Akaiwa et al. Pacific Journal of Mathematics for Industry 2014, 6:10 http://www.pacific-mathforindustry.com/content/6/1/10

ORIGINAL ARTICLE Open Access A tridiagonal construction by the quotient difference recursion formula in the case of multiple eigenvalues Kanae Akaiwa1*, Masashi Iwasaki2, Koichi Kondo3 and Yoshimasa Nakamura1

Abstract In this paper, we grasp an inverse eigenvalue problem which constructs a with specified multiple eigenvalues, from the viewpoint of the quotient difference (qd) recursion formula. We also prove that the characteristic and the minimal polynomials of a constructed tridiagonal matrix are equal to each other. As an application of the qd formula, we present a procedure for getting a tridiagonal matrix with specified multiple eigenvalues. Examples are given through providing with four tridiagonal matrices with specified multiple eigenvalues. Keywords: Quotient difference formula; Tridiagonal matrix; Multiple eigenvalues; Characteristic polynomial; Minimal polynomial

1 Introduction matrix with specified multiple eigenvalues, based on the One of the important problems in is to qd formula. The reason why the sequence of discussions construct matrices with specified eigenvalues. This is an was stopped is expected that multiple- arith- inverse eigenvalue problem which is classified in Struc- metic and symbolic computing around the published year tured Inverse Eigenvalue Problem (SIEP) called in [1]. of Rutishauser’s works for the qd formula were not suf- The main purpose of this paper is to design a procedure ficiently developed. The qd formula, strictly speaking the forsolvinganSIEPinthecasewheretheconstructed differential form of it, for computing tridiagonal eigenval- matrix has tridiagonal form with multiple eigenvalues, ues acts with high relative accuracy in single-precision or through reconsidering the quotient difference (qd) for- double-precision arithmetic [7], while, actually, that serv- mula. It is known that the qd formula has the applications ing for constructing a tridiagonal matrix gives rise to no to computing a continued fraction expansion of power small errors. Thus, the qd formula serving for construct- series [5], zeros of polynomial [3], eigenvalues of a so- ing a tridiagonal matrix is not so worth in single-precision called Jacobi matrix [9] and so on. Though the book [9] or double-precision arithmetic. In recent computers, it is refers to an aspect similar to in the following sections, not difficult to employ not only single or double preci- it gives only an anticipated comment without proof in sion arithmetic but also arbitrary-precision arithmetic or the case of multiple eigenvalues. There is no observa- symbolic computing. In fact, an expression involving only tion about numerical examples for verifying it. The key symbolic quantities achieves exact arithmetic on the sci- point for the purpose is to investigate the Hankel deter- entific computing software such as Wolfram Mathemat- minants appearing in the solution to the qd ica, Maple and so on. Numerical errors frequently occur in formula with the help of the Jordan canonical form. In finite-precision arithmetic, so that a constructed tridiag- this paper, we give our focus on the unsettled case in onal matrix probably does not have multiple eigenvalues order to design a procedure for constructing a tridiagonal without symbolic computing. The resulting procedure in this paper is assumed to be carried out on symbolic *Correspondence: [email protected] computing. 1Graduate School of Informatics, Kyoto University, Yoshida-Hommachi, This paper is organized as follows. In Section 2, we Sakyo-ku, Kyoto 606-8501, Japan first give a short explanation of some already obtained Full list of author information is available at the end of the article

© 2014 Akaiwa et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Akaiwa et al. Pacific Journal of Mathematics for Industry 2014, 6:10 Page 2 of 9 http://www.pacific-mathforindustry.com/content/6/1/10

properties concerning the qd formula. In Section 3, we Theorem 1. ([4], pp. 596, 603, 610) Let F(z) be factor- observe a tridiagonal matrix whose characteristic poly- ized as in (3).Thenitholdsthat nomial is associated with the minimal polynomial of a ( ) H n = 0, s = l + 1, l + 2, ..., n = l + 1, l + 2, .... general matrix through reconsidering the qd formula. s 0 0 (5) The tridiagonal matrix essentially differs from the Jacobi matrix in that it is not always symmetrized. We also Let us assume that discuss the characteristic and the minimal polynomi- (n) = = ... = ... als of a tridiagonal matrix in Section 4. In Section 5, Hs 0, s 1, 2, , l, n 0, 1, .(6) we design a procedure for constructing a tridiagonal Then the qd formula (4) with the initial settings matrix with specified multiple eigenvalues, and then demonstrate four tridiagonal matrices as examples of (n) = (n) = fn+1 = ... e0 0, q1 , n 0, 1, (7) the resulting procedure. Finally, in Section 6, we give fn conclusion. admits the determinant solution (n+1) (n) 2 Some properties for the qd recursion formula ( ) Hs H − q n = s 1 , s = 1, 2, ..., l, n = 0, 1, ..., In this section, we briefly review two theorems in [4] s (n) (n+1) Hs Hs−1 concerning the qd formula from the viewpoint of a gener- (8) ating function, the Hankel determinant and a tridiagonal (n) (n+1) ( ) H + H − matrix. e n = s 1 s 1 , s = 0, 1 ..., l, n = 0, 1, .... (n) (n) ... s (n) (n+1) Let us introduce the Hankel H1 , H2 , Hs Hs { }∞ given in terms of a complex sequence fn 0 as (9)    ···   fn fn+1 fn+s−1  (n) = =   From (9) with (5), it follows that el 0forn 0, 1, fn+1 fn+2 ··· fn+s ( ) ( ) (n)   n n H =   ,(1)....Moreover,itturnsoutthatqs and es for s = l + 1, s  . . .. .   . . . .  l + 2, ... and n = 0, 1, ... are not given in the same form   fn+s−1 fn+s ··· fn+2s−2 as (8) and (9). s = 1, 2, ..., n = 0, 1, ..., Let us introduce s-by-s tridiagonal matrices, ⎛ ⎞ (n) (n) (n) (n) (n) q1 q1 e1 where H− = 0andH = 1forn = 0, 1, ....Moreover, ⎜ ⎟ 1 0 ⎜ ( ) ( ) . ⎟ let F(z) be a generating function associated with { f }∞ as ⎜ n + n .. ⎟ n 0 (n) = ⎜ 1 q2 e1 ⎟ Ts ⎜ ⎟ , (10) ∞ ⎝ .. .. (n) (n) ⎠  . . qs−1es−1 n 2 ( ) ( ) F(z) = fnz = f0 + f1z + f2z +··· .(2) n + n 1 qs es−1 n=0 s = 1, 2, ..., l, n = 0, 1, ... Let us consider that F(z) is a rational function with (n) (n) respect to z with a pole of order l0 ≥ 0 at infinity and finite with the qd variables qs and es .LetIs be the s-by-s poles zk = 0oforderlk for k = 1, 2, ..., L. Then the sum . Then we obtain a theorem for the charac- = + +···+ (n) of the orders of the finite poles is l l1 l2 lL,and teristic polynomial of Tl . F(z) is factorized as Theorem 2. ([4], pp. 626, 635) Let F(z) be factorized as (n) = ... G(z) in (3).LetusassumethatHs satisfies (6).Forn 0, 1, , F(z) = G0(z) + ,(3) it holds that ( − )l1 ( − )l2 ···( − )lL z z1 z z2 z zL     ( ) − l1 − l2 − lL det zI − T n = z − z 1 z − z 1 ··· z − z 1 . where G(z) is a polynomial of degree at most l,andG0(z) l l 1 2 L is a polynomial of degree l0 if l0 > 0, or G0(z) = 0ifl0 = 0. (11) The following theorem gives the determinant solution to the qd recursion formula 3 Tridiagonal matrix associated with general matrix ⎧ In this section, from the viewpoint of the characteristic ⎨ (n+1) + (n+1) = (n) + (n) qs es−1 qs es , and the minimal polynomials, we associate a general M- s = 1, 2, ..., n = 0, 1, ..., (n) ⎩ ( + ) ( + ) ( ) ( ) by-M complex matrix A with a tridiagonal matrix T . q n 1 e n 1 = q n e n , s = 1, 2, ..., n = 0, 1, .... l s s s+1 s Let λ1, λ2, ..., λN be the distinct eigenvalues of A, which (4) are numbered as |λ1|≥|λ2| ≥ ··· ≥ |λN |.Itisnotedthat Akaiwa et al. Pacific Journal of Mathematics for Industry 2014, 6:10 Page 3 of 9 http://www.pacific-mathforindustry.com/content/6/1/10

some of |λ1|, |λ2|, ..., |λN | may equal to each other in the From (15) and (16), we derive the Jordan canonical form case where some of λ1, λ2, ..., λN are negative eigenvalues of A as or complex eigenvalues. Let Mk be the algebraic multiplic- −1 = ity of λk,whereM = M1 +M2 +···+MN . For the identity V AV J (17) M×M matrix IM ∈ R ,letφA(z) = det(zIM − A) be the characteristic polynomial of A,namely, with the nonsingular matrix

M1 M2 MN M×M φA(z) = (z − λ1) (z − λ2) ···(z − λN ) . (12) V = (V1 V2 ··· VN ) ∈ C , (18) { }∞ Let us prepare the sequence fn 0 given by and the block H n fn = w A u, n = 0, 1, ... (13) M×M J = diag(J1, J2, ..., JN ) ∈ C , (19) forsomenonzeroM-dimensional complex vectors u and where w,wherethesuperscriptH denotes the Hermitian trans- ... × pose. Originally, f0, f1, were called the Schwarz con- = ( ··· ) ∈ CM Mk Vk Vk,1 Vk,2 Vk,Mk , (20) stants, but they are usually today called the moments × = ( ( ) ( ) ··· ( )) ∈ CM mk,j or the Markov parameters [2]. Since the matrix power Vk,j vk,j 1 vk,j 2 vk,j mk,j , (21) ∞ ( )n ( ) = Mk×Mk series n=0 zA is a Neumann series (cf. [6]), F z Jk = diag(Jk,1, Jk,2, ..., Jk,M ) ∈ C , (22) ∞ H ( )n | | < ⎛ ⎞ k n=0 w zA u converges absolutely in the disk D : z λ |λ |−1 ( ) = H ( − )−1 k 1 1 . Moreover, we derive F z w IM zA u ⎜ ⎟ ⎜ . ⎟ which implies that F(z) is a rational function with the λ .. × = ⎜ k ⎟ ∈ Cmk,j mk,j ( − ) = Mφ ( −1) Jk,j ⎜ ⎟ . (23) denominator det IM zA z A z as follows. ⎝ . ⎠ .. 1 ˜ G(z) λk F(z) = , (1 − λ z)M1 (1 − λ z)M2 ···(1 − λ z)MN 1 2 N ≥ (14) Without loss of generality, we may assume that mk,1 ≥···≥ mk,2 mk,Mk . ˜ = { ... } ≥ where G(z) is some polynomial with respect to z.It Let mk max mk,1, mk,2, , mk,Mk .Sincemk,1 ˜ ≥ ··· ≥ = is remarkable that the numerator G(z) may have the mk,2 mk,Mk ,itisobviousthatmk mk,1.With same factors as the denominator (1 − λ z)M1 (1 − the help of the Jordan canonical form of A as in (17), we 1 ∞ M M { } λ2z) 2 ···(1−λN z) N .Inotherwords,F(z) has the poles get a proposition for the sequence fn 0 in (13). λ−1 λ−1 ... λ−1 1 , 2 , , N whose orders are equal to or less than M1, M2, ..., MN ,respectively. Proposition 1. Let u be the vector given by the linear Let us introduce the Jordan canonical form of A in order combination of the eigenvectors and the generalized eigen- κ to investigate the poles of F(z) with (13) even in the case vectors of A, namely, for some constants k,j,i, where A has multiple eigenvalues. Let Mk be the geo- M N k mk,j metric multiplicity of λk which indicates the dimension of u = κk,j,ivk,j(i). (24) eigenspace Ker(A − λkIM).ItisnotedthatMk is equal k=1 j=1 i=1 to or less than the algebraic multiplicity Mk. The matrix A has Mk eigenvectors corresponding to λk, and then the ... Moreover, for a vector w,let eigenvectors, denoted by vk,1, vk,2, , vk,Mk ,satisfy M k mk,j Av = λ v , j = 1, 2, ..., M . (15)  k,j k k,j k = κ  H ( − + ) ck,i k,j,i vk,j i i 1 w. (25) j=1 i=i Hereinafter, for j = 1, 2, ..., Mk,letvk,j(1) = vk,j. More- = ... M v ( ) v ( ) ... v ( ) over, for j 1, 2, , k,let k,j 2 , k,j 3 , , k,j mk,j { }∞ denote the generalized eigenvectors associated with the Then, the sequence fn 0 in (13) can be expressed by eigenvectors vk,j(1),wheremk,j is the maximal integer   N mk ( ) ( ) ... ( ) n such that vk,j 1 , vk,j 2 , , vk,j mk,j are linearly indepen- = λn−i+1 fn ck,i k , (26) dent. Of course, m + m +···+m M = M .Then, i − 1 k,1 k,2 k, k k k=1 i=1 the generalized eigenvectors vk,j(2), vk,j(3), ..., vk,j(mk,j) satisfy where the binomial coefficients are 0 if n < i − 1.Also,for suitable u and w,itholdsthat Avk,j(i) = λkvk,j(i) + vk,j(i − 1),

i = 2, 3, ..., mk,j, j = 1, 2, ..., Mk. (16) ck,i = 0, i = 1, 2, ..., mk, k = 1, 2, ..., N. (27) Akaiwa et al. Pacific Journal of Mathematics for Industry 2014, 6:10 Page 4 of 9 http://www.pacific-mathforindustry.com/content/6/1/10

Proof. From V −1AV = J in (17), it holds that An = By writing down two summations, we get VJnV −1. By combining it with (13) and (24), we derive

H n −1 N M    fn = w VJ V u  k = κ H ( ) n λn M m fn k,j,1w vk,j 1 N k k,j 0 k H n −1 k=1 j=1 = κk,j,iw VJ V vk,j(i). (28)       k=1 j=1 i=1 + κ H ( ) n λn + H ( ) n λn−1 k,j,2 w vk,j 2 k w vk,j 1 k  0 1 Let ρ be the column number in which v (i) arranges. k,j,i k,j n n − Then it is obvious that V −1v (i) = e (i) where e (i) + κ wH v (3) λn + wH v (2) λn 1 k,j k,j k,j k,j,3 k,j 0 k k,j 1 k denotes a unit vector such that the ρk,j,ith entry is 1 and    n − the others are 0. Thus, it follows that +wH v (1) λn 2 k,j 2 k M m N k k,j +··· H n fn = κk,j,iw VJ ek,j(i). (29)    n k=1 j=1 i=1 + κ wH v (m ) λn k,j,mk,j k,j k,j 0 k n   Since J is the block diagonal matrix, the matrix J and n − ( )n + wH v (m − 1) λn 1 its small blocks Jk are also so. It also turns out that k,j k,j 1 k (J )n is upper triangle. So, it is worth noting that Jne (i)    k,j k,j − + n H n n mk,j 1 becomes the ρk,j,ith column vector of J and the zero- +···+w v (1) λ . k,j m − 1 k entries arrange in except for its ρk,j,1th, ρk,j,2th, ..., ρk,j,ith k,j rows. The Jordan blocks Jk,j can be decomposed as = λ + Moreover, by paying our attention to the binomial coeffi- Jk,j kImk,j Emk,j , (30) ⎛ ⎞ cients, we can rewrite fn as 01 ⎜ ⎟ ⎜ . ⎟ ⎡ .. ×   ⎜ 0 ⎟ mk,j mk,j N M Em = ⎜ ⎟ ∈ R . (31)  k k,j ⎝ . ⎠ = ⎣ n λn κ H ( ) + κ H ( ) .. 1 fn k k,j,1w vk,j 1 k,j,2w vk,j 2 = 0 = 0 k 1 j 1  +···+κ H ( )  k,j,mk,j w vk,j mk,j It is emphasized that Emk,j is a whose i th    M ≥ k power becomes the zero-matrix O for i mk,j.Thus, n − n + λn 1 κ H ( ) + κ H ( ) (Jk,j) can be expressed as k k,j,2w vk j,1 k,j,3w vk j,2 1 = j 1  mk,j      H n n n−i +1 i −1 +···+κk,j,m w vk,j(mk,j − 1) (J ) = λ (Em ) , (32) k,j k,j i − 1 k k,j i=1 +··· ⎤ 0   M where (E ) = I .Letusintroduceanm - − + k mk,j mk,j k,j + n λn mk,j 1 κ H ( )⎦ dimensional unit vector e(i) which is regarded as a part k k,j,mk,j w vk,j 1  m − 1 ( ) ( )i −1 ( ) = k,j j=1 of ek,j i . Then, by taking account that Emk,j e i ⎡    M e(i − i + 1) in (32), we derive N k mk,j ⎣ n n H = λ κk,j,iw vk,j(i − 1 + 1) i   k   = 0 = = n ( ) = n λn−i +1 ( −  + ) k 1 j 1 i 1 J ek,j i  ek,j i i 1 . (33) − k   M mk,j = i 1 k  i 1 + n λn−1 κ H ( − + ) k k,j,iw vk,j i 2 1   1 = = Since it holds that Vek,j(i − i + 1) = vk,j(i − i + 1),by j 1 i 2 combining it with (29) and (33), we therefore have +···   M m n n−m +1 N k k,j i + λ k,j = κ H ( −  + ) m − 1 k fn k,j,iw vk,j i i 1 k,j ⎤ = j=1 i=1 = M m k 1  i 1 k k,j  × κ H ( − + )⎦ × n λn−i +1 k,j,iw vk,j i mk,j 1 .  k . (34) = = i − 1 j 1 i mk,j Akaiwa et al. Pacific Journal of Mathematics for Industry 2014, 6:10 Page 5 of 9 http://www.pacific-mathforindustry.com/content/6/1/10

≥ H ( −  + ) = H ( −  + ) =  + − From mk mk,j and w vk,j i i 1 vk,j i i 1 w, By letting n n i 1, we derive it follows that     N mk ∞  ⎛ ⎞ − n + i − 1  M ( ) = i 1 (λ )n N mk k mk,j F z ck,iz kz .  i − 1 = ⎝ κ H ( − + ) ⎠ k=1 i=1 n=0 fn k,j,ivk,j i i 1 w = = = =  (39) k 1 i 1 j 1i i n − + × λn i 1. (35) It is noted that, for |z| < 1, i − 1 k   ∞   n + i − 1  1 The exchange of i for i in (35) brings us to (25) and (26). zn = . (40) i − 1 (1 − z)i For example, let us consider the case where the con- n=0 stants κk,j,i are all 1. Then u becomes the sum of all the eigenvectors and generalized eigenvectors. Moreover, let From (39) and (40), it turns out that F(z) converges abso- − | | < |λ |−1 w = V H α in (25) where α is an M-dimensional vector lutely in the disk D : z 1 . Simultaneously, we have  κ H ( − + (36) for z ∈ D. It is obvious that (36) with λN = 0becomes with all the entries 1. Then it holds that k,j,ivk,j i i   (37). Moreover, (36) and (37) immediately lead to the latter 1)w = e (i − i + 1)α = 1. Thus, it is concluded that k,j half concerning the poles of F(z). ck,i = 0. The above discussion suggests that there exists at least a pair of u and w for satisfying (27). Let ψA(z) be the polynomial whose degree is the small- est such that ψ (A) = O.Hereψ (z) is called the Proposition 1 leads to a theorem concerning the gener- A A ( ) = H n minimal polynomial of A. Let us recall here that the max- ating function F z with the moments fn w A u. ... imal dimension of the Jordan blocks Jk,1, Jk,2, , Jk,Mk corresponding to λ is m .So,ψ (z) is representable as Theorem 3. Let F(z) be the generating function with the k k A = H n ( ) moments fn w A u.Then,F z converges absolutely in ψ ( ) = ( − λ )m1 ( − λ )m2 ···( − λ )mN −1 A z z 1 z 2 z N . (41) the disk D : |z| < |λ1| ,andF(z) is expressed as Therefore, we have the main theorem in this section for N mk i−1 ck,iz the relationship between the minimal polynomial of a F(z) = . (36) ( − λ )i = = 1 kz general matrix A and the characteristic polynomial of a k 1 i 1 (n) tridiagonal matrix Tl . Especially, if λN = 0,thenF(z) is expressed as Theorem 4. Let F(z) be given by the generating function mN −1 = H n F(z) =cN,1 + cN,2z +···+cN,m z with the moments fn w A u.Letusassumethat(6) and N λ = λ = ... λ = − (27) hold for suitable u and w.If 1 0, 2 0, , N N1 mk i−1 + ck,iz 0,thenitholdsthat i . (37) (1 − λkz)  k=1 i=1 − (n) = ψ ( ) = ... det zIm Tm A z , n 0, 1, , (42) Let us assume that (27) holds for suitable u and w.If λ = ( ) λ−1 λ−1 ... λ−1 N 0,thenF z has the finite poles 1 , 2 , , N of otherwise, ... the orders m1, m2, , mN , respectively, and the sum of the  ( ) ψ (z) orders is m = m1 +m2 +···+mN .IfλN = 0,thenF(z) has n A det zIm−m − T − = , n = 0, 1, .... N m mN mN the pole of the order mN − 1 at infinity and the finite poles z λ−1 λ−1 ... λ−1 ... (43) 1 , 2 , , N−1 of the orders m1, m2, , mN ,respec- tively,andthesumoftheordersofallthefinitepolesis m − m . Proof. It is remarkable that three integers L, l, lk and a N (n) complex zk associated with the tridiagonal matrix Tl in Proof. By substituting fn in (26) into F(z) in (2), we get Theorem 2 are given in terms of three integers N, m, mk and a complex λk associated with a general matrix A.If     N mk ∞ λN = 0, then it follows from the latter half of Theorem 3 ( ) = n n λn−i+1 = = = = ... = F z ck,i z k that L N, l m, l0, l1 m1, l2 m2, , lN mN and i − 1 −1 k=1 i=1 n=0 zk = λ . So, from (11) and (41), we derive (42). Similarly,     k N mk ∞ if λN = 0, then L = N − 1, l = m − mN , l0 = mN − 1, l1 = n n n−i+1 −1 = c z λ . (38) m1, l2 = m2, ..., lN−1 = mN−1 and z = λ . Thus (11) k,i i − 1 k k k k=1 i=1 n=i−1 and (41) lead to (43). Akaiwa et al. Pacific Journal of Mathematics for Industry 2014, 6:10 Page 6 of 9 http://www.pacific-mathforindustry.com/content/6/1/10

i i Incidentally, the editors in ([9], pp. 444–445) give a in (49) and by taking account that D pl(λk) = D (z − λ )m1 ( − λ )m2 ···( − λ )ml | = = ... simple example with short comments concerning the min- 1 z 2 z l z=λk 0fori 0, 1, , imal polynomial, the Jordan canonical form of A and the mk − 1, we obtain multiple poles of F(z). λ + = ( ) = ... − kpk,i ipk,i−1 Tl pk,i, i 0, 1, , mk 1. 4 Minimal polynomial of tridiagonal matrix (50) In this section, with the help of the Jordan canonical form, we clarify the relationship of the characteristic polynomial Moreover, it follows that (n) of the tridiagonal matrix T to the minimal one. ⎧ l  For simplicity, let us here adopt the following abbrevia- ⎨ (Tl) Pk,0 = λkPk,0, (n) ( ) = λ + tions for matrices Ts , ⎩ Tl Pk,i kPk,i Pk,i−1, (51) ⎛ ⎞ i = 1, 2, ..., mk − 1, u1 v1 ⎜ ⎟ ⎜ .. ⎟ where P = (1/i! )p .Thus,bylettingP = (P P ··· = ⎜ 1 u2 . ⎟ = ... k,i k,i 1,0 1,1 Ts ⎜ ⎟ , s 0, 1, , l, (44) − | ··· − |··· | ··· − ) ∈ ⎝ . . ⎠ P1,m1 1 P2,0 P2,1 P2,m2 1 PL,0 PL,1 PL,mL 1 .. .. v − l×l  ˆ s 1 C ,wehave(Tl) P = PJ. 1 us Here, it remains to prove that P is nonsingular. Of course, P = O since the (i + 1)th row of P is where l = m if λN = 0orl = m − mN if λN = 0. Let k,i   k,i i (λ )/ = = ( ) − λ i = p0(z) = 1andps(z) = det(zIs−Ts) for s = 1, 2, ..., l.Then D pi k i! 1. Let Wk,i Ker Tl kIl for i ... − pl(z) is just the characteristic polynomial of Tl,namely, 1, 2, mk 1, which indicates the generalized eigenspace ( ) λ of Tl corresponding to k. Then it is obvious from (51) m1 m2 mL φT (z) = (z − λ1) (z − λ2) ···(z − λL) , (45) ( ) − λ = ∈ that Tl kIl Pk,0 O and Pk,0 Wk,1. Eq. (51) with = ( ) − λ 2 = ∈ where L = N if λN = 0orL = N − 1ifλN = 0. The i 1 also leads to that Tl kIl Pk,1 O and Pk,1 following proposition gives the Jordan canonical form of Wk,2. Simultaneously, it is observed that Pk,1 ∈/ Wk,1.Let ∈ ( ) = λ the tridiagonal matrix Tl. us assume that Pk,1 Wk,1,namely, Tl Pk,1 kPk,1. Then, from (51), we derive Pk,0 = O, which contradicts Proposition 2. There exists a nonsingular matrix P such with Pk,0 = O. Thus, it follows that Pk,1 ∈/ Wk,1. Similarly, that by induction for i = 2, 3, ..., mk − 1inPk,i,wehave −1  ˆ P (Tl) P = J, (46) Pk,i ∈/ Wk,1, Wk,2, ..., Wk,i, Pk,i ∈ Wk,i+1, ˆ l×l J = diag(J1,1, J2,1, ..., JL,1) ∈ C , (47) i = 1, 2, ..., mk − 1. (52) where J1,1, J2,1, ..., JL,1 are of the same form as (23). From (52), it turns out that Pk,i for i = 0, 1, ..., mk −1and = ... Proof. The characteristic polynomials p0(z), p1(z), ..., k 1, 2, , L are linearly independent. Therefore, it is ( ) concluded that P is nonsingular and the Jordan canonical pl z satisfy  ⎧ form of (Tl) is given by (46). ⎨ zp0(z) = u1p0(z) + p1(z),

zps(z) = vsps−1(z) + us+1ps(z) + ps+1(z), (48) Proposition 2 suggests that the minimal polynomial of ⎩  s = 1, 2, ..., l − 1. (Tl) becomes This is easily derived from the expansion of det(zI − T ) s s ψ (z) = (z − λ )m1 (z − λ )m2 ···(z − λ )mL , (53) by the sth row minors. By taking the 0th, the 1st, ...,the T 1 2 L (mk − 1)th derivatives with respect to z in (48), we get ⎧ which is equal to the characteristic polynomial of Tl in i ( ) + i−1 ( ) = i ( ) + i ( ) = = ··· = = ⎪ zD p0 z iD p0 z u1D p0 z D p1 z , (45). If m1 m2 mL 1, then it is obvious that ⎪ = ... − ⎨ i 0, 1, , mk 1, Tl is diagonalizable. Otherwise, Tl is not diagonalizable. i ( ) + i−1 ( ) ⎪ zD ps z iD ps z This is because multiplicity of roots in minimal polyno- ⎪ = i ( ) + i ( ) + i ( ) ⎩⎪ vsD ps−1 z us+1D ps z D ps+1 z , mial coincides with maximal size of the Jordan blocks. i = 0, 1, ..., mk − 1, s = 1, 2, ..., l − 1, To sum up, we have a theorem for the properties of the (49) tridiagonal matrix Tl. i ( ) ( ) where D ps z denotes the ith derivative of ps z Theorem 5. The minimal polynomial of Tl is equal to = ( i (λ ) i (λ ) with respect to z.Letpk,i D p0 k , D p1 k , the characteristic one. Also, Tl is diagonalizable tridiago- i  l ..., D pl−1(λk)) ∈ C . Then, by substituting z = λk nal matrix if and only if it has no multiple eigenvalues. Akaiwa et al. Pacific Journal of Mathematics for Industry 2014, 6:10 Page 7 of 9 http://www.pacific-mathforindustry.com/content/6/1/10

(n) (n) 5 Procedure for constructing tridiagonal matrix the diagram for getting qs and es by the qd formula and its examples = (0) (4). A procedure for constructing T Tl with the same In this section, based on the discussions in the previ- nonzero eigenvalues as A is therefore as follows. ous sections, we first design a procedure for constructing = λ = = − λ = a tridiagonal matrix with specified multiple eigenvalues. 1: Set l m if N 0 or l m mN if N 0. 2: Choose u and w as in (6) and (27). We next give four kinds of examples for demonstrating H n 3: Compute fn = w A u for n = 0, 1, ...,2l − 1. that the resulting procedure can provide with tridiagonal (n) = = ... − 4: Set e0 0 for n 0, 1, ,2l 3. matrices with multiple eigenvalues. Examples have been (n) 5: Compute q = f + /f for n = 0, 1, ...,2l − 2. carried out with our computer with OS: Mac OS X 10.8.5, 1 n 1 n s = ... l CPU:IntelCorei72GHz,RAM:8GB.Wealsousethe 6: Repeat (a) and (b) for 2, 3, , . (n) = (n+1) + (n+1) − (n) scientific computing software Wolfram Mathematica 9.0. (a) Compute es−1 qs−1 es−2 qs−1 for In every example, all the entries of u are simply set to 1 and n = 0, 1, ...,2l − 2s + 1. (n) = (n+1) (n+1)/ (n) those of w are not artificial. The readers will realize that (b) Compute qs qs−1 es−1 es−1 for the settings of u and w are not so difficult for satisfying (6) n = 0, 1, ...,2l − 2s. and (27). 7: Construct a tridiagonal matrix by arranging (0) (0) ... (0) (0) (0) ... (0) Let us here consider the relationship of five theorems in q1 , q2 , , ql and e1 , e2 , , el−1. the previous sections. Theorem 2 shows that the eigen- (n) According to Theorem 5, the minimal and the character- values of Tl in the tridiagonal form as (10) are equal to the poles of the generating function F(z) and the multi- istic polynomials of the resulting tridiagonal matrix T are plicity of the eigenvalues coincide with the those of the equal to each other. Moreover, T is diagonalizable if and poles of F(z). Theorems 3 and 4 claim that the mini- only if it has no multiple eigenvalues. mal polynomial of a general matrix A, denoted by ψA(z) It is necessary to control the eigenvalues of A for getting H n is just the denominator of F(z) involving fn = w A u, T as a tridiagonal matrix with specified eigenvalues. It is and it coincides with the characteristic polynomial of easy to specify the eigenvalues of the diagonal matrix and (n) φ ( ) thoseoftheJordanmatrix. Tl denoted by T z , except for the factor correspond- ing to zero-eigenvalues. With the help of Theorem 1, we First, in the procedure, let us consider the case where ( ) 0 6×6 thus realize that the nonzero eigenvalues of Tl with the A = diag(2, 2, 2, 1, 1, 1) ∈ R ( ) ( ) ( ) ( ) ( ) ( ) entries involving q 0 , q 0 , ..., q 0 and e 0 , e 0 , ..., e 0 1 2 l 1 2 l−1 which is a diagonal matrix with two eigenvalues 1 and 2 become roots of the minimal polynomial ψA(z) in the case ( ) ( ) ( ) ( ) ( ) ( ) each of multiplicity 3. Obviously, the characteristic and where q 0 , q 0 , ..., q 0 and e 0 , e 0 , ..., e 0 are given by 1 2 l 1 2 l−1 the minimal polynomials are factorized as (z − 1)3(z − 2)3 (n) = the qd formula (4) under the initial settings e0 0and and (z−1)(z−2), respectively. So, the integers l and m are (n) = / = H n = = q1 fn+1 fn with fn w A u. See also Figure 1 for immediately determined as l 2andm 6. Moreover,

Figure 1 The qd diagram for a tridiagonal matrix construction. Akaiwa et al. Pacific Journal of Mathematics for Industry 2014, 6:10 Page 8 of 9 http://www.pacific-mathforindustry.com/content/6/1/10

by letting u = (1, 1, 1, 1, 1, 1) and w = (1, 1, 1, 1, 1, 1), and w = (1, 1, 1, 1, 1, 1, 1, 1) bring us to a tridiagonal we derive a tridiagonal matrix as matrix, which can not be symmetrized,   ⎛ ⎞ 3 1 × 13 1 T = 2 4 ∈ R2 2 ⎜ 4 16 ⎟ 1 3 ⎜ 17 − 13 ⎟ 2 ⎜ 1 4 2 ⎟ ⎜ ⎟ × T = ⎜ 1 − 11 116 ⎟ ∈ R5 5 whose characteristic and minimal polynomials are both ⎜ 26 169 ⎟ ( − )( − ) ⎝ 1232 − 13 ⎠ factorized as z 1 z 2 . The tridiagonal matrix T 1 377 841 is a with the distinct eigenvalues 1 77 1 29 and 2. Next, let us adopt a , which can be whose characteristic and minimal polynomials are both regarded as the Jordan matrix, with eigenvalues 2 of mul- factorized as (z − 2)2(z − 3)3, which is just equal to the tiplicity 6 as A,namely, minimal one of A. The tridiagonal matrix T is not a diag- ⎛ ⎞ onalizable matrix with eigenvalues 2 and 3 of multiplicity 21 ⎜ ⎟ 2 and 3, respectively. ⎜ 21 ⎟ ⎜ ⎟ Finally, let us A be set as the Jordan matrix with complex ⎜ 21 ⎟ 6×6 + − A = ⎜ ⎟ ∈ R , eigenvalues 2 i and 2 i each of multiplicity 2 and distinct ⎜ 21 ⎟ ⎝ ⎠ real eigenvalues 1 and 2, namely, 21 ⎛ ⎞ 2 2 + i 1 ⎜ ⎟ ⎜ 2 + i ⎟ in the procedure. Since the characteristic polynomial of ⎜ ⎟ ⎜ 2 − i 1 ⎟ × A is equal to the minimal one, the integers l and m are A = ⎜ ⎟ ∈ C6 6, = = ⎜ 2 − i ⎟ determined as l m 6. Then the procedure with ⎝ ⎠ u = (1, 1, 1, 1, 1, 1) and w = (1, 1, 0, 1, 0, 1) constructs 2 a tridiagonal matrix, which can not be symmetrized, 1 ⎛ ⎞ 11 3 in this procedure. By taking account that the characteris- ⎜ 4 16 ⎟ ⎜ 11 − 4 ⎟ tic and the minimal polynomials of A are equal to each ⎜ 1 12 9 ⎟ = = ⎜ 10 ⎟ other, let l m 6 in the procedure. Under the settings ⎜ 1 3 ⎟ = ( ) = ( ) ⎜ 3 ⎟ × u 1, 1, 1, 1, 1, 1 and w 1, 1, 1, 1, 1, 1 , the result- T = ⎜ ⎟ ∈ R6 6. ⎜ 10−8 ⎟ ing matrix T is a real tridiagonal matrix, which can not be ⎜ ⎟ ⎜ ⎟ symmetrized, ⎝ 29 − 1 ⎠ 1 8 64 ⎛ ⎞ 11 13 − 19 1 8 ⎜ 6 36 ⎟ ⎜ 1 443 − 1920 ⎟ The characteristic and the minimal polynomials of A and ⎜ 114 361 ⎟ ⎜ − 363 − 209 ⎟ ⎜ 1 ⎟ × T are all the same polynomial with respect to z, which is T = ⎜ 760 1600 ⎟ ∈ R6 6. ( − )6 ⎜ 1 1187 − 240 ⎟ factored as z 2 . So, the tridiagonal matrix T is not ⎜ 440 121 ⎟ ⎝ 37 11 ⎠ diagonalizable. 1 66 36 Let us prepare the Jordan matrix 1 13 ⎛ ⎞ 6 31 ⎜ ⎟ The characteristic and the minimal polynomials of A and ⎜ 31 ⎟ ⎜ ⎟ T are all the same polynomial with respect to z, which is ⎜ 3 ⎟ 2 2 ⎜ ⎟ factorized as (z − 2 + i) (z − 2 − i) (z − 2)(z − 1).So,the ⎜ 31 ⎟ × A = ⎜ ⎟ ∈ R8 8. tridiagonal matrix T is not a diagonalizable matrix with ⎜ 3 ⎟ ⎜ ⎟ the same complex multiple eigenvalues and real distinct ⎜ 3 ⎟ ⎝ ⎠ ones as A. 21 2 6Conclusion The matrix A has multiple eigenvalues such as λ1 = 3, In this paper, we clarify that the qd recursion formula λ2 = 3, λ3 = 3, λ4 = 3, λ5 = 3, λ6 = 3, λ7 = 2, is applicable to constructing a tridiagonal matrix with λ8 = 2. It is noted that |λ1|=|λ2|=|λ3|=|λ4|=|λ5|= specified multiple eigenvalues. We first investigate the |λ6| > |λ7|=|λ8| > 0. The characteristic and the mini- denominator of the generating function associated with mal polynomials of A are factorized as (z−2)2(z−3)6 and the sequence given from two suitable vectors and the pow- (z − 2)2(z − 3)3,respectively.So,letl = 5andm = 8in ers of a general matrix A, through considering the Jordan the procedure. Then, the settings u = (1, 1, 1, 1, 1, 1, 1, 1) canonical form of A. Accordingly, it is observed that the Akaiwa et al. Pacific Journal of Mathematics for Industry 2014, 6:10 Page 9 of 9 http://www.pacific-mathforindustry.com/content/6/1/10

minimal polynomial of A coincides with the characteristic polynomial of a tridiagonal matrix T, denoted by φT (z), m or the polynomial z L φT (z) for the multiplicity mL of the zero-eigenvalues of A. Next, by taking account of the Jor- dan canonical form of T, we show that the characteristic and the minimal polynomials of T are equal to each other. We finally present a procedure for constructing a tridiag- onal matrix with specified multiple eigenvalues, and then give four examples for the resulting procedure.

Acknowledgements The authors would like to thank the reviewer for his/her careful reading and beneficial suggestions. This work is supported by JSPS KAKENHI Grant Number 23654032.

Author details 1Graduate School of Informatics, Kyoto University, Yoshida-Hommachi, Sakyo-ku, Kyoto 606-8501, Japan. 2Department of Informatics and Environmental Sciences, Kyoto Prefectural University, 1-5 Nakaragi-cho, Shimogamo, Sakyo-ku, Kyoto 606-8522, Japan. 3Graduate School of Science and Engineering, Doshisha University, 1-3 Tatara Miyakodani, Kyoto 610-0394, Japan.

Received: 12 November 2013 Revised: 25 March 2014 Accepted: 25 March 2014

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