similar factorization proposition on monomial matrix will be introduced at the end.

2 Preliminary

Let n be a positive integer, P be a square matrix of order n. If P is a binary matrix (every entry in it is either 0 or 1, also called 0-1 matrix or (0, 1) matrix) and there is a unique “1” in its every row and every column, then P is called a permutation matrix . If we substitute the units in a permutation matrix by other non-zero elements, it will be called a monomial matrix or a generalized permutation matrix. There is a reason for the name “permutation matrix”. A permutation matrix P of order n multiplies a matrix T of size n r (from the left side of T ) will result the permutation × of rows of T . If U is a matrix of size t by n, that P acts on U from the right hand side of U will leads to the permutation of columns of U. Let k be a positive integer greater than 1, C be an invertible (0, 1) matrix of order k, if Ck = I (I is the identity matrix of order k) and Ci = I for any i (1

If C1 is a permutation matrix of order n, and there are exact k entries in diagonal being k i 0, (here 2 6 k 6 n), if C = In and C = In for any i (1 6 i

If C2 is a (0, 1) matrix of order n, rank C2 = k, and there are exact k entries in C2 is non-zero, (2 6 k 6 n), if Ck is a diagonal of rank k, too, and Ci is non-diagonal (1 6 i

There are some problems presented naturally:

1. What’s the canonical form of a permutation similarity class?

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

3. If B is the canonical form of the permutation matrix A, how to find the permutation matrix T , such that B = T −1AT ?

Theorem 1. (Decomposition Theorem) For any permutation matrix A of order n, if A is not identical, then there are some generalized cycle matrices Q , Q , , Q of type II 1 2 ··· r and a diagonal matrix Dt of rank t, such that, A = Q1 + Q2 + + Qr + Dt, where the r ··· n non-zero elements in Dt are all ones , rankQi + t = n; 1 6 r 6 , r, Pi (i = 1, 2, i=1 2 , r) and Dt are determined by A. P j k ··· r If the cycle order of Qi is ki, (i = 1, 2, , r), then 2 6 ki 6 n. When A is a cycle ··· i=1 matrix, t = 0, r = 1, k1 = n. P Theorem 2. (Factorization Theorem) For any permutation matrix A of order n, if A is not identical, then there are some generalized cycle matrices P1, P2, , Pr of type I , n ··· such that, A = P P P , where 1 6 r 6 ; r, P (i = 1, 2, , r) are determined 1 2 ··· r 2 i ··· by A. Pi and Pi commute (1 6 i1 = i2 6 rj) .k 1 2 6 r If the cycle order of Pi is ki, (i = 1, 2, , r), then 2 6 ki 6 n. ··· i=1 P Theorem 3. (Similarity Theorem) For any permutation matrix A of order n, there is −1 a permutation matrix T , such that, T AT = diag It, Nk1 , , Nkr , where Nki = 0 1 { ··· } 1 0  ..  1 . is a cycle matrix of order ki in standard form, (i = 1, 2, , r),  . .  ···  .. ..     1 0      n r 2 6 k1 6 k2 6 kr, 0 6 r 6 , 0 6 t 6 n, and ki + t = n. T , t, r, kr are ··· 2 i=1 determined by A. j k P

If A is a identity matrix, then t = n, r = 0. When A is a cycle matrix, t = 0, r = 1, k = n. In this theorem, the quasi-diagonal matrix (or block-diagonal matrices) diag I , 1 { t Nk1 , , Nkr will be called the canonical form of a permutation matrix in permutational similarity··· relation.} 4 Proof 4 4 Proof

Here a proof by linear algebra method is presented.

Proof of Theorem 3

For any permutation matrix A of order n, let A be a linear transformation defined on the vector space Rn with bases = {e , e , , e }, where e = (0 0, 1, 0 0)T, B 1 2 ··· n i ··· ··· i−1 n−i 1 (i = 1, 2, , n), such that A is the matrix of the transformation A in the basis , or ··· B for any vector α Rn with the coordinates x (in the basis ), the| coordinates{z } | of{zA }α is Ax, i.e., A α = ∈Ax. Here the coordinates are written in columnB vectors. B It is clear that the coordinates of e in the basis is (0 0, 1, 0 0)T. Since A is a i B ··· ··· i−1 n−i permutation matrix, Aei is the i’th column of A. | {z } | {z } Let S = {1, 2, , n}, = e i S , a = min S, F = [a ], G = [e ]. (Here F ··· C i ∈ 11 1 11 1 a11 1 and G are sequences, or sets equipped with orders) Of course Ae . If Ae = e , 1  a11 ∈ C a11 6 a11 assume ea12 = Aea11 , then put a12 and ea12 to the ends of the sequences F1 and G1, respectively. If Ae = e , suppose e = Ae , (i.e., Aje = e ), then add a1,j 6 a11 a1,(j+1) a1,j a11 a1,(j+1) a1,(j+1) and ea1,(j+1) to the ends of sequences F1 and G1, respectively (j = 1, 2, ). Since ′ ··· ′ Aei ( i S), there will be a k1 such that Aea1,k = ea11 (otherwise the sequence ∈ C ∀ ∈ 1 2 3 ′ ea11 , Aea11 , A ea11 , A ea11 , will be infinite). Suppose that k1 is the minimal integer ··· ′ k′ k′ 6 6 1 1 6 satisfying this condition (1 k1 n). It is clear that A ea11 = ea11 , A ea1j = ea1j , (1 ′ ′ ′ j 6 k1). It is possible that k1 = 1 or n. So, at last F1 = G1 = k1. Then remove the elements in G from , remove the elements in F from| |S. | | 1 C 1 2 Now, if S = Ø, let a21 = min S, F2 = [a21], G2 = [ea21 ]. It is clear that Aea21 . If i−1 6 i−1 ∈ C A ea = ea , suppose A ea = ea i (i = 2, 3, ), then add a2i and ea i to the ends 21 21 21 2 ′ 2 6 ··· ′ k2 of the sequences F2 and G2, respectively. There will be a k2, such that A ea21 = ea21 (let ′ ′ ′ k2 be the minimal integer satisfying this condition. It is possible that k2 = 1 or n k1). k′ ′ − Obviously, A 2 e = e , (1 6 i 6 k ). Then remove the elements in G from , remove a2i a2i 2 2 C the elements in F2 from S. If S =Ø, continue this step and constructing F3, F4, and G , G , . It will stop since n 6is finite. ··· 3 4 ··· u Assume that we have F1, F2, , Fu and G1, G2, , Gu, such that Fi = { 1, 2, , ··· ··· i=1 ··· u S n }, Gi = { e1, e2, , en }, Fi Fj = Gi Gj = Ø, (1 6 i = j 6 u). There is a i=1 ··· ∩ ∩ 6 possibilityS that u = 1 (when A is a cycle matrix of order n) or n (when A is an identity matrix). Sort F , F , , F by candinality, we will have F ′ 6 F ′ 6 6 F ′ . Then 1 2 ··· u | 1| | 2| ··· | u| 1 Here the regular letter “T” in the upper index means transposition. 2 Otherwise Aea21 G1, since all the elements in G1 are removed from , then there is a k0, s. t. k0 ∈ k0−1 C , A ea11 = Aea21 , of course k0 = 0, so, A ea11 = ea21 as A is invertible, which means that ea21 = k0−1 6 A ea11 is in the set G1. Contradiction. 4 Proof 5

′ ′ ′ ′ sort Gi correspondingly, i.e., Gi = ex x Fi (i = 1, 2, , u). Suppose F1 = F2 ′ { | ∈ } ··· | | | | = = Ft = 1. It is possible that t = n (when A is an identity matrix) or 0. Let ··· | | ′ ′ ′ r = u t. Denote the unique element in Gi by ei (i = 1, 2, , t). Let kj = Gt+j , and − ′ ′ ··· denote the elements in Gt+j by ej,v (j = 1, 2, , r; v = 1, 2, , kj). Hence, the matrix A ··· ··· ′ ′ ′ of the restriction of in the subspace spanned by the bases 0 = { e1, e2, , e t } is I since A e′ = e′ (i = 1, 2, , t), or A = I ; and the matrixD of the restriction··· of t i i ··· D0 D0 t A in the subspace spanned by the bases = { e′ , e′ , , e′ } (j = 1, 2, , r) is Dj j,1 j,2 ··· j,kj ··· 0 1 1 0   .. A ′ ′ Nki = 1 . , a cycle matrix of order kj, as ej,v = ej,v+1, (v = 1, 2, ,  . .  ···  .. ..     1 0   A ′ ′  A A kj 1), and ej,kj = ej,1, i.e., j = jNj (j = 1, 2, , r). So, the matrix of with − ′ ′ ′ ′ ′D D ′ ′ ··· ′ ′ bases = { e1, e2, , et, e1,1, e1,2, , e1,k1 , , er,1, er,2, , er,kr } is B = It Nk1 D N = diag···I , N , , N···. ••• ··· ⊕ ⊕··· ⊕ kr { t k1 ··· kr } Since is a reordering of , so there is a permutation matrix T , such that = T . Then BD = T −1AT . So TheoremB 3 is proved. D B

Proof of Theorem 1

′ With Fi generated above, construct a 0-1 matrix Dt of order n, such that the j’th column t ′ of Dt is the j’th column of A (j Fi ) and the other columns of Dt are 0 vectors. Of ∈i=1 course, Dt is a diagonal matrix ofS rank t, as the j’th column of A is ej (by definition, Aej = ej). Construct a 0-1 matrix Qi ( i = 1, 2, , r ) of order n, such that the j’th ′ ··· column of Qi is the j’th column of A (j Ft+i) and the other columns of Qi are 0 vectors. t r u ∈ As F ′ F ′ = F ={1,2, , n }, and F F = Ø, (1 6 i = i 6 u), i t+i i ··· i1 ∩ i2 1 6 2 i=1  i=1  i=1 S S S S r so every column of A appears exact once in a matrix in the expression Qi + Dt in the i=1 same position as it appears in A, besides, the columns in the same positionP in all the r other addend matrices in the expression Qi + Dt are all 0 vectors, so, i=1 r P

Qi + Dt = A. i=1 X

Now we prove that Qi is a generalized cycle matrix of type II with cycle order ki. Suppose the members in F ′ (i = 1, 2, , r) are a′ , a′ , , a′ . Suppose F ′ = F t+i ··· i,1 i,2 ··· i,ki t+i s for some s (1 6 s 6 u), 3 By the definition of F , we know that Ae ′ = e ′ , (v = 1, s ai,v ai,v+1

3 ′ Then we have a relation about the members in Gt+i and the members in a certain Gs, i.e., ′ e = ea′ , (v =1, 2, , ki), i,v i,v · · · 4 Proof 6

′ ′ ′ ′ 2, , ki 1), Aea = ea , so ea is the ai,v’th column of A. ··· − i,ki i,1 i,v+1 As Qi is made of the 0 vector and ki columns of A. The columns of A are linear inde- ′ ′ pendent, so the rank of Qi is ki. While the ai,v’th column of Qi is the ai,v’th column of ′ ′ ′ ′ ′ A, so Qiea = ea , (v = 1, 2, , ki 1), Qiea = ea , Qiel =0( el Gt+i ). i,v i,v+1 ··· − i,ki i,1 ∀ ∈ D v ki v Therefore Q ea′ = ea′ (v = 1, 2, , ki 1), Q ea′ = ea′ , (so Q is not diagonal as i i,1 i,v+1 ··· − i i,1 i,1 i  v ki v−1 ki−v+1 v−1 Q ea′ = ea′ = ea′ ). Then Q ea′ = Q Q ea′ = Q ea′ = ea′ , (v i i,1 i,v+1 6 i,1 i i,v i i i,v i i,1 i,v = 1, 2, , k ). Hence Qki is a diagonal matrix of rank k . Therefore Q is a generalized ··· i i i i cycle matrices of type II with cycle order ki.

Proof of Theorem 2

(b) a Let Dt = In Dt. Construct a 0-1 matrix Ji (i = 1, 2, , r) of order n, such that the − a ′ ··· a j’th column of Ji is the j’th column of In (j Ft+i) and the other columns of Ji are 0 r ∈ (b) (a) (a) (b) (b) vectors. Let Ji = In Ji . So, Ji + Dt = In. It is obvious that Dt Dt = DtDt − i=1 (a) (b) (b) (a) (bP) (b) (a) (a) (b) = 0, Ji Ji = Ji Ji = 0, QiJi = Ji Qi = 0, and Qi1 Ji2 = Ji2 Qi1 = 0, Qi1 Ji2 = J (b)Q = Q = 0 (1 6 i = i 6 r). Although J (a)J (a) = J (a)J (a) = 0, but J (a)J (b) = i2 i1 i1 6 1 6 2 i1 i2 i2 i1 i1 i2 J (b)J (a) = J (a) = 0. It is not difficult to prove that J (b)J (b) = J (b)J (b) = I J (a) J (a). i2 i1 i1 6 i1 i2 i2 i1 n − i2 − i1 (b) (a) Let Pi = Qi + Ji , so rank Pi = n, In + Qi = Pi + Ji . P P = Q + I J (a) Q + I J (a) = Q + Q + I J (a) J (a), then i1 i2 i1 n − i1 i2 n − i2 i1 i2 n − i1 − i2     r r r r r P = Q + I J (a) = Q + I J (a) = Q + D = A. i i n − i i n − i i t i=1 i=1 i=1 i=1 i=1 Y Y   X X X

We can prove the equality above in another way. It is clear that Q Q = 0, Q D = D Q = 0 (1 6 i = i 6 r). So i1 i2 i1 t t i1 1 6 2 r r

(In + Dt) (In + Qi)= In + Qi + Dt = In + A. (4.1) i=1 i=1 Y X

(a) (a) Since DtQi = QiDt = 0, so DtJi = Ji Dt = 0. By definition, when 1 6 i 6 r, the r t ′ ′ v’th column or the v’th row of Pj with v Ft+i Fi are same as the v’th ∈ i=1 16j6r   jQ=6 i S S column or the v’th row of In, respectively, (since the v’th column or the v’th row of Pj t (1 6 j 6 r, j = i) with v F ′ F ′ are the same as the v’th column or the v’th 6 ∈ t+i i i=1  S S here ea′ Gs. i,v ∈ 4 Proof 7

r (a) row of In), so when Ji is multiplied by Pj (no matter from the left or right), the 16j6r jQ=6 i v’th column and the the v’th row will not change, while the other columns and rows of r r (a) (a) (a) Ji is 0 vector, so Ji Pj = Ji . For the same reason, Dt Pi = Dt. As i=1 16j6r jQ=6 i Q

r r (a) (In + Dt) (In + Qi)=(In + Dt) Pi + Ji i=1 i=1 Q Q   r r r r r   (a)  (a) = (In + Dt) Pi + Ji Pj = (In + Dt) Pi + Ji i=1 i=1 i=1 i=1   6 6    Q P  1 Qj r  Q P   j=6 i  r r r  r r r r  (a)  (a)  (a) = Pi + Ji + Dt Pi + Dt Ji = Pi + Ji + Dt +0= Pi + In, i=1 i=1 i=1 i=1 i=1 i=1 i=1 thatQ is, P Q P Q P Q

r r

(In + Dt) (In + Qi)= Pi + In. (4.2) i=1 i=1 Y Y

r By equations (4.1) and (4.2), we will have Pi + In = In + A, i=1 r Q so Pi = A. i=1 Q Now we prove that Pi is a generalized cycle matrix of type II with cycle order ki. m Since P = Q +J (b), Q J (b) = J (b)Q = 0, so P m = Qm + J (b) = Qm +J (b) ( m Z+), i i i i i i i i i i i i ∀ ∈ k k (b) k then P i = Q i + J . As Q i e ′ = Aki e ′ = e ′ , (v = 1, 2, , k ); Q e =0= i i i i ai,v ai,v ai,v ··· i i l ⇒ k (b) (b) Q i e =0( e G′ ). On the other hand, J e ′ =0(v = 1, 2, , k ), J e = i l ∀ l ∈ D t+i i ai,v ··· i i l ′ ′ ki (b) ki ′ el, ( el Gt+i ). So for any el in , if el Gi, Qi + Ji el = Qi el = el, if el / Gi, ∀ ∈D B ∈ ∈ Qki + J (b) e = J (b)e = e . That means P ki e = Qki + J (b) e = e ( e ). So i i l i l l i l i i l l ∀ l ∈ B ki ki ki ki Pi (e1, e2,  , en)=(e1, e2, , en), or Pi In = In, hence Pi = In. (Actually, Qi = (a) ki··· ki (b) ···(a) (b) m Ji , so Pi = Qi + Ji = Ji + Ji = In.) When 1 6 m

000010 000100  010000  For instance, for the matrix P = , P e = e , P e = e , so F = 1  001000  1 1 5 1 5 1 1    100000     000001    [1, 5], F1 = 2; P1 e2 = e3, P1 e3 =e4, P1 e4 = e2, so F2 = [2, 3, 4], F2 = 3; P1 e6 = e6, | | | | F3 = [6], F3 =1. So P1 is permutationaly similar to the canonical form B1 = diag I1, | | 1 { 0 1  1 0  N , N = , or 2 3}  0 0 1     1 0 0     0 1 0      1 0 1  1 0  P e ; e , e ; e , e , e = e ; e , e ; e , e , e = e ; e , e ; e , e , e . 1 { 6 1 5 2 3 4} { 6 5 1 3 4 2} { 6 1 5 2 3 4}  0 0 1     1 0 0     0 1 0      0 1 0 0 1 0 0 1 0 1  0 1   0 1  As e ; e , e ; e , e , e = e , e , e , e , e , e , let T = , { 6 1 5 2 3 4} { 1 2 3 4 5 6}  0 1  1  0 1       1   1       1 0 0   1 0 0      0 1 000010  0 1 0  1 0 000100 0 1  000010   010000   0 1  then T −1 = T T = , so = 1 1  1   001000   0 1         1   100000   1         1   000001   1 0 0        1  0 1     0 1 1 0  1 0   000010  , ( P = T B T −1).  0 0 1   1  1 1 1 1      1 0 0   1       0 1 0   1          5 On the Number of Permutation Similarity Classes 9

1 0 1 000010 0 1 0 0 1 1 0 000100 0 1  1 0   000010   010000   0 1  = ,  0 0 1   1   001000   0 1           1 0 0   1   100000   1           0 1 0   1   000001   1 0 0   −1        (B1 = T1 P1T1).       

5 On the Number of Permutation Similarity Classes

The number of permutation similarity classes of permutation matrices of order n is the partition number p(n). There is a recursion for p(n), p(n)= p(n 1)+ p(n 2) p(n 5) p(n 7) + + − − − − − − ··· 3k2 k ( 1)k−1p n ± + − − 2 ······   k1 3k2 + k k2 3k2 k = ( 1)k−1p n + ( 1)k−1p n − , (5.1) − − 2 − − 2 Xk=1   Xk=1   (Refer [7], page 55), where √24n +1 1 √24n +1+1 k = − , k = , (5.2) 1 6 2 6     and assume that p(0) = 1. Here x is the floor function, it stands for the maximum integer that is less than or equal to⌊ the⌋ real number x. We may find a famous asymptotic formula for p(n) in references [11] or [1], 1 2 p(n) exp πn1/2 . (5.3) ∼ 4n√3 r3 ! This formula is obtained by Godfrey H. Hardy and Srinivasa Ramanujan in 1918 in the famous paper [3]. (In [2] and [9], we can find two different proofs of this formula. The evaluation of the constants can be found in [8].) Formula 5.3 is very import for analysis in theory. It is very convenient to estimate the value of p(n) especially for ordinary people not majored in mathematics. 4 But the accuracy is not so satisfying when n is small. In [6], several other formulae modified from formula (5.3) is obtained (with high accuracy). Such as

2 exp 3 π√n 1 p(n) + , 1 6 n 6 80. (5.4)  q ′   ≈ 4√3(n + C2(n)) 2       4 Compared with another famous formula in convergent series found by Rademacher in 1937, based on the work of Hardy and Srinivasa Ramanujan, refer [7] or [10]. 6 Result on Monomial Matrix 10 with a relative error less than 0.004%, where 0.4527092482 √n +4.35278 0.05498719946, n =3, 5, 7, , 79; C′ (n)= × − ··· 2 0.4412187317 √n 2.01699 + 0.2102618735, n =4, 6, 8 , 80. ( × − ··· and

2 exp 3 π√n 1 p(n) + , n > 80 (5.5)  q   ≈ 4√3(n + a2√n + c2 + b2) 2       with a relative error less than 5 10−8 when n > 180. Here a = 0.4432884566, b = × 2 2 0.1325096085 and c2 =0.274078.

6 Result on Monomial Matrix

For any monomial matrix M, it can be written as the product of a permutation matrix P and an invertible diagonal matrix D. 5 For the permutation matrix P , there is a −1 permutation matrix T such that T P T = Y is in canonical form diag It, N1, , N as mentioned in Theorem 3. In the expression T −1P T , the permutation{ matrix T···−1 r} changes only the position of the rows, T just changes the position of the columns, neither will change the values of the members, as the non-zero members in M and P share the same positions, so do T −1MT and T −1P T . Suppose the unique non-zero element in the −1 −1 i’th row of T MT is ai, the unique non-zero element in the i’th column of T MT is bi, i = 1, 2, , n. Let D3 = diag { a1, a2, , an }, D4 = diag { b1, b2, , bn }, then −1 ··· ··· ··· T MT = D3Y = YD4.

It It N N  1  −1  1  −1 So M = D1T . T = T . T D2 .. ..      Nr   Nr       It   It  N N  1  −1  1  −1 = TD3 . T = T . D4T . .. ..      Nr   Nr          Acknowledgements

The author would like to express the gratitude to his supervisor Prof. LI Shangzhi from BUAA (Beihang University in China) for his valuable advice.

5 Turn all the non-zero elements in M into 1, then we will have a permutation matrix P . Suppose the unique non-zero elements in the i’th row of M is ci, the unique non-zero elements in the i’th column of M is di, i = 1, 2, , n. Let D1 = diag { c1, c2, , cn }, D2 = diag { d1, d2, , dn }, M = PD2 · · · · · · · · · = D1P . REFERENCES 11 References

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