Spectral Properties of Some Combinatorial Matrices

Spectral Properties of Some Combinatorial Matrices Emrah Kilic TOBB Economics and Technology University Mathematics Department 06560 Sogutozu, Ankara, Turkey e-mail: [email protected]

Gabriela N. St˘ anic˘ a Department of Mathematics Monterey Peninsula College Monterey, CA 93940, USA e-mail: [email protected]

Pantelimon St˘ anic˘ a∗ Naval Postgraduate School Department of Applied Mathematics Monterey, CA 93943, USA e-mail: [email protected] May 9, 2008

Abstract In this paper we investigate the spectra and related questions for various combinatorial matrices, generalizing work by Carlitz, Cooper and Kennedy.

1

Introduction

In [4], R. Peele
and P. St˘ anic˘ a studied
n × n matrices with the (i, j) entry the binomial i−1 i−1 and derived many interesting results on powers of , respectively, n−j coefficient j−1 these matrices. In [6], one of us found that the same is true for a much larger class of what he called netted matrices, namely matrices with entries satisfying a certain type of recurrence among the entries of all 2 × 2 cells.
i−1 Let Rn be the matrix whose (i, j) entries are ai,j = n−j , which satisfy ai,j−1 = ai−1,j−1 + ai−1,j .

(1)

The previous recurrence can be extended for i ≥ 0, j ≥ 0, using the boundary conditions a1,n = 1, a1,j = 0, j 6= n. Remark the following consequences of the boundary conditions and recurrence (1)): ai,j = 0 for i + j ≤ n, and ai,n+1 = 0, 1 ≤ i ≤ n. The matrix Rn was firstly studied by Carlitz [9] who gave explicit forms for the eigenvalues of Rn . Let fn+1 (x) = det (xI − Rn ) be the characteristic polynomial of Rn . ∗

Also Associated to the Institute of Mathematics of Romanian Academy, Bucharest, Romania

1

2 Thus fn (x) =

n+1 X

r(r+1)/2

(−1)

r=0

  n xn−r r F

n r F




denote the Fibonomial coefficient, defined (for n ≥ r > 0) by   n F1 F2 . . . Fn = (F1 F2 . . . Fr ) (F1 F2 . . . Fn−r ) r F

with nn F = n0 F = 1. Carlitz showed that where

fn (x) =

n−1 Y

x − φj φ¯n−j




j=0

√
¯ . . . , φφ¯n−1 , φ¯n . We where φ, φ¯ = 1 ± 5 /2. Thus the eigenvalues of Rn are φn , φn−1 φ, shall give another proof of this result in the next section. In [4] it was proved that the entries of the power Rne satisfy the recurrence (e)

(e)

(e)

(e)

Fe−1 ai,j = Fe ai−1,j + Fe+1 ai−1,j−1 − Fe ai,j−1 ,

(2)

where Fe is the Fibonacci sequence. Closed forms for all entries of Rne were not found, but several results concerning the generating functions of rows and columns were obtained (see [4, 6]). For instance, the entries in the first row and column of Rne are   n−1 (e) a1,j = F n−j F j−1 j − 1 e−1 e (3) (e) n−i i−1 ai,1 = Fe−1 Fe . Further, the generating function for the (i, j)-th entry of the e-th power of a generalization of Rn , namely    i−1 i+j−n−1 n−j Qn (a, b) = a b n−j 1≤i,j≤n is Bn(e) (x, y) =

(Ue−1 + Ue y)(b Ue−1 + y Ue )n−1 . Ue−1 + Ue y − x(Ue + Ue+1 y)

Certainly, Qn (1, 1) = Rn . As an example,  0 0 0  0 0 0   0 0 0 Q6 (a, b) =  3  0 0 b  4  0 b 4ab3 b5 5ab4 10a2 b3

0 0 b2 3ab2 6a2 b2 10a3 b2

0 b 2ab 3a2 b 4a3 b 5a4 b

1 a a2 a3 a4 a5

    .   

Regarding this generalization, in [10], the authors gave the characteristic
polynomial k of Qn (a, b) and the trace of kth power of Qn (a, b, ), that is, tr Qn (a, b) , by using the method of Carlitz [9].

3 Lemma 1. Let gn (x) denote the characteristic polynomial of Qn (a, b) . Thus gn (x) =

n X

i(i+1)/2 i(i−1)/2

(−1)

b

i=0

and

  n xn−i i U

  U kn , tr Qkn (a, b) = Uk

where

n i U




(4)

(5)

stands for the generalized Fibonomial coefficient, defined by

  n U1 U2 . . . Un = , r U (U1 U2 . . . Ur ) (U1 U2 . . . Un−r )

for n ≥ i > 0, where nn U = n0 U = 1. One of us extended in [6] some of these results to netted matrices, say Rnα,β,γ,δ , whose entries ai,j , i ≥ 0, j ≥ 0 satisfy (for i ≥ 1, j ≥ 1) δai,j = αai−1,j + βai−1,j−1 + γai,j−1 ,

(6)

with the boundary conditions βai,0 + γai+1,0 = 0, for all 1 ≤ i ≤ n − 1

(7)

δai+1,n+1 − αai,n+1 = 0, for all 1 ≤ i ≤ n − 1.

(8)

It was shown in [6] that the entries of any power of such a matrix will satisfy a similar recurrence, and generating functions for these powers were found. In the case of the Fibonacci sequence, the generating function of the i-th row of Rne is X (e) (e) ri (x) = ai,j xj−1 = (Fe + Fe+1 x)i−1 (Fe−1 + Fe x)n−i j≥1

and, if e > 1, the generating function for the j-th column of Rne is (e) cj (x)

(1)

 =

Fe+1 x − Fe Fe−1 − Fe x

j−1

"  # j−1    n X Fe−1 n Fe (Fe−1 − Fe x) s 1+ . Fe−1 − Fe x s Fe−1 (Fe+1 x − Fe ) s=1

where cj (x) = (1 − x)j−1−n xn−j . matrix Qn (a, b) was introduced as  a generalization  The     of the Fibonacci matrix Q = 1 F 0 1 n−1 which has the property that Qn · = . The matrix Qn (a, b) displays 1 1 0 Fn a similar property on the sequence Un+1 = aUn + bUn−1 , that is, any e-th power of Qn (a, b) multiplied by a fixed vector gives an n-tuple of consecutive terms of the sequence U = (Uk )k . We refer to [6, 10] for more details.

4

2

Spectra of Rn and Qn (a, b)

In [4], the authors proposed a conjecture on the eigenvalues, which was proven independently in [1] and the unpublished manuscript [7]. In this section we give the proof of the conjecture from [7] and we find the eigenvectors of Rn . In [4, 6], it was shown that the inverse of Rn is the matrix    −1 n+i+j+1 n − i Rn = (−1) , j − 1 1≤i,j≤n and, in general, the inverse of Qn (a, b) is    −1 n+i+j+1 n+1−i−j i−n n − i Qn (a, b) = (−1) a b . j − 1 1≤i,j≤n

(9)

We define Kn to be the matrix with (i, j)-entry δi,n−j+1 (the Kronecker symbol), that is, Kn is a permutation matrix having 1 on the secondary diagonal and 0 elsewhere. √

√

Theorem 2. Let φ = 1+2 5 , φ¯ = 1−2 5 be the golden section and its conjugate. The eigenvalues of Rn are: 1. {(−1)k+i φ2i−1 , (−1)k+i φ¯2i−1 }i=1,...,k , if n = 2k. 2. {(−1)k } ∪ {(−1)k+i φ2i , (−1)k+i φ¯2i }i=1,...,k , if n = 2k + 1. ¯ . . . , φφ¯n−1 , φ¯n . Equivalently, the eigenvalues of Rn are φn , φn−1 φ, Proof. First, we show that Rn is a permutation matrix away from Ln , namely Rn · Kn = Ln ⇐⇒ Rn = Ln · Kn .

(10)

It is a trivial matter to prove Kn2 = In , which will give the equivalence. It suffices to show the second identity, which follows easily, since an entry in Ln · Kn is  n  X i−1 k=1

k−1

 δk,n−j+1 =

 i−1 . n−j

Now, denote by An , the matrix obtained by taking absolute values of entries of Rn−1 . We show that Rn has the same characteristic polynomial (eigenvalues) as An , namely we prove their similarity, Kn · Rn · Kn = An . (11) Since Kn · Rn · Kn = Kn · Ln (by (10)), to show (11) it suffices to prove that Kn · Ln = 
 n−i . Therefore, we need j−1 i,j

n X k=1

which is certainly true.

    k−1 n−i δi,n−k+1 = , j−1 j−1

5 We use a result of [3] to show that An in turn is similar to Dn , the diagonal matrix whose diagonal entries are the elements in the eigenvalues set listed in decreasing order according to size of the absolute value. For instance, for n = 4,  3  α 0 0 0  0 −α 0 0   D4 =   0 0 −β 0  0 0 0 β3 Our claim will be proved if one can show that An is similar to Dn , and this will follow from [3]. We sketch here that argument for the convenience of the reader. Define an array bn,m by bn,0 = 1 for all n ≥ 0 bn,m = 0 for all m > n bn,m = bn−1,m−1 (−1)m Certainly, |bn,m | =

Fn for all m ≤ n Fm

Fn Fn−1 · · · Fn−m+1 . Let Cn = (ci,j )i,j , where Fm · · · F1   if i = 1, . . . , n − 1 ci,i+1 = 1 cn,j = −bn,n+1−j if j = 1, . . . , n   ci,j = 0 otherwise.

We observe that, in fact, Cn is the companion matrix
of the polynomial with coefficients j−1 n−j n−i bn,n+1−j . Let Xn be the matrix with entries j−1 Fi−2 Fi−1 . It turns out that the eigenvector matrix En of An , with columns vectors listed in decreasing order of absolute value of the corresponding eigenvalues, normalized so that the last row is made up of all 1’s, satisfies Xn En = Vn , where Vn is the Vandermonde matrix, which is the eigenvector matrix of Cn with eigenvectors listed in decreasing order of the absolute values of the corresponding eigenvalues. Also, Xn An Xn−1 = Cn and En−1 An En = Dn . Our theorem follows. Easily we deduce Corollary 3. The eigenvectors matrix of Rn , say Wn , with eigenvectors listed in decreasing order of the absolute values of the corresponding eigenvalues, is Wn = Kn En = Kn Xn−1 Vn . Proof. We showed that Kn Rn Kn = An and En−1 An En = Dn . It follows that (En−1 Kn )Rn (Kn En ) = Dn , which together with Xn En = Vn , proves the corollary.

6 Example 4. For n = 4, the eigenvectors matrix is   −α3 α β −β 3  α2 β2  − 13 β − 13 α  W4 =   −α − 1 α2 − 1 β 2 −β  3 3 1 1 1 1 We present below the first few characteristic polynomials of Qn (a, b) (for 1 ≤ n ≤ 7) and we ask whether one can find their roots as in Theorem 2: x2 − ax − b;
−(b + x) −xa2 + b2 + x2 − 2bx ;

b3 − axb − x2 xa3 + 3bxa + b3 − x2 ;


b2 − x −xa4 − 4bxa2 + b4 + x2 − 2b2 x b4 + 2xb2 + a2 xb + x2 ;


− b5 − 3axb2 − a3 xb − x2 b5 + axb2 − x2 xa5 + 5bxa3 + 5b2 xa + b5 − x2 . The argument that proves the next theorem is similar to the proof of Theorem 2, and so, we omit it. √

√

2 ¯ = a− a2 +4b be the roots of the polynomial x2 − ax − b. Theorem 5. Let λ = a+ a2 +4b , λ 2 The eigenvalues of Qn (a, b) are: ¯ 2i−1 }i=1,...,k , if n = 2k. 1. {(−b)k−i λ2i−1 , (−b)k−i λ ¯ 2i }i=1,...,k , if n = 2k + 1. Equivalently, the eigenvalues 2. {(−b)k } ∪ {(−b)k−i λ2i , (−b)k−i λ ¯ . . . λλ ¯ n−1 , λ ¯n. of Qn (a, b) can also be written in the following compact form λn , λn−1 λ,

Since Rn , Qn (a, b) are column justified triangular matrices, the determinants of our matrices are easy to compute, or we can use the previous result. Corollary 6. We have det(Rn ) = (−1)bn/2c and det(Qn (a, b)) = (−1)bn/2c bn(n−1)/2 .

3

Arithmetic progressions and spectra of other combinatorial matrices

Let the sequences {un }, {vn } be defined by un = aun−1 + bun−2 vn = avn−1 + bvn−2 , for n > 1, where u0 = 0, u1 = 1, and v0 = 2, v1 = a, respectively. Let α, β be the roots of the associated equation x2 − ax − b = 0. The next lemma appears in [11]. Lemma 7. For k ≥ 1 and n > 1, ukn = vk uk(n−1) + (−1)k+1 bk uk(n−2) vkn = vk vk(n−1) + (−1)k+1 bk vk(n−2) .

(12)

7
Using the sequence vk , we define the n × n matrix Hn vk , bk as follows:     n−j  i − 1  i+j−n−1 k k Hn vk , b = vk − (−b) . n−j 1≤i,j≤n
Observe that Hn v1 , b1 = Qn (a, b) . As an example,  0 0 0 0 0 k  0 0 0 0 − (−b)   0 0 0 b2k −2 (−b)k vk  H6 (vk , bk ) =  3k  0 0 − (−b) 3b2k vk −3 (−b)k vk2   0 b4k −4 (−b)3k vk 6b2k vk2 −4 (−b)k vk3 − (−b)5k 5b4k vk −10 (−b)3k vk2 10b2k vk3 −5 (−b)k vk4

1 vk vk2 vk3 vk4 vk5

     .   

As in equation (9), we can easily find the inverse of the matrix Hn , namely    n−i −1 k j+1 −k(n−i) n+1−i−j Hn (vk , b ) = (−1) (−b) v . j−1 i,j It is well known that for n ≥ −1, un+1 =

X r  a2r−n bn−r . n − r r

(13)

We generalize this identity next. Lemma 8. For k > 0 and n ≥ −1,    n−r uk(n+1) X r = vk2r−n − (−b)k . uk n−r r Proof. (Induction on n) If n = 1, then equations (12) and (13) give the identity. Suppose that the result is true for n − 1 and n. Then uk(n+1) = vk ukn + (−b)k+1 uk(n−1)   n−1−r X r = vk vk2r−n+1 − (−b)k n−1−r r   n−2−r X r − (−b)k vk2r−n+2 − (−b)k n−2−r r      n−1−r X r r = + vk2r−n+2 − (−b)k n−1−r n−2−r r    n−1−r X r+1 = vk2r−n+2 − (−b)k n−1−r r    n−r X r = vk2r−n − (−b)k . n − r r

8 Lemma 9. Let k, m > 0 and 0 ≤ r ≤ n. Then     ukm x − (−b)m u(k−1)m r u(k+1)m x − (−b)m ukm n−r um um   X n − rn − r1  n − rk−1 ... = r2 rk r1 r ,r ,...,r 1

2

k

kn−r−2r1 −···−2rk−1 −rk × vm

(− (−b)m )r1 +r2 +...+rk xn−rk .

(14)

Proof. (Induction on k) For k = 1, the proof follows from u2n = un vn and X n − r m n−r n−r−s r vm (− (−b)m )s xn−s x (vm x − (−b) ) = s s (see [10]). We assume that the equality holds for some positive integer k. If we replace
vm − (−b)m x−1 by x and then multiply by xn , the left side of this equation becomes     u(k+1)m x − (−b)m ukm r u(k+2)m x − (−b)m u(k+1)m n−r . um um After some simplifications, if we expand the right side of this equation, then we get      X n − r n − r1 n − rk ... r1 r2 rk+1 r ,r ,...,r 1

2

k+1

k(n+1)−r−2r1 −···−2rk −rk+1 ×vm

(− (−b)m )r1 +r2 +···+rk+1 xn−rk+1 .

The proof is complete. Lemma 10. For all m > 0,  u   knm = tr Hnm vk , bk . uk Proof. If we multiply both sides of the equation (14) by xr and summing over r, we get    n  X ukm x − (−b)m u(k−1)m r u(k+1)m x − (−b)m ukm n−r r x um um r=0   X n − rn − r1  n − rk−1 = ... r1 r2 rk r ,r ,...,r 1

2

k

kn−r−2r1 −···−2rk−1 −rk ×vm

(− (−b)m )r1 +r2 +···+rk xn+r−rk ,

where the coefficient of xn on the right side is tr Hnm vk , bk . The coefficient of xn on the left side of this equation without its denominator is X rn − r
r−s t uskm − (−b)m u(k−1)m u(k+1)m (− (−b)m ukm )s s t r+s+t=n   X r n − r
r−s n−r−s = uskm − (−b)m u(k−1)m u(k+1)m (− (−b)m ukm )s . s s r+s≤n

9 For easy writing, denote this last expression by cn . Then ∞ X

cn xn

n=0 ∞ X

  X n − r 
n−r−s r m r r−s 2s r+s u(k+1)m x = (− (−b) ) u(k−1)m ukm x s s n=r+s r,s=0 ∞   X
−s−1 r r−s r+s = (− (−b)m )r u(k−1)m u2s 1 − u(k+1)m x km x s r,s=0   ∞ X
−s−1 X r
r−s 2s = (− (−b)m )s u2s x 1 − u x (− (−b)m ) u(k−1)m x (k+1)m km s s=0

=

∞ X

r≥s

2s (− (−b)m )s u2s 1 − u(k+1)m x km x


−s−1

1 + (−b)m u(k−1)m x


−s−1

s=0

=

1

1 − u(k+1)m x 1 + (−b)m u(k−1)m x 1 −

1

=

1

. m 2 1 − u(k+1)m x 1 + (−b) u(k−1)m x + (−b)m u2s km x

2 (−(−b)m )s u2s km x (1−u(k+1)m x)(1+(−b)m u(k−1)m x)

(15)

Here by the Binet formula for {un } , we note that for all integers k > 0, ukn − (−b)k uk(n−2) = vk(n−1) uk and −

ukn uk(n−2) − u2k(n−1)

!

u2k

= (−b)k(n−2) .

Using (16) and (17), we write the right side of the equation (15) as 1

=

1 − u(k+1)m x 1





2 1 + (−b) u(k−1)m x + (−b)m u2s km x

. 1 − vmk x + (−b)mk x2

m k = Thus, cn = ukmn uk , and so, tr Hn vk , b

(16)

m

uknm uk .

The proof is complete.


Theorem 11. The eigenvalues of Hn vk , bk are αkn , αk(n−1) β k , . . . , αk β k(n−1) , β kn .

(17)

10 Proof. Let
fn (x) = det xI − Hn vk , bk Hn vk , bk . Then by Lemma 10, 0

fn (x) fn (x)

=

= =

=

and λ0 , λ1 , . . . , λn−1 denote the eigenvalues of

n−1 m X X 1 −m−1 = x λm j x − λj

n−1 X j=0 ∞ X





m=0



m x−k−1 tr Hn−1 vk , bk

k=0 ∞ X m=0 ∞ X

x−m−1

x

−m−1

m=0

= Thus

fn (x) =



uk(n−1)m uk n−1 X

αjkm β k(n−j−1)m

j=0

n−1 X j=0

j=0



1 x−

n−1 Y

αjk β (n−j−1)k

.

x − αjk β (n−j−1)k



j=0

and so the eigenvalues of Hn


vk , bk are

αk(n−1) , αk(n−2) β k , . . . , αk β k(n−2) , β k(n−1) .

Next, we give an expression of the characteristic polynomial of Hn in terms of Gaussian binomials. Fix k and let rn := ukn . Define the k-Fibonomial coefficients by nno r1 r2 . . . rn = , m u,k (r1 r2 . . . rm ) (r1 r2 . . . rn−m )   where nn u,k = n0 u,k = 1. n
n Clearly, when k = 1, then m is reduced to the Fibonomial coefficient m . u,k u Theorem 12. fn (x) =

n−1 Y

jk (n−j−1)k

x−α β



=

j=0

n X

(−1)i (−b)ki(i−1)/2

i=0

nno i

xn−i .

u,k

Proof. Here we can use the following familiar identity n−1 Y j=0

n hni
X 1 − qj x = (−1)i q i(i−1)/2 xi , i q i=0

(18)

11 where hni i

q


(1 − q n ) . . . 1 − q n−i+1 , = (1 − q i ) . . . (1 − q)

are the usual q-binomial coefficients (Gaussian binomials). If we replace q by (β/α)k we find that for all positive fixed integers k, hni nno → αki(i−n) . i q i u,k Thus (18) becomes n−1 Y

1− α

−1

j=0

n nno
kj  X β x = (−1)i β ki(i−1)/2 αki(i+1)/2−nki xi . i u,k i=0

Replacing x by αk(n−1) x, we get n−1 Y

1−α

(n−j−1)k kj

β x



=

j=0

=

n X i=0 n X i=0

(−1)i (αβ)ki(i−1)/2 (−1)i (−b)ki(i−1)/2

nno i nno i

xi

u,k

xi .

u,k

Finally replacing x by x−1 gives us n−1 Y j=0

n  X nno 1 − α(n−j−1)k β kj x = (−1)i (−b)ki(i−1)/2 xn−i . i u,k i=0

Acknowledgements. P.S. acknowledges NPS Foundation and RIP funding support from Naval Postgraduate School.

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