mathematics
Article Characteristic Polynomials and Eigenvalues for Certain Classes of Pentadiagonal Matrices
María Alejandra Alvarez 1,* , André Ebling Brondani 2 , Francisca Andrea Macedo França 2 and Luis A. Medina C. 1
1 Departamento de Matemáticas, Facultad de Ciencias Básicas, Universidad de Antofagasta, Antofagasta 1240000, Chile; [email protected] 2 Departamento de Matemática, Instituto de Ciências Exatas, Universidade Federal Fluminense, Volta Redonda 27213-145, RJ, Brazil; [email protected] (A.E.B.); [email protected] (F.A.M.F.) * Correspondence: [email protected]
Received: 13 May 2020; Accepted: 28 June 2020; Published: 1 July 2020
Abstract: There exist pentadiagonal matrices which are diagonally similar to symmetric matrices. In this work we describe explicitly the diagonal matrix that gives this transformation for certain pentadiagonal matrices. We also consider particular classes of pentadiagonal matrices and obtain recursive formulas for the characteristic polynomial and explicit formulas for their eigenvalues.
Keywords: pentadiagonal matrix; symmetric pentadiagonal matrix; eigenvalue; characteristic polynomial
MSC: 15B99; 15A18
1. Introduction Tridiagonal and pentadiagonal matrices appear in several areas of mathematics and engineering, specially involving linear systems of differential equations. In [1], the authors give necessary and sufficient conditions for a matrix to be diagonally similar to a symmetric matrix. For tridiagonal matrices, the explicit construction of the diagonal matrix involved in this similar transformation is given in [2]. There are several results concerning different types of tridiagonal matrices and the obtaining of their eigenvalues and eigenvectors, see for instance [3] and [4] and the references therein. In the case of pentadiagonal matrices, there are many articles concerning algorithms for solving systems of equations associated with them. Among these works, we mention [5–12]. Furthermore, there are some results for particular cases of pentadiagonal matrices. In [13], the author gives a recurrence formula for the determinant of pentadiagonal matrices A = (Ai,j), such that Ai,j 6= 0 for |i − j| = 1. In [14], an algorithm is given to find the determinant of pentadiagonal matrices satisfying Ai,i+2 6= 0. In [15], the author shows that the characteristic polynomial for such matrices is the product of two polynomials given in terms of Chebyshev polynomials. In [16], the authors study pentadiagonal Toeplitz matrices and give determinantal identities for the symmetric and skew-symmetric cases. In [17] and [18], banded Toeplitz matrices are studied and, in particular, results on particular banded pentadiagonal Toeplitz matrices are obtained. Since every 3 × 3 matrix is a pentadiagonal one, it is clear that not every pentadiagonal matrix is similar to a symmetric matrix. In this work, we consider two classes of pentadiagonal matrices and obtain recursive formulas for the characteristic polynomials and explicit formulas for the eigenvalues of these classes of pentadiagonal matrices. The paper is organized as follows: in Section2, we present two classes of pentadiagonal matrices and show explicitly that these matrices are similar to symmetric pentadiagonal ones. In Section3 we obtain recursive formulas for the characteristic polynomials of this
Mathematics 2020, 8, 1056; doi:10.3390/math8071056 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 1056 2 of 12 type of matrices and a result regarding their eigenvalues. In Section4, we consider special subclasses of pentadiagonal matrices and, using the results of Section3, we show their eigenvalues and provide some results on the nullity and spectral radius of these matrices. Let Mn(R) be the set of square matrices of order n with real entries and let S ⊂ Mn(R) be the set of symmetric matrices. We denote the (i, j) entry of a matrix A by Ai,j, and in some cases, to avoid confusion, we will use the notation [A]i,j. The highest integer lower than or equal to x will be denoted by bxc. A matrix A ∈ Mn(R) is called a pentadiagonal matrix if Ai,j = 0 whenever |i − j| > 2. The class of pentadiagonal matrices, denoted by Pn, consists of matrices of the form a1 b1 d1 0 . . . 0 . . c a b d .. . 1 2 2 2 . . . ...... e1 c2 0 A = . ...... 0 . . . . dn−2 ...... . . . . . bn−1 0 . . . 0 en−2 cn−1 an We will distinguish two subclasses of pentadiagonal matrices, namely:
n − 1 C = A ∈ Pn : b = c = d − = e − = 0, for i = 1, 2, . . . , 1 2i 2i 2i 1 2i 1 2 and n j n ko C = A ∈ Pn : b − = c − = d = e = 0, for i = 1, 2, . . . , . 2 2i 1 2i 1 2i 2i 2
For a real symmetric matrix A = (Ai,j) of order n, the non-directed graph G(A) associated to A consists of vertices {1, 2, ... , n} and edges {i, j} for which i 6= j and Ai,j Aj,i 6= 0. The matrix A is acyclic if G(A) has no cycles.
2. Pentadiagonal Matrices Similar to Symmetric Matrices
In [1], the authors give necessary and sufficient conditions for a matrix A = (Ai,j) to be diagonally similar to a symmetric matrix, namely
Ai,j 6= 0 implies Aj,i 6= 0, (1) and for any sequence of integers i1,..., ir such that 1 ≤ ik ≤ n, k = 1, . . . , r the following holds
= Ai1,ir Air,ir−1 ... Ai3,i2 Ai2,i1 Ai1,i2 Ai2,i3 ... Air−1,ir Air,i1 . (2)
The construction of the diagonal matrix for the case of tridiagonal matrices is done in [2]. For acyclic matrices is given in [19] and [20]. In the following theorem we impose conditions to −1 matrices A ∈ C1 ∪ C2 and give the explicit construction of a diagonal matrix D such that DAD is a symmetric pentadiagonal matrix.
Theorem 1. Let A ∈ C1 ∪ C2. If any of the following conditions holds:
(i)A ∈ C1 satisfies bici > 0 or bi = ci = 0, for i = 1, . . . , n − 1; (ii)A ∈ C2 satisfies diei > 0 or di = ei = 0, for i = 1, . . . , n − 2, then A is similar to a symmetric pentadiagonal matrix.
Proof. Under assumptions (i) or (ii), A satisfies conditions (1) and (2), therefore the existence of D is guaranteed. If bici > 0 (respectively diei > 0) then the sign of bi and ci (respectively di and ei) are Mathematics 2020, 8, 1056 3 of 12
the same. Furthermore, if bi = 0 (respectively if di = 0) then ci = 0 (respectively ei = 0). We define σi (respectively τi) in the following manner: 1, if b > 0; 1, if d > 0; i i σi = 0, if bi = 0; and τi = 0, if di = 0; −1, if bi < 0, −1, if di < 0.
We will consider each condition separately. Assume A satisfies condition (i). We will prove that A is similar to a symmetric pentadiagonal matrix Rn of order n of the form: • If n is even:
√ √a1 σ1 b1c1 0√ 0 ...... 0 σ1 b1c1 a2 0 τ2 d2e2 0 √ 0 0 a3 σ3 b3c3 0 0 √ √ √ 0 τ2 d2e2 σ3 b3c3 a 0 τ d e 0 . . . 0 4 4 4 4 Rn = . , (3) ...... ...... . . √ . 0 0√ 0 √an−1 σn−1 bn−1cn−1 0 . . . 0 τn−2 dn−2en−2 σn−1 bn−1cn−1 an • If n is odd: ! Rn−1 0 Rn = , where Rn−1 is of the form (3). 0 an
Let D = diag{α1, α2, ... , αn} be the diagonal matrix whose diagonal entries are given, recursively, by:
α1 = 1; s b2k−1 n α2k = α2k−1, for k = 1, . . . , 2 ; c2k−1 s r d2k c2k+1 n α2k+1 = α2k, for k = 1, . . . , 2 − 1; e2k b2k+1 αn = 1, if n is odd.
In the previous formula: s d2j−2 • If b2j−1 = c2j−1 = 0 for some j, then we replace α2j = α2j−1 = α2j−2. e2j−2 s c2j+1 • If d2j = e2j = 0 for some j, then we replace α2j+1 = α2j. b2j+1
−1 −1 Consider now X = DAD . Then X = (Xi,j) = (αi Ai,jαj ), where • If i is odd, then a , if j = i; a , if j = i; i i √ Ai,j = bi, if j = i + 1; and [Rn]i,j = σi bici, if j = i + 1; 0, elsewhere, 0 elsewhere.
• If i is even, then p ei−2, if j = i − 2; τi−2 di−2ei−2, if j = i − 2; p c , if j = i − 1; σ b c , if j = i − 1; i−1 i−1 i−1 i−1 Ai,j = ai, if j = i; and [Rn]i,j = ai, if j = i; √ di, if j = i + 2; τi diei, if j = i + 2; 0, elsewhere, 0, elsewhere. Mathematics 2020, 8, 1056 4 of 12
Then we obtain:
(a) If i is odd, then r −1 ci −1 p Xi,i = ai = [Rn]i,i, and Xi,i+1 = αibiαi+1 = αibi αi = σi bici = [Rn]i,i+1. bi
(b) If i is even, then
s s r −1 bi−1 di−2 ci−1 −1 p Xi,i−2 =αiei−2αi−2 = αi−2ei−2αi−2 = τi−2 di−2ei−2 = [Rn]i,i−2, ci−1 ei−2 bi−1 s −1 bi−1 −1 p Xi,i−1 =αici−1αi−i = αi−1ci−1αi−1 = σi−1 bi−1ci−1 = [Rn]i,i−1, ci−1 −1 Xi,i =αiaiαi = [Rn]i,i, r r s −1 ci+1 ei bi+1 −1 p Xi,i+2 =αidiαi+2 = αidi αi = τi diei = [Rn]i,i+2. bi+1 di ci+1
Therefore X = Rn and A is similar to a symmetric pentadiagonal matrix. Assume now that A satisfies condition (ii). We will prove that A is similar to a symmetric matrix Tn of the form:
• If n is odd:
√ a1 0 τ1 √d1e1 0 ...... 0 0 a σ b c 0 0 √ √2 2 2 2 √ τ d e σ b c a 0 τ d e 0 1 1 1 2 2 2 3 3 √ 3 3 0 0 0 a4 σ4 b4c4 0 0 . . . 0 ...... Tn = ...... (4) ...... ...... ...... ...... √ 0 0√ 0 √an−1 σn−1 bn−1cn−1 0 . . . 0 τn−2 dn−2en−2 σn−1 bn−1cn−1 an
• If n is even: ! Tn−1 0 Tn = where Tn−1 is of the form (4). 0 an
Let D = diag{β1, β2,..., βn} be the diagonal matrix defined by:
β1 = 1; s r d2k−1 c2k j n−1 k β2k = β2k−1, for k = 1, . . . , 2 ; e2k−1 b2k s b2k j n−1 k β2k+1 = β2k, for k = 1, . . . , 2 ; c2k βn = 1, if n is even.
In the previous formula: s d2j−1 • If b2j = c2j = 0 for some j, then we replace β2j+1 = β2j = β2j−1. e2j−1 s c2j • If d2j−1 = e2j−1 = 0 for some j, then we replace β2j = β2j−1. b2j
−1 Analogously to case (i), we obtain DAD = Tn and the proof is complete. Mathematics 2020, 8, 1056 5 of 12
Corollary 1. If A is a pentadiagonal matrix under the hypothesis of Theorem1, then all the eigenvalues of A are real and A is diagonalizable.
Example 1. Consider the pentadiagonal matrix,
1 1 0 0 0 0 3 2 0 −2 0 0 − − 0 0 1 2 0 0 A = . 0 −1 −5 3 0 4 0 0 0 0 2 7 0 0 0 5 7 0
Since A satisfies the conditions of Theorem1, it is similar to the symmetric pentadiagonal matrix √ √1 3 0√ 0 0 0 3 2 0 − 2 0 0 √ − − 0√ 0 √1 10 0√ 0 B = , 0 − 2 − 10 3 0 20 0 0 0√ 0 2 7 0 0 0 20 7 0
( r r r r r ) 1 5 2 8 8 by the diagonal matrix D = diag 1, , , , , . 3 3 3 15 15
Remark 1. The pentadiagonal matrices of Theorem1 are, in general, not similar by permutation to tridiagonal matrices. In fact, the graph associated to a pentadiagonal matrix considered in Theorem1 is a caterpillar, while the graph associated to a tridiagonal matrix is a path. Since a caterpillar is, in general, not a path, the problem of finding a symmetric matrix similar to a given pentadiagonal matrix one, is not equivalent to the related problem regarding tridiagonal matrices.
3. On Characteristic Polynomials In this section we give recursive formulas for the characteristic polynomials of the considered matrices. Let An ∈ C1 ∩ S and Bn ∈ C2 ∩ S, i.e.,
a1 b1 0 0 0 0 ... 0 a1 0 d1 0 0 0 ... 0 b1 a2 0 d2 0 0 ... 0 0 a2 b2 0 0 0 ... 0 0 0 a3 b3 0 0 ... 0 d b2 a3 0 d3 0 ... 0 1 0 d2 b3 a4 0 d4 ... 0 0 0 0 a4 b4 0 ... 0 0 0 0 0 a5 b5 ... 0 0 0 d3 b4 a5 0 ... 0 An = 0 0 0 d4 b5 a6 ... 0 and Bn = 0 0 0 0 0 a6 ... 0 , (5) ...... ...... ...... ...... . . . . 0 0 0 0 0 0 ... an−2 0 dn−2 0 0 0 0 0 0 ... an−2 0 dn−2 0 0 0 0 0 0 ... 0 an−1 bn−1 0 0 0 0 0 0 ... 0 an−1 bn−1 0 0 0 0 0 0 ... dn−2 bn−1 an 0 0 0 0 0 0 ... dn−2 bn−1 an where the entries bn−1 and dn−2 are equal to zero according to the parity of the order of each matrix. Mathematics 2020, 8, 1056 6 of 12
Proposition 1. Let P0(λ) = 1. The characteristic polynomial Pn(λ) of An is given by the recursion formula:
P1(λ) = (a1 − λ), 2 P2(λ) = (a2 − λ)(a1 − λ) − b1, and ( (a − λ)P (λ) if n is odd, n ≥ 3 ( ) = n n−1 Pn λ 2 2 (an − λ)(an−1 − λ) − bn−1 Pn−2(λ) − dn−2(an−1 − λ)(an−3 − λ)Pn−4(λ) if n is even, n ≥ 4.
Proof. By computing the determinant along the last row, we obtain
2 2 Pn(λ) = (an − λ)Pn−1(λ) − bn−1Pn−2(λ) − dn−2(an−1 − λ)Pn−3(λ).
Clearly, if k is odd, we obtain that Pk(λ) = (ak − λ)Pk−1(λ). By replacing in the previous formula, we obtain the result.
Analogously, we obtain:
Proposition 2. The characteristic polynomial Qn(λ) of Bn is given by the recursion formula:
Q1(λ) = a1 − λ,
Q2(λ) = (a2 − λ)(a1 − λ), 2 2 Q3(λ) = (a1 − λ)(a2 − λ)(a3 − λ) − b2(a1 − λ) − d1(a2 − λ), and ( (a − λ)Q (λ) if n is even, n ≥ 4 ( ) = n n−1 Qn λ 2 2 (an − λ)(an−1 − λ) − bn−1 Qn−2(λ) − dn−2(an−1 − λ)(an−3 − λ)Qn−4(λ), if n is odd, n ≥ 5.
The following two corollaries show that the symmetric pentadiagonal matrices in C1 or C2 have exactly n distinct eigenvalues.
n Corollary 2. Let An ∈ C1 ∩ S such that ai = a for i = 1 ... n. If b2j−1d2k 6= 0 for j = 1 ... b 2 c and n−2 k = 1 . . . b 2 c, then the eigenvalues of An are all simple.
n Proof. First of all, notice that P2k(a) 6= 0 for 0 ≤ k ≤ b 2 c. If P2k(a) = 0 for some k, this would imply 2 that P2j(a) = 0 for 1 ≤ j < k, but P2(a) = −b1 6= 0, contradicting the assumption. Since for a real symmetric matrix, the eigenvalues of a submatrix of lower order interlace the eigenvalues of the matrix, then the zeros of P2k−1(λ) interlace the zeros of P2k(λ). Assume now that P2k(λ) and P2k−1(λ) have a common zero µ. If 0 = P2k−1(µ) = (a − µ)P2k−2(µ) we must have µ = a or P2k−2(µ) = 0. By the previous, µ = a is not a zero of P2k(λ), therefore µ 6= a and P2k−2(µ) = 0. In this 2 case, 0 = P2k(µ) = −d2k−2(a − µ) P2k−4(µ) and then P2k−4(µ) = 0. This implies that P2j(µ) = 0 for 2 2 1 ≤ j < k, but P4(µ) = −d2(a − µ) P0(µ) 6= 0 contradicting the assumption, therefore P2k(λ) and P2k−1(λ) do not have a common zero and then all the eigenvalues of An are simple.
n−2 Corollary 3. Let Bn ∈ C2 ∩ S such that ai = a for i = 1 ... n. If b2jd2k−1 6= 0 for j = 1 ... b 2 c and n k = 1 . . . b 2 c, then the eigenvalues of Bn are all simple.
Proof. Follows analogously to the proof of Corollary2.
Remark 2. As we mentioned before, there are some works concerning the obtaining of formulas for the characteristic polynomial, eigenvalues and eigenvectors of a pentadiagonal matrix A = (Ai,j). In [13], the author consider matrices satisfying Ai,j 6= 0 for |i − j| = 1. In [14], the matrices are constrain to the condition Ai,i+2 6= 0. As one can easily see, the classes C1 and C2 are not contained in the previous cases. Mathematics 2020, 8, 1056 7 of 12
4. Spectrum of Special Classes of Pentadiagonal Matrices Here we apply the previous results to special classes of pentadiagonal matrices to obtain explicit formulas of the eigenvalues and some results concerning them.
a 4.1. The Class C1
Consider the families of subclasses of C1, given by: ai = a, for i = 1, . . . , n; a n C1 = An ∈ C1 : b2i−1 = c2i−1 = b ≥ 0, for i = 1, . . . , 2 ; . n d2i = e2i = d ≥ 0, for i = 1, . . . , 2 − 1.
a So, the elements of C1 are of the form:
a b 0 0 . . . 0 . . b a 0 d .. . . . . ! ...... 0 0 0 A2k 0 A2k = and A2k+1 = . ...... 0 a 0 . . . . d ...... . . . . . b 0 . . . 0 d b a
a 0 Remark 3. Every matrix A ∈ C1 can be decomposed as A = a I +B where B ∈ C1 . Then every eigenvalue of A is of the form λi(A) = a + λi(B) for i = 1, . . . , n. a 0 Therefore, in order to obtain the spectra of C1, it is enough to study the class C1 .
In view of this, we present in the next theorem the characteristic polynomial for matrices in this class.
0 Theorem 2. Let Pn(λ) be the characteristic polynomial of An ∈ C1 .
k b 2 c l k − l 2 2 k−2l 2l (i) If n = 2k, then P2k(λ) = ∑(−1) (λ − b ) (dλ) . l=0 l (ii) If n = 2k + 1, then P2k+1(λ) = −λP2k(λ).
Proof. By Proposition1 we obtain that
2 2 2 2 P2k(λ) = (λ − b )P2k−2(λ) − d λ P2k−4(λ).
2 2 2 2 Let r = λ − b and s = d λ . Then P2k(λ) can be obtained as
k k P2k(λ) = αx1 + βx2,
2 where x1, x2 are the roots of x − rx + s = 0, this is √ r ± r2 − 4s x = . 1,2 2 √ Set r = 2 s cos θ. Then √ √ √ √ 2 s cos θ ± 4s cos2 θ − 4s 2 s cos θ ± i2 s sin θ √ x = = = se±iθ = dλe±iθ. 1,2 2 2 Mathematics 2020, 8, 1056 8 of 12
The scalars α and β are obtained by replacing the formula with the initial conditions
2 2 P0(λ) = 1, P2(λ) = λ − b .
Then √ √ √ 1 = α + β, 2 s cos θ = α s [cos θ + i sin θ] + β s [cos θ − i sin θ] , and we obtain that 1 cos θ 1 cos θ α = 1 − i and β = 1 + i . 2 sin θ 2 sin θ Therefore 1 cos θ 1 cos θ P (λ) = 1 − i (dλ)keikθ + 1 + i (dλ)ke−ikθ, 2k 2 sin θ 2 sin θ and by reducing this expression, we obtain
sin((k + 1)θ) P (λ) = (dλ)k . (6) 2k sin θ Now, by considering the recursion formula
b m−1 c 2 m − 1 − l sin(mθ) = sin θ ∑ (−1)l 2m−2l−1(cos θ)m−2l−1, (7) l=0 l
λ2 − b2 and since cos θ = , we obtain that 2dλ
k b 2 c l k − l 2 2 k−2l 2l P2k(λ) = ∑(−1) (λ − b ) (dλ) . (8) l=0 l
As we mentioned in the proof of Proposition1, P2k+1(λ) = −λP2k(λ) and the result follows.
0 Corollary 4. Consider An ∈ C1 . Then ( (−1)kb2k if n = 2k, det(An) = 0 if n = 2k + 1.
Now we are ready to stablish the main result of this section:
0 Theorem 3. The eigenvalues of An ∈ C1 of order n = 2k, are given by the following formulas:
s s πj 2 2 πj 2 πj 2 2 πj 2 µ = d cos + d cos + b and µ + = d cos − d cos + b , j k + 1 k + 1 k j k + 1 k + 1 for j = 1, . . . , k.
Proof. Consider equation (6). Then P2k(λ) = 0 if and only if sin((k + 1)θ) = 0 or λ = 0. 1 i Set now t 2 = e θ. Then
k+1 − k+1 k+ 1 − 1 t 2 − t 2 t 1 − 1 t 2 − t 2 t − 1 sin((k + 1)θ) = = k+1 and sin θ = = 1 . 2i 2it 2 2i 2it 2 Therefore sin((k + 1)θ) 1 tk+1 − 1 0 = = sin θ tk+1 t − 1 Mathematics 2020, 8, 1056 9 of 12
2πj i k+1 and all the roots of this equation are the (k + 1)-th roots of unit distinct of 1. Let ξj = e for 1 1 2πj 2 iθj 2 i k+1 πj j = 1, . . . , k be these roots. Then e = ξj = e and θj = k+1 for j = 1, . . . , k. λ2 − b2 Since, by the proof of Theorem2, cos θ = , we obtain that λ2 − (2d cos θ)λ − b2 = 0, and 2dλ then √ 2d cos θ ± 4d2 cos2 θ + 4b2 λ = . 2 By replacing,
r 2d cos πj ± 4d2 cos2 πj + 4b2 s k+1 k+1 πj πj λ = = d cos ± d2 cos2 + b2. 2 k + 1 k + 1
Therefore, all the eigenvalues of P2k(λ) are given by:
s s πj 2 2 πj 2 πj 2 2 πj 2 µ = d cos + d cos + b and µ + = d cos − d cos + b (9) j k + 1 k + 1 k j k + 1 k + 1 for j = 1, . . . , k. Since we have obtained 2k roots of P2k(λ), we conclude that these are all the eigenvalues of An.
Remark 4. λ = 0 is a root of P2k(λ) if and only if b = 0. In fact, if λ = 0 is a root of P2k(λ) then 0 = P2m(0) 2 for every m < k. Since P2(0) = −b then we must have b = 0. Equation (9) says that the condition b = 0 is necessary in order to obtain λ = 0 as an eigenvalue of an even order matrix An.
Corollary 5. If b = 0 in An, then the multiplicity of λ = 0 is 2m, if n = 4m; 2m + 1, if n = 4m + 1; mult(0) = 2m + 2, if n = 4m + 2; 2m + 3, if n = 4m + 3.
πj Proof. If b = 0 and n = 2k, we obtain by equation (9), that µ = 2 cos and µ + = 0 for j k + 1 k j πj k + 1 j = 1, ... , k. Since cos = 0 if and only if (k + 1) is even and j = , we obtain multiplicity k + 1 2 k for n = 2(2m) and multiplicity k + 1 for n = 2(2m + 1). If n is odd, the eigenvalues of An are λ = 0 and the (n − 1) eigenvalues of An−1. Then the result follows. s π π Corollary 6. The spectral radius of A ∈ C0 is ρ = d cos + d2 cos2 + b2. n 1 k + 1 k + 1
s πj πj Proof. Consider f (j) = d cos + d2 cos2 + b2. Then k + 1 k + 1
πj π d sin k+1 k+1 f 0(j) = − f (j) r 2 2 πj 2 d cos k+1 + b Mathematics 2020, 8, 1056 10 of 12
πj π By Remark4, f (j) = 0 if and only if b = 0 and in this case f 0(j) = −2d sin 6= 0 for k + 1 k + 1 j = 1, ... , k. Therefore the critical points of f (j) are in j = 1 and j = k. If b 6= 0, we conclude by the previous analysis, that j = 1 and j = k are the critical points. Moreover, since f (1) > 0 and f (k) < 0, we obtain that f (j) has a maximum value at j = 1, and the proof is complete.
Example 2. If d = 0 in An of even order n = 2k, we obtain a tridiagonal matrix and the eigenvalues are b and −b, both with multiplicity k.
πj Example 3. Consider k = 5. Then θ = for j = 1, . . . , 5. Therefore the eigenvalues of A , are: j 6 10
√ ! s √ √ √ √ 3 3 d 3 + 3d2 + 4b2 d 3 − 3d2 + 4b2 µ =d + d2 + b2 = , µ = , 1 2 4 2 6 2 s √ √ 1 1 d + d2 + 4b2 d − d2 + 4b2 µ =d + d2 + b2 = , µ = , 2 2 4 2 7 2 √ 2 µ3 = b = b, µ8 = −b, s √ √ 1 1 −d + d2 + 4b2 −d − d2 + 4b2 µ =d − + d2 + b2 = , µ = , 4 2 4 2 9 2 √ ! s √ √ √ √ 3 3 −d 3 + 3d2 + 4b2 −d 3 − 3d2 + 4b2 µ =d − + d2 + b2 = , µ = . 5 2 4 2 10 2
0 4.2. On the Class C2 a Analogously to the class C1, one can define the families of subclasses of C2, given by: ai = a for i = 1, . . . , n; j k a = = ≥ = n−1 C = Bn ∈ C2 : b2i c2i b 0 for i 1, . . . , 2 ; . 2 j k n−1 d2i−1 = e2i−1 = d ≥ 0 for i = 1, . . . , 2 .
a Then elements of C2 are of the form:
a 0 d 0 . . . 0 . . 0 a b 0 .. . . . . ! ...... d b 0 B2k−1 0 B2k−1 = and B2k = . ...... 0 a 0 . . . . d ...... . . . . . b 0 . . . 0 d b a
0 a We will only consider the class C2 , since the general case C2 follows by Remark3. Using Propositions1 and2 and replacing the coefficients by b and d properly, we obtain the following recursion formula:
2 Q2k+1(λ) = − λP2k(λ) + λd P2k−2(λ), (10) Q2k+2(λ) = − λQ2k+1(λ). Mathematics 2020, 8, 1056 11 of 12
By equations (10) and (6), we obtain k k−1 sin((k + 1)θ) sin(kθ) Q + (λ) = λ d λ −λ + d . 2k 1 sin θ sin θ
d sin(kθ) Therefore, Q + (λ) has one root λ = 0 and the remaining are given by λ = . 2k 1 sin((k + 1)θ) Furthermore, by the proof of Theorem2, λ2 − (2d cos)λ − b2 = 0. These together imply
d2 sin2(kθ) − 2d2 cos θ sin(kθ) sin((k + 1)θ) − b2 sin2((k + 1)θ) = 0.
A straightforward computation shows that
d2 sin(kθ) sin((k + 2)θ) + b2 sin2((k + 1)θ) = 0.
1 i Again, if t 2 = e θ, then we obtain
(tk − 1)(tk+2 − 1) b2 = − , (tk+1 − 1)2 d2 and k−1 k+1 ∑ ti ∑ tj i=0 j=0 b2 = − . k d2 ∑ ti i=0 Therefore
k−1 k−1 ∑ (s + 1)(b2 + d2)ts + (kd2 + (k + 1)b2)tk + ∑ (s + 1)(b2 + d2)t2k−s = 0. (11) s=0 s=0
If ti, i = 1, . . . , 2k, are the roots of equation (11), then
1 k dt 2 (t − 1) λ = i i = k i k+1 , 1, . . . 2 (ti − 1) are the remaining roots of Q2k+1(λ).
Remark 5. If b 6= 0 or d 6= 0 in Bn, the multiplicity of λ = 0 is ( 1, if n is odd; mult(0) = 2, if n is even.
5. Conclusions In this work, we have considered certain pentadiagonal matrices that are diagonally similar to symmetric pentadiagonal matrices. We provided recursive formulas for the characteristic polynomial of symmetric pentadiagonal matrices, and for particular subclasses we have given explicit formulas for the eigenvalues.
Author Contributions: All authors have contributed equally to this work. All authors have read and agreed to the published version of the manuscript. Funding: This research was funded by Fondo Puente de Investigación de Excelencia—FPI-18-02 and Coloquio de Matemática from Universidad de Antofagasta. Mathematics 2020, 8, 1056 12 of 12
Acknowledgments: The authors thank the referees for their comments and suggestions which improved the presentation of this paper. The researchers A.E. Brondani and F.A.M. França thank the hospitality of Departamento Matemática of Universidad de Antofagasta where this paper was started. M.A. Alvarez and L. Medina thank the support of Vicerrectoría de Investigación, Innovación y Postgrado from Universidad de Antofagasta. Conflicts of Interest: The authors declare no conflict of interest.
References
1. Parter, S.V.; Youngs, J.W.T. The symmetrization of matrices by diagonal matrices. J. Math. Anal. Appl. 1962, 4, 102–110. [CrossRef] 2. Rozsa, R. On periodic continuants. Linear Algebra Appl. 1969, 2, 267–274. [CrossRef] 3. Losonczi, L. Eigenvalues and eigenvectors of some tridiagonal matrices. Acta Math. Hung. 1992, 60, 309–332. [CrossRef] 4. da Fonseca, C.M.; Kowalenko, V. Eigenpairs of a family of tridiagonal matrices: three decades later. Acta Math. Hung. 2020, 160, 376–389. [CrossRef] 5. Askar, S.S.; Karawia, A.A. On Solving Pentadiagonal Linear Systems via Transformations. Math. Probl. Eng. 2015, 2015, 232456. [CrossRef] 6. Batista, M. A method for solving cyclic block penta-diagonal systems of linear equations. arXiv 2008, arXiv:0806.3639V5. 7. Jia, J.T.; Sogabe, T. A novel algorithm for solving quasi penta-diagonal linear systems. J. Math. Chem. 2013, 51, 881–889. [CrossRef] 8. Karawia, A.A. A computational algorithm for solving periodic pentadiagonal linear systems. Appl. Math. Comput. 2006, 174, 613–618. 9. Lv, X.-G.; Le, J. A note on solving nearly pentadiagonal linear systems. Appl. Math. Comput. 2008, 204, 707–712. 10. El-Mikkawy, M.; El-Desouky Rahmo. Symbolic algorithm for inverting cyclic pentadiagonal matrices recursively—Derivation and implementation. Comput. Math. Appl. 2010, 59, 1386–1396. [CrossRef] 11. Navon, I.M. A periodic pentadiagonal systems solver. Commun. Appl. Numer. Methods 1987, 3, 63–69. [CrossRef] 12. Nguetchue, S.N.N.; Abelman, S. A computational algorithm for solving nearly penta-diagonal linear systems. Appl. Math. Comput. 2008, 203, 629–634. [CrossRef] 13. Sweet, R.A. A recursive relation for the determinant of a pentadiagonal matrix. Commun. ACM 1969, 12, 330–332. [CrossRef] 14. Hadj, A.D.A.; Elouafi, M. On the characteristic polynomial, eigenvectors and determinant of a pentadiagonal matrix. Appl. Math. Comput. 2008, 198, 634–642. 15. Elouafi, M. On formulae for the determinant of symmetric pentadiagonal Toeplitz matrices. Arab. J. Math. 2018, 7, 91–99. [CrossRef] 16. Andelic, M.; da Fonseca, C.M. Some determinantal considerations for pentadiagonal matrices. Linear Multilinear Algebra 2019.[CrossRef] 17. Trench, W.F. On the Eigenvalue Problem for Toeplitz Band Matrices. Linear Algebra Appl. 1985, 64, 199–214. [CrossRef] 18. Ekström, S.-E.; Serra-Capizzano, S. Eigenvalues and eigenvectors of banded Toeplitz matrices and the related symbols. Numer. Linear Algebra Appl. 2018, 25, e2137. [CrossRef] 19. Berman, A.; Hershkowitz, D. Matrix diagonal stability and its applications. SIAM J. Algebr. Discre. Methods 1983, 4, 377–382. [CrossRef] 20. Berman, A.; Hershkowitz, D. Characterization of acyclic d-stable matrices. Linear Algebra Appl. 1984, 58, 17–31. [CrossRef]
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