2維有限型的子移位之混合性質

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(1)國立交通大學應用數學系碩士論文. 2 維有限型的子移位之混合性質 The mixing property of 2-dimensional subshift of finite type. 研究生：蔡佩君指導老師：林松山. 教授. 中華民國九十四年六月.

(2) 2 維有限型的子移位之混合性質 The mixing property of 2-dimensional subshift of finite type. 研究生：蔡佩君. Student：Pei-Jiun Tsai. 指導教授：林松山. Advisor：Song-Sun Lin. 國立交通大學應用數學系碩士論文. A Thesis Submitted to Department of Applied Mathematics College of Science National Chiao Tung University in Partial Fulfillment of the Requirements for the Degree of Master in Applied Mathematics June 2005 Hsinchu, Taiwan, Republic of China. 中華民國九十四年六月.

(3) 2 維有限型的子移位之混合性質. 學生：蔡佩君. 指導老師：林松山教授. 國立交通大學應用數學系(研究所)碩士班. 摘. 要. 在這篇論文中，討論 n 階置換矩陣 Α n 的原始性質。而這些主題與 2 維有限型的移位之混合性質有關。我們的目的是給定 2 階置換矩陣 Α 2 的某些必備條件，進而証得矩陣 Α n 的原始性質。這些結果可以被提供去研究有安全符號的矩陣 Α n 之原始性質。在這篇論文中，我們也檢查一些與有限型的矩陣子移位 Λ = Ω(Α2 ) 是弱擴張混合相關的例子，對於這些例子，我們都可証得所有 n ≥ 2 的 n 階矩陣 Α n 是原始的。最後，我們描述一些與 Α n 的原始性質有關的結果。. i.

(4) The mixing property of 2-dimensional subshift of finite type Student：Pei-Jiun Tsai. Advisors：Song-Sun Lin. Department﹙Institute﹚of Applied Mathematics National Chiao Tung University. ABSTRACT. In this paper, the primitivity of n-th order transition matrices Α n defined on Ζ 2×n are studied, this topics related to the mixing property of 2-dimensional shift. of finite type. Our propose is to give some necessary conditions for Α 2 to guarantee the primitivity of Α n . The results can be applied to study the primitivity of Α n which has safe symbol. In the paper, we also check some examples related to the matrix subshift of finite type Λ = Ω(Α 2 ) be extensively weak mixing, and for these examples, we all show that Α n is primitive for all n ≥ 2 . Finally, we describe some results related to the primitivity of Α n .. ii.

(5) 誌. 謝. 這篇論文的完成必須感謝許多協助與支持我的人。首先，我想要感謝的人就是林松山教授，您兩年來耐心的指導與勉勵，讓我得以順利完成此篇論文，此外在學問與待人處世方面，對我也有很大的啟發。謹此，致上我最誠摰的敬意與謝意。還有要感謝班榮超學長和林吟衡學姊，常常給予適時的幫助，告訴我該如何著手，也會提供一些想法來解決問題，尤其是班榮超學長，從他身上看到了對數學的熱情，豐富的知識，在在令人佩服。. 接著，我要感謝我最親愛的父母親蔡進先生和陳素霞女士，因為他們偉大無私的父母愛，讓我得以成長茁壯至今。在求學的過程中，他們總是在背後默默的支持著我，給我最大的空間，讓我做任何我想做的事，讓我可以無後顧之憂的學習。另外，也謝謝哥哥、姊姊和弟弟，在我需要幫助時，永遠是第一個伸出雙手，深濃的兄弟姐妹之情，是支持我向前的原動力。謹此，獻上我最真誠感恩的心，因為沒有他們，就沒有現在的我。. 最後，感謝我在梧棲國小的同事們，香吟、耀仁、婉娥、亞民、靜宜、雪甄、至芸…，在兩年的碩士班生涯中，給我幫助與鼓勵，讓我能兼顧工作與學業。在我心情低落時，大家陪我一起喝茶、聊天；在我茫然不知所措時，大家給我開導與建議，因為有你們，豐富了我的學習路程，謝謝你們，你們永遠是我最要好的支持者。. iii.

(6) 目. 錄. 中文提要英文提要誌謝目錄一、二、 2.1 2.2 三、四、 4.1 4.2 4.3. ………………………………………………………………… ………………………………………………………………… ………………………………………………………………… ………………………………………………………………… Introduction ………………………………………………… Preliminaries ………………………………………………… Definitions …………………………………………………… s n , Sn , Rn and C n ………………………………………… Main Theorem ……………………………………………… Examples for safe symbol existing Case …………………… Find H and V ……………………………………………… Construct H2 and V2 ………………………………………… Checking all cases in section4.2 can let Α n be primitive for all n ≥ 2 ……………………………………………………… 4.4 Conclusion …………………………………………………… References …………………………………………………………………. iv. i ii iii iv 1 4 4 5 7 15 15 16 22 27 31.

(7) 1. Introduction. Many systems have been studied as models for spatial pattern formation in biology, chemistry, engineering and physics. Lattices play important roles in modeling underlying spatial structures. We mention some works arising in biology([1], [3], [22], [24], [25], [28], [29], [30]), chemical reaction and phase transitions([4], [10], [15], [16], [17], [19], [23], [32], [36]), image processing and pattern recognition([11], [13], [15], [16], [18], [20], [21], [26], [35]), as well as materials science([12], [14], [27]). In Lattice Dynamical Systems(LDS), especially Cellular Neural Networks(CNN), the complexity of the set of all global patterns has received considerable attention in recent years([2], [5], [6], [9]). One of the interesting problem comes from the statistic mechanism, is d-demensional shift of …nite type, state as follows, given a list of patterns with shape F 2 Zd , consider the set n o d (1.1) X = XL = x 2 AZ for all n 2 Zd ; and n (x) F 2 L where A is a …nite set, we call it symbol, and without loss of generality, F is d-demensional cube, i.e., F = f(n1 ; :::; nd ) j 1 nk k; 8k = 1; :::; d g, many invariants related to the shift of …nite will discussed likewise in [31], e.g., the topological entropy, measure-theoretical entropy, variational principle, mixing property, and extension problem. Unfortunately, unlike the one dimensional case, it’s extremely di¢ culty to compute and check those invariants, for example, only a very few example of entropy of 2-dimensional shift of …nite type can be computed explicitly, also for mixing property. In this Paper we start to study the mixing property of d-dimension shift of …nite type, and we focus on d=2. In [7], the authors construct a …nite approximation scheme of higher dimensional shift of …nite type, and call it the series of transition matrices in multi-dimensional lattice model in Z2 , we are going to use the structure of such transition matrices to study the mixing property of higher dimensional shift of …nite type. We …rst recall some results in [7], which are crucial in this study. For simplicity, we only consider two symbols which are given on 2 2 lattice Z2 2 . We begin with a consideration of given horizontal transition matrix 1 0 h11 h12 h13 h14 B h21 h22 h23 h24 C C H2 = B (1.2) @ h31 h32 h33 h34 A h41 h42 h43 h44 1.

(8) which is related to a set of admissible local patterns on Z2 hij 2 f0; 1g for 1 The associated vertical transition matrix 0 v11 v12 B v21 v22 V2 = B @ v31 v32 v41 v42. i; j. 2. , and (1.3). 4. V2 is de…ned by 1 v13 v14 v23 v24 C C v33 v34 A v43 v44. (1.4). In 2-dimensional shift of …nite type, one can immediate construct the H2 according to the list of pattern with shape F = f(n1 ; n2 ) j 1 ni 2; 8i = 1; 2g . In [7], H2 and V2 possess the following property to each other 0 1 v11 v12 v21 v22 B v13 v14 v23 v24 C H2;1 H2;2 C H2 = B (1.5) @ v31 v32 v41 v42 A = H2;3 H2;4 ; v33 v34 v43 v44 and. 0. h11 B h13 V2 = B @ h31 h33. h12 h14 h32 h34. h21 h23 h41 h43. 1 h22 h24 C C= h42 A h44. V2;1 V2;2 V2;3 V2;4. (1.6). :. The recursive formula for n-th order horizontal transition matrices Hn de…ned on Z2 n has been obtained [7] by the following procedure: 0 1 v11 Hk;1 v12 Hk;2 v21 Hk;1 v22 Hk;2 B v13 Hk;3 v14 Hk;4 v23 Hk;3 v24 Hk;4 C C Hk+1 = B (1.7) @ v31 Hk;1 v32 Hk;2 v41 Hk;1 v42 Hk;2 A v33 Hk;3 v34 Hk;4 v43 Hk;3 v44 Hk;4 whenever. Hk =. Hk;1 Hk;2 Hk;3 Hk;4. for k 2. The number of all admissible patterns de…ned on Zm be generated from H2 is now de…ned by m n. (H2 ) = Hnm 1 = the summation of all entries in Hnm 2. (1.8) n. which can (1.9). 1.

(9) The quantitative properties of Hn for n 2 are interesting problem in matrix theory and combinatorial dynamics, the most important one is the primitive property, in matrix analysis, the primitivity of a nonnegative matrix will guarantee the positivity of the maximal eigenvalue of a given matrix, and according to the discussion above, if some H2 is induced from some 2dimensional shift of …nite type, then primitivity of H2 demonstrate the shift is mixing. And some interesting dynamics will appear therein, for example, the periodic orbits is dense, and there exists a unique measure of maximal entropy. Thus, it give rise to the study the primitivity of Hn ; 8n 2: The di¢ culties of this study is that the size of Hn grows exponentially, i.e., Hn 2 M2n 2n ; then it’s of nature and interesting to ask that which kind of su¢ cient conditions will guarantee the primitivity for Hn . To overcome this problem, the powerful tool sn ; Sn ; Rn and Cn will be introduced, thus we obtain some checkable conditions of H2 to guarantee the primitivity for Hn ; 8n 2: The paper is organized as follows. Section 2 introduce some de…nitions, sn ; Sn ; Rn and Cn , the main result and proof will presented in section 3, section 4 included some examples of 2 symbols in Z2 and some results related to the primitivity of Hn ; 8n 2:. 3.

(10) 2. Preliminaries. 2.1. De…nitions. In this section, we give some standard de…nitions in matrix analysis related to our study. As mentioned in the introduction, horizontal transition matrix H2 and vertical transition matrix V2 are related to each other. However, in application, usually it is better working on one matrix then the other. Therefore, we use A2 and B2 to replace H2 and V2 throughout this paper, i.e., if A2 = H2 then B2 = V2 and if A2 = V2 then B2 = H2 : Therefore, for simplicity, only A2 is stated herein. First, we de…ne the non-compressible property for a matrix. De…nition 1 We say a matrix A 2 Mn column and row of A are all zero.. n. (Z) is non-compressible if no. In the other word ,if A is compressible, i.e., at least one column or row of A is all zero. Next, we follow the notation from [7] to denote the recursive formulae for n-th order transition matrices An de…ned on Z2 n (or Zn 2 ) ; by An = (An 1 )2n. 1. 2n. E2n. 1. 2. 2n. A2;1 A2;2 A2;3 A2;4. 2. ; 2n. 1. 2n. (2.1). 1. for n > 2; where A2;1 A2;2 A2;3 A2;4. 0. b11 B b13 =B @ b31 b33. b12 b14 b32 b34. and A2; 2 M2 2 (Z), 8 2 f1; 2; 3; 4g : Then, we de…ne the rn and cn below: De…nition 2 If A 2 Mn. n. b21 b23 b41 b43. 1 b22 b24 C C = A2 b42 A b44. (Z) ; we de…ne. r (A) = fi j A (i; j) = 0; 8j g and c (A) = fi j A (j; i) = 0; 8j g ;. and from (2:1), we denote rn and cn for given A2 as rn (A2 ) = r (An ) and cn (A2 ) = c (An ) : 4.

(11) Therefore, we de…ne the safe symbol of a matrix, and the conception of safe symbol comes from [34]. De…nition 3 If A 2 Mn. n. (Z) ; we say index i is a safe symbol if. A (j; i) = 1; 8j 2 f1; :::; ng nr (A) ; and A (i; j) = 1; 8j 2 f1; :::; ng nc (A) : 0 1 1 1 1 0 B 1 0 1 0 C C Example 4 If A2 = B @ 1 1 0 0 A ; then from De…nition2 and 3; for 0 0 0 0 n = 2; we have r2 = f4g c2 = f4g and i = 1 is a safe symbol.. 2.2. sn ; Sn ; Rn and Cn. In this section, sn ; Sn ; Rn and Cn are introduced, these four concepts are de…ned in, and is crucial for our study. First, we de…ne the sn : De…nition 5 If A = (aij )ni;j=1 2 Mn n (Z) has a safe symbol i; we denote s (A) = aii , and from (2:1), we denote sn for given A2 as sn (A2 ) = s (An ) : Next, for (2:1) ; we de…ne the Sn ; Rn and Cn below. De…nition 6 From (2:1), if An 2 M2n 2n (Z) has a safe symbol; we denote S (An ) = A2; S , for S 2 f1; 2; 3; 4g ; be the 2 2 block that s (An ) is inside and we denote Sn for given A2 as Sn (A2 ) = S (An ) : De…nition 7 From (2:1), if An 2 M2n. 2n. (Z) has a safe symbol; we de…ne. R (An ) = fA2; j A2; and S (An ) are on the same rowg ; C (An ) = fA2; j A2; and S (An ) are on the same columng. and we denote Rn and Cn for given A2 as Rn (A2 ) = R (An ) and Cn (A2 ) = C (An ) : 5.

(12) Next, from (2:1) and De…nition7; if A2 has a safe symbol, i.e., s2 exists, we de…ne the proposition below. b 1 b 2 ; 8 2 f1; 2; 3; 4g ; we say A2; has b 3 b 4 property R if it satis…ed one of the follow situations: 8i 2 f1; 2g nr (A2; ) ; Proposition 8 For A2; =. (1) (2) (3) (4). if if if if. s2 s2 s2 s2. = = = =. b11 ; b14 ; b41 ; b44 ;. then then then then. A2; A2; A2; A2;. (i; 1) = 1; 8 (i; 2) = 1; 8 (i; 1) = 1; 8 (i; 2) = 1; 8. 2 f1; 3g ; 2 f1; 3g ; 2 f2; 4g ; 2 f2; 4g ;. and we say A2; has property C if it satis…ed one of the follow situations: 8j 2 f1; 2g nc (A2; ) ; = b11 ; then A2; (1; j) = 1; 8 2 f1; 2g ; = b14 ; then A2; (2; j) = 1; 8 2 f1; 2g ; = b41 ; then A2; (1; j) = 1; 8 2 f3; 4g ; = b44 ; then A2; (2; j) = 1; 8 2 f3; 4g : 0 1 0 1 1 1 b11 b12 b21 b22 B b13 b14 b23 b24 C B A2;1 A2;2 C B 1 0 1 Example 9 If A2 = B @ b31 b32 b41 b42 A = A2;3 A2;4 = @ 1 1 0 0 0 0 b33 b34 b43 b44 then A2 has a safe symbol. Therefore, from De…nition5, 6, 7 and Proposition9; for n = 2; we have s2 = b11 ; (1) (2) (3) (4). if if if if. s2 s2 s2 s2. S2 = A2;. s. = A2;1 =. 1 1 1 0. ;. R2 = fA2;1 ; A2;2 g C2 = fA2;1 ; A2;3 g : and A2; has property R; 8 A2; has property C; 8 6. 2 f1; 3g 2 f1; 2g. 1 0 0 C C; 0 A 0.

(13) 3. Main Theorem. In this section, we will formulate the main theorems of our study and proof. First, we de…ne the primitive property for a matrix. De…nition 10 Let A 2 Mn n (Z) is called primitive if there exists an integer k 1 such that Ak En n (full matrix), and let (A) be the minimum number of such k, i.e., (A). min k : Ak. En. n. :. In this paper, we use the "generalized" primitive property for a matrix, i.e., let A 2 Mn n (Z) is called "generalized" primitive if there exists an integer k 1 such that Ak (i; j). 1; 8i 2 f1; :::; ng nr (A) ; 8j 2 f1; :::; ng nc (A). Before proving the main Theorem, we show lemma …rst. Lemma 11 If A 2 Mn n (Z) has at least one safe symbol then A is primitive. Proof. By the de…nition of primitive, it su¢ ces to show that there exists an integer k 1 such that Ak (i; j). 1; 8i 2 f1; :::; ng nr (A) ; 8j 2 f1; :::; ng nc (A) :. Since A 2 Mn n (Z) has at least one safe symbol, we let A = (aij ) index i = m be a safe symbol. Then. (3.1) n i;j=1. and. amj = 1; 8j 2 f1; :::; ng nc (A) ;. (3.2). ajm = 1; 8j 2 f1; :::; ng nr (A) :. (3.3). and Indeed, let A2 = ( ij. n ij ) i;j=1 ,. then. = ai1 a1j + ai2 a2j + ::: + aim amj + ::: + ain anj :. (3.4). From (3:2) and (3:3) ; 8i 2 f1; :::; ng nr (A) ; 8 j 2 f1; :::; ng nc (A) ; we get aim amj = 1 7. (3.5).

(14) and from (3:4) and (3:5) ; we …nd, 8i 2 f1; :::; ng nr (A) ; j 2 f1; :::; ng nc (A) ; (3.6). 1:. ij. I.e., Ak (i; j). 1; 8i 2 f1; :::; ng nr (A) ; 8j 2 f1; :::; ng nc (A) :. Thus from (3:1) ; k = 2 is chose as we want. This complete the proof of Lemma11. 1 0 b11 b12 b21 b22 B b13 b14 b23 b24 C A2;1 A2;2 C Next, we give A2 and write it as A2 = B @ b31 b32 b41 b42 A = A2;3 A2;4 b33 b34 b43 b44 where A2; 2 M2 2 (Z), 8 2 f1; 2; 3; 4g : And we follow the recursive formulae for n-th order transition matrices An from (2:1) : Then we prove the main Theorem of this paper. Theorem 12 If A2 2 M4 4 (Z) satis…es the following properties (1) A2 and A3 have safe symbols. (2) There exist sequences f k gqk=0 such that for all k = 0; :::; q and q we have (a) k 2 f1; 4g ; (b) there exists 0 m q; such that m = 0 or m+1 = 1 ; (c) s2 = b 0 1 and s3 = b 0 1 b 1 2 = s2 b 1 2 ; Then An is primitive for all n 2:. 2;. Proof. We prove it by induction, since A2 has a safe symbol and s2 = b 0 1 ; then A2 (s2 ; s2 ) = 1; (3.7) A2 (s2 ; j) = 1; 8j 2 1:::22 nc2 ;. (3.8). A2 (i; s2 ) = 1; 8i 2 1:::22 nr2 :. (3.9). S2 = A2; 0 ;. (3.10). and Therefore R2 = fA2; j. 2 f1; 2g or f3; 4gg ; where A2; has property C. 8. (3.11). ,.

(15) and C2 = fA2; j. 2 f1; 3g or f2; 4gg ; where A2; has property R:. Since A3 has a safe symbol and s3 = b. 0 1. b. 1 2. = s2 b. 1 2. (3.12). ; then. A3 (s3 ; s3 ) = 1;. (3.13). A3 (s3 ; j) = 1; 8j 2 1:::23 nc3 ;. (3.14). A3 (i; s3 ) = 1; 8i 2 1:::23 nr3 :. (3.15). S3 = A2; 1 ;. (3.16). and Therefore R3 = fA2; j. 2 f1; 2g or f3; 4gg ; where A2; has property C. (3.17). 2 f1; 3g or f2; 4gg ; where A2; has property R:. (3.18). and C3 = fA2; j. Next, we show that s4 exists, such that A4 (s4 ; s4 ) = 1; A4 (s4 ; j) = 1; 8j 2 f1:::24 g nc4 ; and A4 (i; s4 ) = 1; 8i 2 f1:::24 g nr4 : By (2:1) ; we perform A3 and A4 for given A2 as follows A3 =. A3;1 A3;2 A3;3 A3;4. ;. where A3;. = 0. for. b 1 A2;1 b 2 A2;2 b 3 A2;3 b 4 A2;4. b B b = B @ b b. 1 b11 1 b13 3 b31 3 b33. b b b b. 1 b12 1 b14 3 b32 3 b34. b b b b. 2 b21 2 b23 4 b41 4 b43. 2 f1; 2; 3; 4g ; and. A4 =. A4;1 A4;2 A4;3 A4;4 9. ;. b b b b. 2 b22 2 b24 4 b42 4 b44. 1. C C; A.

(16) where b 1 A3;1 b 2 A3;2 b 3 A3;3 b 4 A3;4. A4; = 0. b B b =B @ b b. for. 1 b11 A2;1. b b b b. 1 b13 A2;3 3 b31 A2;1 3 b33 A2;3. 1 b12 A2;2. b b b b. 1 b14 A2;4 3 b32 A2;2 3 b34 A2;4. 2 b21 A2;1. b b b b. 2 b23 A2;3 4 b41 A2;1 4 b43 A2;3. 2 b22 A2;2 2 b24 A2;4 4 b42 A2;2 4 b44 A2;4. 1. C C; A. (3.19). 2 f1; 2; 3; 4g : Next, we use the conditions (2)-(a)~(2)-(b) to take the di¤erent s4 in the follow situations: (1) If 1 = 0 ; then we take s4 = b. 0 0. b. 0 0. b. 0 0. = s3 b. 0 0. :. so S2 = S3 = S4 = A2; 0 ; and R2 = R3 = R4 ; C2 = C3 = C4 ; (2) If. 0. 6=. 1. and. 1. 6=. s4 = b. 2. (i:e:; 0 1. b. 1 0. then we take. 2. =. 0) ;. b. 0 1. = s3 b. 0 1. :. 1 1. :. so S2 = A2; 0 ; S3 = A2; 1 ; S4 = A2; 0 ; and R2 = R4 6= R3 ; C2 = C4 6= C3 ; (3) If. 0. 6=. 1. and. 2. =. s4 = b. 1;. then we take. 0 1. b. 1 1. b. 10. 1 1. = s3 b.

(17) so S2 = A2; 0 ; S3 = S4 = A2; 1 ; and R2 = 6 R3 ; R3 = R4 ; C2 = 6 C3 ; C3 = C4 ; Then from (3:7) and (3:13) ; we get A4 (s4 ; s4 ) = 1; from (3:14) ; (3:19) and (3:11) or (3:17) ; we get A4 (s4 ; j) = 1; 8j 2 f1:::24 g nc4 ; and from (3:15) ; (3:19) and (3:12) or (3:18) ; we get A4 (i; s4 ) = 1; 8i 2 f1:::24 g nr4 , i.e., A4 has at least one safe symbol. Now, we assume An 2 and An 1 have safe symbols and sn 2 = sn 3 b 0 1 and sn 1 = sn 3 b 0 1 b 1 2 = sn 2 b 1 2 , in the same fashion of proof of A4 has at least one safe symbol, since An 2 has a safe symbol and s2 = sn 3 b 0 1 ; then An 2 (sn 2 ; sn 2 ) = 1; (3.20) An. 2. (sn 2 ; j) = 1; 8j 2 1:::2n. 2. ncn 2 ;. (3.21). An. 2. (i; sn 2 ) = 1; 8i 2 1:::2n. 2. nrn 2 :. (3.22). and Therefore Sn Rn. 2. (3.23). = A2; 0 ;. 2. = fA2; j. 2 f1; 2g or f3; 4gg ; where A2; has property C. (3.24). 2. = fA2; j. 2 f1; 3g or f2; 4gg ; where A2; has property R: (3.25). and Cn. Since An. 1. has a safe symbol and sn An. 1. 1. = sn 3 b. 0 1. b. 1 2. = sn 2 b. 1 2. ; then (3.26). (sn 1 ; sn 1 ) = 1;. An. 1. (sn 1 ; j) = 1; 8j 2 1:::2n. 1. ncn 1 ;. (3.27). An. 1. (i; sn 1 ) = 1; 8i 2 1:::2n. 1. nrn 1 :. (3.28). and. 11.

(18) Therefore Sn Rn. 1. (3.29). = A2; 1 ;. 1. = fA2; j. 2 f1; 2g or f3; 4gg ; where A2; has property C. (3.30). 1. = fA2; j. 2 f1; 3g or f2; 4gg ; where A2; has property R: (3.31). and Cn. Next, we show that sn exists, such that An (sn ; sn ) = 1; An (sn ; j) = 1; 8j 2 f1:::2n g ncn ; and An (i; sn ) = 1; 8i 2 f1:::2n g nrn : By (2:1) ; we perform An 1 and An for given An 2 as follows An. 1. An An. =. 1;1 1;3. An An. 1;2. ;. 1;4. where An. 1;. = 0. for. b 1 An b 3 An. b B b = B @ b b. 2;1 2;3. b 2 An b 4 An. 1 b11 An 3;1 1 b13 An 3;3 3 b31 An 3;1 3 b33 An 3;3. b b b b. 2;2 2;4. 1 b12 An 3;2 1 b14 An 3;4 3 b32 An 3;2 3 b34 An 3;4. b b b b. 2 b21 An 3;1 2 b23 An 3;3 4 b41 An 3;1 4 b43 An 3;3. b b b b. 2 b22 An 3;2 2 b24 An 3;4 4 b42 An 3;2 4 b44 An 3;4. 2 f1; 2; 3; 4g ; and An =. An;1 An;2 An;3 An;4. 1. C C; A. ;. where An; = 0. for. b B b =B @ b b. b 1 An b 3 An. b 2 An b 4 An. 1;1 1;3. 1 b11 An 2;1 1 b13 An 2;3 3 b31 An 2;1 3 b33 An 2;3. b b b b. 1;2 1;4. 1 b12 An 2;2 1 b14 An 2;4 3 b32 An 2;2 3 b34 An 2;4. b b b b. 2 b21 An 2;1 2 b23 An 2;3 4 b41 An 2;1 4 b43 An 2;3. b b b b. 2 b22 An 2;2 2 b24 An 2;4 4 b42 An 2;2 4 b44 An 2;4. 1. C C; A. (3.32). 2 f1; 2; 3; 4g : Next, we use the conditions (2)-(a)~(2)-(b) to take the di¤erent sn in the follow situations: 12.

(19) (1) If. 1. =. 0;. then we take sn = sn 3 b. 0 0. b. 0 0. b. 0 0. = sn 1 b. 0 0. :. so Sn. = Sn. 2. 1. = Sn = A2; 0 ;. and Rn Cn (2) If. 0. 6=. 1. and. 1. 6=. = Rn = Cn. 2 2. 2. (i:e:;. 2. b. 1 0. sn = sn 3 b. 0 1. 1 1. = b. = Rn ; = Cn ; 0) ;. 0 1. then we take. = sn 1 b. 0 1. :. 1 1. :. so Sn 2 = A2; 0 ; Sn 1 = A2; 1 ; Sn = A2; 0 ; and Rn Cn (3) If. 0. 6=. 1. and. 2. =. 2. 1;. sn = sn 3 b. = Rn 6= Rn 1 ; = Cn = 6 Cn 1 ;. 2. then we take 0 1. b. 1 1. b. 1 1. = sn 1 b. so Sn Sn. 2 1. = A2; 0 ; = Sn = A2; 1 ;. and Rn Cn. 2 2. 6= Rn 1 ; Rn 6 = Cn 1 ; Cn 13. 1 1. = Rn ; = Cn ;.

(20) Then from (3:20) and (3:26) ; we get An (sn ; sn ) = 1; from (3:27) ; (3:32) and (3:24) or (3:30) ; we get An (sn ; j) = 1; 8j 2 f1:::2n g ncn ; and from (3:28) ; (3:32) and (3:25) or (3:31) ; we get An (i; sn ) = 1; 8i 2 f1:::2n g nrn , i.e., An has at least one safe symbol. Therefore, Lemma11 is applied to show An is primitive. This complete the proof of Theorem12.. 14.

(21) 4. Examples for safe symbol existing Case. For the section, we will show some examples in [33] to check primitivity of An ; 8n 2. In [33], the authors prove the Theorem related to the extensively weak mixing property. We will use the conditions of the extensively weak mixing property to …nd H and V: Then we use some results in [33] to construct horizontal transition matrix H2 and vertical transition matrix V2 : And to …nd the cases that if A2 = H2 or A2 = V2 can all let An be primitive for all n 2:. 4.1. Find H and V. We …rst state the Theorem related to the extensively weak mixing property in [33] . Theorem 13 (Theorem 5.8 of [33]) Suppose = (A1 ; :::; Av ) is a subshift of …nite type and each Ai is a p p matrix. Then is extensively weak mixing if and only if for all 2 f 1; 1gv there exists n satisfying 1. n. p2. 2p + 2. (4.1). and (. )n. A1 1 :::A(v v )n > 0:. (4.2). For simplicity, we only consider two symbols which are given on 2 2 lattice Z2 2 : So we just suppose = (H; V ) ; p = 2 and = 1: Since = 1; in [33], we know that the matrix subshift = (H; V ) is of …nite type if and only if HV = V H: So by the above statement, we can get three conditions below, such that = (H; V ) is extensively weak mixing. Condition 1 HV = V H Condition 2 1 n 2 Condition 3 H n V n > 0 From the above conditions, we can …nd eleven cases for H and V: 1 1 Case 1 H = V = 1 0 0 1 Case 2 H = V = 1 1 15.

(22) Case 3. H=V =. 1 1 1 1. 1 1 1 0 ,V = 1 0 0 1 1 0 1 1 Case 5 H = ,V = 0 1 1 0 0 1 1 0 Case 6 H = ,V = 1 1 0 1 1 0 0 1 Case 7 H = ,V = 0 1 1 1 1 1 1 0 Case 8 H = ,V = 1 1 0 1 1 0 1 1 Case 9 H = ,V = 0 1 1 1 1 1 0 1 Case 10 H = ,V = 1 1 1 0 0 1 1 1 Case 11 H = ,V = 1 0 1 1 Next, we will take these cases of H and V to construct horizontal transition matrix H2 and vertical transition matrix V2 : Case 4. 4.2. H=. Construct H2 and V2. we consider …rst some results in [7], which are crucial to the constructs of H2 and V2 : We begin with 1 2 column pattern hi ; hi =. u2 u1. or. u2 u1. (4.3). and (4.4). i = 1 + 2u1 + u2 :. A 2 2 pattern U = (u 1 2 ) can now be obtained by a horizontal direct sum of two 1 2 pattern, i.e., hi1 i2. hi1. hi2 u12 u22 u11 u21 16. or. u12 u11. u22 ; u21. (4.5).

(23) where ik = 1 + 2uk1 + uk2 ; 1. k. (4.6). 2:. Therefore, the complete set of all 16(= 22 2 ) 2 2 patterns in 2 2 can be listed by a 4 4 matrix H2 = (hi1 i2 ) with 2 2 pattern hi1 i2 as its entries in. 0 0 1 0 0 1 1 1. 0 B B B B B B B B B B B B B B B B @. 0 0. 1 0. 0 1. 1 1. 0 0 0 0. 0 1 0 0. 0 0 0 1. 0 1 0 1. 1 0 0 0. 1 1 0 0. 1 0 0 1. 1 1 0 1. 0 0 1 0. 0 1 1 0. 0 0 1 1. 0 1 1 1. 1 0 1 0. 1 1 1 0. 1 0 1 1. 1 1 1 1. 1 C C C C C C C C C C C C C C C C A. (4.7). Similarly, a 2 2 pattern can also be viewed as a vertical direct sum of two 2 1 patterns, i.e., vj1 j2 vj1 vj2 ; (4.8) where vjl =. u1l u2l. or u1l. u2l ;. (4.9). 2:. (4.10). and jl = 1 + 2u1l + u2l ; 1. 17. l.

(24) A4. 4 matrix V2 = (vj1 j2 ) can also be obtained for. 0 0. 0 1. 1 0. 1 1. 0 B B B B B B B B B B B B B B B B @. 2 2;. 0 0. 0 1. 1 0. 1 1. 0 0 0 0. 0 1 0 0. 1 0 0 0. 1 1 0 0. 0 0 0 1. 0 1 0 1. 1 0 0 1. 1 1 0 1. 0 0 1 0. 0 1 1 0. 1 0 1 0. 1 1 1 0. 0 0 1 1. 0 1 1 1. 1 0 1 1. 1 1 1 1. From above, H2 can also be represented by vj1 j2 as 0 1 v11 v12 v21 v22 B v13 v14 v23 v24 C C H2 = B @ v31 v32 v41 v42 A : v33 v34 v43 v44. i.e., we have. 1 C C C C C C C C C C C C C C C C A. (4.11). (4.12). In (4:12) ; the indices j1 j2 are arranged by two Z-map successively, as 0 1 1 ! 2 @ A; . (4.13) 3 ! 4. i.e., the path from 1 to 4 in (4:13) is Z shaped and is then called a Z-map. More precisely, H2 can be decomposed by H2 =. V2;1 V2;2 V2;3 V2;4. V2;K =. vk1 vk2 vk3 vk4. (4.14). and :. (4.15). Where, H2 is arranged by a Z-map (V2;K ) in (4:14) and each V2;K is also arranged by a Z-map (vkl ) in (4:15) : Therefore, the indices of v in (4:12) consist of two Z-map. 18.

(25) Then, we use the above mention to get the value of H2 and V2 . We …rst let 0 1 a1 a2 a3 a4. 0 1. H=. ; V =. 0 1. 0 1 b1 b3 b2 b4. (4.16). where ai ; bj 2 f0,1g ; for i; j = 1 to 4.. So by (4:3) , (4:9). h1 =. 0 1 = b1 ; h2 = = b2 ; 0 0. h3 =. 0 1 = b3 ; h4 = = b4 1 1. (4.17). (4.18). and v1 =. 0 0 = a1 ; v 2 = 0 1 = a2 ;. v3 =. 1 0 = a3 ; v 4 = 1 1 = a4. (4.19). But by (4:7) and (4:11) ; if we want to get the value of H2 and V2 , we must consider the value of H and V at the same time. Therefore, 0 1 h11 h12 h13 h14 B h21 h22 h23 h24 C C H2 = B @ h31 h32 h33 h34 A h41 h42 h43 h44 0 1 a1 a1 b 1 b 1 a2 a1 b 1 b 2 a1 a2 b 1 b 3 a2 a2 b 1 b 4 B a3 a1 b 2 b 1 a4 a1 b 2 b 2 a3 a2 b 2 b 3 a3 a2 b 2 b 4 C C = B (4.20) @ a1 a3 b 3 b 1 a2 a3 b 3 b 2 a1 a4 b 3 b 3 a2 a4 b 3 b 4 A a3 a3 b 4 b 1 a4 a3 b 4 b 2 a3 a4 b 4 b 3 a4 a4 b 4 b 4 and. V2. 0. v11 v12 B v21 v22 = B @ v31 v32 v41 v42 0 a1 a1 b 1 b 1 B a1 a2 b 1 b 3 = B @ a1 a3 b 3 b 1 a1 a4 b 3 b 3. v13 v23 v33 v43. 1 v14 v24 C C v34 A v44. a2 a1 b 1 b 2 a2 a2 b 1 b 4 a2 a3 b 3 b 2 a2 a4 b 3 b 4 19. a3 a1 b 2 b 1 a3 a2 b 2 b 3 a4 a3 b 4 b 1 a3 a4 b 4 b 3. 1 a4 a1 b 2 b 2 a3 a2 b 2 b 4 C C a4 a3 b 4 b 2 A a4 a4 b 4 b 4. (4.21).

(26) From above, we can also …nd 0 v11 v12 v21 B v13 v14 v23 H2 = B @ v31 v32 v41 v33 v34 v43 and. 0. 1 v22 v24 C C= v42 A v44. V2;1 V2;2 V2;3 V2;4. (4.22). 1 h11 h12 h21 h22 B h13 h14 h23 h24 C H2;1 H2;2 C V2 = B (4.23) @ h31 h32 h41 h42 A = H2;3 H2;4 h33 h34 h43 h44 Therefore we get V2 for given H2 : Next, for those eleven cases of H and V in section4.1: We also use the introduced method above to get eleven cases below for H2 and V2 : 1 1 Case 1 If H = V = 1 0 0 1 0 1 1 1 1 0 1 1 1 0 B 1 0 1 0 C B 1 0 1 0 C C B C then H2 = B @ 1 1 0 0 A ; V2 = @ 1 1 0 0 A : 0 0 0 0 0 0 0 0 0 1 Case 2 If H = V = 1 1 0 1 0 1 0 0 0 0 0 0 0 0 B 0 0 1 1 C B 0 0 1 1 C C B C then H2 = B @ 0 1 0 1 A ; V2 = @ 0 1 0 1 A : 0 1 1 1 0 1 1 1 1 1 Case 3 If H = V = 1 1 0 1 0 1 1 1 1 1 1 1 1 1 B 1 1 1 1 C B C C ; V2 = B 1 1 1 1 C : then H2 = B @ 1 1 1 1 A @ 1 1 1 1 A 1 1 1 1 1 1 1 1 1 1 1 0 Case 4 If H = ,V = 1 0 0 1 0 1 1 0 1 0 0 0 1 0 0 1 C B B 0 0 0 0 C C ; V2 = B 0 1 0 0 C : then H2 = B @ 0 0 1 0 A @ 0 0 0 0 A 0 0 0 0 1 0 0 0 20.

(27) If H = 0 1 0 B 0 1 then H2 = B @ 0 0 0 0 Case 5. If H = 0 0 0 B 0 0 then H2 = B @ 0 0 1 0 Case 6. Case 7 If H = 0 0 B 0 then H2 = B @ 0 0. Case 8 If H = 0 1 B 0 then H2 = B @ 0 1. Case 9 If H = 0 1 B 0 then H2 = B @ 0 0. 0 1 0 0. 0 0 0 0. 0 1 0 0. Case 10 If H = 0 0 0 B 0 1 then H2 = B @ 0 1 0 0. 1 0 1 1 ,V = 0 1 1 0 1 0 0 0 1 0 C B 0 0 C 0 0 ;V =B 1 0 A 2 @ 0 0 0 0 1 0 0 1 1 0 ,V = 1 1 0 1 0 1 0 1 0 0 C B 0 1 0 0 C ;V =B 0 0 A 2 @ 0 0 0 1 0 0 1 0 0 1 ,V = 0 1 1 1 1 0 0 0 0 0 C B 0 0 C 0 0 ; V2 = B A @ 1 0 0 0 0 1 1 0 1 1 1 0 ,V = 1 1 0 1 1 0 0 1 1 0 C B 0 0 C 0 1 ; V2 = B A @ 0 0 0 0 0 1 0 0 1 0 1 1 ,V = 0 1 1 1 1 0 0 0 1 0 B 0 0 0 0 C C; V = B 1 0 A 2 @ 0 0 0 1 1 0 1 1 0 1 ,V = 1 1 1 0 1 0 0 0 0 0 B 0 0 1 0 C C; V = B 1 0 A 2 @ 0 1 0 0 1 0 21. 0 0 0 0. 1 1 0 C C: 0 A 0. 0 0 1 0. 1 0 0 C C: 0 A 1. 0 0 0 0. 1 1 0 C C: 0 A 1. 0 0 1 0. 1 0 0 C C: 0 A 1. 0 0 0 0. 1 1 0 C C: 0 A 1. 0 1 0 0. 1 1 0 C C: 0 A 0.

(28) 0 1 1 1 Case 11 If H = ,V = 1 0 1 1 0 1 0 0 0 0 1 0 0 0 B 0 0 1 0 C B 0 1 1 C B then H2 = B @ 0 1 0 0 A ; V2 = @ 0 1 1 1 0 0 0 0 0 0 Next, we want to check that for these cases, if An is primitive for all n 2:. 4.3. 1 0 0 C C: 0 A 0 A2 = H2 or A2 = V2 ; then. Checking all cases in section4.2 can let An be primitive for all n 2. We separate these cases into four subsections. 4.3.1. s2 = b11 or b44 and one column and one row of A are all zero. We consider the case1 and case2 in section4.2. 0 1 0 b11 b12 b21 b22 1 1 B b13 b14 b23 b24 C B 1 0 C B Example 14 Consider A2 = B @ b31 b32 b41 b42 A = H2 = @ 1 1 0 0 b33 b34 b43 b44 By (2:1) ; it is easily 0 checked that 1 1 1 1 0 1 1 0 0 B 1 0 1 0 1 0 0 0 C B C B 1 1 0 0 1 1 0 0 C B C B 0 0 0 0 0 0 0 0 C C (1) A2 and A3 = B B 1 1 1 0 0 0 0 0 Chave safe symbols. B C B 1 0 1 0 0 0 0 0 C B C @ 0 0 0 0 0 0 0 0 A 0 0 0 0 0 0 0 0 (2) there exists 0 = 1; 1 = 1; 2 = 1. 1 1 0 0. and s2 = b11 ; s3 = b11 b11 = s2 b11 such that (2)-(a)~(2)-(c) of Theorem12 hold, then Theorem12 is applied to show that An is primitive for all n 2: 22. 1 0 0 C C. 0 A 0.

(29) 0. b11 B b13 Example 15 Consider A2 = B @ b31 b33 By (2:1) ; it is easily 0 checked that 0 0 0 0 B 0 0 0 0 B B 0 0 0 0 B B 0 0 0 0 (1) A2 and A3 = B B 0 0 0 0 B B 0 0 1 1 B @ 0 0 0 1 0 0 1 1 (2) there exists 0 = 4; 1. b12 b14 b32 b34 0 0 0 0 0 0 0 0. b21 b23 b41 b43 0 0 1 1 0 0 1 1. = 4;. 0 0 0 1 0 1 0 1 2. 1 0 0 0 b22 C B 0 0 b24 C = H2 = B A @ 0 1 b42 b44 0 1 0 0 1 1 0 1 1 1. 1. 1 0 0 1 1 C C. 0 1 A 1 1. C C C C C Chave safe symbols. C C C C A. =4. and s2 = b44 ; s3 = b44 b44 = s2 b44 such that (2)-(a)~(2)-(c) of Theorem12 hold, then Theorem12 is applied to show that An is primitive for all n 2: 4.3.2. A is full matrix. We consider the case3 in section4.2. 0 b11 b12 B b13 b14 Example 16 Consider A2 = B @ b31 b32 b33 b34 By (2:1) ; it is easily 0 checked that 1 1 1 1 1 B 1 1 1 1 1 B B 1 1 1 1 1 B B 1 1 1 1 1 (1) A2 and A3 = B B 1 1 1 1 1 B B 1 1 1 1 1 B @ 1 1 1 1 1 1 1 1 1 1 23. b21 b23 b41 b43. 1 0 b22 1 1 1 C B b24 C 1 1 1 = H2 = B @ 1 1 1 b42 A b44 1 1 1. 1 1 1 1 1 1 1 C C 1 1 1 C C 1 1 1 C Chave safe symbols. 1 1 1 C C 1 1 1 C C 1 1 1 A 1 1 1. 1 1 1 C C. 1 A 1.

(30) (2) there exists 0. = 1;. 1. = 1;. 2. =1. and s2 = b11 ; s3 = b11 b11 = s2 b11 such that (2)-(a)~(2)-(c) of Theorem12 hold, then Theorem12 is applied to show that An is primitive for all n 2: 4.3.3. s2 = b11 or b44 and two column and row of A are all zero. We consider the case4~case9 in section4.2. Example 17 Consider 0 A2. 1 b22 b24 C C b42 A b44 0 1 0 0 B 0 0 0 = one of H2 and V2 = B @ 0 0 0 1 0 0. b11 B b13 = B @ b31 b33. b12 b14 b32 b34. By (2:1) ; it is easily 0 checked that 1 0 0 B 0 0 0 B B 0 0 0 B B 0 0 0 (1) A2 and A3 = B B 0 0 0 B B 0 0 0 B @ 0 0 0 1 0 0 (2) there exists 0 = 1;. b21 b23 b41 b43. 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 1. 1 1 0 C C: 0 A 0. 1 0 0 1 0 0 0 C C 0 0 0 C C 0 0 0 C Chave safe symbols. 0 0 0 C C 0 0 0 C C 0 0 0 A 0 0 1. = 1;. 2. =1. and s2 = b11 ; s3 = b11 b11 = s2 b11 such that (2)-(a)~(2)-(c) of Theorem12 hold, then Theorem12 is applied to show that An is primitive for all n 2: 24.

(31) Example 18 Consider 0 A2. 1 b22 b24 C C b42 A b44 0 0 B 0 = one of H2 and V2 = B @ 0 1 b11 B b13 = B @ b31 b33. b12 b14 b32 b34. By (2:1) ; it is easily 0 checked that 0 0 0 B 0 0 0 B B 0 0 0 B B 0 0 0 (1) A2 and A3 = B B 0 0 0 B B 0 0 0 B @ 0 0 0 1 0 0 (2) there exists 0 = 4;. b21 b23 b41 b43. 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 1. 0 0 0 0 0 0 0 0. = 4;. 0 0 0 0 0 0 0 0 2. 0 0 0 0 1 0 0 0 0 0 0 1. 1 0 1 0 0 C C: 0 0 A 0 1. 1. C C C C C Chave safe symbols. C C C C A. =4. and s2 = b44 ; s3 = b44 b44 = s2 b44 such that (2)-(a)~(2)-(c) of Theorem12 hold, then Theorem12 is applied to show that An is primitive for all n 2: Example 19 Consider 0 A2. 1 b22 b24 C C b42 A b44 0 1 0 0 B 0 0 0 = one of H2 and V2 = B @ 0 0 0 1 0 0. b11 B b13 = B @ b31 b33. b12 b14 b32 b34. b21 b23 b41 b43. 25. 1 1 0 C C: 0 A 1.

(32) By (2:1) ; it is easily 0 checked that 1 0 0 B 0 0 0 B B 0 0 0 B B 0 0 0 (1) A2 and A3 = B B 0 0 0 B B 0 0 0 B @ 0 0 0 1 0 0 (2) there exists 0 = 1;. 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 1. 1 0 0 1 0 0 0 C C 0 0 0 C C 0 0 0 C Chave safe symbols. 0 0 0 C C 0 0 0 C C 0 0 0 A 0 0 1. = 1;. 2. =1. and s2 = b11 ; s3 = b11 b11 = s2 b11 such that (2)-(a)~(2)-(c) of Theorem12 hold, then Theorem12 is applied to show that An is primitive for all n 2: 4.3.4. s2 = b14 ; s3 = b14 b44 and two column and row of A are all zero. We consider the case10 and case11 in section4.2. Example 20 Consider 0 A2. 1 b22 b24 C C b42 A b44 0 1 0 0 0 0 B 0 1 1 0 C C = one of H2 and V2 = B @ 0 1 1 0 A: 0 0 0 0. b11 B b13 = B @ b31 b33. b12 b14 b32 b34. b21 b23 b41 b43. By (2:1) ; it is easily checked that. 26.

(33) 0. B B B B B (1) A2 and A3 = B B B B B @. 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0. 0 0 1 0 0 1 0 0. 0. = 1;. 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0. (2) there exists. 1. 1 0 0 0 0 0 0 C C 1 0 0 C C 0 0 0 C Chave safe symbols. 0 0 0 C C 1 0 0 C C 0 0 0 A 0 0 0. = 4;. 2. =1. and s2 = b14 ; s3 = b14 b41 = s2 b41 such that (2)-(a)~(2)-(c) of Theorem12 hold, then Theorem12 is applied to show that An is primitive for all n 2:. 4.4. Conclusion. In the section we describe some Remarks and examples related to the primitivity of An : Remark 21 For the examples in the section4.3; we can …nd all cases for H2 and V2 which satisfying the Theorem5.8 related to the extensively weak mixing property in [33] can all let An be primitive for all n 2: Therefore, we can cover the Theorem5.8 in [33], i.e., the Theorem12 in our paper can be used to show more examples that An is primitive therein for all n 2 than the Theorem5.8 in [33]. Remark 22 Observing the examples in the section4.3; we …nd one of H and V 2. 1 1 1 1. ;. 1 1 1 0. ;. 0 1 1 1. ;. In fact, if H and V satisfy one of the follow situations, and using the introduced method to get H2 and V2 ; then for A2 = one of H2 and V2 ; Theorem12 is applied to show that An is primitive for all n 2: (1) one of H and V = E; and the other 2 = fOg ; (2) one of H and V = G; and the other 2 fU; L; I; T1 ; T2 ; K1 g ; (3) one of H and V = G0 ; and the other 2 fU; L; I; T3 ; T4 ; K4 g ; 27.

(34) (4) one of H and V 2 fU; L; Ig ; and the other 2 fK1 ; K2 ; K3 ; K4 g ; (5) one of H and V 2 fT1 ; T2 g ; and the other 2 fK1 g ; (6) one of H and V 2 fT3 ; T4 g ; and the other 2 fK4 g ; (7) H = V and H; V 2 fK1 ; K2 ; K3 ; K4 g , where E =. 1 1 1 1. ;O =. 0 0 0 0. ;G =. U =. 1 1 0 1. ;L =. 1 0 1 1. ;I =. T1 =. 1 0 1 0. ; T2 =. 1 1 0 0. ; T3 =. K1 =. 1 0 0 0. ; K2 =. 0 1 0 0. ; K3 =. 1 1 1 0. 0 1 1 1. ; G0 =. 1 0 0 1. ;J =. 0 1 0 1. 0 1 1 0. ;. 0 0 1 1. ; T4 =. 0 0 1 0. ; K4 =. 0 0 0 1. Next, we give one example to show the statement given above. Example 23 (from(1)) If H=E= then. Consider. 0. 1 B 1 H2 = B @ 1 0 0. b11 B b13 A2 = B @ b31 b33. b12 b14 b32 b34. 1 1 1 0. 1 1 1 1 1 1 1 0. b21 b23 b41 b43. ;V = G =. 1 0 0 1 1 B 1 0 0 C C;V = B 0 A 2 @ 1 1 0 1 0. 1 1 1 0 1 1 0 0. 1 1 0 C C: 0 A 0. 1 0 b22 1 1 1 C B b24 C 1 1 1 = H2 = B A @ b42 1 1 1 b44 0 0 0. By (2:1) ; it is easily checked that. 28. ;. 1 0 0 C C: 0 A 0. ; :.

(35) 0. B B B B B (1) A2 and A3 = B B B B B @. 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0. 0 0 0 0 0 0 0 0. 1 1 1 0 1 1 0 0. (2) there exists. 0. = 1;. 1. 1 1 0 0 1 0 0 C C 1 0 0 C C 0 0 0 C Chave safe symbols. 1 0 0 C C 1 0 0 C C 0 0 0 A 0 0 0. = 1;. 2. =1. and s2 = b11 ; s3 = b11 b11 = s2 b11 such that (2)-(a)~(2)-(c) of Theorem12 hold, then Theorem12 is applied to show that An is primitive for all n 2: Remark 24 If A2 is not constructed from H and V; then Theorem12 is also applied to show that An is primitive for all n 2: we give one example follow. Example 25 (Simpli…ed Golden Mean) Consider 0 1 0 b11 b12 b21 b22 1 1 B b13 b14 b23 b24 C B 1 0 C B A2 = B @ b31 b32 b41 b42 A = @ 1 0 b33 b34 b43 b44 0 0 By (2:1) ; it is easily 0 checked that 1 1 1 B 1 0 0 B B 1 0 0 B B 0 0 0 (1) A2 and A3 = B B 1 1 0 B B 1 0 0 B @ 0 0 0 0 0 0 (2) there exists 0 = 1;. 0 0 0 0 0 0 0 0. 1 1 0 0 0 0 0 0 1. 1 0 0 C C: 0 A 0. 1 1 0 0 0 0 0 C C 0 0 0 C C 0 0 0 C Chave safe symbols. 0 0 0 C C 0 0 0 C C 0 0 0 A 0 0 0. = 1;. 29. 1 0 0 0. 2. =1.

(36) and s2 = b11 ; s3 = b11 b11 = s2 b11 such that (2)-(a)~(2)-(c) of Theorem12 hold, then Theorem12 is applied to show that An is primitive for all n 2:. 30.

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