The Fibonacci Identities of Orthogonality

Transcription

1 The Fibonacci Identities of Orthogonality Kyle Hawins, Ursula Hebert-Johnson and Ben Mathes January 14, 015 Abstract In even dimensions, the orthogonal projection onto the two dimensional space of second order recurrence sequences is particularly nice: it is a scaled Hanel 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 ran one Lucas perturbation of a scaled Lucas Hanel matrix, from which more Fibonacci identities are derived 1 The Fibonacci projection Let R n denote the n dimensional vector space of all real n-tuples, and let R n (, l) denote the vector subspace consisting of the (s i ) R n for which s i s i 1 + ls i, for i 3,,n (See [7] for a delightful article on general Fibonacci sequences) The element of R n (, l) whose first two coordinates are 0 and 1 (respectively) will be denoted (f i ) and is called the general Fibonacci sequence, and the sequence beginning with and is denoted (l i ) 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 wors, in [5] it was shown that, when n is even, the orthogonal projection onto R n (, 1) 0 AMS 000 Classification numbers: Primary: 15B36 Secondary: 65F35 0 Key Words: Orthogonal projection, Fibonacci numbers, Fibonacci identities 1

2 is the Hanel matrix f n+1 f n+ f 1 f 0 f n+ f n+3 f 0 f 1 f 1 f 0 3 f 0 f 1 1 (A proof of this in the special case of R n (1, 1) may now be found in []) We begin with a new proof of this fact Let E denote the matrix E It follows that for all integers m E m fm 1 f m f m f m+1, Lemma 1 For any integer t and any nonnegative integer s, we have s i0 E t+4i f (s+1) E t+s Proof It suffices to prove the scalar identity s i0 f t+4i f (s+1) f t+s, and we proceed via induction on s Since f, the identity holds when s 0, so we assume the induction hypothesis s 1 f t+4i f s f t+(s 1) i0 Substituting this in for the first s summands of s i0 f t+4i, we see that the equality holds s f t+4i f (s+1) f t+s i0

3 if and only if The equality f s f t+s + f t+4s f (s+1) f t+s E t+4s E t+s E s yields the identity f t+4s f s f t+s 1 + f s+1 f t+s from which we derive f s+ f t+s f s f t+s + f s+1 f t+s f s f t+s + f s f t+s 1 + f s+1 f t+s f s f t+s + f t+4s Let H f denote the n n matrix f n+1 f n+ f 1 f 0 f n+ f n+3 f 0 f 1 H f f 1 f 0 3 f 0 f 1 1 We will call H f the central Fibonacci Hanel matrix Replacing f i with l i yields H l,thecentral Lucas Hanel matrix Theorem For even n, thematrix H f is the orthogonal projection onto R(, 1) Proof It is clear that H f is selfadjoint, and its range is R(, 1), so we need only prove H f is idempotent, which is equivalent to proving H f H f Write n m and notice that E n+ E n+4 E E 0 E n+4 E n+6 E 0 E H f E E 0 E n 6 E n 4 E 0 E E n 4 E n E (i+j) (m+1) m i,j1 3

4 Using our lemma, it follows that Hf m r1 E(i+r) (m+1) E (r+j) (m+1) m i,j1 m r1 E(i+j) 4m+4(r 1) m i,j1 m 1 f m fn fn H f s0 E(i+j) 4m+4s m i,j1 E (i+j) 4m+(m 1) m E (i+j) (m+1) m i,j1 i,j1 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 H f leads immediately to the identity f i+j n+1 n 1 fn f i+ n+1 f +j n+1, for all even n and n +1 i, j n 1 This proves the following 0 Corollary 3 Assume that n is even and n +1 i, j n 1 We have f i u f j u f i+j n+1 n 1 u0 Assume that n is even and s (s i ) n 1 i0 R(, 1), which we will thin of as a column vector We then let s denote the transpose of s, whichisa row vector We define s ( s n 1,s n, s n 3, s 1,s 0 ) (( 1) i+1 s n i 1 ) n 1 i0, and observe that this is also an element of R(, 1), and {s, s } forms an orthogonal basis of R(, 1) for each non-zero s R(, 1) Normalizing, and denoting u s s and v s, then, by elementary linear s algebra, the orthogonal projection onto R(, 1) is the matrix uu + vv Equating this matrix with H f establishes the following identity (Note that s s 1 (s n 1s n s 0 s 1 )) 4

5 Theorem 4 Assume that n is even Then, for all 0 i, j n 1 we have ie we have s n 1 s n s 0 s 1 (s i s j +( 1) i+j s n i 1 s n j 1 ) f i+j n+1 (s i s j +( 1) i+j s n i 1 s n j 1 ) (s n 1 s n s 0 s 1 )f i+j n+1 The preceding identity has many famous identities as special cases Example 1 Tae s to be the Fibonacci sequence itself, and let i j 0 The result is that 1 f n 1 f n+1 for all even n In other words, f q f q for odd subscripts q Example Tae s to be the Fibonacci sequence again, and this time let i 1 and j 0 The result is that 1 ( 1 ) f n+ for all even n In other words, f q f q for all even subscripts q Example 3 With s the Fibonacci sequence again, let i n/ and j n/ to obtain the identity f q + f q 1 f q 1 for all positive integers q Let φ denote the positive solution of the equation x x 10, which we call the -golden ratio We let ψ denote the negative solution Verify that φψ 1, from which we see that when s (1,φ,,φ n 1 ) and t (1,ψ,,ψ n 1 ), we have another orthogonal basis {s, t} of R(, 1) Of course, in this case t is just a multiple of s We equate the entries of the matrix 1 s ss + 1 t tt with those of the projection in Theorem to obtain the following 5

6 Theorem 5 With n even we have 1 φ 1 φ n φi+j + 1 ψ 1 ψ n ψi+j f i+j n+1 for all 0 i, j n 1 Alternatively, this can be written φ i+j+1 φ n 1 + ψi+j+1 ψ n 1 f i+j n+1 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 φ n 1 + ψn+1 ψ n 1 Using the fact that n is even and φψ 1, one calculates φ n+1 φ n 1 + ψn+1 ψ n (φ n ψ n )φ ψ, 1 from which the familiar Binet formula appears as 1 (φ n ψ n )φ ψ The Lucas Projection In this section we consider odd values o, withn m + 1 We now let f and l denote the central Fibonacci and Lucas sequences, which are the transposes of (f m,,f 0,,f m ) and (l m,,l 0,,l m ), respectively A computation reveals that f f mf m+1 and l l ml m+1 Recalling that f 0 0, f i ( 1) i+1 f i, and l i ( 1) i l i, you see that m i m f i l i m f i l i + f 0 l 0 + i1 6 m f i l i 0, i1

7 so the pair {f, l} forms an orthogonal basis of R(, 1) One expression for the orthogonal projection onto R(, 1) is then given by ff + ll f m f m+1 l m l m+1 Our second expression for this projection gives it as a linear combination of the two Lucas matrices H l and ll Theorem 6 The orthogonal projection onto R(, 1) is the matrix ( 1) m+1 (4 + ll + H l )f m f m+1 l m l m+1 Proof The Hanel matrix H l has eigenvectors f and l with corresponding eigenvalues (4+ )f mf m+1 and lml m+1 (respectively) It follows that H l (4+ )f mf m+1 f mf m+1 ff + lml m+1 l ml m+1 ll (4+ ) ff + 1 ll, and, letting P denote our projection, we compute P f mf m+1 ff + l ml m+1 ll f mf m+1 ( (4+ ) H l 1 (4+ ) ll )+ l ml m+1 ll Simplifying the above equation while using the identity l m l m+1 (4 + )f m f m+1 ( 1) m gives the expression ( 1) m+1 P (4 + ll + H l )f m f m+1 l m l m+1 Corollary 7 I is odd with n m +1, then f i f j + l il j ( 1) m+1 f m f m+1 l m l m+1 (4 + l i l j + l i+j )f m f m+1 l m l m+1 for all m i, j m 7

8 Example 5 The identity l m l m+1 (4 + )f m f m+1 ( 1) m 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 (4 + )f m + l m l m 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(, 1) and H s is the corresponding central Hanel matrix, then H s is diagonalizable with eigenvectors in R(, 1) corresponding to non-zero eigenvalues If s is ran two (as most of the H s are), then choosing an orthonormal eigenbasis {u, v} of R(, 1) gives us an expression for both H s and P : P uu + vv and H s α uu + β vv One can then choose to solve for one of the orthogonal projections in the expression for H s and substitute into the expression for P, giving P in terms of H s and the other projection, for example P ( 1 α H s β α vv )+vv Let us express P in terms of the central Fibonacci Hanel matrix H f illustrate this method Recall that H f is defined by to H f (f i+j ) m ij m Verify that H f (f) fmf m+1 l and H f (l) lml m+1 f It is convenient to write a fmf m+1 f and b lml m+1 l, from which we see that H f ( abf + al) ab( abf + al), 8

9 and H f ( abf al) ab( abf al) Since H f is symmetric, it is reassuring to chec that the eigenvector abf +al is orthogonal to abf al As f is orthogonal to l, we get abf + al abf al ab f + a l f 4 l We let u ( abf+al) f l and v ( abf al),withα ab f l f l and β ab f l, in the equation above to prove the following Theorem 8 Assume that n is odd, and n m +1 The orthogonal projection onto R(, 1) is the matrix (p ij ), where p ij f l f i+j +v i v j and i, j m,, m f l f i+j + 1 f f i f j + 1 l l i l j 1 f l (f il j + l i f j ) Example 7 For all i, j m,, m one has f i+j f i l j + l i f j, which can be seen following from Theorem 8 by remembering that and hence p ij so f l f i+j 1 f l (f il j + l i f j ) References p ij 1 f f if j + 1 l l il j, f l f 1 i+j + p ij f l (f il j + l i f j ), [1] M Bahsi and S Sola, On the spectral norms of Hanel matrices with Fibonacci and Lucas numbers, Selcu J Appl Math, 1 no 1 (011),

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