The Fibonacci Identities of Orthogonality
The Fibonacci Identities of Orthogonality Kyle Hawkins, Ursula Hebert-Johnson and Ben Mathes January 14, 2015 Abstract In even dimensions, the orthogonal projection onto the two dimensional space of second order recurrence sequences is particularly nice: it is a scaled Hankel matrix whose entries consist of the classical Fibonacci sequence. A new proof is given of this result, and new Fibonacci identities are derived from it. Examples are given showing that familiar Fibonacci identities can be viewed as special cases. We show that the projection in odd dimensions can be written as a rank one Lucas perturbation of a scaled Lucas Hankel matrix, from which more Fibonacci identities are derived.
1
The Fibonacci projection
Let Rn denote the n dimensional vector space of all real n-tuples, and let Rn (k, l) denote the vector subspace consisting of the (si ) ∈ Rn for which si = ksi−1 + lsi−2 , for i = 3, . . . , n. (See [7] for a delightful article on general Fibonacci sequences.) The element of Rn (k, l) whose first two coordinates are 0 and 1 (respectively) will be denoted (fi ) and is called the general Fibonacci sequence, and the sequence beginning with 2 and k is denoted (li ) and is called the general Lucas sequence. Matrices with Fibonacci and Lucas entries were studied in [1],[3],[4],[5],[6], [8],[9], and [10],. Motivated by these works, in [5] it was shown that, when n is even, the orthogonal projection onto Rn (k, 1) 0 0
AMS 2000 Classification numbers: Primary: 15B36 Secondary: 65F35 Key Words: Orthogonal projection, Fibonacci numbers, Fibonacci identities.
1
is the Hankel matrix k fn
f−n+1 f−n+2 f−n+2 f−n+3 .. .. . . f−1 f0 f0 f1
... ... .. .
f−1 f0 .. .
f0 f1 .. .
. . . fn−3 fn−2 . . . fn−2 fn−1
.
(A proof of this in the special case of Rn (1, 1) may now be found in [2].) We begin with a new proof of this fact. Let E denote the 2 × 2 matrix � � 0 1 E= . 1 k It follows that E
m
=
�
fm−1 fm fm fm+1
�
,
for all integers m. Lemma 1 For any integer t and any nonnegative integer s, we have s �
E t+4i =
i=0
Proof.
f2(s+1) t+2s E . k
It suffices to prove the scalar identity s �
ft+4i =
i=0
f2(s+1) ft+2s , k
and we proceed via induction on s. Since f2 = k, the identity holds when s = 0, so we assume the induction hypothesis s−1 � i=0
ft+4i =
f2s f . k t+2(s−1)
� Substituting this in for the first s summands of si=0 ft+4i , we see that the equality holds s � f2(s+1) ft+4i = ft+2s k i=0
2
if and only if f2s ft+2s−2 + kft+4s = f2(s+1) ft+2s . The equality E t+4s = E t+2s E 2s yields the identity ft+4s = f2s ft+2s−1 + f2s+1 ft+2s from which we derive f2s+2 ft+2s = f2s ft+2s + kf2s+1 ft+2s = f2s ft+2s−2 + kf2s ft+2s−1 + kf2s+1 ft+2s = f2s ft+2s−2 + kft+4s � Let Hf denote the n × n matrix
Hf =
f−n+1 f−n+2 f−n+2 f−n+3 .. .. . . f−1 f0 f0 f1
... ... .. .
f−1 f0 .. .
f0 f1 .. .
. . . fn−3 fn−2 . . . fn−2 fn−1
.
We will call Hf the central Fibonacci Hankel matrix. Replacing fi with li yields Hl , the central Lucas Hankel matrix. Theorem 2 For even n, the matrix R(k, 1).
k fn Hf
is the orthogonal projection onto
Proof. It is clear that fkn Hf is selfadjoint, and its range is R(k, 1), so we need only prove fkn Hf is idempotent, which is equivalent to proving Hf2 =
fn Hf . k
Write n = 2m and notice that −n+2 E E −n+4 . . . E −2 E0 E −n+4 E −n+6 . . . E 0 E2 . . . .. .. .. .. .. Hf = . . 0 n−6 n−4 E −2 E ... E E E0 E2 . . . E n−4 E n−2 3
� �m . = E 2(i+j)−2(m+1) i,j=1
Using our lemma, it follows that ��m � 2(i+r)−2(m+1) E 2(r+j)−2(m+1) m Hf2 = r=1 E i,j=1 � ��m 2(i+j)−4m+4(r−1) m = E ��r=1 �m i,j=1 m−1 2(i+j)−4m+4s = s=0 E i,j=1 f2m � 2(i+j)−4m+2(m−1) �m = k E i,j=1 � �m = fkn E 2(i+j)−2(m+1) i,j=1 = fkn Hf � The theme of this paper is the interplay between the linear algebraic equations of orthogonality and Fibonacci identities. To illustrate with an example, observe that the idempotence of Hf leads immediately to the identity n−1 k k2 � fi+j−n+1 = 2 fi+k−n+1 fk+j−n+1 , fn fn k=0
for all even n and −n + 1 ≤ i, j ≤ n − 1. This proves the following.
Corollary 3 Assume that n is even and −n + 1 ≤ i, j ≤ n − 1. We have k
n−1 �
fi−u fj−u = fn fi+j−n+1 .
u=0
Assume that n is even and s = (si )n−1 i=0 ∈ R(k, 1), which we will think ∗ of as a column vector. We then let s denote the transpose of s, which is a row vector. We define s⊥ ≡ (−sn−1 , sn−2 , −sn−3 . . . , −s1 , s0 )∗ = ((−1)i+1 sn−i−1 )n−1 i=0 , and observe that this is also an element of R(k, 1), and {s, s⊥ } forms an orthogonal basis of R(k, 1) for each non-zero s ∈ R(k, 1). Nor⊥ s malizing, and denoting u = ||s|| and v = ||ss⊥ || , then, by elementary linear algebra, the orthogonal projection onto R(k, 1) is the matrix uu∗ + vv ∗ . Equating this matrix with fkn Hf establishes the following identity. (Note that ||s||2 = ||s⊥ ||2 = k1 (sn−1 sn − s0 s−1 ).) 4
Theorem 4 Assume that n is even. Then, for all 0 ≤ i, j ≤ n − 1 we have k k (si sj + (−1)i+j sn−i−1 sn−j−1 ) = fi+j−n+1 sn−1 sn − s0 s−1 fn i.e. we have (si sj + (−1)i+j sn−i−1 sn−j−1 )fn = (sn−1 sn − s0 s−1 )fi+j−n+1 . The preceding identity has many famous identities as special cases. Example 1 Take s to be the Fibonacci sequence itself, and let i = j = 0. The result is that k k f2 = f−n+1 fn−1 fn n−1 fn for all even n. In other words, fq = f−q for odd subscripts q. Example 2 Take s to be the Fibonacci sequence again, and this time let i = 1 and j = 0. The result is that k k (−fn−2 fn−1 ) = f−n+2 fn−1 fn fn for all even n. In other words, fq = −f−q for all even subscripts q. Example 3 With s the Fibonacci sequence again, let i = n/2 and j = n/2 to obtain the identity 2 fq2 + fq−1 = f2q−1 for all positive integers q. Let φ denote the positive solution of the equation x2 − kx − 1 = 0, which we call the k-golden ratio. We let ψ denote the negative solution. Verify that φψ = −1, from which we see that when s = (1, φ, . . . , φn−1 )∗ and t = (1, ψ, . . . , ψ n−1 )∗ , we have another orthogonal basis {s, t} of R(k, 1). Of course, in this case t is just a multiple of s⊥ . We equate the entries of the matrix 1 1 ss∗ + tt∗ 2 ||s|| ||t||2 with those of the projection in Theorem 2 to obtain the following. 5
Theorem 5 With n even we have 1 − φ2 1 − ψ2 k i+j φ + ψ i+j = fi+j−n+1 1 − φ2n 1 − ψ 2n fn for all 0 ≤ i, j ≤ n − 1. Alternatively, this can be written φi+j+1 ψ i+j+1 k + = fi+j−n+1 2n 2n φ −1 ψ −1 fn Example 4 Letting j = n − 1 and i = 1 in the equation above yields a slightly convoluted formulation of the Binet equation: 1 φn+1 ψ n+1 = 2n + 2n . fn φ −1 ψ −1 Using the fact that n is even and φψ = −1, one calculates � n+1 � φ ψ n+1 + (φn − ψ n ) = φ − ψ, φ2n − 1 ψ 2n − 1 from which the familiar Binet formula appears as 1 n (φ − ψ n ) = φ − ψ. fn
2
The Lucas Projection
In this section we consider odd values of n, with n = 2m + 1. We now let f and l denote the central Fibonacci and Lucas sequences, which are the transposes of (f−m , . . . , f0 , . . . , fm ) and (l−m , . . . , l0 , . . . , lm ), respectively. A computation reveals that ||f ||2 =
2fm fm+1 2lm lm+1 and ||l||2 = . k k
Recalling that f0 = 0, f−i = (−1)i+1 fi , and l−i = (−1)i li , you see that m �
i=−m
f i li = −
m �
f i li + f 0 l0 +
i=1
m � i=1
6
fi li = 0,
so the pair {f , l} forms an orthogonal basis of R(k, 1). One expression for the orthogonal projection onto R(k, 1) is then given by k k ff∗ + ll∗ . 2fm fm+1 2lm lm+1 Our second expression for this projection gives it as a linear combination of the two Lucas matrices Hl and ll∗ . Theorem 6 The orthogonal projection onto R(k, 1) is the matrix � � k k(−1)m+1 ∗ ll + Hl . (4 + k 2 )fm fm+1 lm lm+1 Proof. The Hankel matrix Hl has eigenvectors f and l with correspond2 ing eigenvalues (4+k )fkm fm+1 and lm lkm+1 (respectively). It follows that (4+k2 )fm fm+1 k ∗ k 2fm fm+1 ff 2 (4+k ) ∗ ff + 12 ll∗ 2
Hl = =
+
lm lm+1 k ∗ k 2lm lm+1 ll
,
and, letting P denote our projection, we compute P
=
k ∗ 2fm fm+1 ff
=
k 2 2fm fm+1 ( (4+k2 ) Hl
+
k ∗ 2lm lm+1 ll
−
1 ll∗ ) (4+k2 )
+
k ∗ 2lm lm+1 ll
.
Simplifying the above equation while using the identity lm lm+1 − (4 + k 2 )fm fm+1 = 2(−1)m k gives the expression k P = 2 (4 + k )fm fm+1
�
� k(−1)m+1 ∗ ll + Hl . lm lm+1 �
Corollary 7 If n is odd with n = 2m + 1, then � � fi fj li lj 2 k(−1)m+1 + = li lj + li+j fm fm+1 lm lm+1 (4 + k 2 )fm fm+1 lm lm+1 for all −m ≤ i, j ≤ m. 7
Example 5 The identity lm lm+1 − (4 + k 2 )fm fm+1 = 2(−1)m k used in the derivation of the above is recovered when we let i = j = 0, when i = j = 1 and again when we let one of the indices equal zero, and the other equal m. This suggests to us that this identity is intrinsically an orthogonality relation. Example 6 the identity
Let i = j = m in the equation above and simplify to obtain 2 2 (4 + k 2 )fm + lm = 2 l2m .
There is nothing sacred about the Lucas numbers in the above expression for our projection, aside from the fact that the resulting formula is the simplest we have found. Indeed, if s is any element of R(k, 1) and Hs is the corresponding central Hankel matrix, then Hs is diagonalizable with eigenvectors in R(k, 1) corresponding to non-zero eigenvalues. If s is rank two (as most of the Hs are), then choosing an orthonormal eigenbasis {u, v} of R(k, 1) gives us an expression for both Hs and P : P = uu∗ + vv∗ and Hs = α uu∗ + β vv∗ . One can then choose to solve for one of the orthogonal projections in the expression for Hs and substitute into the expression for P , giving P in terms of Hs and the other projection, for example 1 β P = ( Hs − vv∗ ) + vv∗ . α α Let us express P in terms of the central Fibonacci Hankel matrix Hf to illustrate this method. Recall that Hf is defined by Hf = (fi+j )m ij=−m . fm fm+1 l and Hf (l) = lm lkm+1 f . It is convenient k 2 and b = lm lkm+1 = ||l|| 2 , from which we see that
Verify that Hf (f ) = a=
fm fm+1 k
=
||f ||2 2
√ √ √ Hf ( abf + al) = ab( abf + al),
8
to write
and
√ √ √ Hf ( abf − al) = − ab( abf − al).
Since Hf is symmetric, it is reassuring to check that the eigenvector √ is orthogonal to abf − al. As f is orthogonal to l, we get
√
abf +al
√ √ ||f ||4 ||l||2 || abf + al||2 = || abf − al||2 = ab||f ||2 + a2 ||l||2 = . 2 √ √ √ √ √ +al) 2( abf −al) We let u = 2(||f ||abf and v = , with α = ab = ||f ||||l|| and 2 ||l|| 2 ||f ||2 ||l|| √ β = − ab =
−||f ||||l|| , 2
in the equation above to prove the following.
Theorem 8 Assume that n is odd, and n = 2m + 1. The orthogonal projection onto R(k, 1) is the matrix (pij ), where pij
=
2 ||f ||||l|| fi+j
+ 2vi vj
=
2 ||f ||||l|| fi+j
+
1 ff ||f ||2 i j
+
1 ll ||l||2 i j
−
1 ||f ||||l|| (fi lj
+ li f j )
and i, j = −m, . . . , m. Example 7
For all i, j = −m, . . . , m one has 2fi+j = fi lj + li fj ,
which can be seen following from Theorem 8 by remembering that pij = and hence pij = so
2 ||f ||||l|| fi+j
=
1 1 fi fj + li lj , 2 ||f || ||l||2
2 1 fi+j + pij − (fi lj + li fj ), ||f ||||l|| ||f ||||l||
1 ||f ||||l|| (fi lj
+ li fj ).
References [1] M. Bahsi and S. Solak, On the spectral norms of Hankel matrices with Fibonacci and Lucas numbers, Selcuk J. Appl. Math., 12 no. 1 (2011), 71-76. 9
[2] O. Bretscher, Linear Algebra with Applications, Fifth Edition, Pearson Education, New Jersey. [3] D. Bozkurt and M. Akbulak, On the norms of Toeplitz matrices involving Fibonacci and Lucas numbers, Hacet. J. Math. Stat., 37 (2)(2008), 89-95. [4] J. Dixon, B. Mathes, and D. Wheeler, An application of matricial Fibonacci identities to the computation of spectral norms, Rocky Mt. J. Math., to appear. [5] E. Dupree and B. Mathes, Singular values of k-Fibonacci and k-Lucas Hankel Matrices, Int. J. Contemp. Math. Sciences, to appear. [6] A. Ipek, On the spectral norms of circulant matrices with classical Fibonacci and Lucas numbers entries, Appl. Math. Comput., 217 (2011), 6011-6012. [7] D. Kalman and R. Mena, The Fibonacci numbers – exposed, Mathematics Magazine, vol. 76, no. 3 (2003). [8] S. Shen, On the norms of Toeplitz matrices involving k-Fibonacci and k-Lucas numbers, Int. J. Contemp. Math. Sciences, 7 (8) (2012), 363368. [9] S. Shen, J. Cen, On the spectral norms of r-circulant matrices with the k-Fibonacci and k-Lucas numbers, Int. J. Contemp. Math. Sciences, 5 (12) (2010), 569-578. [10] S. Solak, On the norms of circulant matrices with the Fibonacci and Lucas numbers, Appl. Math. Comput., 160 (2005), 125-132.
10
1
The Fibonacci projection
Let Rn denote the n dimensional vector space of all real n-tuples, and let Rn (k, l) denote the vector subspace consisting of the (si ) ∈ Rn for which si = ksi−1 + lsi−2 , for i = 3, . . . , n. (See [7] for a delightful article on general Fibonacci sequences.) The element of Rn (k, l) whose first two coordinates are 0 and 1 (respectively) will be denoted (fi ) and is called the general Fibonacci sequence, and the sequence beginning with 2 and k is denoted (li ) and is called the general Lucas sequence. Matrices with Fibonacci and Lucas entries were studied in [1],[3],[4],[5],[6], [8],[9], and [10],. Motivated by these works, in [5] it was shown that, when n is even, the orthogonal projection onto Rn (k, 1) 0 0
AMS 2000 Classification numbers: Primary: 15B36 Secondary: 65F35 Key Words: Orthogonal projection, Fibonacci numbers, Fibonacci identities.
1
is the Hankel matrix k fn
f−n+1 f−n+2 f−n+2 f−n+3 .. .. . . f−1 f0 f0 f1
... ... .. .
f−1 f0 .. .
f0 f1 .. .
. . . fn−3 fn−2 . . . fn−2 fn−1
.
(A proof of this in the special case of Rn (1, 1) may now be found in [2].) We begin with a new proof of this fact. Let E denote the 2 × 2 matrix � � 0 1 E= . 1 k It follows that E
m
=
�
fm−1 fm fm fm+1
�
,
for all integers m. Lemma 1 For any integer t and any nonnegative integer s, we have s �
E t+4i =
i=0
Proof.
f2(s+1) t+2s E . k
It suffices to prove the scalar identity s �
ft+4i =
i=0
f2(s+1) ft+2s , k
and we proceed via induction on s. Since f2 = k, the identity holds when s = 0, so we assume the induction hypothesis s−1 � i=0
ft+4i =
f2s f . k t+2(s−1)
� Substituting this in for the first s summands of si=0 ft+4i , we see that the equality holds s � f2(s+1) ft+4i = ft+2s k i=0
2
if and only if f2s ft+2s−2 + kft+4s = f2(s+1) ft+2s . The equality E t+4s = E t+2s E 2s yields the identity ft+4s = f2s ft+2s−1 + f2s+1 ft+2s from which we derive f2s+2 ft+2s = f2s ft+2s + kf2s+1 ft+2s = f2s ft+2s−2 + kf2s ft+2s−1 + kf2s+1 ft+2s = f2s ft+2s−2 + kft+4s � Let Hf denote the n × n matrix
Hf =
f−n+1 f−n+2 f−n+2 f−n+3 .. .. . . f−1 f0 f0 f1
... ... .. .
f−1 f0 .. .
f0 f1 .. .
. . . fn−3 fn−2 . . . fn−2 fn−1
.
We will call Hf the central Fibonacci Hankel matrix. Replacing fi with li yields Hl , the central Lucas Hankel matrix. Theorem 2 For even n, the matrix R(k, 1).
k fn Hf
is the orthogonal projection onto
Proof. It is clear that fkn Hf is selfadjoint, and its range is R(k, 1), so we need only prove fkn Hf is idempotent, which is equivalent to proving Hf2 =
fn Hf . k
Write n = 2m and notice that −n+2 E E −n+4 . . . E −2 E0 E −n+4 E −n+6 . . . E 0 E2 . . . .. .. .. .. .. Hf = . . 0 n−6 n−4 E −2 E ... E E E0 E2 . . . E n−4 E n−2 3
� �m . = E 2(i+j)−2(m+1) i,j=1
Using our lemma, it follows that ��m � 2(i+r)−2(m+1) E 2(r+j)−2(m+1) m Hf2 = r=1 E i,j=1 � ��m 2(i+j)−4m+4(r−1) m = E ��r=1 �m i,j=1 m−1 2(i+j)−4m+4s = s=0 E i,j=1 f2m � 2(i+j)−4m+2(m−1) �m = k E i,j=1 � �m = fkn E 2(i+j)−2(m+1) i,j=1 = fkn Hf � The theme of this paper is the interplay between the linear algebraic equations of orthogonality and Fibonacci identities. To illustrate with an example, observe that the idempotence of Hf leads immediately to the identity n−1 k k2 � fi+j−n+1 = 2 fi+k−n+1 fk+j−n+1 , fn fn k=0
for all even n and −n + 1 ≤ i, j ≤ n − 1. This proves the following.
Corollary 3 Assume that n is even and −n + 1 ≤ i, j ≤ n − 1. We have k
n−1 �
fi−u fj−u = fn fi+j−n+1 .
u=0
Assume that n is even and s = (si )n−1 i=0 ∈ R(k, 1), which we will think ∗ of as a column vector. We then let s denote the transpose of s, which is a row vector. We define s⊥ ≡ (−sn−1 , sn−2 , −sn−3 . . . , −s1 , s0 )∗ = ((−1)i+1 sn−i−1 )n−1 i=0 , and observe that this is also an element of R(k, 1), and {s, s⊥ } forms an orthogonal basis of R(k, 1) for each non-zero s ∈ R(k, 1). Nor⊥ s malizing, and denoting u = ||s|| and v = ||ss⊥ || , then, by elementary linear algebra, the orthogonal projection onto R(k, 1) is the matrix uu∗ + vv ∗ . Equating this matrix with fkn Hf establishes the following identity. (Note that ||s||2 = ||s⊥ ||2 = k1 (sn−1 sn − s0 s−1 ).) 4
Theorem 4 Assume that n is even. Then, for all 0 ≤ i, j ≤ n − 1 we have k k (si sj + (−1)i+j sn−i−1 sn−j−1 ) = fi+j−n+1 sn−1 sn − s0 s−1 fn i.e. we have (si sj + (−1)i+j sn−i−1 sn−j−1 )fn = (sn−1 sn − s0 s−1 )fi+j−n+1 . The preceding identity has many famous identities as special cases. Example 1 Take s to be the Fibonacci sequence itself, and let i = j = 0. The result is that k k f2 = f−n+1 fn−1 fn n−1 fn for all even n. In other words, fq = f−q for odd subscripts q. Example 2 Take s to be the Fibonacci sequence again, and this time let i = 1 and j = 0. The result is that k k (−fn−2 fn−1 ) = f−n+2 fn−1 fn fn for all even n. In other words, fq = −f−q for all even subscripts q. Example 3 With s the Fibonacci sequence again, let i = n/2 and j = n/2 to obtain the identity 2 fq2 + fq−1 = f2q−1 for all positive integers q. Let φ denote the positive solution of the equation x2 − kx − 1 = 0, which we call the k-golden ratio. We let ψ denote the negative solution. Verify that φψ = −1, from which we see that when s = (1, φ, . . . , φn−1 )∗ and t = (1, ψ, . . . , ψ n−1 )∗ , we have another orthogonal basis {s, t} of R(k, 1). Of course, in this case t is just a multiple of s⊥ . We equate the entries of the matrix 1 1 ss∗ + tt∗ 2 ||s|| ||t||2 with those of the projection in Theorem 2 to obtain the following. 5
Theorem 5 With n even we have 1 − φ2 1 − ψ2 k i+j φ + ψ i+j = fi+j−n+1 1 − φ2n 1 − ψ 2n fn for all 0 ≤ i, j ≤ n − 1. Alternatively, this can be written φi+j+1 ψ i+j+1 k + = fi+j−n+1 2n 2n φ −1 ψ −1 fn Example 4 Letting j = n − 1 and i = 1 in the equation above yields a slightly convoluted formulation of the Binet equation: 1 φn+1 ψ n+1 = 2n + 2n . fn φ −1 ψ −1 Using the fact that n is even and φψ = −1, one calculates � n+1 � φ ψ n+1 + (φn − ψ n ) = φ − ψ, φ2n − 1 ψ 2n − 1 from which the familiar Binet formula appears as 1 n (φ − ψ n ) = φ − ψ. fn
2
The Lucas Projection
In this section we consider odd values of n, with n = 2m + 1. We now let f and l denote the central Fibonacci and Lucas sequences, which are the transposes of (f−m , . . . , f0 , . . . , fm ) and (l−m , . . . , l0 , . . . , lm ), respectively. A computation reveals that ||f ||2 =
2fm fm+1 2lm lm+1 and ||l||2 = . k k
Recalling that f0 = 0, f−i = (−1)i+1 fi , and l−i = (−1)i li , you see that m �
i=−m
f i li = −
m �
f i li + f 0 l0 +
i=1
m � i=1
6
fi li = 0,
so the pair {f , l} forms an orthogonal basis of R(k, 1). One expression for the orthogonal projection onto R(k, 1) is then given by k k ff∗ + ll∗ . 2fm fm+1 2lm lm+1 Our second expression for this projection gives it as a linear combination of the two Lucas matrices Hl and ll∗ . Theorem 6 The orthogonal projection onto R(k, 1) is the matrix � � k k(−1)m+1 ∗ ll + Hl . (4 + k 2 )fm fm+1 lm lm+1 Proof. The Hankel matrix Hl has eigenvectors f and l with correspond2 ing eigenvalues (4+k )fkm fm+1 and lm lkm+1 (respectively). It follows that (4+k2 )fm fm+1 k ∗ k 2fm fm+1 ff 2 (4+k ) ∗ ff + 12 ll∗ 2
Hl = =
+
lm lm+1 k ∗ k 2lm lm+1 ll
,
and, letting P denote our projection, we compute P
=
k ∗ 2fm fm+1 ff
=
k 2 2fm fm+1 ( (4+k2 ) Hl
+
k ∗ 2lm lm+1 ll
−
1 ll∗ ) (4+k2 )
+
k ∗ 2lm lm+1 ll
.
Simplifying the above equation while using the identity lm lm+1 − (4 + k 2 )fm fm+1 = 2(−1)m k gives the expression k P = 2 (4 + k )fm fm+1
�
� k(−1)m+1 ∗ ll + Hl . lm lm+1 �
Corollary 7 If n is odd with n = 2m + 1, then � � fi fj li lj 2 k(−1)m+1 + = li lj + li+j fm fm+1 lm lm+1 (4 + k 2 )fm fm+1 lm lm+1 for all −m ≤ i, j ≤ m. 7
Example 5 The identity lm lm+1 − (4 + k 2 )fm fm+1 = 2(−1)m k used in the derivation of the above is recovered when we let i = j = 0, when i = j = 1 and again when we let one of the indices equal zero, and the other equal m. This suggests to us that this identity is intrinsically an orthogonality relation. Example 6 the identity
Let i = j = m in the equation above and simplify to obtain 2 2 (4 + k 2 )fm + lm = 2 l2m .
There is nothing sacred about the Lucas numbers in the above expression for our projection, aside from the fact that the resulting formula is the simplest we have found. Indeed, if s is any element of R(k, 1) and Hs is the corresponding central Hankel matrix, then Hs is diagonalizable with eigenvectors in R(k, 1) corresponding to non-zero eigenvalues. If s is rank two (as most of the Hs are), then choosing an orthonormal eigenbasis {u, v} of R(k, 1) gives us an expression for both Hs and P : P = uu∗ + vv∗ and Hs = α uu∗ + β vv∗ . One can then choose to solve for one of the orthogonal projections in the expression for Hs and substitute into the expression for P , giving P in terms of Hs and the other projection, for example 1 β P = ( Hs − vv∗ ) + vv∗ . α α Let us express P in terms of the central Fibonacci Hankel matrix Hf to illustrate this method. Recall that Hf is defined by Hf = (fi+j )m ij=−m . fm fm+1 l and Hf (l) = lm lkm+1 f . It is convenient k 2 and b = lm lkm+1 = ||l|| 2 , from which we see that
Verify that Hf (f ) = a=
fm fm+1 k
=
||f ||2 2
√ √ √ Hf ( abf + al) = ab( abf + al),
8
to write
and
√ √ √ Hf ( abf − al) = − ab( abf − al).
Since Hf is symmetric, it is reassuring to check that the eigenvector √ is orthogonal to abf − al. As f is orthogonal to l, we get
√
abf +al
√ √ ||f ||4 ||l||2 || abf + al||2 = || abf − al||2 = ab||f ||2 + a2 ||l||2 = . 2 √ √ √ √ √ +al) 2( abf −al) We let u = 2(||f ||abf and v = , with α = ab = ||f ||||l|| and 2 ||l|| 2 ||f ||2 ||l|| √ β = − ab =
−||f ||||l|| , 2
in the equation above to prove the following.
Theorem 8 Assume that n is odd, and n = 2m + 1. The orthogonal projection onto R(k, 1) is the matrix (pij ), where pij
=
2 ||f ||||l|| fi+j
+ 2vi vj
=
2 ||f ||||l|| fi+j
+
1 ff ||f ||2 i j
+
1 ll ||l||2 i j
−
1 ||f ||||l|| (fi lj
+ li f j )
and i, j = −m, . . . , m. Example 7
For all i, j = −m, . . . , m one has 2fi+j = fi lj + li fj ,
which can be seen following from Theorem 8 by remembering that pij = and hence pij = so
2 ||f ||||l|| fi+j
=
1 1 fi fj + li lj , 2 ||f || ||l||2
2 1 fi+j + pij − (fi lj + li fj ), ||f ||||l|| ||f ||||l||
1 ||f ||||l|| (fi lj
+ li fj ).
References [1] M. Bahsi and S. Solak, On the spectral norms of Hankel matrices with Fibonacci and Lucas numbers, Selcuk J. Appl. Math., 12 no. 1 (2011), 71-76. 9
[2] O. Bretscher, Linear Algebra with Applications, Fifth Edition, Pearson Education, New Jersey. [3] D. Bozkurt and M. Akbulak, On the norms of Toeplitz matrices involving Fibonacci and Lucas numbers, Hacet. J. Math. Stat., 37 (2)(2008), 89-95. [4] J. Dixon, B. Mathes, and D. Wheeler, An application of matricial Fibonacci identities to the computation of spectral norms, Rocky Mt. J. Math., to appear. [5] E. Dupree and B. Mathes, Singular values of k-Fibonacci and k-Lucas Hankel Matrices, Int. J. Contemp. Math. Sciences, to appear. [6] A. Ipek, On the spectral norms of circulant matrices with classical Fibonacci and Lucas numbers entries, Appl. Math. Comput., 217 (2011), 6011-6012. [7] D. Kalman and R. Mena, The Fibonacci numbers – exposed, Mathematics Magazine, vol. 76, no. 3 (2003). [8] S. Shen, On the norms of Toeplitz matrices involving k-Fibonacci and k-Lucas numbers, Int. J. Contemp. Math. Sciences, 7 (8) (2012), 363368. [9] S. Shen, J. Cen, On the spectral norms of r-circulant matrices with the k-Fibonacci and k-Lucas numbers, Int. J. Contemp. Math. Sciences, 5 (12) (2010), 569-578. [10] S. Solak, On the norms of circulant matrices with the Fibonacci and Lucas numbers, Appl. Math. Comput., 160 (2005), 125-132.
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