Pascal Eigenspaces and Invariant Sequences of the First or Second Kind

Transcription

1 Pascal Eigenspaces and Invariant Sequences of the First or Second Kind I-Pyo Kim a,, Michael J Tsatsomeros b a Department of Mathematics Education, Daegu University, Gyeongbu, 38453, Republic of Korea b Department of Mathematics and Statistics, Washington State University, Pullman, WA 99164, USA Abstract An infinite real sequence a n } is called an invariant sequence of the first (resp, second ind if a n = n ( n = ( 1 a (resp, a n = ( =n n ( 1 a We review and investigate invariant sequences of the first and second ind, and study their relationships using similarities of Pascal-type matrices and their eigenspaces Keywords: Invariant sequence, Pascal matrix, Eigenvalue, Eigenvector 1 MSC: 15B18, 11B39, 11B65 1 Introduction Inverse relations play an important role in combinatorics [11] The binomial inversion formula, which states that for sequences a n } and b n } (n =, 1,,, a n = n = ( n ( 1 b if and only if b n = n = ( n ( 1 a, (11 is a typical inverse relation considered in [6, 8, 1, 14, 15, 16] Specifically, (11 motivated Sun [14] to investigate the following sequences Definition 11 Let a n } (n =, 1,, be a sequence such that ( 1 s 1 a n = n = ( n ( 1 a ; s = 1 or s = (1 5 We refer to a n } as an invariant sequence (when s = 1 or an inverse invariant sequence (when s = of the first ind Research supported by Daegu University Research Grant 13 Corresponding author addresses: imipyo@daeguacr (I-Pyo Kim, tsat@mathwsuedu (Michael J Tsatsomeros Preprint submitted to Journal of LATEX Templates September 1, 17

2 Several examples of invariant sequences of the first ind can be found in [14], including 1 n }, nf n 1}, L n }, ( 1 n B n } (n, where F 1 = and F n }, L n }, and B n } are the Fibonacci sequence, Lucas sequence, and Bernoulli numbers [7], respectively In this paper, we will establish (see Lemma 1 the modified binomial inversion formula, that is, a n = =n ( ( 1 b if and only if b n = n Motivated by (13, we will introduce and consider the following sequences =n Definition 1 Let a n } (n =, 1,, be a sequence such that ( 1 s 1 a n = =n ( ( 1 a (13 n ( ( 1 a ; s = 1 or s = (14 n We refer to a n } as an invariant sequence (when s = 1 or an inverse invariant sequence (when s = of the second ind 1 15 Naturally arising are the questions of existence, identification, and construction of (inverse invariant sequences of the second ind, as well as the problem of characterizing such sequences and examining their relationship to their counterparts of the first ind Invariant sequences, which are also called self-inverse sequences in [15], have indeed been studied by several authors [6, 8, 14, 15] They are naturally connected to involutory (also nown as involution or self-invertible matrices [9] and to Riordan involutions [5] Involutory matrices find use in numerical methods for differential equations [, 9] They are also useful in cryptography, information theory, and computer security by providing convenient encryption and decryption methods [1] Motivated by Shapiro s open questions [1], Riordan involutions have been intensely investigated as a combinatorial concept [4, 5] In this paper, we investigate invariant sequences by means of the eigenspaces of P D and P T D, where P is the Pascal matrix and D an infinite diagonal matrix with alternating diagonal entries in 1, 1} (see Sections, 3 In fact, P D and P T D are involutory matrices and P D is a Riordan involution Our investigation follows the ideas and connections of invariant sequences to the eigenspaces of P D and P T D developed in Choi et al [6] This will allow us to associate (inverse invariant sequences of the first and second ind, as well as identify and construct such sequences (Section 4 Notation and preliminaries 5 The following notation and conventions are used throughout the manuscript The infinite matrices in this paper have infinite numbers of rows i and columns, with i,, 1,, }

3 E λ (A denotes the eigenspace of a (finite or infinite matrix A corresponding to its eigenvalue λ For a matrix A with columns A ( =, 1,, [ and ] with denoting the vector of zeros in R, A denotes the matrix whose th column is A, where is by convention vacuous For a matrix A, its (possibly infinite row and column index sets are J and K, respectively For J J, K K, let A(J K denote the matrix obtained from A by deleting rows in J and columns in K, and let A[J K ] denote the matrix A(J K, where J = J \J, K = K \K For brevity, write A( K and A(J in place of A( K and A(J, respectively Further, for m, n, 1,, }, we let A m,n = A[, 1,,, m}, 1,,, n}]; A m,m is abbreviated by A m The binomial coefficient ( i choose is denoted by ( i with the convention that it equals when i < or < [ (i P = ] (i, =, 1,, denotes the (infinite Pascal matrix D = diag(1, 1, 1, 1, Infinite real sequences x n } are identified with the infinite dimensional real vector space R consisting of column vectors x = [x, x 1, x, ] T Notice that as a consequence of the binomial inversion formula (11, we have that [ P 1 = DP D = ( 1 i ( i ] (i, =, 1,, Thus (P D 1 = P D and (11 can be converted [6] into a vector equation for x = [a, a 1, a, ] T and y = [b, b 1, b, ] T R, as follows: P Dx = y if and only if P Dy = x (1 Lemma 1 Let P and D be the Pascal matrix and the diagonal matrix defined above, and let x, y R Then P T Dx = y if and only if P T Dy = x, ( and the modified binomial inversion formula (13 holds 45 Proof As P 1 = DP D, we have (P T D 1 = D(P T 1 = P T D As a consequence, ( holds Letting x = [a, a 1, a, ] T and y = [b, b 1, b, ] T implies (13 Let F = [F, F 1, F, ] T and L = [L, L 1, L, ] T denote the vectors in R whose entries are the members of the Fibonacci and Lucas sequences, respectively; that is F =, F 1 = 1, F n = F n 1 + F n (n, L =, L 1 = 1, L n = L n 1 + L n (n 3

4 x The generating functions of F and L are h 1 (x = 1 x x and h (x = x 1 x x, respectively [3, 7] The following fact is nown, however, we include a proof for completeness Lemma P D F = F and P D L = L 5 Proof Let g(x = 1 x 1 x and f(x = 1 x For =, 1,,, the generating function of the th column of P D is g(xf(x [13] Thus the generating function of P DF is [g(x, g(xf(x, g(xf(x, ][F, F 1, F, ] T = g(x(f + F 1 f(x + F f(x + = g(xh 1 (f(x = h 1 (x, which implies that P D F = F The proof of P D L = L is similar That is, 1 and 1 are eigenvalues of P D and consequently of P T D In fact, these are the only eigenvalues of P D and P T D; see [6] The corresponding eigenspaces are infinite dimensional Indeed, if we consider the Pascal-type matrices P and Q constructed via the Pascal matrix P and the matrix [ ] 1 T Q = P +, P then, as shown in [6], the columns of [ T P ] and form bases for E 1 (P D and E 1 (P D, respectively The following observation follows directly from the definitions and properties mentioned above: Observation 3 The entries of x R form an invariant sequence of the first ind if and only if x E 1 (P D; Q 55 an inverse invariant sequence of the first ind if and only if x E 1 (P D; an invariant sequence of the second ind if and only if x E 1 (P T D; an inverse invariant sequence of the second ind if and only if x E 1 (P T D Based on Observation 3, our goal is to study the eigenspaces E λ (P D and E λ (P T D (λ 1, 1} and discover their relationships Our approach entails showing the existence of an infinite invertible matrix N such that N(P T DN 1 = (P T 1 D 1 (P T 1 D 1 (P T 1 D 1 (3 6 and D(N 1 T D (P D DN T D = (P 1 D 1 (P 1 D 1 (P 1 D 1, (4 which are infinite direct sums of copies of P1 T D 1 and P 1 D 1, respectively This result will be applied to characterize E λ (P D and E λ (P T D Extending the wor in [6], we will also show that the columns of P T and Q T ( form bases for E λ (P T D This will indeed allow us to investigate the relationships between invariant sequences of the first and second ind 4

5 3 The Eigenspaces of P T D and P D Let B = [b i ] (i, =, 1,, be the matrix defined by ( 1 i, if i, b i =, if i >, (31 and let J(a denote the infinite Jordan bloc of the form a 1 a 1 O a 1 a 1 O a (3 It readily follows that B 1 = J(1 65 In the next two lemmas, we will construct an infinite matrix N and its inverse M = N 1, which will give rise to similarity transformations of P T D and P D into direct sums as in (3 and (4 Lemma 31 Let m be a positive integer and let N (m = H (m H (m 1 H (1 with H (l = I l B (l = 1,,, m, where B is the matrix defined in (31 Then lim N (m = N = [n i ] m is the infinite matrix defined by n = 1, n = and n i = for i, = 1,,, and n ( 1 i( i 1 + i i =, if 1 i, i 1, if i > 1 Proof Let m be a positive integer and let N (m = H (m H (m 1 H (1 = [n m i ] with nm = 1, n m = and n m i = for i, = 1,, We will prove that n m ( 1 i( + i i =, if i,, if i > by induction on m, where = i 1, if i = 1,,, m; = 1,,, m 1, if i = m + 1, m +, ; = 1,, The claim is clear for m = 1 For m =, by the construction of H (1 and H (, we have n ( 1 i( + i i =, if i,, if i >, 5

6 where = i 1, if i = 1,, 3, 4; = 1,,, 1, if i = 5, 6, ; = 1,,, since for each pair of indices i and with i = 3, 4, and = 1,,, n i = ( l ( l ( 1 l i ( 1 l = ( 1 i l=i l=i ( 1 + i = ( 1 i 1 Let now m 3 Then by the construction of H (m, we have n m i = nm 1 i for all i = 1,,, m and all = 1,, By the induction hypothesis, n m i = ( 1 i( + i, if i,, if i >, where because = n m i = i 1, if i = 1,,, m; = 1,,, m 1, if i = m + 1, m +, ; = 1,,, ( m + l ( 1 l i ( 1 l m l=i ( m + l = ( 1 i m l=i ( m 1 + i = ( 1 i m 1 (33 for each pair of indices i and with i = m 1, m, and = 1,, Thus, by (33, we obtain lim m N (m = N = [n i ] given by n ( 1 i( i 1 + i i =, if i, i 1, if i > for each i, = 1,, Clearly, we have n = 1, n = and n i construction of H (l (l = 1,,, and the proof is complete = for i, = 1,, by the The difference sequence a = [ a, a 1, a, ] T of a sequence a = [a, a 1, a, ] T is defined by a i = a i+1 a i for each i =, 1,, Let a = [ a, a 1, a, ] T ( =, 1,, be the th difference sequence defined inductively by a = ( 1 a, where a = a The infinite matrix a a 1 a a a 1 a a a 1 a 6

7 is called the difference matrix of a It is well nown [3] that for each n =, 1,,, ( ( ( ( n n n n a n = a + a + a + + n a, (34 1 n which is used in the proof of the following lemma Lemma 3 Let M = [m i ] be the matrix with m = 1, m = and m i and ( m i = i, if i, (i, = 1,,, if i > = for i, = 1,,, 7 Then M = N 1, where N is the matrix defined in Lemma 31 Proof Let n i denote the (i, entry of N (i, =, 1,, We would lie to show that n il m l = δ i, l= the Kronecer delta For i = or =, there is nothing to show, so let i 1 and 1 If i = resp i >, then clearly n il m l = l= i ( i 1 n il m li = ( i i 1 = 1 resp l=i n il m l = So it is enough to show that if = i + r with r >, then l= n il m i 1 l = Let = [ and (+r ( consider the sequence z =, +r 1 (, +r (,, ( 1 r 1 +1 ( ] T, ( 1 r,, We can construct the difference matrix A = [a i ] having z as its first column as follows: A = ( +r ( +r 1 ( 1 r ( +r 1 ( +r 1 ( r 1 ( +r ( +r 3 ( 1 r Y ( 1 r ( +1 1 ( 1 r ( ( 1 r 1( 1 ( 1 r 1( 1 ( 1 r( ( 1 r( 1 1 ( 1 r( ( 1 r ( + ( 1 r 1( +1 l=, 7

8 where Y is the (r + 1 (r + 1 matrix given by Y = Thus a = for = + 1,, + r Since + 1 i+r i 1+r+r = + r, by using (34 we obtain where = i 1 n il m l = l= = i+r ( ( + l i n il m l = ( 1 l i t i + r l l=i t ( ( + r s t ( 1 r+s = ( 1 r a t =, s l=i s= i+r and t =, completing the proof Let M (m = U (1 U ( U (m, where U (l = I l J(1 (l = 1,,, m where J(1 is the matrix defined in (3 Then, by Lemmas 31 and 3, we have lim m M (m = M, which is the matrix defined in Lemma 3 In fact we have N =

9 and M = N 1 = We can now state and prove the similarity transformations of P D and P T D claimed in (3 and (4 [ (i Theorem 33 Let P = ] (i, =, 1,, and D = diag(1, 1, 1, 1, Then, 75 (a N(P T DM = (P T 1 D 1 (P T 1 D 1 (P T 1 D 1, (b (DM T D (P D (DN T D = (P 1 D 1 (P 1 D 1 (P 1 D 1, where P 1 resp D 1 are the leading submatrices of P resp D, and N resp M are the matrices defined in Lemmas 31 resp 3 Proof (a For each = 1,,, let H ( = I B and U( = I J(1, where B is the matrix defined in (31 Let m, n be arbitrary positive integers with n = m + 1 First, we will show by induction on m that ( H(m H (m 1 H (1 P T D ( U (1 U ( U (m = Z (P T D, where Z is an n n matrix such that Z = m }} (P T 1 D 1 (P T 1 D 1 (P T 1 D 1 When m = 1, the first row of H (1 P T DU (1 is clearly [1, 1,,, ] Since for a pair of indices i and with i, = 1,,, (( ( ( ( 1 (H (1 P T D i = ( ( 1 = ( 1, i i + 1 i 1 the second row of H (1 P T DU (1 is [, 1, 1, 1, 1, ]U (1 = [, 1,,,, ] For a pair of indices i and with i,, we have ( ( ( 1 (H (1 P T DU (1 i = ( ( 1 = ( 1, i 1 i 1 i and so H (1 P T DU (1 = (P T 1 D 1 (P T D By induction on m, it follows N (m P T DU (m = m }} (P T 1 D 1 (P T 1 D 1 (P T 1 D 1 (P T D, 9

10 where N (m = H (m H (m 1 H (1 and M (m = U (1 U ( U (m Since m was an arbitrary positive integer, NP T DM = (P T 1 D 1 (P T 1 D 1 (P T 1 D 1, where N = lim m N (m and M = lim m M (m (b It follows directly from (a that (DM T DP D(DN T D = (P 1 D 1 (P 1 D 1 (P 1 D 1, completing the proof 8 Let e i denote the ith column of the identity matrix I (i =, 1, Then E λ (P T D and E λ (P D (λ 1, 1} can be characterized via Theorem 33 as follows: Theorem 34 Let N resp M = N 1 be the matrices in Lemma 31 resp Lemma 3 Then the following hold: (a Me, Me, Me 4, } is a basis for E 1 (P T D 85 (b M(e + e 1, M(e + e 3, } is a basis for E 1 (P T D (c DN T D(e + e 1, DN T D(e + e 3 } is a basis for E 1 (P D (d DN T De 1, DN T De 3, DN T De 5, } is a basis for E 1 (P D 9 Proof (a Let B (1 = e, e, e 4, } and x = [x, x 1, x, ] T E 1 (NP T DM Then, by Theorem 33, we have ( [ ] (NP T 1 DM Ix = x =, i=1 which implies that x i = t i, x i+1 = for each i =,, 4, and t i R So B (1 spans E 1 (NP T DM and since e, e, e 4, are linearly independent, B (1 is a basis for E 1 (NP T DM Therefore Me, Me, Me 4, } is a basis for E 1 (P T D (b Let B ( 1 = e + e 1, e + e 3, } and y = [y, y 1, y, ] T E 1 (NP T DM Then, by Theorem 33, we have ( [ ] (NP T 1 DM + Iy = y =, i=1 [ ] [ ] y i 1 which implies that = s i for each i =,, 4, and s i R So B ( 1 spans y i+1 E 1 (NP T DM and since e + e 1, e + e 3, are linearly independent, B ( 1 is a basis for E 1 (NP T DM Therefore M(e + e 1, M(e + e 3, } is a basis for E 1 (P T D Clauses (c and (d can be proven similarly 1

11 Consider now P T = The following is a matrix expression of Theorem 34; the last two clauses appeared in [6] [ ] 1 T [ (i Corollary 35 Let Q = P + and D = diag(( 1, ( 1 1,, where P = P ] (i, =, 1,, Then the following hold: 95 (a The columns of P T form a basis for E 1 (P T D (b The columns of Q T ( form a basis for E 1 (P T D [ ] (c The columns of T P form a basis for E 1 (P D (d The columns of Q form a basis for E 1 (P D Proof (a Let (i, be a pair of indices i and with i, The ith component of Me equals ( }} i when i, and equals, otherwise; Me = [,,, ( (, ( 1,,,,, ] T is the th column of P T, which implies (a (b The ith component of M(e +e +1 equals ( ( ( i + i 1 = +1 ( i + i 1, when i +1, and equals otherwise; }} ( ( ( M(e + e +1 = [,,,,,,,,, ] T }} ( ( ( + [,,,,,,,,, ] T 1 [ ] T is the th column of P T 1 T ( + ( = Q T (, which implies (b Using the fact P T that for a pair of indices i and with i 1, ( 1 (DN T D i = ( 1 i ( 1 i + i 1 ( 1, clauses (c and (d can be proven similarly 11

12 1 4 Invariant sequences of two inds: Relations and examples We begin by some basic examples of (inverse invariant sequences Example 41 It follows from Corollary 35 (c and (d that the Fibonacci sequence F is an inverse invariant sequence of the first ind and the Lucas sequence L is an invariant sequence of the first ind The ith row of P T (i =, 1, is i }} ( i [,,, ( ( i i 1 i,,, 1, ( i,,, ], (41 15 from which we can get that J(F is an invariant sequence of the second ind and J(L is an inverse invariant sequence of the second ind Recall that J( is the infinite Jordan bloc with in the main diagonal, which is the matrix defined in (3 In the following theorem, we provide a general mechanism for transforming invariant sequences into inverse invariant sequences, and vice versa 11 Theorem 4 Let J(λ denote the matrix defined in (3 Then the following hold: (a If x is an invariant sequence of the second ind, then (J(1 + J( T x is an inverse invariant sequence of the second ind (b If x is an inverse invariant sequence of the second ind, then J( 1 J(x is an invariant sequence of the second ind (c If x is an invariant sequence of the first ind, then ( J(J( 1 T x is an inverse invariant 115 sequence of the first ind (d If x is an inverse invariant sequence of the first ind, then (J( 1 + J(x is an invariant sequence of the first ind Proof (a If x is an invariant sequence of the second ind, then by Corollary 35 (a, there exists b R such that P T b = x Let L = [l i ] (i, =, 1,, be the infinite lower triangular matrix defined by ( 1 i+, if i, l i =, if i < Then LP T ( = P T because for a pair of indices i and with i,, the (i, entry of LP T ( equals ( ( ( ( ( 1 i+ + ( 1 i ( 1 i++i = 1 i i 1

13 1 when i, and equals otherwise, which coincides with the (i, entry of P T Thus, [ ] LQ T ( = P T + L T P T From the fact that L 1 = J(1 T and by Corollary 35 (b, it follows that [ ] Q T ( b = J(1 T P T b + T P T (4 b = J(1 T x + [, x T ] T = (J(1 + J( T x is an inverse invariant sequence of the second ind (b Let x be an inverse invariant sequence of the second ind From (4 and J(J(1 T = J(1, it readily follows that J( 1 J(Q T ( = P T and so by Corollary 35 (b, we conclude that J( 1 J(x is an invariant sequence of the second ind 1 T T (c Since Q = P + T, we obtain J( T Q = J( T P 1 T + P T For nonnegative P integers i and with i, =, 1,,, since ( ( i i + i ( i i ( i+ = i+1 i++1, it implies that ([ ] [ ] [ ] T Q T P T P Ω =, where Ω is the infinite (, 1-matrix with 1 s everywhere on and [ ] [ ] below its main diagonal Since J( 1 T = Ω 1, we get (I J( 1 T 1 T T =, which implies that ( J(J( 1 T 15 x is an inverse invariant sequence of the first ind Clause (d easily follows from (c similarly Q P Let τ 1 = 1+ 5 and τ = 1 5 It is well nown that F n = 1 5 τ n τ n and L n = τ n 1 + τ n where F n and L n are the nth terms of F and L, respectively (n =, 1,, [3] Since J(F is an invariant sequence of the second ind by (41, it follows from Theorem 4 (a that J(1 T J(F + J( T J(F = J(1 T J(F + F is an inverse invariant sequence of the second ind In fact, the nth term (n =, 1,, of J(1 T J(F + F is 1 τ1 n+ 1 τ n+ + 1 τ1 n 1 τ n = L n+1, which implies that J(1 T J(F + F = J(L On the other hand, since for a pair of nonnegative integers i and with i, =, 1,,, (J( 1 ( 1 i ( 1 i = i+1, if i,, if i >, 13

14 and the n-th term of J( 1 J( L is ( 1 (1/ +1 (τ1 ++n + τ ++n = 1 τ1 n+1 1 τ n+1 = F n+1, 5 5 = we have J( 1 J( L = J(F, which is an invariant sequence of the second ind by Theorem 4 (b, as stated in Example 41 It directly follows from Theorem 4 (d that (J(+J( 1F, namely L, is an invariant sequence of the first ind Since for a pair of nonnegative integers i and with i, =, 1,,, ( J( 1 ( 1 i = i+1, if i,, if i >, we obtain ( J( 1 T [, L T ] T = F, which is an inverse invariant sequence of the first ind by Theorem 4 (c, because for n =, 1,,, n 1 n 1 (1/ n L = (1/ n ((τ 1 + (τ = 1 τ n 1 τ n = F n 5 5 = = 13 The sequence B = (B, B 1, B, T defined by B = 1 and n ( n+1 = B = (n 1 comprises the Bernoulli numbers and DB is an invariant sequence of the first ind [14], which also follows directly from the fact that P DDB = DB A new inverse invariant sequence of the first ind from the Bernoulli numbers B is provided next See Table 1 for explicit members of these sequences Corollary 43 Let B = (B, B 1, B, T be the Bernoulli numbers Then the sequence K = (K, K 1, K, T defined by n 1 K =, K n = (1/ n ( 1 B (n = 1,, is an inverse invariant sequence of the first ind Proof It follows from Theorem 4 (c that ( J( 1 T J( T DB = ( J( 1 T [, DB T ] T = 135 is an inverse invariant sequence of the first ind Notice now that the sequence K in the statement is indeed equal to ( J( 1 T [, DB T ] T By Theorem 4 (d, we get DB = (J(+J( 1K, since the first component of (J(+J( 1K is clearly ( 1 B, and for each positive integer i with i = 1,,, ith component of (J(+J( 1K is (1/ i ( 1 B + i 1 = i (1/ i+1 ( 1 B = ( 1 i B i = By Corollary 35 and Theorem 4, we can directly get more (inverse invariant sequences of the first and second ind as follows: 14

15 Table 1: (Inverse invariant sequences of the first ind associated with the Bernoulli numbers n B DB 1 1 K Corollary 44 For a positive integer n, let Φ n = S ln S ln 1 S l3 S l S l1, where P T, if l i = 1, S li = J(1 T + J( T, if l i = 1, and let Φ n = S ln Sln 1 S l3 Sl Sl1 where J( S 1 J(, if l i = 1, li = Q T (, if l i = 1 for i = 1,,, n Then for x R, we have the following: 14 (a If n is odd (even and l i = ( 1 i+1 for i = 1,,, n, then Φ n x is an invariant (inverse invariant sequence of the second ind (b If n is odd (even and l i = ( 1 i for i = 1,,, n, then Φ n x is an inverse invariant (invariant sequence of the second ind Example 45 It follows from (41 and Corollary 44 (a that for x = [x, x 1, x, ] T R, Φ x = y is an inverse invariant sequence of the second ind, where y = [y, y 1, y, ] T with y i = i t= i (( t + i 1 t ( t + 1 x t (i =, 1,, i t For example, let x = [,,,,,,, 1,,, ] T This results to the inverse invariant sequence of the second ind y = [y, y 1, y, ] T, where ( ( 7 8 y i = + (i =, 1,, i 1 7 i 7 That is, y = = y 6 =, y 16 = y 17 = =, and by direct calculation, one can compute the nonzero components y 7, y 8,, y 15 ; eg, y 1 = ( ( = 77 Corollary 46 For a positive integer n, let Ψ n = T ln T ln 1 T l3 T l T l1, where T li = Q, if l i = 1, ( J( 1 T J( T, if l i = 1 15

16 and let Ψ n = T ln Tln 1 T l3 Tl Tl1, where J( + J( 1, if l i = 1, [ ] T li = T, if l i = 1 P 145 for i = 1,,, n Then for x R, we have the following: (a If n is odd (even and l i = ( 1 i+1 for i = 1,,, n, then Ψ n x is an invariant (inverse invariant sequence of the first ind (b If n is odd (even and l i = ( 1 i for i = 1,,, n, then Ψ n x is an inverse invariant (invariant sequence of the first ind [ ] T Example 47 For i = 1,,, the ith row of is [ ( ( i 1, i ( i 1 1,,,,, ] It follows i 1 P that for x = [x, x 1, x, ] T R, Ψ x = y is an invariant sequence of the first ind, where by Corollary 46 (b, y = (y, y 1, y, T satisfies i (( i 1 t y i = + t 1 t= ( i t t x t For example, let x = [,,,,,,, 1,,, ] T This results to the invariant sequence of the first ind y = [y, y 1, y, ] T, where ( ( i 1 7 i 7 y i = + (i = 1,, Thus y = y 1 = = y 13 = and one can find by direct calculation, the nonzero components y 14, y 15, ; eg, y 14 = ( ( = In the next two theorems, we obtain direct relationships among (inverse invariant sequences of the first ind and (inverse invariant sequences of the second ind 155 Theorem 48 Let x and y be, respectively, either (i an invariant sequence of the first ind and an inverse invariant sequence of the second ind, or (ii an inverse invariant sequence of the first ind and an invariant sequence of the second ind Then x T Dy = 16 Proof From P Dx = λx and P T Dy = λy, we have x T DP T Dy = λx T Dy, which implies that λx T Dy = for λ 1, 1} The following theorem is a useful tool for proving the final theorem Theorem 49 For every positive integer n, 16

17 (a the columns of (P +D n are invariant sequences of the first ind, and the columns of (P D n are inverse invariant sequences of the first ind 165 (b the columns of (P T + D n are invariant sequences of the second ind, and the columns of (P T D n are inverse invariant sequences of the second ind Proof Let n be a positive integer Then and P D(P + D n = P D(P + D(P + D n 1 = (P + D n P T D(P T D n = P T D(P T D(P T D n 1 = (P T D n, which respectively imply that each column of (P +D n is an invariant sequence of the first ind, and each column of (P T D n is an inverse invariant sequence of the second ind The other assertions of the theorem follow similarly 17 The following theorem is a form of converse of Theorem 48 Theorem 41 Let x i R (i =, 1,,, y R \ }, and let X = [x, x 1, x, ] Then for each i =, 1,,, 175 (a if x i T Dy = and X = P + D (X = P D, then y is an inverse invariant (invariant sequence of the second ind, and x i is an invariant (inverse invariant sequence of the first ind (b if x i T Dy = and X = P T + D (X = P T D, then y is an inverse invariant (invariant sequence of the first ind and x i is an inverse invariant (invariant sequence of the second ind 18 Proof If x T i Dy = and X = P + D, then since X T Dy = (P T + DDy =, we have P T Dy = y and by Theorem 49, P Dx i = x i Thus y is an inverse invariant sequence of the second ind and x i is an invariant sequence of the first ind This proves the first case of part (a For the case of X = P D and part (b, the results can be shown similarly References 185 [1] B Acharya, SK Panigrahy, SK Patra, G Panda, Image encryption using advanced Hill Ciper Algorithm, Int J Recent Trends in Engineering, 1(1 ( [] AL Andrew, Eigenvectors of certain matrices, Linear Algebra Appl, 7 ( [3] RA Brualdi, Introductory combinatorics, Fourth ed, Prentice-Hall, Upper Saddle River, 4 17

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