Advances in Pure Mathematics, 2011, 1, 105-117 doi:10.4236/apm.2011.14023 Published Online July 2011 (http://www.SciRP.org/journal/apm)
Left Eigenvector of a Stochastic Matrix
Sylvain Lavalle´e Departement de mathematiques, Universite du Quebec a Montreal, Montreal, Canada E-mail: [email protected] Received January 7, 2011; revised June 7, 2011; accepted June 15, 2011
Abstract
We determine the left eigenvector of a stochastic matrix M associated to the eigenvalue 1 in the commu- tative and the noncommutative cases. In the commutative case, we see that the eigenvector associated to the eigenvalue 0 is (,,NN1 n ), where Ni is the ith principal minor of NMI= n , where In is the 11 identity matrix of dimension n . In the noncommutative case, this eigenvector is (,P1 ,Pn ), where Pi is the sum in aij of the corresponding labels of nonempty paths starting from i and not passing through i in the complete directed graph associated to M .
Keywords: Generic Stochastic Noncommutative Matrix, Commutative Matrix, Left Eigenvector Associated To The Eigenvalue 1, Skew Field, Automata
1. Introduction stochastic free field and that the vector 11 (,,PP1 n ) is fixed by our matrix; moreover, the sum 1 It is well known that 1 is one of the eigenvalue of a of the Pi is equal to 1, hence they form a kind of stochastic matrix (i.e. the sum of the elements of each noncommutative limiting probability. row is equal to 1) and its associated right eigenvector is These results have been proved in [1] but the proof the vector (1,1, ,1)T . But, what is the left eigenvector proposed in this paper are completely different. associated to 1? Indeed for the major part the proof in these two In the commutative case, this eigenvector is the vector instances, we use elementary operations on the rows and
(,,)M1 M n , where M i is the i th principal minor the columns of the matrix. of the matrix. This formula is known in probability theory; it amounts to finding the stationary distribution 2. Commutative Case of the finite Markov chain whose transition matrix is the stochastic irreducible matrix. The commutative case is well known in probability In the noncommutative case, we must involve inverses theory. Indeed, we calculate the limit probability of a of elements of the skew field and as these may be finite Markov chain by replacing the stochastic matrix undefined, we take a generic noncommutative stochastic M by M I, where I is the identity matrix of the matrix: this is the matrix ()aij of noncommuting vari- appropriate dimension. For this, we use the Markov ables aij subject only to the stochastic identities; i.e. the chain tree Theorem, where this calculus is expressed in sum of each row equals one. terms of spanning trees. The Markov chain tree theorem We work in the free field generated by these variables gives a formula for the stationary distribution of a finite (in the sense of Paul Cohn), which we call the stochastic Markov chain. Equivalently, this formula gives a row free field. Considering the complete digraph on the set vector fixed by a matrix fixing (1,1, ,1)T . This
{1, ,n }, let M i be the set of paths from i to i . Let theorem is attributed to Kirchoff by Persi Diaconis, who
Pi be the paths starting from i and not passing gives a probabilistic proof of it (see [2] p. 443 and 444). through i again. We identify Pi with the noncom- See also [3,4]. mutative power series which is equal to the sum of all the The proof uses only the definition of the determinant words in Pi and we still denote this series by Pi . Next which involves the principal minors. 1 we show that the elements Pi can be evaluated in the Proposition 1 Let M be a stochastic matrix and
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NMI==() nijijb 1,n; i.e. for all in=1, , , Proof. By definition of N , we have det N = 0 . n We show that j=1bij =0. Then (,,NN1 n ), where Ni is the i th (,NN ,, N )=0N principal minor of N , is the left eigenvector associated 12 n to the eigenvalue 0. We verify this Equation with the first column of N .
Nb111 Nbnn1
bb22 23 b2n bb11 12 b1,n 1
bb32 33 b3n bb21 22 b2,n 1 = bb11 n1
bbnn23 b nn bbnn1,1 1,2 b nn 1, 1 n bb1121,1jn b j=2 bb b 22 23 2n n
bb32 33 b3n bb2222,1jn b = b11 j=2 bn1 bb b nn23 nn n bbnj1, n1,2bnn 1, 1 j=2
bb22 23 b 2n bb12 12 b 1,n 1
bb32 33 b 3n bb22 22 b 2,n 1 = bb11 n 1
bbnn23 b nn bbnn1,2 1,2 b nn 1, 1 =0
bb1,nn 1 12 b 1, 1 bb1121,1nn b
bb2,nn 1 22 b 2, 1 bb2222,1nn b bbnn11
bbnn1, 1 n 1,2 bnn 1, 1 bbnn1, n 1,2 b nn 1, 1 =0
bb22 23 b2n bb1121,1nn b
bb32 33 b3n bb2222,1nn b = bb11 n1
bbnn23 b nn bbnn1, n 1,2 b nn 1, 1
bb22 23 b2n bb12 13 b1n
bb32 33 b3n n2 bb22 23 b2n =(1b11 ) bn1
bbnn23 b nn bbnn1,2 1,3 b nn 1,
bb22 23 b2n bb12 13 b1n
bb32 33 b3n n1 bb22 23 b2n =(1)bb11 n1
bbnn23 b nn bbnn1,2 1,3 b nn 1, =detN =0
Remark 2 If M is reducible, then this proposition is irreducible and stochastic, then its adjoint will be of the asserting that the zero vector ia an eigenvector. If M is form (1, ,1)uT , where uT is a null vector. Te pro-
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position follows from the cofactor formula for the adjoint. that its characteristic series wL wA is ratio- nal. We shall identify a language and its characteristic 3. Noncommutative Case series. This is exposed in [5] or [6].
In [1], the authors have prove in two manners that are 3.2. Free Field actually similar, Theorem 9. In the first, results from variable-length prefix codes are required; and the second Let be a field. If the multiplicative group of is deals with general variable-length codes, not necessarily commutative, then is a commutative field. Else, is called skew field. The ring of rational formal series in prefix. Moreover, in Appendix 2 of [1], we see how the rat theory of quasideterminants may be used to obtain these noncommutative variables A is not a skew results on noncommutative matrices. field. However, skew fields containing A do exist. The major part of the new proof of this Theorem One of them is the free field (in the sense of Cohn), involves only the elementary operations on the rows and denoted . The free field is the noncommutative on the columns of the stochastic matrix. Therefore, we analogue of the field of fractions of the ring of commu- need the following results. tative polynomials. There are several constructions of the free field: Amitsur [7], Bergman [8], Malcolmson [9], 3.1. Languages and Series Cohn [10] and [11]. A square matrix of order n is full if it can’t expressed Let A be a finite alphabet and A* be the free monoid as a product of matrices nn 1 by nn1 . generated by A . A language is a subset of a free Theorem 3 ([11], Thm 4.5.8.) Each full matrix M monoid A* . A language is rational if it is obtained from with coefficients in X is invertible in the free field. finite languages by the operations (called rational) union, A square matrix of order n is called hollow if it product (concatenation) and star. The product of two contains a pq block of zeros such that pqn 1. languages LL12 is ww12 w 1 L 1, w 2 L 2 , and the star 001 of L is LwwwLn* =, 0= Ln . Ra- Example 4 The matrix 001 is hollow since 1 ni n0 tional languages may be obtained by using only unam- 111 biguous rational operations; these are: disjoint union, there is a 22 block of 0 and 4>3. unambiguous product (meaning that if wLL 12, then Proposition 5 ([11], Prop. 4.5.4) A hollow matrix is w has a unique factorization ww= 12w, wii L) and not full. * the star L restricted to languages which are free Proposition 6 Take 1n , M nn and * submonoids of A . n1 , where is a skew field. We suppose that A formal series is an element of the -algebra of M noncommutative series A , where A is a set of M is invertible. Then is invertible if and 0 noncommuting variables. A rational series is an element 1 of the smallest subalgebra of A , which contains only if M 0 . the -algebra of noncommutative polynomials A , Proof. 1 and which is closed under the operation () Suppose that cM= 0, then the inverse M * n 1 of is SS ==1 S S n=0 0 which is defined if S has zero constant term. We MMcMMc11111 1 denote by A rat the -algebra of rational series. cM11 c 1 Let L be a rational language. Since L may be obtained by unambiguous rational expressions, it follows Indeed, this matrix is the right inverse of M since
M MM111111 c MMc 11 1 0 cM c MM()(111111 M c M cM MM 11 c) c1 = 1111 11 ()MMcM Mc 11 11 1 1 1cM cM c c 10 = MMcMMc111111 = 01 ==cc
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This matrix is also the left inverse of M since
MMcMMc111111 M 11 1 cM c 0 ()()MMcMMMc1 111 11 MMcM 1 111 = 11 1 11 ()cMMc () cM 1Mc11 Mc 11 M 1 Mc 11 M 1 10 = =c = 11 11 cc cM 01 =c
M In other words, the sum of each line of S is equal to () Suppose that is invertible. Let 1; hence S is a stochastic matrix. We call S a gene- 0 ric noncommutative stochastic matrix. The algebra over M be its inverse. Then we have generated by its coefficients is a free a ssociative algebra, since it is isomorph to the algebra aiij , j. Indeed, we ca n eliminate the a with the stochastic 0 ii relations (1). We denote this alge bra by a /(1) . ij M Hence, ther e is a corresponding free field called = 00 stochastic free field denoted S . 1 3.4. Paths 0 Hence M == 0 and =1. It follows Consider the set of nonempty paths in the complete directed graph with a set of vertices 1, ,n starting 0 from i and not passing through i ; we denote by P that i the sum in a of all the corresponding words. It is 1 ij = M classically a rational series, and thus defines an element == M 1 1 of the free field .
1 ab M 0 Example 8 M = . The graph is cd From this proposition, we deduce th e follo wing result. Corollary 7 Let , M and be matrices with M coefficients in A where M is full. If is 0
M not full (in particul ar, if is equivalent by then 0 ** elementary operations on the lines and on the columns to PbdPc12=1 , =1 a 1 a hollow matrix) then M =0 in the free field. We shall also consider rational expressions over any skew field D , and say that such an expression is 3.3. Generic Stochastic Matrices evaluable in D if it can be evaluated without inversion of 0. I f the elements of D appea ring in the rationa l Let M = a be a generic noncommutative matrix; ij 1,ij n expression are actua lly in a subring R of D , we say that the expression is over R . i.e. the aij are noncommutative var iables. We denote rat the corresponding free field. We associate to M There is a cano ni cal embedding of A into , which can be seen as follows: let S be any rational the matrix S : it’s exactly the same matrix of which the * 1 coefficients satisfy the stochastic identities series, we replace in it the operation T by 1T ; then one obtains a rational expression in , which is n evaluable in and represents the image of S under ina=1, , , ij =1 (1) rat j=1 the embedding A . Thus, each rational
Copyright © 2011 SciRes. APM S. LAVALL E´E 109 language and each rational series is naturally an element n PLi=,=1,, n, of the free field. See [12]. iij j=1 By Theorem 1 of [1], the Pi can be evaluated in the stochastic free field. where We can now state the main result of this paper. 1, if i= j Theorem 9 Let M = aij be a generic nn 1,ij n Lij = La =,a La ifi j ik kj ij ik kj stochastic noncommutative matrix. Let Pi be defined as kk=1 =1,k j above. Then
Let Si be the system composed of the n Equa- 11 11 P11,, PPPnnM = ,, (2) tions Pi , LLLiiiii1,1,1,, , ,, Lin. Multip lying to the 1 11 left by Pi each equation of Si , we obtain the system In other words, PP1 ,, n is the left eig envector n associated to the eigenvalue 1 o f M . 11 1= PPLiii ij, =1, ,n Example 10 (Continued.) Let be S be the m atrix jji1, T : M with ab =1 and cd =1. We show that i n PL1= Pa1 PLa1 iij iij iikkj 11 11 kki1, P12,=,PPPM 12 1 Let QPLij = i ij , then we have From this system, we deduce the two following equa- n tions: 1 11 1 1= PQiii j, =1, ,n Pa12 P 1=d P1 (3) jji1, T : i n P111=aPdP1 (4) QPa= 1 Qa 12 2 ij i ij ik kj kki1, From the Equation (3), we have This system transforms into the following system 11 1 Pa12 P1= d P 1 n 1 1= PQiniij , =1, , 11 jji1, PdPa211=1 Ui : n 1 ** 0=Paiijijjj Q a1 Qaikkj dP21= aP (5) kki1, , kj dcaabd*1=1*** We show that n d**d1= da * aa ** 1 ad * 11 Pakk11 P k =1 ** ** daad= Define Hence, Equation (3) is satisfied. From the Equation (4), n RPa= 1 P1 (6) we have kk11 k =1 111 PaPdP121 =2 We show that R =0. So Equation (6) is converted
11 into the following Equation Pa121=1 d n 11 ** RP11111= a Pakk 0 (7) aP12= dP k =2 Since we again obtain Equation (5), it follows that Consider the system of the n2 1 Equations: Equa-
Equation (4) is also satisfied. tion (7) and the n systems UU1 ,, n . We have the Proof of Theorem 9. matrical representation
Let M = a be the matrix of the automaton 11 ij 1,ij n RP,,,,,,,,,,112131221232 Q Q Qnn P Q Q Q , and P be the language of the labels of the paths i PQ1,,, Q E =λ starting from i and not passing through i . We have nn1,1 nn
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aamm12 a mmmmmn,1,1 a a aa1 a a a 11 12 1,mm 1 1, 1 1n aa21 22 1 a2,mm 1 a2, 1 a2n A = m aa a1 a a m1,1 m 1,2 mm 1, 1 mm 1, 1 mn 1, aam1,1 m 1,2 a mm 1, 1 a mm 1, 1 1 a mn 1, aann,1 ,2 a nmnmnn, 1 a, 1 a1 where mn=1, , and =(0, ,0,1, ,1). E is a ntimes square matrix of order n2 1. We have RE=, 1
T E =(1,0, ,0) and F = . The order of this 0 n2 times matrix is n2 2 . Moreover, with the stochastic identities, we have where the Bi are defined by
aamm12 a m ,m1 a m ,m1 a mn n aa11km2 a1,11 a,1m a 1n kk=1, 1 n aa21 2kma2, 1 a2, m 1 a2n kk=1, 2 n Bm = am1,1 am1,2 aam1,km1,m 1 a m 1,n kk=1, 2, km1 n aam1,1 m1,2 amm 1, 1 amk 1, amn 1, kkm=1, 1 n aa a aann,1 ,2 nm,1 nm,1 nk kkn=1,
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To show that R =0, we show that the matrix F is Step 2. We want a nn block of 0 in the right upper hollow. For this, we apply elementary operations on the corner of F1 . For this, we consider the j - th row of rows and on the colum ns of F until F contains a F1 (correspo nding to the j - th row of B1 ). By s t block of 0 such that stn2 3. construction of B1 , the first index of the elements of the j - th row of this block is j . Now, we consid er the
row which passes in the block B2 such that this first index of its elements is j . Such a row exists by
construction of B2 . We repeat this operation wi th the row which have the first index equals to j in the blocks
BB3 ,, n . We add each one of these rows to the j - th
row of F1 . It follows that the n last elements of the j first rows of this new matrix are all equal to 1. From each row of the j first rows, we subtract the last row of F . 1 We obtain the following matrix Step 1. We eliminate the first rowe and the last column of F . We obtain the square matrix F1 of order n2 1 (stn2 2 ).
where n aa11k2 a1,11ma1,ma 1n kk=1, 1 n aa21 2kmaaa2, 1 2,m 1 2 n Cm = kk=1, 2 n aa aa a nn,1 ,2 nmnm,, 1 , 1 nk kkn=1,
mn=2,, . We reduce F2 in removing the last line LLiknknmk (1)= L L(1)n, to obtain the matrix and the nth column if we start from the end. We 2 have the following matrix F3 of order n , ( s tn 1).
where, for all mn=2, , , DB=,=,=dbdmmbm bm m ij1,ijn m ij ij ij nj Step 3. Consider the rows Li of F3 , where in> and in 0mod . We want that the (1)n last ele- We remove the (1)n rows Li where in> and ments of these rows to be equals to 0. Let ik= nm , ni and the (1n ) last columns of F4 to obtain the 2 where 01mn, then we apply the transformation matrix F5 of order ( st n n2 ).
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( st n2 n2 ).
where Dm is the matrix obtained from Dm in remo- ving the last row. where B1 is the block obtained from B1 by removing 1 the first column. We find a nn(1 )2 1 block of 0 Step 4. Denote by Cm the first column of the matrix located in the right upper corner and Cm and by dm the column of F5 , which is the ex- 1 1 tension of C . We want C to be transformed into a 2 22 m m nn11 =nn 2 nn 1=dimF vector of 0, mn=2,, . For this, we apply the 8 It follows that F is a hollow. From Corollary 7, operation dccj 23 cn , where ci is the it h column of F . 1n 11 11 3 RE==k =1Pakk1 P1 =0. Hence P1 ,, Pn Finally we want that the first n elements of the firs t is the left eigenvector associated to 1. column of F5 to be all 0; for this, we apply the aa11 12a 13 transformation cc12 cn . We obtain the matrix F6 2 2 Example 11. Let M = aaa be a generic of order nn1 (stnn2 ) 21 22 23 a31aa 32 33 noncommutative matrix in w hich the variables satisfy the
stochastic ide ntities aaaii123 i=1 , i = 1,2,3 . We associate the a utomato n
where Cm is the ma trix obt ained from Cm by removing the first column.
Step 5. We want each Cm will be transformed into a q matrix of 0. To each colum n q of Cm , denoted Cm
q T correspon ding to the column dqm=C ,*,,* of F . By construction of C (hence of C ), there exists 6 m m Let P be the language recognized by the automaton T 1 a column b of F such that b =,*,,C q *. q 6 qm
Hence, for all q , we calculate dbqq . We obtain the 2 2 matrix F7 of order nn1 (stn n2 ).
Let L be the language of words which start from i ij to j . We have PL=, L L where Step 6. Permuting the fir st and the nth column of 1111213 2 L11 =1 F7 , we obtain the matrix F8 of or der n n1
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LaLaLa12= 12 12 22 13 32 Permuting the index of the equations of U1 , we obtain the two following systems LaLaLa13 = 13 12 23 13 33 1 We have the system 1= PQQ22123 PLL1 1 12 13 UPaQaQ2 :0= 2 21 12 11 1 23a 31 SL:=aLaLa 1 1 121212221332 0=Pa Qa Q a 1 22321132333 LaLaLa13= 13 12 23 13 33 1 1 1= PQQ33132 Multiplying to left each equation of S1 by P1 , we 1 obtain the system UPaQaQ3:0= 3 31 31 111 32a 21 1= PPLPL11 1 1 1112113 0=Pa33231123222 Qa Q a 1 111 1 TPLP1:= 1 12 1 aPLaPLa12 1 12 22 1 13 32 111 111 To show that PPP123,,M = PPP 123 ,,, it 111 1 111 1 PL1 13= Pa 1 13 PLa 1 12 23 PLa 1 13 33 suffices to prove that Pa111 Pa 2213311 Pa= P. 1 Define Define QPLij = 1 ij , then RPaPaPaP= 1111 1= PQQ1 1112213311 11213 then 1 111 TQ1:= 12 PaQaQa 1 12 12 22 13 32 RP1112213311= a Pa Pa 0 1 QPaQaQa1311312231333= Under the matrical system, we have
We obtain 11 RP,1,,,,,,,, QQP QQP QQ E =λ 1 1 12 13 2 21 23 3 31 32 1= PQQ11213 1 where UPaQa1 :0= 1 12 12 22 1 Q13a 32 λ = 0,0,0,0,0,0,0,1,1,1 0=Pa1 Qa Q a 1 11312231333 and 1000000000 1 aa a 0000100 11 12 13 0aa 1 0000100 22 23 0aa32 33 10000100 aa00a 00010 E = 21 21 23 000aa11 1 13 00010 000aa 100010 31 33 aa310000 31a 32 001 aa 0000011 1 12 001 00000aa21 22 1001
We have RE= 1 , where T tities equal to = 1,0,0,0,0,0,0,0,0,0 and E with stochastic iden- 1000000000 aa a a 0000100 12 13 12 13 0aa a 0000100 21 23 23 0aaa32 31 32 0000100 aa00 a00010 21 21 23 00 0aa12 13 a13 0 0010 00 0aaa 0 0010 31 31 32 aa31 0000 31 a32 001 00000aa a 001 12 13 12 00000aaa21 21 23 001
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E Let F = , we show that F is hollow. 0
10000000001 aa12 13 a 12 a13 00001000 0aa a 00001000 21 23 23 0aaa 00001000 32 31 32 aa00 a000100 21 21 23 F = 00 0aa12 13 a13 0 00100 00 0aaa31 31 32 0 00100 aa0000 a0010 31 31 32 00000aa a 0010 12 13 12 00000aaa21 21 23 0010 00000001110
aa12 13 a 12 a13 0000100 0aa21 23 a 23 0000100 0aaa 0000100 32 31 32 aa00a 00010 21 21 23
00 0aa12 13 a13 0 0010 F1 = 00 0a31 aa31 32 0 0010 aa31 0 0 0 0 31 a32 001 00 0 0 0aa a 001 12 13 12 00 0 0 0aaa 001 21 21 23 0000000111
LLLL158102491036710,, LLLL LLLL ,
aa12 13 a 12 a13 aa12 13 a 13 aa12 13 a 12 000 aaaa21 21 23 23 a21 a23 a21 aa21 23 000 aaaaaaaaa 000 31 32 31 32 31 31 32 31 32 a00aa00010 21 21 23 00 0aa12 13 a 13 0 0010 F2 = 00 0aaa31 31 32 0 0010 aa31 0000 31 a32 001 00000aa a 001 12 13 12 00000aaa 001 21 21 23 0000000111
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aa12 13 a 12 a13 aa12 13 a 13 aa12 13 a 12 00 aaaa21 21 23 23 a21 a23 a21 aa21 23 00 aaaaaaaaa 00 31 32 31 32 31 31 32 31 32 aaa00 0010 21 21 23 F31=0 0 0aa213 a13 0 0 10 00 0aaa31 31 32 0 010 aa31 000031 a32 01 00 0 0 0aa a 01 12 13 12 00 0 0 0a21 aa21 23 01
LL46,,, LL 56 LL 79 LL 89,
aa12 13 a 12 a13 aa12 13 a13 aa12 13 a12 00 aaaa21 21 23 23 a21 a23 a21 aa21 23 00 aaaaa aaa a00 31 32 31 32 31 31 32 31 32 aaaaaa00 0 000 21 21 31 23 31 32 F4=0 0 0aaaaaa121331133132 0 0 00 00 0aaa31 31 32 0 010 aaa31 00 0 0 31 21 a32aa 21 23 00 00 0 0 0aaaaaa 00 12 13 21 12 21 23 00 0 0 0 aaa21 21 23 01
aa12 13 a 12 a13 aa12 13 a13 aa12 13 a12 aaaa21 21 23 23 a21 a23 a21 aa21 23 aaaaa aaa a 31 32 31 32 31 31 32 31 32 F =00 aaaaaa00 52 1 2131233132 00 0aaaaaa 0 0 12 13 31 13 31 32 aa31 00 0 0 31a 21a 32a 21a 23 00 0 0 0aaaa12 13 21 12aa 21 23
CCC123,, CCC 423 CCC 623 ,
000aa12 13 a13 a12 000aa a a aa 21 23 23 23 21 23 000aaa32 31 32 aa31 32 a32 F621=00aa2131233132aaaa0 0 00 0aaaaaa 0 0 12 13 31 13 31 32 aa31 00 0 0 31a 21a 32a 21a 23 00 0 0 0 aaaaaa12 13 21 12 21 23
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CC53,, CC 72
00aa12 13 000 00aa21 23 a 23 000 00aaa 000 32 31 32 F =00aaaaaa0 0 721 2131233132
00 0aaaaaa12 13 31 13 31 32 0 0 aa31 00 0 0 31a 21a 32a 21a 23 00 0 0 0 aaaaaa12 13 21 12 21 23
CC13
aa13 12 00000 aaa23 21 23 00000 aa a 00000 31 32 32 F =0 0 aaaaaa 0 0 8 212131233132
000aaaaaa12 13 31 13 31 32 0 0 00aa31 0 0 31a 21a 32a 21a 23 0000 0aaaaaa12 13 21 12 21 23
We have a 35 block of zero with 8>7. It follows column. Again, we show that F is hollow, and hence, that F is hollow. By Corollary 7, by Corollary 7, we have R = 0 . This proves Equation 1111 R ==Pa111Pa221Pa3311 P 0. (8). This proof allo ws us to prove the next result. Example 13. (Continued.)
Theorem 12. Let Pi be defined as above, then in 11 11 * we have PP1211= PP 1 ad by Equation (5)
11*1 * 1 n ==1=PPbdP bdPP=1 1 11 1 11 Pi =1 (8) i=1 Proof. Define 4. Conclusions
n 1 R =1 Pi One of the contribution of this article is the using of i=1 elementary operations on the lines and the columns. This we will show that method provides a new way to obtain some results in
n skew field theory with a minimum of knowledge of this 1 R Pi =1 theory. Moreover, Theorems 9 and 12 are proved exactly i=1 the same way. We replace the first column of the matrix E by the vector 5. Acknowledgements
T 1,1,0, ,0,1,0, ,0, ,1,0, ,0 I am obliged to thank to Christophe Reutenauer for his and the first element of λ by 1. Denoted E and λ comments, corrections and suggestions.
E these two matrices. Consider the matrix F = . 6. References 0 We repeat each step of the proof of Theorem 9, except [1] S. Lavall´ee, D. Perrin, C. Reutenauer and V. Retakh, for the last operation of step 4, which concerns the first “Codes and Noncommutative Stochastic Matrices,” To
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