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 Copyright © 2011 SciRes. APM 106 S. LAVALL E´E 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- Copyright © 2011 SciRes. APM S. LAVALL E´E 107 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 Copyright © 2011 SciRes.