Spectral Properties of Anti-Heptadiagonal Persymmetric

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0 ...... 0 αn βn γn dn . ..  . . αn−1 βn−1 γn−1 dn−1 cn−1  . . .  . .. .. β γ d c b   n−2 n−2 n−2 n−2 n−2   . . . .   ......   . γn−3 dn−3 cn−3 bn−3 an−3   . . . . .   ......   . dn−4 cn−4 bn−4 an−4 0   . .   ......   ......     ......   0 α5 β5 γ5 d5 . . . . .     ......   α4 β4 γ4 d4 c4 . . . .     .. .. .   β3 γ3 d3 c3 b3 . . .     .. .   γ2 d2 c2 b2 a2 . .     d1 c1 b1 a1 0 ...... 0      is called anti-heptadiagonal. Throughout, we shall consider the following n × n anti-heptadiagonal persymmetric Hankel matrix

0 ...... 0 a b cd . .  . .. a b c dc  . . .  . .. .. b c d cb     . . . .   ......   . c d c ba   . . . . .   ......   . d c b a 0   ......  Hn =  ......  . (1.1)  ......     ......   0 a b c d . . . . .     ......   ab c d c . . . .     .. .. .   bc d c b . . .     . .   cd c b a .. .     dc b a 0 ...... 0      2 Auxiliary tools

Consider the class of complex matrices deﬁned by

A M C n := An ∈ n( ):[An]k−1,ℓ + [An]k+1,ℓ = [An]k,ℓ−1 + [An]k,ℓ+1 , k,ℓ =1, 2,...,n n o where it is assumed [An]n+1,ℓ = [An]k,n+1 = [An]0,ℓ = [An]k,0 = 0 for all k,ℓ. We begin by retrieve a pair of results stated in [1] (see Propositions 2.1 and 2.2), announced here for complex matrices. The proofs do not suﬀer any changes from the original ones, whence we will omit the details.

2 Lemma 1 (a) The class An is the algebra generated over C by the n × n matrix

0 1 0 ...... 0 .. .  1 0 1 . .  .. .. .  0 1 0 . . .     ......  Ωn =  ......  . (2.1)    . . .   . .. .. 0 1 0     . .   . .. 1 0 1     0 ...... 0 1 0      ⊤ (b) If An ∈ An and a is its ﬁrst row then

n−1 k An = ωk+1Ωn k X=0 ⊤ where Ωn is the n × n matrix (2.1) and ω = (ω1,ω2,...,ωn) is the solution of the upper triangular ω system Tn = a, with [Tn]0,0 := 1, [Tn]0,ℓ := 0 for all 1 6 ℓ 6 n,

[T ] + [T ] , 1 6 k 6 ℓ 6 n [T ] := n k−1,ℓ−1 n k+1,ℓ−1 . n k,ℓ 0, otherwise

Given a positive integer n, Sn+2 will denote the (n + 2) × (n + 2) symmetric, involutory and orthogonal matrix deﬁned by

2 kℓπ [S ] := sin . (2.2) n+2 k,ℓ n +3 n +3 r Our second auxiliary result makes useful Lemma 1 above and provides us an orthogonal diagonalization for the following (n + 2) × (n + 2) complex matrix

0 ...... 0 a b c − a d − b . .  . .. abc dc − a  . . .  . .. .. bcd c b     . . . .   ......   . cdc b a   . . . . .   ......   . dcb a 0   ......  Hn =  ......  . (2.3) +2  ......     ......  b  0 a bcd . . . . .     ......   a b cdc . . . .     .. .. .   b c dcb . . .     . .   c − ad cba .. .     d − b c − a b a 0 ...... 0      Lemma 2 Let n ∈ N, a,b,c,d ∈ C and

nkπ (n+1)kπ (n+2)kπ λk = −2a cos n+3 − 2b cos n+3 − 2c cos n+3 − d cos(kπ) , k =1, 2,...,n +2. (2.4) h i h i 3 If Hn+2 is the (n + 2) × (n + 2) matrix (2.3) then

b Hn+2 = Sn+2Λn+2Sn+2 where Λn+2 = diag (λ1, λ2,...,λn+2) andb Sn+2 is the (n + 2) × (n + 2) matrix (2.2).

Proof. Suppose n ∈ N and a,b,c,d ∈ C. Since Hn+2 ∈ An+2 and its ﬁrst row is

⊤ a = (0,...,b0,a,b,c − a, d − b) we have, from Lemma 1,

2 3 Hn+2 = (d − 2b)In+2 + (c − 3a)Ωn+2 + b Ωn+2 + a Ωn+2 Jn+2 where Jn+2 is the (bn + 2) × (n + 2) exchange matrix (that is, the matrix such that [Jn+2]k,n−k+3 =1 for every 1 6 k 6 n + 2 with the remaining entries zero). Using the spectral decomposition

n+2 ℓπ Ω = 2cos s s⊤, n+2 n +3 ℓ ℓ ℓ X=1 where 2 ℓπ n+3 sin n+3  q 2 2ℓπ  n+3 sin n+3 sℓ =  q .   .     2 (n+1)ℓπ   n sin n   +3 +3   q h i  (i.e. the ℓth column of Sn+2 in (2.2)) and the decomposition

2 3 n+3 Jn+2 = Sn+2 diag (−1) , (−1) ,..., (−1) Sn+2 (see Lemma 1 of [8]) with Sn+2 given by (2.2), it follows

n+2 ℓπ 2 ℓπ 3 ℓπ ⊤ Hn+2 = (d − 2b)+2(c − 3a)cos +4b cos +8a cos sℓ sℓ Jn+2 ( n +3 n +3 n +3 ) ℓ=1 . X 2 3 n+3 2 3 n+3 b = Sn+2 diag (−1) λ1, (−1) λ2,..., (−1) λn+2 diag (−1) , (−1) ,..., (−1) Sn+2 = Sn+2 diag (λ1, λ2,...,λn+2) Sn+2 The proof is completed.

The decomposition below plays a central role in the study of the spectral properties of anti- heptadiagonal persymmetric Hankel matrices. The statement is split in two cases of n ∈ N: n is an even number and n is an odd number. For the case in which n is even, we shall take the same steps as in [1] (see pages 106 and 107); the case in which n is odd, this procedure must be improved and we shall follow closely the bordering technique of Fasino (see [5], pages 303 and 304).

Lemma 3 Let n ∈ N, a,b,c,d ∈ C, λk, k =1, 2,...,n +2 be given by (2.4) and Hn the n × n matrix (1.1). (a) If n is even then

⊤ ⊤ diag (λ3, λ5,...,λn+1)+ λ1uu O Hn = RnPn ⊤ PnRn (2.5a) O diag (λ , λ ,...,λn)+ λn vv 2 4 +2

4 where 3π 2π sin( n+3 ) sin( n+3 ) π π sin( n+3 ) sin( n+3 )

 5π   4π  sin( n+3 ) sin( n+3 ) π π  sin( n )   sin( n )  u =  +3  , v =  +3  , (2.5b)  .   .   .   .   (n+1)π   nπ   sin[ n+3 ]   sin( n+3 )   π   π   sin( n )   sin( n )   +3   +3      Rn is the n × n matrix given by

2 (k + 1)(ℓ + 1)π [R ] = sin (2.5c) n k,ℓ n +3 n +3 r and Pn is the n × n permutation matrix deﬁned by

1 if ℓ =2k − n − 1 or ℓ =2k [Pn]k,ℓ = . (2.5d) ( 0 otherwise

(b) If n is odd then

⊤ diag (λ3, λ5,...,λn, λn+2)+ λ1uu O ⊤ ⊤ Hn = RnPn ⊤ Pn Rn O diag (λ , λ ,...,λn− )+ λn vv 2 4 1 +1 (2.6a) where 3π sin n ( +3 ) 2π π sin( n+3 ) sin n ( +3 ) (n+1)π sin[ n+3 ]  5π  sin n ( +3 )  4π  π sin( n+3 ) sin n  ( +3 )  (n+1)π    sin[ n ]  u =  .  , v =  +3  , (2.6b)  .   .   (n+1)π   .  sin n  [ +3 ]   (n−1)π   π   sin[ n+3 ]  sin n  ( +3 )   (n+1)π     sin[ n ]   1   +3        Rn is the n × n matrix given by

2 (k+1)(ℓ+1)π n+3 sin n+3 if ℓ

[Pn]k,ℓ =  1 if k = n and ℓ = (n + 1)/2 . (2.6d)  0 otherwise  Proof. Consider n ∈ N, a,b,c,d ∈ C and λk, k =1, 2,...,n + 2 given by (2.4).

(a) Supposing n even and Sn+2 given by (2.2), we have

θ x⊤ θ Sn+2 =  x Rn y  θ y⊤ −θ     5 2 π where θ := n+3 sin n+3 , q 2 2π 2 2π n+3 sin n+3 (−1) n+3 sin n+3  q   q  2 sin 3π (−1)2 2 sin 3π x =  n+3 n+3  and y =  n+3 n+3  . (2.7)  .   .   q .   q .   .   .       2 (n+1)π   n 2 (n+1)π   n+3 sin n+3   (−1) n+3 sin n+3       q h i   q h i  From Lemma 2, we obtain Hn+2 = Sn+2 diag (λ1, λ2,...,λn+2) Sn+2 and since Hn is obtained by deleting the ﬁrst and last row and column of the matrix Hn+2 in (2.3), it follows b −1 −1 ⊤ −1 −1 ⊤ Hn = Rn diag(λ2, λ3,...,λn+1)+ λ1Rn x Rnb x + λn+2Rn y Rn y Rn h i with x + y R−1x = − , (2.8) n 2θ x − y R−1y = − . (2.9) n 2θ The odd components of (2.8) and the even components of (2.9) are zero; moreover, the kth even component of (2.8) and the kth odd component of (2.9) is

(k+1)π sin n+3 − . h π i sin n+3 −1 −1 ⊤ −1 −1 ⊤ Permuting rows and columns of diag(λ2, λ3,...,λn+1)+ λ1Rn x Rn x + λn+2Rn y Rn y according to (2.5d) yields (2.5a). (b) Consider n odd and the (n + 2) × (n + 2) orthogonal matrix Sn+2 := Sn+2Kn+2, where Sn+2 is the matrix (2.2) and Kn+2 is the (n + 2) × (n + 2) permutation matrix b In 0 0 ⊤ Kn+2 :=  0 0 1  .  0⊤ 1 0      ⊤ Using Lemma 2, we have Hn+2 = Sn+2 diag (λ1, λ2,...,λn, λn+2, λn+1) Sn+2. Putting

b b θ z⊤ η b Sn+2 =  x Rn y  θ w⊤ −η   b   2 π 2 (n+1)π where θ := n+3 sin n+3 , η := n+3 sin n+3 , q q h i

6 2 2π 2 4π n+3 sin n+3 (−1) n+3 sin n+3  q   q  2 sin 3π (−1)2 2 sin 6π x =  n+3 n+3  , y =  n+3 n+3   .   .   q .   q .   .   .       2 (n+1)π   n 2 2(n+1)π   n+3 sin n+3   (−1) n+3 sin n+3       q 2 h 2π i   q 2 h 2π i  n+3 sin n+3 (−1) n+3 sin n+3 (2.10)  q   q  2 sin 3π (−1)2 2 sin 3π  n+3 n+3   n+3 n+3   .   .  z =  q .  , w =  q .   .   .       2 sin nπ   (−1)n−1 2 sin nπ   n+3 n+3   n+3 n+3       q   q   2 (n+2)π   2 (n+2)π   n+3 sin n+3   n+3 sin n+3       q h i   q h i  and Rn given by (2.6c), we get

−1 −1 ⊤ −1 −1 ⊤ ⊤ Hn = Rn diag(λ2, λ3,...,λn, λn+2)+ λ1Rn x Rn x + λn+1Rn y Rn y Rn h i with z + w R−1x = − , (2.11) n 2θ z − w R−1y = − (2.12) n 2η since Hn is obtained by deleting the first and last row and column of the matrix Hn+2. The first n − 1 odd components of (2.11) are zero; additionally, the even components of (2.12) are zero as well as its last component. In the first n − 1 components of (2.11), the kth even componentb is

(k+1)π sin n+3 − , h π i sin n+3 and the kth odd component of (2.12) is

(k+1)π sin n+3 − . h (n+1)π i sin n+3 h i Permuting rows and columns of

−1 −1 ⊤ −1 −1 ⊤ diag(λ2, λ3,...,λn, λn+2)+ λ1Rn x Rn x + λn+1Rn y Rn y according to (2.6d) we obtain (2.6a) establishing the thesis.

ℓ+1 Remark Let us observe that for n odd the condition Sn+2 = (−1) Sn+2 , ℓ =1, 2,...,n+2 n+2,ℓ 1,ℓ is not satisﬁed (unlike the case in which n is evenh andi we have [S h] i= (−1)ℓ+1 [S ] , b n+2 nb+2,ℓ n+2 1,ℓ ℓ = 1, 2,...,n + 2). This foreshadowed that the bordering technique can be founded taking account only the four corners of the orthogonal matrix (in order to get det(Rn) 6= 0) and discarding any relation between the remaining entries of its ﬁrst and last rows.

7 3 Spectral properties of Hn 3.1 Eigenvalue localization

The decomposition presented in Lemma 3 allows us to express the eigenvalues of Hn as the zeros of explicit polynomials.

Theorem 1 Let n ∈ N, a,b,c,d ∈ C, λk, k =1, 2,...,n+2 be given by (2.4) and Hn the n×n matrix (1.1).

(a) If n is even then the eigenvalues of Hn are precisely the zeros of

n n 2 (2j+1)π n 2 2 2 sin n+3 f(t)= (t − λ2j+1) + (t − λ1) (t − λ2m+1) (3.1a) 2h π i j=1 j=1 sin n+3 m=1 Y X mY=6 j and n n 2 2jπ n 2 2 2 sin n+3 g(t)= (t − λ2j ) + (t − λn+2) (t − λ2m). (3.1b) 2 π j=1 j=1 sin n+3 m=1 Y X mY=6 j (b) If n is odd then the eigenvalues of Hn are precisely the zeros of

n+1 n+1 2 (2j+1)π n+1 2 2 2 sin n+3 f(t)= (t − λ2j+1) + (t − λ1) (t − λ2m+1) (3.2a) 2h π i j=1 j=1 sin n+3 m=1 Y X mY=6 j and n−1 n−1 2 2jπ n−1 2 2 2 sin n+3 g(t)= (t − λ2j ) + (t − λn+1) (t − λ2m). (3.2b) 2 (n+1)π j=1 j=1 sin n+3 m=1 Y X mY=6 j h i Proof. Let n ∈ N, a,b,c,d ∈ C and λk, k =1, 2,...,n + 2 be given by (2.4).

(a) Consider n even and Rn the matrix deﬁned by (2.5c). We have

2 ⊤ ⊤ Rn = In − xx − yy

2 16 4 π where x, y are given by (2.7) and det(Rn)= (n+3)2 sin n+3 6= 0 (see [1], page 106), so that −2 ⊤ ⊤ −1 Rn = In − xx − yy . Setting

Σ n := diag (λ , λ ,...,λ )+ λ uu⊤ 2 3 5 n+1 1 Γ n := diag (λ , λ ,...,λ )+ λ vv⊤ 2 2 4 n n+2 with u, v given by (2.5b), we obtain

det(tIn − Hn)=

n 2 −2 ⊤ Σ O = [det(Rn)] det tR − P 2 Pn n n O Γ n 2 n 2 −2 ⊤ Σ O = det R det tPnR P − 2 n n n O Γ n 2 8 n 2 ⊤ ⊤ ⊤ −1 Σ O = det R det t Pn In − xx − yy P − 2 n n O Γ n 2 2 π −1 4 sin ( n ) I n − +3 uu⊤ O Σ n O 2 2 n+3 2 = det Rn det t 2 π −  4 sin n n   n ( +3 ) ⊤  O Γ O I − n vv 2 2 +3      n 1 ⊤  I − ⊤ n+3 uu O 2 u u− 2 π n 4 sin ( n ) Σ O R2 +3 2 = det n det t n 1 ⊤ − n   O I − ⊤ n vv  O Γ  2 v v− +3 2 2 π 4 sin ( n )   +3       2 t ⊤ = det R det tI n − diag (λ , λ ,...,λn ) − λ + uu · n 2 3 5 +1 1 ⊤ n+3   u u − 2 π   4 sin ( n )  +3     t  ⊤ det tI n − diag (λ , λ ,...,λn) − λn + vv 2 2 4 +2 ⊤ n+3   v v − π   4 sin2  ( n+3 )    (see [9], page 88 and [10], page 19) where Pn the permutation matrix (2.5d). Thus, supposing t 6= λ2k+1 n u⊤u n+3 for each k =1,..., 2 and noting that − 2 π = −1, we get 4 sin ( n+3 )

t ⊤ det tI n − diag (λ , λ ,...,λn ) − λ + uu = 2 3 5 +1 1 ⊤ n+3   u u − π   4 sin2  ( n+3 )  n  2 (2j+1)π n  2 2  sin n+3  = 1 − (λ1 − t) (t − λ2j+1)  h 2 i π  (3.3) j=1 (t − λ2j+1) sin j=1  X n+3  Y n n 2 (2j+1)π n 2 2 2  sin n+3  = (t − λ2j+1) + (t − λ1) (t − λ2m+1). 2h π i j=1 j=1 sin n+3 m=1 Y X mY=6 j Since the determinant is a continuous function, relation (3.3) holds for all t. Analogously, we have ⊤ n+3 v v − 2 π = −1 and 4 sin ( n+3 )

t ⊤ det tI n − diag (λ , λ ,...,λn) − λn + vv = 2 2 4 +2 ⊤ n+3   v v − π   4 sin2  ( n+3 )  n  2 2jπ n  2 2  sin n+3  = 1 − (λn+2 − t) (t − λ2j )  2 π  j=1 (t − λ2j ) sin j=1 X n+3 Y n n 2jπ  n 2 2 2 2 sin n+3 = (t − λ2j ) + (t − λn+2) (t − λ2m) 2 π j=1 j=1 sin n+3 m=1 Y X mY=6 j for all t, so that the characteristic polynomial of Hn is

4 n n 2 (2j+1)π n π 2 2 2 16 sin n+3 sin n+3 det(tIn − Hn)= (t − λ2j+1) + (t − λ1) · (t − λ2m+1) · (n + 3)2  2h π i  j=1 j=1 sin n+3 m=1  Y X mY=6 j   

9 n n 2 2jπ n 2 2 2 sin n+3 (t − λ2j ) + (t − λn+2) (t − λ2m) .  2 π  j=1 j=1 sin n+3 m=1 Y X mY=6 j   

(b) Supposing n odd and Rn the matrix deﬁned by (2.6c), it follows

⊤ ⊤ ⊤ Rn Rn = In − zz − ww where z and w are given in (2.10). Hence,

⊤ ⊤ ⊤ det(Rn Rn) = det(In − zz − ww ) (z⊤w)2 = (1 − z⊤z) 1 − w⊤w − 1 − z⊤z 16 π (n + 1)π = sin2 sin2 6=0 (n + 3)2 n +3 n +3 (see [13], page 70) and ⊤ −1 ⊤ ⊤ −1 Rn Rn = In − zz − ww . Putting

⊤ Σ n+1 := diag (λ3, λ5,...,λn, λn+2)+ λ1uu 2 ⊤ Γ n−1 := diag (λ2, λ4,...,λn−1)+ λn+1vv 2 with u, v given by (2.6b), we obtain det(tIn − Hn)=

Σ n+1 O −1 ⊤ −1 2 ⊤ ⊤ = det(Rn) det tRn (Rn ) − Pn Pn det(Rn ) O Γ n−1 " 2 # !

Σ n+1 O ⊤ ⊤ ⊤ −1 2 = det Rn Rn det tPn (Rn Rn) Pn − O Γ n−1 " 2 #! −1 Σ n+1 O ⊤ ⊤ ⊤ ⊤ 2 = det Rn Rn det t Pn In − zz − ww Pn − O Γ n−1 " 2 #!

2 π −1 4 sin ( n+3 ) ⊤ I n+1 − uu O Σ n+1 O ⊤ 2 n+3 2 = det R Rn det t 2 (n+1)π − n 4 sin n−   [ n+3 ] ⊤  O Γ 1  n−1 " # O I − n+3 vv 2  2  ⊤     = det Rn Rn ·

n 1 ⊤ I +1 − ⊤ n+3 uu O 2 u u− 2 π n+1 4 sin n Σ O ( +3 ) 2 det t 1 −   n− ⊤  n−  O I 1 − v⊤v n+3 vv O Γ 1 2 − (n+1)π " 2 # 4 sin2   [ n+3 ]       ⊤ t ⊤ = det Rn Rn det tI n+1 − diag (λ3, λ5,...,λn, λn+2) − λ1 + uu · 2 ⊤ n+3   u u − 2 π   4 sin ( n )  +3     t  ⊤ det tI n−1 − diag (λ2, λ4,...,λn−1) − λn+1 + vv 2 ⊤ n+3   v v − n π   4 sin2 ( +1)  [ n+3 ]      10 (see [9], page 88 and [10], page 19) where Pn the permutation matrix (2.6d). Thus, admitting t 6= λ2k+1 n+1 for each k =1,..., 2 , we have

t ⊤ det tI n+1 − diag (λ3, λ5,...,λn, λn+2) − λ1 + uu = 2 ⊤ n+3   u u − π   4 sin2  ( n+3 )  n−1 2 (2j+1)π  n−1 2 2  sin n+3 1  = 1 − (λ1 − t) + (t − λn+2) (t − λ2j+1)   2 π t − λn  j=1 (t − λ2j+1) sin +2 j=1  X n+3  Y n+1 2 (2j+1)π n+1  (3.4) 2 2  sin n+3  = 1 − (λ1 − t) (t − λ2j+1)  h 2 i π  j=1 (t − λ2j+1) sin j=1  X n+3  Y n+1 n+1 2 (2j+1)π n+1 2 2 2  sin n+3  = (t − λ2j+1) + (t − λ1) (t − λ2m+1) 2h π i j=1 j=1 sin n+3 m=1 Y X mY=6 j ⊤ n+3 provided that u u − 2 π = −1. Since the determinant is a continuous function, relation (3.4) 4 sin ( n+3 ) ⊤ n+3 is valid for every t. Analogously, we have v v − 2 (n+1)π = −1 and 4 sin [ n+3 ]

t ⊤ det tI n−1 − diag (λ2, λ4,...,λn−1) − λn+1 + vv = 2 ⊤ n+3   v v − n π   4 sin2 ( +1)  ( n+3 )  n−1  n−1 2 2jπ n−1 2 2 2  sin n+3 = (t − λ2j ) + (t − λn+1) (t − λ2m) 2 (n+1)π j=1 j=1 sin n+3 m=1 Y X mY=6 j h i for all t, whence the characteristic polynomial of Hn is

n−1 16 π (n + 1)π 2 det(tI − H )= sin2 sin2 (t − λ ) (t − λ )+ n n (n + 3)2 n +3 n +3 n+2 2j+1 ( j=1 Y n−1 2 (2j+1)π n−1 n−1 2 2 2 (λn+2 − t) sin n+3 (t − λ1) (t − λ2j+1)+ (t − λ2j+1) · 2 π " j=1 sin n+3 m=1 j=1 #) X mY=6 j Y n−1 n−1 2 2jπ n−1 2 2 2 sin n+3 (t − λ2j ) − (λn+1 − t) (t − λ2m) .  2 (n+1)π  j=1 j=1 sin n+3 m=1   Y X mY=6 j  h i The proof is completed.  

The previous theorem leads us to the following separation result for the eigenvalues of real matrices Hn.

Corollary 1 Let n ∈ N, a,b,c,d ∈ R, λk, k = 1, 2,...,n +2 be given by (2.4) and Hn the n × n matrix (1.1). (a) If n is even and:

11 i. λτ(1) 6 λτ(3) 6 ... 6 λτ(n+1) are arranged in nondecreasing order by some bijection τ then

λτ(2k−1) 6 µk 6 λτ(2k+1) (3.5a)

for any k = 1,..., n , where µ ,...,µ n are the zeros of (3.1a). Moreover, if λ are 2 1 2 τ(2k−1) all distinct then the kth inequalities (3.5a) are strict, and λτ(2k−1) = λτ(2k+1) if and only if λτ(2k−1) = µk = λτ(2k+1).

ii. λσ(2) 6 λσ(4) 6 ... 6 λσ(n+2) are arranged in nondecreasing order by some bijection σ then

λσ(2k) 6 νk 6 λσ(2k+2) (3.5b)

for all k = 1,..., n , where ν ,...,ν n are the zeros of (3.1b). Furthermore, if λ are all 2 1 2 σ(2k) distinct then the kth inequalities (3.5b) are strict, and λσ(2k) = λσ(2k+2) if and only if λσ(2k) = νk = λσ(2k+2).

(b) If n is odd and:

i. λτ(1) 6 λτ(3) 6 ... 6 λτ(n) 6 λτ(n+2) are arranged in nondecreasing order by some bijection τ then λτ(2k−1) 6 µk 6 λτ(2k+1) (3.6a) n+1 for every k = 1,..., , where µ1,...,µ n+1 are the zeros of (3.2a). Moreover, if λτ(2k−1) are 2 2 all distinct then the kth inequalities (3.5a) are strict, and λτ(2k−1) = λτ(2k+1) if and only if λτ(2k−1) = µk = λτ(2k+1).

ii. λσ(2) 6 λσ(4) 6 ... 6 λσ(n+1) are arranged in nondecreasing order by some bijection σ then

λσ(2k) 6 νk 6 λσ(2k+2) (3.6b)

n−1 for each k = 1,..., , where ν1,...,ν n−1 are the zeros of (3.2b). Furthermore, if λσ(2k) 2 2 are all distinct then the kth inequalities (3.6b) are strict, and λσ(2k) = λσ(2k+2) if and only if λσ(2k) = νk = λσ(2k+2).

Proof. Since all assertions can be proven in the same way, we only prove i. of (a). Let n ∈ N be even 6 6 n and consider τ(2ξ − 1)=1 for some 1 ξ 2 +1. According to Theorem 1, we have

n n 2 (2j+1)π n 2 2 2 sin n+3 f(t)= (t − λ2j+1) + (t − λ1) (t − λ2m+1) 2h π i j=1 j=1 sin n+3 m=1 Y X mY=6 j n n 2 τ(2j−1)π n 2 +1 2 +1 2 +1 sin n+3 = t − λτ(2j−1) + t − λτ(2ξ−1) t − λτ(2m−1) h2 π i j=1 j=1 sin n+3 m=1 jY=6 ξ Xj=6 ξ mY=6 ξ,j 6 n 6 6 so that f λτ(2k−1) f λτ(2k+1) 0 for all k = 1,..., 2 . Hence, λτ(2k−1) µk λτ(2k+1). If λ are all distinct then λ <µk 6 λ or λ 6 µk < λ . In the ﬁrst case, if τ(2k−1) τ(2 k−1) τ(2k+1) τ(2k−1) τ(2k+1) µk = λτ(2k+1) then f λτ(2k+1) = 0, implying λτ(2k+1) = λτ(2ζ−1) for some ζ 6= k +1 which contradicts the hypothesis; the case λτ(2k−1) 6 µk < λτ(2k+1) is analogous. Hence, the kth inequalities (3.5a) are strict.

Remark Let us observe that, in the assumptions of Corollary 1, if λτ(1) > 0 and λσ(2) > 0 then Hn is positive deﬁnite.

12 3.2 Eigenvectors

From Lemma 3 we can get also an explicit representation for the eigenvectors of Hn.

Theorem 2 Let n ∈ N, a,b,c,d ∈ R, λk, k =1, 2,...,n+2 be given by (2.4) and Hn the n×n matrix (1.1).

(a) If n is even, u, v are given in (2.5b), Rn is the n × n matrix (2.5c), Pn is the n × n permutation matrix (2.5d),

i. µ ,...,µ n are the zeros of (3.1a) and λ , λ ,...,λ are all distinct then 1 2 1 3 n+1

3π sin( n+3 ) π k (µ −λ3) sin( n+3 )  5π  sin( n+3 ) π (µk−λ ) sin  5 ( n+3 )   .  n ⊤  .  ⊤ I + uu O   R P 2 (n+1)π n n n ⊤  sin  (3.7a) O I + vv  [ n+3 ]  2 π  (µk−λn ) sin   +1 ( n+3 )     0     .   .       0    n is an eigenvector of Hn associated to µk, k =1,..., 2 . ii. ν ,...,ν n are the zeros of (3.1b) and λ , λ ,...,λ are all distinct then 1 2 2 4 n+2

0 .  .  0    2π  sin( n+3 ) n ⊤  π  ⊤ I + uu O  (νk −λ ) sin  R P 2 2 ( n+3 ) n n n ⊤   (3.7b) O I + vv 4π 2  sin   ( n+3 )  π  (νk −λ ) sin   4 ( n+3 )   .   .   .   sin nπ   ( n+3 )  π  (νk−λn) sin   ( n+3 )    n is an eigenvector of Hn associated to νk, k =1,..., 2 .

(b) If n is odd, u, v are given in (2.6b), Rn is the n × n matrix (2.6c), Pn is the n × n permutation matrix (2.6d),

i. µ1,...,µ n+1 are the zeros of (3.1a) and λ1, λ3,...,λn, λn+2 are all distinct then 2

13 3π sin( n+3 ) π k (µ −λ3) sin( n+3 )  5π  sin( n+3 ) π (µk−λ ) sin  5 ( n+3 )   .   .   .  ⊤ n π I n+1 + uu O  sin ( +1)  2  [ n+3 ]  RnPn ⊤ π (3.8a) O I n−1 vv  (µk−λn) sin  " + #  ( n+3 )  2    1   µk−λn+2     0     .   .       0  n+1   is an eigenvector of Hn associated to µk, k =1,..., 2 .

ii. ν1,...,ν n−1 are the zeros of (3.1b) and λ2, λ4,...,λn+1 are all distinct then 2

0 .  .  0    sin 2π  ⊤ ( n+3 ) n+1  (n+1)π  I + uu O k 2  (ν −λ2) sin[ n+3 ]  RnPn ⊤   (3.8b) O I n−1 + vv  sin 4π  " 2 # ( n+3 )  (n+1)π   (νk −λ ) sin   4 [ n+3 ]   .   .   .  n− π  sin ( 1)   [ n+3 ]  (n+1)π  (νk −λn− ) sin   1 [ n+3 ]    n−1 is an eigenvector of Hn associated to νk, k =1,..., 2 . Proof. Since both assertions can be proven in the same way, we only prove (a). Let n ∈ N be even.

i. Supposing λ1, λ3,...,λn+1 all distinct, Corollary 1 guarantees that the zeros of (3.1a) are all simple. Setting

Υ n := diag(λ , λ ,...,λ ) 2 3 5 n+1 ∆ n := diag(λ , λ ,...,λ ) 2 2 4 n

we can rewrite the matricial equation (µkIn − Hn)q = 0 as

µ I n − Υ n + (µ − λ )uu⊤ O ⊤ k 2 2 k 1 RnPn PnRnq = 0 (3.9) O µ I n − ∆ n + (µ − λ )vv⊤ " k 2 2 k n+2 #

with u, v given by (2.5b), Rn deﬁned in (2.5c) and Pn the permutation matrix (2.5d). Thus,

µ I n − Υ n + (µ − λ )uu⊤ q = 0, k 2 2 k 1 1 ⊤ µ I n − ∆ n + (µ − λ )vv q = 0, k 2 2 k n+2 2 q 1 = P R q q n n 2

14 that is,

− q = α(µ I n − Υ n ) 1u, 1 k 2 2

q2 = 0

for α 6= 0 and −1 α(µkI n − Υ n ) u q = R−1P⊤ 2 2 n n 0 is a nontrivial solution of (3.9), and so an eigenvector of Hn associated to the eigenvalue µk. Since

−1 ⊤ −2 ⊤ Rn Pn = RnRn Pn ⊤ ⊤ −1 ⊤ = Rn In − xx − yy Pn ⊤ ⊤ ⊤ ⊤ −1 (3.10) = RnP n Pn In − xx − yy Pn I n + uu⊤ O ⊤ 2 = RnP , n O I n + vv⊤ 2 where x, y are given by (2.7), the result follows choosing α = 1.

ii. Considering λ2, λ4,...,λn+2 all distinct, the zeros of (3.1b) are all simple according to Corollary 1. Hence,

ν I n − Υ n + (ν − λ )uu⊤ O ⊤ k 2 2 k 1 RnPn PnRnq = 0 O ν I n − ∆ n + (ν − λ )vv⊤ " k 2 2 k n+2 # leads to −1 ⊤ 0 q = Rn Pn −1 , α(νkI n − ∆ n ) v 2 2 for α 6= 0, which is an eigenvector of Hn associated to the eigenvalue νk. The conclusion follows from (3.10).

Remark Recall that if n is even and some λ2k−1 is a root of (3.1a) (and a fortiori an eigenvalue of Hn) then λ2k−1 = λ2ℓ−1 for a suitable ℓ 6= k according to Corollary 1, so that

n ⊤ ⊤ I + uu O q1 RnP 2 , n O I n + vv⊤ 0 2 where u, v are given in (2.5b), Rn is the n × n matrix (2.5c), Pn is the n × n permutation matrix (2.5d), (2ℓ−1)π (2k−1)π sin n+3 ik − sin n+3 iℓ if k,ℓ 6=1 h i h i q1 =  i ,  ℓ if k =1  ik if ℓ =1 and i denotes the kth column of I n , is an eigenvector of H associated to λ ; if a λ is root of k  2 n 2k−1 2k (3.1b) then λ2k = λ2ℓ for some ℓ 6= k and

I n + uu⊤ O 0 ⊤ 2 RnPn ⊤ , O I n + vv q2 2

15 with u, v given by (2.5b), Rn the n × n matrix (2.5c), Pn the n × n permutation matrix (2.5d) and

2ℓπ 2kπ n+2 sin n+3 ik − sin n+3 iℓ if k,ℓ 6= 2  q2 = i n+2  ℓ if k = 2   n+2 ik if ℓ = 2   is an eigenvector of Hn associated to λ2k. Analogously, if n is odd and some λ2k−1 is a root of (3.2a) then λ2k−1 = λ2ℓ−1 for a suitable ℓ 6= k and

⊤ I n+1 + uu O q 2 1 RnPn ⊤ , O I n−1 + vv 0 " 2 # where u, v are given in (2.6b), Rn is the n × n matrix (2.6c), Pn is the n × n permutation matrix (2.6d), (2ℓ−1)π (2k−1)π sin n+3 ik − sin n+3 iℓ if k,ℓ 6=1 h i h i q1 =  i ,  ℓ if k =1  ik if ℓ =1  and ik denotes the kth column of I n+1 , is an eigenvector of Hn associated to λ2k−1; if a λ2k is a root  2 of (3.2b) then λ2k = λ2ℓ for some ℓ 6= k and

⊤ I n+1 + uu O 0 2 RnPn ⊤ , O I n−1 + vv q2 " 2 # with u, v given by (2.6b), Rn the n × n matrix (2.6c), Pn the n × n permutation matrix (2.6d),

2ℓπ 2kπ n+1 sin n+3 ik − sin n+3 iℓ, if k,ℓ 6= 2 q 2 =  i , if k = n+1  ℓ 2  n+1 ik, if ℓ = 2  and ik denotes the kth column of I n−1 , is an eigenvector of Hn associated to λ2k. 2

4 Integer powers of Hn

The decomposition in Lemma 3 can still be used to compute explicitly the inverse of Hn.

Theorem 3 Let n ∈ N, a,b,c,d ∈ C, λk, k =1, 2,...,n+2 be given by (2.4) and Hn the n×n matrix (1.1). If λk 6=0 for every k =1, 2,...,n +2, Hn is nonsingular and: (a) n is even then n −1 ⊤ U O H = RnP 2 PnRn n n O V n 2 where Rn is the n × n matrix (2.5c), Pn is the n × n permutation matrix (2.5d),

n n ⊤ −1 −1 −1 λ1 U = I + uu diag λ3 , λ5 ,...,λn+1 − u⊤ −1 −1 −1 u · 2 2 1+λ1 diag(λ ,λ ,...,λn ) 3 5 +1 (4.1a) − − − ⊤ − − − ⊤ diag λ 1, λ 1,...,λ 1 uu diag λ 1, λ 1,...,λ 1 I n + uu 3 5 n+1 3 5 n+1 2 16 and

n n n ⊤ −1 −1 −1 λ +2 V = I + vv diag λ2 , λ4 ,...,λn − v⊤ −1 −1 −1 v · 2 2 1+λn+2 diag(λ ,λ ,...,λn ) 2 4 (4.1b) − − − − diag λ 1, λ 1,...,λ−1 vv⊤diag λ 1, λ 1,...,λ−1 I n + vv⊤ 2 4 n 2 4 n 2 with u, v given by (2.5b). (b) n is odd then

U n+1 O −1 2 ⊤ ⊤ Hn = RnPn Pn Rn O V n−1 " 2 # where Rn is the n × n matrix (2.6c), Pn is the n × n permutation matrix (2.6d),

⊤ − − − − λ n n 1 1 1 1 1 U +1 = I +1 + uu diag λ3 , λ5 ,...,λn , λn+2 − u⊤ −1 −1 −1 −1 u · 2 2 1+λ1 diag(λ ,λ ,...,λn ,λn ) 3 5 +2 (4.2a) −1 −1 −1 −1 ⊤ −1 −1 −1 −1 ⊤ diag λ , λ ,...,λn , λn uu diag λ , λ ,...,λn , λn I n+1 + uu 3 5 +2 3 5 +2 2 and

⊤ − − − λn+1 n− n− 1 1 1 V 1 = I 1 + vv diag λ2 , λ4 ,...,λn−1 − v⊤ −1 −1 −1 v · 2 2 1+λn+1 diag(λ ,λ ,...,λn− ) 2 4 1 (4.2b) −1 −1 −1 ⊤ −1 −1 −1 ⊤ diag λ , λ ,...,λn− vv diag λ , λ ,...,λn− I n−1 + vv 2 4 1 2 4 1 2 with u, v given by (2.6b).

Proof. Consider n ∈ N even, a,b,c,d ∈ C λk 6= 0, k = 1, 2,...,n + 2 and Hn nonsingular. According to Sherman-Morrison-Woodbury formula (see [10], page 19 or [9], page 424), we have

−1 ⊤ −1 −1 ⊤ −1 −1 ⊤ −1 Υ n + λ1uu = Υ n − λ1 1+ λ1u Υ n u Υ n uu Λ n , 2 2 2 2 2 where Υ n := diag (λ , λ ,...,λ ), u is given in (2.5a) and U n is the matrix (4.1a). Similarly, we 2 3 5 n+1 2 obtain −1 ⊤ −1 −1 ⊤ −1 −1 ⊤ −1 ∆ n + λn+2vv = ∆ n − λn+2 1+ λn+2v ∆ n v ∆ n vv ∆ n , 2 2 2 2 2 where ∆ n := diag (λ , λ ,...,λ ), v is given in (2.5a) and V n is the matrix (4.1b). Hence, the 2 2 4 n 2 decomposition of Lemma 3, 8.5b of [9] (page 88), identity (3.10) and

−1 −2 PnRn = PnRn Rn ⊤ ⊤ −1 = Pn In − xx − yy Rn I n + uu⊤ O 2 = PnRn, O I n + vv⊤ 2 where x, y are given by (2.7), establish the thesis in (a). The proof of (b) is analogous and so will be omitted.

Remark Note that if n is even, λk 6= 0 for all k =2, 3,...,n + 1,

n 2 (2j+1)π n 2 2jπ 2 2 λ sin n+3 λ sin n+3 1 6= −1 and n+2 6= −1 2 π λh j i 2 π λ j sin j=1 2 +1 sin j=1 2 n+3 X n+3 X 17 then Hn is nonsingular. Similarly, if n is odd, λk 6= 0 for all k =2, 3,...,n,n + 2,

n+1 (2j+1)π n−1 2jπ 2 2 2 2 λ sin n+3 λ sin n+3 1 6= −1 and n+1 6= −1 2 π λh j i 2 (n+1)π λ j sin j=1 2 +1 sin j=1 2 n+3 X n+3 X h i then Hn is a nonsingular matrix.

Let us point out that the previous Theorems 1 and 2 lead to an immediate eigenvalue decomposition with orthogonal eigenvector matrix for any real anti-heptadiagonal persymmetric Hankel matrix having all the eigenvalues distinct. Indeed, admitting Hn in (1.1) real with its eigenvalues all diﬀerent it is well- known that the set of its eigenvectors is orthogonal (see, for instance, [6] page 109); hence, normalizing each of them we can obtain an orthonormal set, and so an orthogonal eigenvector matrix (see [18], page 257). However, it is not guaranteed, in general, that the eigenvalues of Hn are all diﬀerent even assuming the distinctness of every λ2k−1 and all λ2k, as the following examples show: 01000 1 0 1 0 10000 0 1 0 1 H = or H =  00000  . 4  1 0 1 0  5  00001   0 1 0 1       00010        It is possible to establish an alternative diagonalization for Hn avoiding its eigenvalues. The following auxiliary result is the central key in its accomplishment.

Lemma 4 Let n ∈ N, a,b,c,d ∈ C, λk, k =1, 2,...,n +2 be given by (2.4) and Hn the n × n matrix (1.1). (a) If n is even and:

i. u is given in (2.5b) then the eigenvalues of

2 π 4 sin n+3 diag (λ , λ ,...,λ )+ diag (λ − λ , λ − λ ,...,λ − λ ) uu⊤ (4.3a) 3 5 n+1 n +3 1 3 1 5 1 n+1 are the zeros of the polynomial

n n n 2 4 2 (2j + 1)π 2 F (t)= (t − λ )+ sin2 (λ − λ ) (t − λ ). (4.3b) 2j+1 n +3 n +3 2j+1 1 2m+1 j=1 j=1 m=1 Y X mY=6 j

Moreover, if λ , λ ,...,λ are all real and distinct then the eigenvalues φ , φ ,...,φ n of (4.3b) 1 3 n+1 1 2 2 are all simple and 3π (λ3−λ1) sin( n+3 ) λ3−φk (λ −λ ) sin 5π  5 1 ( n+3 )  f := λ5−φk (4.3c) k  .   .   (n+1)π  λn −λ  ( +1 1) sin[ n+3 ]     λn+1−φk   n  is an eigenvector associated to φk, k =1,..., 2 . ii. v is given in (2.5b) then the eigenvalues of

2 π 4 sin n+3 diag (λ , λ ,...,λ )+ diag (λ − λ , λ − λ ,...,λ − λ ) vv⊤ (4.3d) 2 4 n n +3 n+2 2 n+2 4 n+2 n

18 are the zeros of the polynomial

n n n 2 4 2 2jπ 2 G(t)= (t − λ )+ sin2 (λ − λ ) (t − λ ). (4.3e) 2j n +3 n +3 2j n+2 2m j=1 j=1 m=1 Y X mY=6 j

Furthermore, if λ , λ ,...,λ are all real, nonzero and diﬀerent then the eigenvalues ψ , ψ ,...,ψ n 2 4 n+2 1 2 2 of (4.3e) are all simple and 2π n (λ2−λ +2) sin( n+3 ) λ2−ψk π (λ −λn ) sin 4  4 +2 ( n+3 )  g := λ4−ψk (4.3f) k  .   .   nπ  λn−λn  ( +2) sin( n+3 )     λn−ψk   n  is an eigenvector associated to ψk, k =1,..., 2 .

(b) If n is odd and:

i. u is given in (2.6b) then the eigenvalues of

2 π 4 sin n+3 diag (λ , λ ,...,λ , λ )+ diag (λ − λ , λ − λ ,...,λ − λ , λ − λ ) uu⊤ 3 5 n n+2 n +3 1 3 1 5 1 n 1 n+2 (4.4a) are the zeros of the polynomial

n+1 n+1 n+1 2 4 2 (2j + 1)π 2 F (t)= (t − λ )+ sin2 (λ − λ ) (t − λ ) (4.4b) 2j+1 n +3 n +3 2j+1 1 2m+1 j=1 j=1 m=1 Y X mY=6 j

Moreover, if λ1, λ3,...,λn, λn+2 are all real and distinct then the eigenvalues φ1, φ2,...,φ n+1 of 2 (4.4b) are all simple and 3π (λ3−λ1) sin( n+3 ) λ3−φk (λ −λ ) sin 5π  5 1 ( n+3 )  λ5−φk  .  fk :=  .  (4.4c) nπ  (λn−λ ) sin   1 ( n+3 )   λn−φk   (n+2)π  λn −λ  ( +2 1) sin[ n+3 ]     λn+2−φk   n  is an eigenvector associated to φk, k =1,..., 2 . ii. v is given in (2.6b) then the eigenvalues of

2 (n+1)π 4 sin n+3 diag (λ , λ ,...,λ )+ diag (λ − λ , λ − λ ,...,λ − λ ) vv⊤ 2 4 n−1 nh+3 i n+1 2 n+1 4 n+1 n−1 (4.4d) are the zeros of the polynomial

n−1 n−1 n−1 2 4 2 2jπ 2 G(t)= (t − λ )+ sin2 (λ − λ ) (t − λ ). (4.4e) 2j n +3 n +3 2j n+1 2m j=1 j=1 m=1 Y X mY=6 j

19 Furthermore, if λ2, λ4,...,λn−1, λn+1 are all real and diﬀerent then the eigenvalues ψ1, ψ2,...,ψ n−1 2 of (4.4e) are all simple and

2π n (λ2−λ +1) sin( n+3 ) λ2−ψk π (λ −λn ) sin 4  4 +1 ( n+3 )  g := λ4−ψk (4.4f) k  .   .   (n−1)π  λn− −λn  ( 1 +1) sin[ n+3 ]     λn−1−ψk    n−1 is an eigenvector associated to ψk, k =1,..., 2 . Proof. Since both assertions can be proven in the same way, we only prove (a). Consider n ∈ N, a,b,c,d ∈ C and λk, k = 1, 2,...,n + 2 given by (2.4). The proof follows the same steps of the ones displayed in Section 3.

i. Setting

4 sin2 π n+3 ⊤ Φ n := diag (λ3, λ5,...,λn+1)+ diag (λ1 − λ3, λ1 − λ5,...,λ1 − λn+1) uu 2 n +3 with u given in (2.5b), we have

det(tI n − Φ n ) = det [diag (t − λ ,t − λ ,...,t − λ )] · 2 2 3 5 n+1 2 π 4 sin ( n ) λ −λn 1 − +3 u⊤diag λ1−λ3 , λ1−λ5 ,..., 1 +1 u n+3 t−λ3 t−λ5 t−λn+1 n n n 2 2 2 4 2 (2j+1)π = (t − λ2j+1)+ n+3 sin n+3 (λ2j+1 − λ1) (t − λ2m+1) =: F (t). j=1 j=1 m=1 Y X h i mY=6 j

Supposing λ1, λ3,...,λn+1 all real, distinct and arranged in increasing order by some bijection ϕ, i.e. λϕ(1) < λϕ(3) <...<λϕ(n+1), we get

n n n 2 +1 2 +1 2 +1 4 2 ϕ(2j−1)π F (t)= t − λϕ(2j−1) + n+3 sin n+3 λϕ(2j−1) − λϕ(2ϑ−1) t − λϕ(2m−1) j=1 j=1 m=1 jY=6 ϑ jX=6 ϑ h i mY=6 ϑ,j

6 6 n for some ϑ satisfying ϕ(2ϑ − 1)=1,1 ϑ 2 + 1 which yields F λϕ(2k−1) F λϕ(2k+1) < 0 for n all k = 1,..., . Hence, the zeros φ1, φ2,...,φ n of F (t) are all simple. From Theorem 5 of [3] 2 2 (see [3], page 41), it follows that fk in (4.3c) is an eigenvector of Φ n associated to the eigenvalue n 2 φk, k =1,..., 2 . ii. Analogously, putting

4 sin2 π n+3 ⊤ Ψ n := diag (λ2, λ4,...,λn)+ diag (λn+2 − λ2, λn+2 − λ4,...,λn+2 − λn) vv 2 n +3 with v given in (2.5b), we obtain

det(tI n − Ψ n ) = det [diag (t − λ ,t − λ ,...,t − λ )] · 2 2 2 4 n 2 π 4 sin ( n ) λn −λ λn −λ λn −λn 1 − +3 v⊤diag +2 2 , +2 4 ,..., +2 v n+3 t−λ2 t−λ4 t−λn 20 n n n 2 2 2 4 2 2jπ = (t − λ2j )+ n+3 sin n+3 (λ2j − λn+2) (t − λ2m) =: G(t). j=1 j=1 m=1 Y X mY=6 j

Assuming λ2, λ4,...,λn all real, distinct and arranged in increasing order by some bijection ρ, that is λρ(2) < λρ(4) <...<λρ(n), we have

n n n 2 +1 2 +1 2 +1 4 2 ρ(2j)π G(t)= t − λρ(2j) + n+3 sin n+3 λρ(2j) − λρ(2υ) t − λρ(2m) j=1 j=1 m=1 jY=6 υ jX=6 υ h i mY=6 υ,j 6 6 n for some υ such that ρ(2υ)= n +2, 1 υ 2 + 1 which implies G λρ(2k) G λρ(2k+2) < 0 for n all k = 1,..., . Thus, the zeros ψ1, ψ2,...,ψ n of G(t) are all simple. Hence, Theorem 5 of [3] 2 2 (see [3], page 41), ensures that gk in (4.3f) is an eigenvector of Ψ n associated to the eigenvalue n 2 ψk, k =1,..., 2 .

Next, we announce a diagonalization for Hn in (1.1) discarding its eigenvalues.

Theorem 4 Let n ∈ N, a,b,c,d ∈ R, λk, k = 1, 2,...,n +2 be given by (2.4) and Hn the n × n anti-heptadiagonal persymmetric Hankel matrix (1.1).

(a) If n is even, λ1, λ3,...,λn+1 are all distinct, λ2, λ4,...,λn+2 are all distinct then

−1 F n O F n O ⊤ 2 n n 2 −1 Hn = RnPn diag φ1, φ2,...,φ , ψ1, ψ2,...,ψ −1 PnRn (4.5a) O G n 2 2 O G n 2 " 2 # where φ , φ ,...,φ n are the zeros of (4.3b), ψ , ψ ,...,ψ n are the zeros of (4.3e), 1 2 2 1 2 2

(2k+1)π (λ2k+1 − λ1) sin n+3 F n := (4.5b) 2  λ2k+1 − φhℓ i  k,ℓ  2kπ  (λ2k − λn+2) sin n+3 G n := (4.5c) 2  λ2k − ψℓ  k,ℓ   Rn is the n × n matrix (2.5c) and Pn is the n × n permutation matrix (2.5d).

(b) If n is odd, λ1, λ3,...,λn, λn+2 are all distinct, λ2, λ4,...,λn−1, λn+1 are all distinct then

−1 F n+1 O F n+1 O 2 2 ⊤ −1 Hn = RnPn diag φ1, φ2,...,φ n+1 , ψ1, ψ2,...,ψ n−1 −1 Pn Rn O G n−1 2 2 O G n− " # " 1 # 2 2 (4.6a) where φ1, φ2,...,φ n+1 are the zeros of (4.4b), ψ1, ψ2,...,ψ n−1 are the zeros of (4.4e), 2 2

(2k+1)π (λ2k+1 − λ1) sin n+3 F n+1 := (4.6b) 2  λ2k+1 − φhℓ i k,ℓ  2kπ  (λ2k − λn+1) sin n+3 G n−1 := (4.6c) 2  λ2k − ψℓ  k,ℓ   Rn is the n × n matrix (2.6c) and Pn is the n × n permutation matrix (2.6d).

21 Proof. Consider n ∈ N, a,b,c,d ∈ R and λk, k =1, 2,...,n + 2 given by (2.4). (a) According to Lemma 3 and

2 π 4 sin ( n ) I n − +3 uu⊤ O 2 ⊤ 2 n+3 PnRnPn = 2 π 4 sin ( n )  O I n − +3 vv⊤  2 n+3   we get

⊤ ⊤ diag (λ3, λ5,...,λn+1)+ λ1uu O 2 ⊤ −1 Hn = RnPn ⊤ PnRnPn PnRn O diag (λ , λ ,...,λn)+ λn vv 2 4 +2 2 π 4 sin ( n ) diag (λ , λ ,...,λ )+ λ uu⊤ I n − +3 uu⊤ ⊤ 3 5 n+1 1 2 n+3 = RnP n  O  O 2 π −1 4 sin ( n ) PnR diag (λ , λ ,...,λ )+ λ vv⊤ I n − +3 vv⊤ n 2 4 n n+2 2 n+3  2 π  4 sin ( n+3 ) ⊤ ⊤ diag (λ3, λ5,...,λn+1)+ diag (λ1 − λ3, λ1 − λ5,...,λ1 − λn+1) uu = RnPn n+3 " O O −1 2 π PnR . 4 sin ( n+3 ) ⊤ n diag (λ2, λ4,...,λn)+ n+3 diag (λn+2 − λ2, λn+2 − λ4,...,λn+2 − λn) vv # From Lemma 4, − F n diag φ , φ ,...,φ n F 1 2 1 2 2 n and − G n diag ψ , ψ ,...,ψ n G 1 2 1 2 2 n are eigenvalue decompositions for (4.3a) and (4.3d), respectively (see [6], page 85). Thus,

−1 F n O F n O ⊤ 2 n n 2 −1 Hn = RnPn diag φ1, φ2,...,φ , ψ1, ψ2,...,ψ −1 PnRn . O G n 2 2 O G n 2 " 2 # The proof of (b) is analogous being so omitted.

Setting

3π 5π (n + 1)π Ξ n := diag (λ3 − λ1) sin , (λ5 − λ1) sin ,..., (λn+1 − λ1) sin 2 n +3 n +3 n +3 and 1 Q n := 2 λ − φ 2k+1 ℓ k,ℓ we have F n = Ξ n Q n . Straight computations permit us to verify that the inverse of the alternant 2 2 2 matrix Q n is 2 n n 2 2 (φk − λ2j+1) (φj − λ2ℓ+1)  j=1 j=1  n j=6 k −1 2 Q n = (−1) Qn n Q . 2  2 2     (φk − φj ) (λ2ℓ+1 − λ2j+1)  j=1 j=1     jQ=6 k jQ=6 ℓ k,ℓ  

22 Hence,

n n 2 2 (φk − λ2j+1) (φj − λ2ℓ+1)  j=1 j=1  − n j=6 k −1 −1 1 2 Q Q F n = Q n Ξ n = (−1) n n (4.7) 2 2 2  2 2   (2ℓ+1)π  (λ2ℓ+1 − λ1) sin n+3 (φk − φj ) (λ2ℓ+1 − λ2j+1)  j=1 j=1     h i jQ=6 k jQ=6 ℓ k,ℓ   Similarly,

n n 2 2 (ψk − λ2j ) (ψj − λ2ℓ)  j=1 j=1  n j=6 k −1 2 G n = (−1) Q n Q n (4.8) 2  2 2   2ℓπ  (λ2ℓ − λn+2) sin n+3 (ψk − ψj ) (λ2ℓ − λ2j )  j=1 j=1     jQ=6 k jQ=6 ℓ k,ℓ  n+1 n+1  2 2 (φk − λ2j+1) (φj − λ2ℓ+1) j=1 j=1 n   −1 +1 j=6 k n 2 Q Q F +1 = (−1) n+1 n+1 (4.9) 2    (2ℓ+1)π 2 2  (λ2ℓ+1 − λ1) sin (φk − φj ) (λ2ℓ+1 − λ2j+1)  n+3   j=1 j=1   h i jQ=6 k jQ=6 ℓ k,ℓ  n−1 n−1  2 2 (ψk − λ2j ) (ψj − λ2ℓ) j=1 j=1 n−   −1 1 j=6 k n− 2 Q Q G 1 = (−1) n−1 n−1 . (4.10) 2  2 2   2ℓπ  (λ2ℓ − λn+1) sin (ψk − ψj ) (λ2ℓ − λ2j )   n+3   j=1 j=1   jQ=6 k jQ=6 ℓ k,ℓ   The ultimate statement below gives us a formula to compute integer powers of Hn in (1.1) only at the expense of the zeros of F (t) and G(t) presented in Lemma 4.

Corollary 2 Let n,m ∈ N, a,b,c,d ∈ R, λk, k = 1, 2,...,n +2 be given by (2.4) and Hn the n × n anti-heptadiagonal persymmetric Hankel matrix (1.1).

(a) If n is even, λ1, λ3,...,λn+1 are all distinct, λ2, λ4,...,λn+2 are all distinct then

F n O m ⊤ 2 m m m m m m Hn = RnPn diag φ1 , φ2 ,...,φ n , ψ1 , ψ2 ,...,ψ n O G n 2 2 2 −1 (4.11) F n O I n + uu⊤ O 2 2 −1 ⊤ PnRn O G n O I n + vv " 2 # 2 where φ , φ ,...,φ n are the zeros of (4.3a), ψ , ψ ,...,ψ n are the zeros of (4.3e), F n is the n × n 1 2 2 1 2 2 2 2 2 n n −1 n n −1 n n matrix (4.5b), G n is the × matrix (4.5c), F n is the × matrix (4.7), G n is the × matrix 2 2 2 2 2 2 2 2 2 (4.8), Rn is the n × n matrix (2.5c), Pn is the n × n permutation matrix (2.5d) and u, v are given by (2.5b). Moreover, if Hn is nonsingular then (4.11) holds for any integer m.

23 (b) If n is odd, λ1, λ3,...,λn, λn+2 are all distinct, λ2, λ4,...,λn−1, λn+1 are all distinct then

F n+1 O m 2 m m m m m m Hn = RnPn diag φ1 , φ2 ,...,φ n+1 , ψ1 , ψ2 ,...,ψ n−1 O G n−1 2 2 " 2 # (4.12) −1 ⊤ F n+1 O I n+1 + uu O 2 2 P⊤R⊤ −1 n− ⊤ n n O G n−1 O I 1 + vv " 2 # " 2 # where φ1, φ2,...,φ n+1 are the zeros of (4.3a), ψ1, ψ2,...,ψ n−1 are the zeros of (4.3e), F n+1 is the 2 2 2 − n+1 n+1 n− n−1 n−1 1 n+1 n+1 2 × 2 matrix (4.6b), G 1 is the 2 × 2 matrix (4.6c), F n+1 is the 2 × 2 matrix (4.9), 2 2 −1 n−1 n−1 G n−1 is the 2 × 2 matrix (4.10), Rn is the n × n matrix (2.6c), Pn is the n × n permutation 2 matrix (2.6d) and u, v are deﬁned in (2.6b). Furthermore, if Hn is nonsingular then (4.12) is valid for every integer m.

Proof. According to Theorem 4, it suﬃces to observe that for n even

n ⊤ −1 I + uu O PnR = 2 PnRn, n O I n + vv⊤ 2 and for n odd, ⊤ I n+1 + uu O ⊤ −1 2 ⊤ ⊤ Pn Rn = ⊤ Pn Rn . O I n−1 + vv " 2 #

Remark It is worth pointing out that all results presented here for matrices Hn having the form (1.1) are still valid for anti-pendiagonal persymmetric Hankel matrices or anti-tridiagonal persymmetric Hankel matrices, that is to say, for matrices Hn in (1.1) such that a =0 or a = b = 0.

Acknowledgements

This work is a contribution to the Project UID/GEO/04035/2013, funded by FCT - Funda¸c˜ao para a Ciˆencia e a Tecnologia, Portugal.

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