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Evaluation of Spectrum of 2-Periodic Tridiagonal-Sylvester Matrix

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Evaluation of Spectrum of 2-Periodic Tridiagonal-Sylvester Matrix Evaluation of spectrum of 2-periodic tridiagonal-Sylvester matrix Emrah KILIÇ AND Talha ARIKAN Abstract. The Sylvester matrix was …rstly de…ned by J.J. Sylvester. Some authors have studied the relationships between certain or- thogonal polynomials and determinant of Sylvester matrix. Chu studied a generalization of the Sylvester matrix. In this paper, we introduce its 2-period generalization. Then we compute its spec- trum by left eigenvectors with a similarity trick. 1. Introduction There has been increasing interest in tridiagonal matrices in many di¤erent theoretical …elds, especially in applicative …elds such as numer- ical analysis, orthogonal polynomials, engineering, telecommunication system analysis, system identi…cation, signal processing (e.g., speech decoding, deconvolution), special functions, partial di¤erential equa- tions and naturally linear algebra (see [2, 4, 5, 6, 15]). Some authors consider a general tridiagonal matrix of …nite order and then describe its LU factorizations, determine the determinant and inverse of a tridi- agonal matrix under certain conditions (see [3, 7, 10, 11]). The Sylvester type tridiagonal matrix Mn(x) of order (n + 1) is de- …ned as x 1 0 0 0 0 n x 2 0 0 0 2 . 3 0 n 1 x 3 .. 0 0 6 . 7 Mn(x) = 6 . .. .. .. .. 7 6 . 7 6 . 7 6 0 0 0 .. .. n 1 0 7 6 7 6 0 0 0 0 x n 7 6 7 6 0 0 0 0 1 x 7 6 7 4 5 1991 Mathematics Subject Classi…cation. 15A36, 15A18, 15A15. Key words and phrases. Sylvester Matrix, spectrum, determinant. Corresponding author. 1 2 EmrahKILIÇANDTalhaARIKAN and Sylvester [12] gave its determinant as n det Mn(x) = (x + n 2k): k=0 Y Askey [1] showed two ways to compute the determinant of Mn(x), one matrix-theoretic and another based on orthogonal polynomials. Also he explored their connection to orthogonal polynomials. For the relationships between orthogonal polynomials and other determinants of Sylvester type matrices related to Krawtchouk, Hahn and Racah polynomials, we refer to [14]. Holtz [8] showed how the determinants in [12] can be evaluated by left eigenvectors of corresponding matrices coupled with a simple similarity trick. Chu [13] generalized the Sylvester matrix by adding a new parameter as shown x 1 0 n x + y 2 2 . 3 n 1 x + 2y .. Mn(x; y) = 6 . 7 6 .. .. n 1 7 6 7 6 2 x + (n 1)y n 7 6 7 6 0 1 x + ny 7 6 7 4 5 and by using the method which Holtz uses in [8], evaluated its deter- minant as n ny n 2k det M (x; y) = (x + + 4 + y2); n 2 2 k=0 Y p via the generalized Fibonacci sequences. In this paper, we consider a new generalization of tridiogonal-Sylvester matrix. Then we compute its spectra and also determinant. 2. A Periodic Tridiagonal-Sylvester Matrix We de…ne 2-period Sylvester matrix of order (n + 1) as follows : x 1 0 n y 2 2 . 3 n 1 x .. An(x; y) = 6 . 7 ; 6 .. .. n 1 7 6 7 6 2 an 1 (x; y) n 7 6 7 6 0 1 a (x; y) 7 6 n 7 4 5 2-periodictridiagonal-Sylvestermatrix 3 where x if n is even, a (x; y) = n y if n is odd. If we take x = y, then the matrix An(x; x) gives the Sylvester matrix Mn (x) : K¬l¬ç[9] studied the case y = x. In this paper, our main purpose is to prove determinant formula for the matrix An(x; y): n=2 x (xy 4t2) if n is even, t=1 det A (x; y) = 8 n > Q > n=2 < b c(xy (2t + 1)2) if n is odd. > t=0 > Q We will frequently denote:> the matrix An(x; y) by An and, an (x; y) by an: 1 1 1 1 2 2 Let 1 = (x+y)+ and 2 = (x+y) where = (x y) + (2n) : 2 2 2 2 For the matrix An of order n + 1 with odd n, de…ne the vectors with (n + 1) dimension: p + (y x)+ (y x)+ (y x)+ z := 1 1 1 2n 2n 2n and (y x) (y x) (y x) z := 1 1 1 : 2n 2n 2n For the matrix A of order n+1 with even n; de…ne the vectors with n n + 1 dimension: + (y x)+ (y x)+ (y x)+ s := 1 1 1 1 2n 2n 2n and (y x) (y x) (y x) s := 1 1 1 1 : 2n 2n 2n We need the following results: Lemma 1. For odd n > 0; the matrix An has the eigenvalues 1 and + 2 with the corresponding left eigenvectors z and z; respectively. + + Proof. To prove the claim, it is su¢ cient to show z An = 1z and th zAn = 2z. From the de…nition of An, we should prove that the k components of zAn are z0x + z1n = z01;2 for k = 0; zn 1n + zny = zn1;2 for k = n; and for 0 < k < n; kzk 1 + akzk + (n k)zk+1 = zk1;2; where an is de…ned as before. 4 EmrahKILIÇANDTalhaARIKAN For the case k = 0; we get 1 1 1 x + n [(y x) ] = (x + y) = 1;2; 2n 2 2 as claimed. Now we consider the case k = n and examine the equality + + z An = 1z : So we get (y x) + (x + y) + (2.1) z+ = n 1 2n 2 1 = (y2 xy + y + 2n2) 2n y (2.2) = n + ((y x) + ) 2n + + = zn 1n + zn y; as claimed. To complete the proof, we show the last case 0 < k < n. Now we examine this case under two conditions: For even k; + + + kzk 1 + xzk + (n k)zk+1 1 1 = k ((y x) + ) + x + (n k) ((y x) + ) 2n 2n 1 = x + n ((y x) + ) 2n 1 1 = (x + y) + = = z+ ; 2 2 1 k 1 and for odd k; 1 kzk 1 + yzk + (n k)zk+1 = k + y ((y x) + ) + (n k) 2n 1 = n + y ((y x) + ) ; 2n which, by the equations (2.1) and (2.2), equals 1 1 + ((y x) + ) ((x + y) + ) = z 1: 2n 2 k + + So the proof is completed for the case z An = 1z . The other case, zAn = 2z; can be similarly shown. Lemma 2. For even n > 0; the matrix An has the eigenvalues 1 and + 2 with the corresponding to the left eigenvectors s and s; respec- tively. Proof. The proof can be done similar to the proof of the previous Lemma. 2-periodictridiagonal-Sylvestermatrix 5 For odd n > 0; we de…ne a (n + 1) (n + 1) matrix Tn as + + . + + + z0 z1 . z2 zn 1 zn 2 . 3 z0 z1 . z2 zn 1 zn Tn = 6 7 ; 6 . 7 6 . 7 6 7 6 . 7 6 0(n 1) 2 . In 1 7 6 7 4 5 where 0(n 1) 2 is the zero matrix of order (n 1) 2 and In is the identity matrix of order n: We can obtain inverse of the matrix Tn as follows: (x y)+ (x y) . 2 2 . 1 0 1 0 1 0 . 2 n n . 0 1 0 1 0 1 3 1 6 7 Tn = 6 . 7 : 6 0(n 1) 2 . 7 6 7 6 . 7 6 . In 1 7 6 . 7 6 . 7 6 7 4 5 So we can see that the matrix An is similar to the matrix En := 1 TnAnTn via the matrix Tn as shown . 1 0 . 02 (n 1) . 2 0 2 . 3 1 TnAnTn = 6 7 ; 6 n(n 1) n (n 1) . 7 6 . 7 6 7 6 . 7 6 0(n 2) 2 . Wn 1 7 6 7 4 5 where the matrix Wn 1 of order (n 1) is given by x 4 n 0 1 n 0 1 n n 2 y 4 0 0 2 . . 3 0 n 3 x 5 .. 6 . .. .. .. .. .. 7 Wn 1 = 6 . 7 : 6 7 6 . .. 7 6 . 3 y n 1 0 7 6 . 7 6 . ... 7 6 . 2 x n 7 6 0 0 1 y 7 6 7 4 5 Considering the 2 2 principal submatrix of En, it is clearly seen that 1 and 2 are two eigenvalues of En: 6 EmrahKILIÇANDTalhaARIKAN We focus on the matrix An for odd n up to now. By consider the matrix An for even n; we de…ne a matrix Yn of order (n + 1) as shown: + + . + + + s0 s1 . s2 sn 1 sn . 2 s0 s1 . s2 sn 1 sn 3 Y = ; n 6 7 6 7 6 7 6 . 7 6 0(n 1) 2 . In 1 7 6 7 4 5 where 0(n 1) 2 and In are de…ned as before. 1 We also obtain the inverse matrix Yn in the form x y+ x y . . 1 0 1 0 1 2 2 n n . 1 2 . 0 1 0 1 0 3 Yn = : 6 7 6 . 7 6 0(n 1) 2 . In 1 7 6 7 4 1 5 So the matrix An is similar to matrix Dn := YnAnYn via the matrix Yn is given by . 1 0 . 02 (n 1) . 2 0 2 . 3 1 YnAnYn = 6 7 ; 6 n(n 1) n (n 1) . 7 6 . 7 6 7 6 . 7 6 0(n 2) 2 . Qn 1 7 6 7 4 5 where the matrix Qn 1 of order (n 1) is given by x 4 n 0 1 n 0 1 n 0 n 2 y 4 0 0 2 .
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