Fibonacci Identities through Matrix Method

Thotsaporn “Aek” ThanatipanondaMahidol University International College

December 25, 2018

1 Introduction

We unified all the generalized Fibonacci numbers in the past by using the matrixmethod. This method is by no mean new, it has been used extensive by many authorand was a chapter in the book Fibonacci and Lucas Number with Applications byKoshy, [11]. However, many papers did not use this matrix method to show theiridentities. It is common that we have a simpler way to show the old identities usingthe properties of the matrix. We also successfully find the new generalization of thenumbers in the past. The outline of the paper are:

Section 2: Apply the matrix method to the known identities.

Section 2.1: On Fibonacci identities from couple papers, Harman [7], Demirturk andKesking [4], Benjamin and Quinn [1] and Bloom [3] .

Section 2.2: On identities of Gaussian Fibonacci numbers

GFn = GFn−1 +GFn−2,

where GF0 = a+ bi and GF1 = c+ di from Jordon [10] .

Section 2.3: On identities of Fibonacci Quaternions

Qn = Fn + iFn+1 + jFn+2 + kFn+3

by Iyer [8] and Halici [6].

Section 3: We consider the matrix of the form[Tn+1,m Tn,mTn,m Tn−1,m

]=

[1 11 0

]n [a bb a− b

]m.

Case a = 1, b = i is the work of Berzsenyi, Gaussian Fibonacci Numbers, [2].Case a = 1, b = 2 is the generalization of Iyer, [8], we mentioned in section 2.We investigated the results of the general form in section 3.3 and 3.4.

Section 4: We deal with the numbers arise from the more general form of matrixmultiplication, i.e.[

Tn+1,m,p Tn,m,p

Tn,m,p Tn−1,m,p

]=

[1 11 0

]n [a bb a− b

]m [A BB A−B

]p.

Section 5: We consider the numbers that arise from multiplication of the two 3-by-3matrices, i.e. tribonacci numbers, trucas numbers. We also consider the identitiesfrom general form in section 5.3.

Section 6: Another direction of the generalization of Complex Fibonacci numbers byHarman, [7]:

G(n+ 2,m) = G(n+ 1,m) +G(n,m),

G(n,m+ 2) = G(n,m+ 1) +G(n,m),

where G(0, 0) = 0, G(1, 0) = 1, G(0, 1) = i, G(1, 1) = 1 + i.

Then we generalize this idea.

2 One Matrix Multiply, P nR

We will use the matrix to define several known generalized Fibonacci numbers andprove their identities.

In this section, let a 2-by-2 matrix P and R be in the form:

P :=

[1 11 0

]and R :=

[a bb a− b

].

We consider the matrix of the form P nR.Note that P n = FnP + Fn−1I, i.e. P 2 = P + I.

2.1 Fibonacci numbers

We define Fibonacci numbers, Fn, by the following matrix:[Fn+1 Fn

Fn Fn−1

]:=

[1 11 0

]nNote that the matrix is well defined.

2.1.1 Identities

The matrix form gives rise to the more compact form of “many” known identities.

2

1. F2 = 1, F1 = 1 and F0 = 0. By substitute n = 1 in the definition.

2. Fn+2 = Fn+1 + Fn.Using the fact that P 2 = P + I then read off from the entry in of matrixP n+2 = P n+1 + P n.

3.n∑

j=1

Fj = Fn+2 − 1.

By the fact that∑n

j=1 Pj =

P (I − P n)

I − P=P (I − P n)

−P−1= P n+2 − P 2.

4.n∑

j=1

F2j−1 = F2n.

By the fact that∑n

j=1 P2j−1 =

P (I − P 2n)

I − P 2= P 2n − I.

i.e. F7+F5+F3+F1 = F8. From the matrix identity P 7+P 5+P 3+P+I = P 8.

5.n∑

j=1

F4j =3F4n+5 − 4F4n+4 − 3

5. Proof can be done in similar fashion.

6. Fn+1Fm+1 + FnFm = Fn+m+1.

Consider the top right entry of the following:[Fn+1 Fn

Fn Fn−1

]·[Fm+2 Fm+1

Fm+1 Fm

]= P nPm+1 = P n+m+1.

Special case m = n− 1:

F2n = Fn+1Fn + FnFn−1 = Fn(Fn+1 + Fn−1) = F 2n+1 − F 2

n−1.

7.∑k

j=1 F4j−2 = F 22k and

∑kj=1 F4j = F 2

2k+1

This can be shown by substituting n = 2j − 1 and n = 2j respectively inthe previous result then do the telescoping sum.

8. (Eq 3.6 of Harman, [7])

Fn+2k+1Fm+2k − Fn+2kFm+2k+1 = Fn+1Fm − FnFm+1.

This can be shown by using the definition[Fn+2k+1 Fm+2k+1

Fn+2k Fm+2k

]=

[1 11 0

]2k·[Fn+1 Fm+1

Fn Fm

]Then take the determinant both sides.

3

9. (Theorem 6 of Demirturk and Kesking, [4])

Fn+kFm+k − FkFn+m+k = (−1)kFnFm,

This can be shown by using the definition[Fn+k Fk

Fn+m+k Fm+k

]=

[Fk+1 Fk

Fm+k+1 Fm+k

]·[

Fn 0Fn−1 1

]=

[1 0

Fm+1 Fm

]·[Fk+1 Fk

Fk Fk−1

]·[

Fn 0Fn−1 1

]=

[1 0

Fm+1 Fm

]·[

1 11 0

]k·[

Fn 0Fn−1 1

]Then take the determinant both sides.

10. (Theorem 6 of Demirturk and Kesking, [4])

Fn+kFm − FkFn+m = (−1)kFnFm−k,

Substitute m by m− k in the previous result.

11. (Theorem 5.1 of Demirturk and Kesking, [4])

Ln+m+kLk − Ln+kLm+k = 5(−1)kFnFm,

This can be shown by using the definition[Ln+m+k Ln+k

Lm+k Lk

]=

[Ln+k+1 Ln+k

Lk+1 Lk

]·[

Fm 0Fm−1 1

]=

[Fn+1 Fn

1 0

]·[Lk+1 Lk

Lk Lk−1

]·[

Fm 0Fm−1 1

]=

[Fn+1 Fn

1 0

]·[

1 11 0

]k [1 22 −1

]·[

Fm 0Fm−1 1

]Then take the determinant both sides.

12. (Theorem 5.2 of Demirturk and Kesking, [4]. substitute m with m+ k)

Fn+m+kLk − Ln+kFm+k = (−1)kLmFn,

This can be shown by using the definition[Fn+m+k Ln+k

Fm+k Lk

]=

[Fn Fn−10 1

]·[Fm+k+1 Lk+1

Fm+k Lk

]=

[Fn Fn−10 1

]·[Fm+k+1 Fm+k

Fm+k Fm+k−1

]·[

1 L−m+1

0 L−m

]=

[Fn Fn−10 1

]·[

1 11 0

]m+k

·[

1 L−m+1

0 L−m

].

Then take the determinant both sides.

4

13. (Finite sum, Theorem 2.1.2 of Demirturk, [5])

n∑j=0

Fmj+k =Fk − Fmn+m+k + (−1)m(Fmn+k − Fk−m)

1 + (−1)m − Lm

.

Consider the (2,1) entry of P k + Pm+k + P 2m+k + · · ·+ Pmn+k :

P k + Pm+k + P 2m+k + · · ·+ Pmn+k =P k(P (n+1)m − I)

Pm − I

=P k(P (n+1)m − I)((−1)mP−m − I)

(Fm+1Fm−1 − (Fm+1 + Fm−1) + 1− F 2m)

=P k − P nm+m+k + (−1)m(P nm+k − P k−m)

(−1)m + 1− Lm

.

14. (Finite sum, Theorem 2.1.1 of Demirturk, [5])

n∑j=0

Lmj+k =Lk − Lmn+m+k + (−1)m(Lmn+k − Lk−m)

1 + (−1)m − Lm

.

The proof is similar to the previous.

15. (Binomial Sum, Lemma 2.1 of Demirturk and Kesking, [4])

n∑j=0

(n

j

)ajbn−jFj+k = (−1)k+1

n∑j=0

(n

j

)(−a)j(a+ b)n−jFj−k.

Consider the upper right hand corner of P k(aP + bI)n:

On one hand: P k(aP + bI)n =∑j

(n

j

)ajbn−jP j+k.

On the other hand:

P k(aP + bI)n = P k((a+ b)I + aP−1)n =∑j

(n

j

)aj(a+ b)n−jP−j+k.

16. (Eq 3.8 of Harman, [7])

2k∑j=1

Fn+jFm+j = Fn+2k+1Fm+2k − Fn+1Fm.

We can show this by induction on k.The base case when k = 0 is true.Induction step: let S(k) =

∑2kj=1 Fn+jFm+j.

S(k)− S(k − 1) = Fn+2kFm+2k + Fn+2k−1Fm+2k−1

= (Fn+2k+1 − Fn+2k−1)Fm+2k + Fn+2k−1(Fm+2k − Fm+2k−2)

= Fn+2k+1Fm+2k − Fn+2k−1Fm+2k−2.

5

17. (Eq 3.9 of Harman, [7])

2k+1∑j=1

Fn+jFm+j = Fn+2k+2Fm+2k+1 − FnFm+1

This could be done similarly to 3.8.

18. (Eq 3.7 of Harman, [7])

2k∑j=1

Fn+jFm+j =1

2(Fn+2kFm+2k+1 + Fn+2k+1Fm+2k − Fn+1Fm − FnFm+1) .

This could also be done similarly to 3.8.

2.1.2 Sum of Products of Fibonacci

This section we consider∑

i+j=n FiFj,∑

i+j+k=n FiFjFk, ... which was asked byBloom, [3], in 1996.

Definition. Let hi,j be defined ash0,j = 0 if j ≥ 0.hj,j = 1 if j ≥ 1.hi,j = 1 if i > j.hi,j = hi,j−1 + hi,j−2 + hi−1,j−1 if i ≥ 1 and j ≥ 2.

Theorem 2.1.1.h1,n = Fn.

Proof. It is obvious from the definition.

Theorem 2.1.2.h2,n =

∑i,j>0i+j=n

FiFj.

Proof. Let H2,n :=

[h2,n+1 h2,nh2,n h2,n−1

]and R :=

[1 00 0

].

By definition of hi,j, we have that

H2,n+1 = H2,nP + P nR

= (H2,n−1P + P n−1R)P + P nR

= H2,n−1P2 + P n−1RP + P nR

= H2,n−2P3 + P n−2RP 2 + P n−1RP + P nR

= · · ·

=n∑

k=0

P n−kRP k.

6

Then we note that

[Fn−k+1 Fn−kFn−k Fn−k−1

] [1 00 0

] [Fk+1 Fk

Fk Fk−1

]=

[Fn−k+1 0Fn−k 0

] [Fk+1 Fk

Fk Fk−1

]=

[Fn−k+1Fk+1 Fn−k+1Fk

Fn−kFk+1 Fn−kFk

].

Thereforen∑

k=0

P n−kRP k =

[ ∑nk=0 Fn+1−kFk+1

∑nk=0 Fn+1−kFk∑n

k=0 Fn−kFk+1

∑nk=0 Fn−kFk

]

Theorem 2.1.3.h3,n =

∑i,j,k>0

i+j+k=n

FiFjFk.

Proof. Similar to the above theorem

H3,n+1 = H3,nP +H2,nR

= (H3,n−1P +H2,n−1R)P +H2,nR

= H3,n−1P2 +H2,n−1RP +H2,nR

= H3,n−2P3 +H2,n−2RP

2 +H2,n−1RP +H2,nR

= · · ·

=n∑

k=0

H2,n−kRPk.

Then note that

H2,n−kRPk =

[ ∑n−kj=0 Fn−k−jFj+1

∑n−kj=0 Fn−k−jFj∑n−k

j=0 Fn−1−k−jFj+1

∑n−kj=0 Fn−1−k−jFj

][Fk+1 Fk

0 0

]

=

[ ∑n+1−kj=1 Fn+1−k−jFjFk+1

∑n+1−kj=1 Fn+1−k−jFjFk∑n+1−k

j=1 Fn−k−jFjFk+1

∑n+1−kj=1 Fn−k−jFjFk

]Therefore

n∑k=0

H2,n−kRPk =

[ ∑nk=0

∑n+1−kj=1 Fn+1−k−jFjFk+1

∑nk=0

∑n+1−kj=1 Fn+1−k−jFjFk∑n

k=0

∑n+1−kj=1 Fn−k−jFjFk+1

∑nk=0

∑n+1−kj=1 Fn−k−jFjFk

]

=

[ ∑n+1k=1

∑n+2−kj=1 Fn+2−k−jFjFk

∑nk=1

∑n+1−kj=1 Fn+1−k−jFjFk∑n+1

k=1

∑n+2−kj=1 Fn+1−k−jFjFk

∑nk=1

∑n+1−kj=1 Fn−k−jFjFk

]

=

[ ∑i,j,k=n+2 FiFjFk

∑i,j,k=n+1 FiFjFk∑

i,j,k=n+1 FiFjFk

∑i,j,k=n FiFjFk

]We then conclude the result.

7

The main theorem can be done by follow these proofs recursively.

Theorem 2.1.4.

hm,n =∑ai>0∑ai=n

m∏i=1

Fai .

2.1.3 Binomial Sum of Fibonacci and Lucas Numbers

The following results are Lemma 3 in Demirturk and Kesking, [4].

Let Rb be a 2-by-2 matrix in the form Rb :=

[1 bb 1− b

].

Proposition 2.1.5.

Fmn+k =∑j

(n

j

)F jmF

n−jm−1Fj+k,

and

Lmn+k =∑j

(n

j

)F jmF

n−jm−1Lj+k.

Proof.

Pmn+kRb = (Pm)nP kRb = (FmP + Fm−1I)nP kRb =

(∑j

(n

j

)F jmF

n−jm−1P

j

)P kRb

=∑j

(n

j

)F jmF

n−jm−1P

j+kRb.

Then set b = 0 for the first equation and b = 2 for the second equation.

The followings are from the book, Proofs that really count by Benjamin and Quinn,page 144, [1].

Here we let R :=

[1 22 −1

].

1. (V69)

5nF2n−1 =2n∑j=0

(2n

j

)F2j−1.

Consider R2nP 2n−1 = P−1(PR)2n = P−1(P 2 + I)2n = P−1∑j

(2n

j

)P 2j.

2. (V73)

5n−1L2n =2n∑j=0

(2n

j

)F 2j , n ≥ 1.

(I can’t proof using matrix method.)

8

2.2 Gaussian Fibonacci Numbers

Let a 2-by-2 matrix R be in the form:

R :=

[1 ii 1− i

].

We define Fibonacci numbers, GFn, by the following matrix:[GFn+1 GFn

GFn GFn−1

]:=

[1 11 0

]n·[

1 ii 1− i

].

Note:

1. The matrix is well defined.

2. It is obvious that GFn+1 = GFn + GFn−1 for n ≥ 1. Moreover P and Rcommute.

3. GFn = Fn + iFn−1 by comparing the bottom left entry of P nR.

4. PR2 =

[1 + 2i 0

0 1 + 2i

].

2.2.1 Identities

The matrix form gives rise to the more compact form of “many” known identities.Most of the identities are from the paper by Jordan, [10].

1. GF1 = 1 and GF0 = i.

2. GFn = Fn + iFn−1.

3.∑n

j=0GFj = GFn+2 − 1.

From 1 + P + P 2 + ...+ P n =P n+1 − 1

P − 1= P n+2 − P.

4. GFn+1GFn−1 −GF 2n = (−1)n(2− i).

Take determinant both sides of the definition.

5. GFn(GFn+1+GFn−1) = (1+2i)F2n−1.(

Or GF 2n+1 −GF 2

n−1 = (1 + 2i)F2n−1.)

Consider the top right entry of the following:[GFn+1 GFn

GFn GFn−1

]·[GFn+1 GFn

GFn GFn−1

]= (P nR)2 = P 2n−1(PR2).

9

6. GFn+1GFm+1 +GFnGFm = (1 + 2i)Fn+m.

Consider the top right entry of the following:[GFn+1 GFn

GFn GFn−1

]·[GFm+2 GFm+1

GFm+1 GFm

]= (P nR)(Pm+1R) = P n+m(PR2).

Special case m = n: GF 2n+1 +GF 2

n = (1 + 2i)F2n.

7.∑n

j=1GF2j = (1 + 2i)F 2

n + (−1)ni− i.Case when n is even:

n∑j=1

GF 2j =

n∑j=1

(Fj + iFj−1)2 by identity (2)

=n∑

j=1

(F 2j − F 2

j−1 + 2iFjFj−1)

= F 2n + 2iF 2

n by identity (5) of previous section

= (1 + 2i)F 2n .

Case when n is odd:

n∑j=1

GF 2j =

n∑j=1

(Fj + iFj−1)2 by identity (2)

=n∑

j=1

(F 2j − F 2

j−1 + 2iFjFj−1)

= F 2n + 2i(F 2

n − 1) by identity (6) of previous section

= (1 + 2i)F 2n − 2i.

8. GF−n = (−1)niGFn+1.Consider P−nR = (P n)−1R = (R−1P−1P n+1R)−1R= (P n+1R)−1PR2 = (1 + 2i)(P n+1R)−1.

9.∑n

j=1GF2j−1 = GF2n − i.Consider the geometric sum PR + P 3R + · · ·+ P 2n−1R.

10.∑n

j=1GF2j = GF2n+1 − 1.

Consider the geometric sum P 2R + P 4R + · · ·+ P 2nR.

11.∑2n

j=1(−1)jGFj = GF2n−1 − 1 + i.

Consider the geometric sum −PR + P 2R + · · ·+ P 2nR.

12.∑n

j=1(−1)jGFj = (−1)nGFn−1 − 1 + i.

Consider the geometric sum −PR + P 2R + · · ·+ P nR.

10

2.3 Fibonacci Quaternions

Let a 2-by-2 matrix R be in the form:

R :=

[1 + i+ 2j + 3k i+ j + 2ki+ j + 2k 1 + j + k

]and let

P :=

[1 11 0

]We define Qn and Ln as following:

Definition (Fibonacci Quaternions).[Qn+1 Qn

Qn Qn−1

]=

[1 11 0

]n·[

1 + i+ 2j + 3k i+ j + 2ki+ j + 2k 1 + j + k

].

Definition (Lucas numbers).[Ln+1 Ln

Ln Ln−1

]=

[1 11 0

]n·[

1 22 −1

].

Again need to show that Qn and Ln is well defined. (This can be done at the begin-ning once and for all.)

We will show some identities from the paper by Iyer, Some results on FibonacciQuaternions, [8].

2.3.1 Identities from Iyer

1. Qn = Fn + iFn+1 + jFn+2 + kFn+3.Ln = 2Fn+1 − Fn.

Fn =1

15(Qn+1(−4− 3I + J − 2K) +Qn(7 + 4I − 3J +K))

Fn =1

5(2Ln+1 − Ln).

All these can be obtained from the definitions.

2. Identity (17)

Qn − iQn+1 − jQn+2 − kQn+3 = 3Ln+3.Note that I + iP + jP 2 + kP 3 = R and I − iP − jP 2 − kP 3 = R∗

(R with conjugate entries) and R ·R∗ = 3

[L4 L3

L3 L2

].

3. Identity (18)

Qn−1Qn+1 −Q2n = (−1)n (2Q1 − 3k) .

which can be obtained by taking the determinant on both sides of the definition.Also note that Q−1Q1 −Q2

0 = 2Q1 − 3k.

11

4. Identity (19)

Q2n +Q2

n−1 = 2Q2n−1 − 3L2n+2.

Multiply P−1R to R +R∗ = 2I to get(P−1R)R = 2P−1R− P−1R ·R∗ which is[

Q0 Q−1Q−1 Q−2

]·[Q1 Q0

Q0 Q−1

]= 2

[Q0 Q−1Q−1 Q−2

]− 3

[L3 L2

L2 L1

]

then multiply

[1 11 0

]2non both sides.

5. Identity (20)

Qn+1Qn +QnQn−1 = Q2n+1 −Q2

n−1 = (2Q2n − 3L2n+3) + 2(−1)n+1(Q0 − 3k).

Multiply R to R +R∗ = 2I to get R2 = 2R−R ·R∗ which is

[Q1 Q0

Q0 Q−1

]2= 2

[Q1 Q0

Q0 Q−1

]− 3

[L4 L3

L3 L2

]

then multiply

[1 11 0

]2non both sides.

6. Identity (21)

QnQn+1 +Qn−1Qn−2 = (6FnQn−1 − 9F2n+2) + 2(−1)n+1(Q−1 − 3k).

Note: Fn+1Fn + Fn−1Fn−2 = F2n − F 2n−1. (This one I can prove)

7. Identity (22)

Qn−1Qn+3 −Q2n+1 = (−1)n (2 + 4i+ 6j + k) .[

Qn+3 Qn

Qn+2 Qn−1

]=

[1 11 0

]n·[Q3 Q0

Q2 Q−1

]Take determinant on both sides to get:Qn+3Qn−1 −Qn+2Qn = (−1)n(Q3Q−1 −Q2Q0).

Also [Qn+2 Qn+1

Qn+1 Qn

]=

[1 11 0

]n·[Q2 Q1

Q1 Q0

]

12

Take determinant on both sides to get:Qn+2Qn −Q2

n+1 = (−1)n(Q2Q0 −Q21).

Add the two equations to get:Qn+3Qn−1 −Q2

n+1 = (−1)n(Q3Q−1 −Q21).

8. Identity (23)

Qn+2Qn−2 −Qn+1Qn−1 = 2(−1)n(−2 + I − J − 6K).Consider [

Qn+2 Qn−1Qn+1 Qn−2

]=

[1 11 0

]n·[Q2 Q−1Q1 Q−2

]Then take determinant on both sides to get:Qn+2Qn−2 −Qn+1Qn−1 = (−1)n(Q2Q−2 −Q1Q−1).

9. Identity (24)

QnQn+1 +Qn−2Qn−3 = 4Q2n−2 − 6L2n+1 + 2(−1)n+1(i+ j − k).

10. Identity (25)

Q2n+1 +Q2

n−1 = 6Fn+1Qn−1 − 9F2n+3 + 2(−1)n+1(1− i− k).

11. Identity (26)

Qn+r + (−1)rQn−r = LrQn.We want to show[

Qn+1+r Qn+r

Qn+r Qn−1+r

]+

[Qn+1−r Qn−rQn−r Qn−1−r

]= Lr

[Qn+1 Qn

Qn Qn−1

]when r is even. And[

Qn+1+r Qn+r

Qn+r Qn−1+r

]−[Qn+1−r Qn−rQn−r Qn−1−r

]= Lr

[Qn+1 Qn

Qn Qn−1

]when r is odd.

These can be shown byCase 1: r is even. Let r = 2k.P n+2kR + P n−2kR = (P 2k + P−2k) · P nR = L2kP

nR.

Case 2: r is odd. Let r = 2k + 1.P n+2k+1R− P n−2k−1R = (P 2k+1 − P−2k−1) · P nR = L2k+1P

nR.

13

Note that P 2k + P−2k = L2k and P 2k+1 − P−2k−1 = L2k+1 can be easily shownusing an induction on k.i.e. P 2k + P−2k = P 2k−1 − P−2k+1 + P 2k−2 + P−2k+2 = L2k−1 + L2k−2 = L2k.

2.3.2 Identities from Halici, [6]

1. Corollary 2(a)

n∑i=0

Qi = Qn+2 −Q1.

Same as Jor(1).

2. Corollary 2(b)

n∑i=0

Q2i = Q2n+1 −Q−1.

(1 + P 2 + P 4 + · · ·+ P 2n)R =(P 2n+2 − 1)R

P= P 2n+1R− P−1R.

3. Corollary 2(c)

n−1∑i=0

Q2i+1 = Q2n −Q0.

Same proof.

4. Theorem 3.5, equation (3.13)n∑

i=0

(n

i

)Qi = Q2n.∑(

ni

)P iR = (P 1 + P 0)nR = P 2nR.

5. Theorem 3.5, equation (3.14)n∑

i=0

(−1)i(n

i

)Qi = (−1)nQ−n.

Similar proof.

2.4 P nR In General

We define Sn as following:

Definition. [Sn+1 Sn

Sn Sn−1

]=

[1 11 0

]n·[a bb a− b

].

It is possible to work with other hyper complex numbers other than Quaternions.However, for example, Cayley numbers are non-associative. With some extra care,we should be able to make it works.

14

2.5 Conclusion

This section serves like a warm-up problems. We use it to learn about the standardtechnique of matrix-form. The results may not very interesting and have been donebefore. Nonetheless, we mention it for completeness.

3 Two Matrix Multiply, P nRm

In this section we let a 2-by-2 matrix R be in the form:

R :=

[a bb a− b

]and consider the identities from matrices of the form P nRm.

3.1 Gaussian Fibonacci Numbers

Motivated by paper by Berzsenyi, Gaussian Fibonacci numbers, [2]. We define Fn,m

as following:

Definition. [Fn+1,m Fn,m

Fn,m Fn−1,m

]=

[1 11 0

]n·[

1 ii 1− i

]m.

We call Fn,m as Gaussian Fibonacci numbers. Here we give a simpler proof than inthe original paper. Some new identities have been discovered as well.

Corollary 3.1.1.Fn,m = Fn−1,m + Fn−2,m,

Fn,m = iFn+1,m−1 + (1− i)Fn,m−1,

Fn,m = Fn,m−1 + iFn−1,m−1,

Fn,m = Fn+1,0F0,m + Fn,0F−1,m.

for all n ∈ Z,m ∈ N with the initial conditions: F0,0 = 0, F1,0 = 1.

Proof. From matrix multiplication and compare the entry on both sides.

Corollary 3.1.2 (Generalize of the first Corollary).

Fn+s,m+t = Fn+1,mFs,t + Fn,mFs−1,t.

Proof. Notice the following matrix multiplication then compare the entry on bothsides. [

Fn+1,m Fn,m

Fn,m Fn−1,m

]·[Fs+1,t Fs,t

Fs,t Fs−1,t

]=

[Fn+s+1,m+t Fn+s,m+t

Fn+s,m+t Fn+s−1,m+t

].

15

Some identities from the original paper which now can be done easily from theproperty of matrix multiplication.

1. iFn+1,m − Fn,m+1 + (1− i)Fn,m = 0, (monodiffric)From the definition[

Fn+1,m+1 Fn,m+1

Fn,m+1 Fn−1,m+1

]=

[Fn+1,m Fn,m

Fn,m Fn−1,m

]·[

1 ii 1− i

].

Then compare the 12 entries.

2. Fn,m =m∑k=0

(m

k

)ikFn−k,0, m ≥ 0

Consider

P nRm = P n−m(PR)m

= P n−m(P + i)m

= P n−mm∑k=0

(m

k

)ikPm−k

=m∑k=0

(m

k

)ikP n−k.

This implies the statement.

3. Fm,2m = 0,

4. Fm+1,2m = (1 + 2i)m,

For (3) and (4), since PR2 = (1 + 2i)I, we have

PmR2m = (1 + 2i)mI.

Then (3) can be obtained by comparing the entry (1,2) on both sides. And (4)can be obtained by comparing the entry (1,1) on both sides.

5. Fn,2m = Fm+1,2mFn−m,0 = (1 + 2i)mFn−m,0,

Consider

P nR2m = P n−m(PmR2m)

= (1 + 2i)mP n−m.

Then compare entry (1,2) on both sides.

16

6. Fn,2m+1 = (1 + 2i)m (Fn−m,0 + iFn−m−1,0) .

Consider

P nR2m+1 = P n−mR(PmR2m)

= (1 + 2i)mP n−mR.

Then compare entry (2,1) on both sides.

Corollary 3.1.3 (Generalize Cassini’s identity).

Fn+1,mFn−1,m − F 2n,m = (−1)n(2− i)m.

Proof. By taking determinant on both sides of our definition.

Vajda’s identity generalize Cassini’s identity.

Theorem 3.1.4 (Generalize Vajda’s identity).

Fn+i,m+kFn+j,m+l − Fn,mFn+i+j,m+k+l = (−1)n(2− i)mFi,kFj,l.

Proof. Take the determinant on both sides of the following equation:

[Fn+i,m+k Fn,m

Fn+i+j,m+k+l Fn+j,m+l

]=

[Fn+1,0 Fn,0

Fn+j+1,l Fn+j,l

]·[

Fi,m+k F0,m

Fi−1,m+k F−1,m

]

=

([1 0

Fj+1,l Fj,l

]·[Fn+1,0 Fn,0

Fn,0 Fn−1,0

])·([

F1,m F0,m

F0,m F−1,m

]·[

Fi,k 0Fi−1,k 1

])

=

([1 0

Fj+1,l Fj,l

]·[

1 11 0

]n)·([

1 ii 1− i

]m·[

Fi,k 0Fi−1,k 1

]).

3.2 Generalize Lucas Numbers: a = 1 and b = 2

Definition. [Sn+1,m Sn,m

Sn,m Sn−1,m

]=

[1 11 0

]n [1 22 −1

]m.

Note:

1. Sn,0 = Fn and Sn,1 = Ln.

2. S1,m = Inverse Binomial Mean transform of the Fibonacci sequence, A074872.

3. P 2 = P + 1, R2 = 5I

17

4. P nR = LnP + Ln−1I

Theorem 3.2.1. Sn,m = 2Sn+1,m−1 − Sn,m−1 = Sn,m−1 + 2Sn−1,m−1

Proof. [Sn+1,m Sn,m

Sn,m Sn−1,m

]=

[1 22 −1

] [Sn+1,m−1 Sn,m−1Sn,m−1 Sn−1,m−1

].

Corollary 3.2.2.Ln = Fn + 2Fn−1.

Proof. Set m = 1 in the theorem and compare the top right entries.

Proposition 3.2.3. Sn,2m = 5mFn and Sn,2m+1 = 5mLn.

Proof. From R2 = 5I.

Corollary 3.2.4. S0,2m = 0 and S0,2m+1 = 2 · 5m.

The following theorem and corollary are also a special case of theorem in the nextsection.

Theorem 3.2.5.

Sn,m =∑j

(m

j

)2jFn−j.

Proof.

P nRm = P n−m(PR)m = P n−m(P+2)m = P n−m∑j

(m

j

)2jPm−j =

∑j

(m

j

)2jP n−j.

Corollary 3.2.6.

5mFn =∑j

(2m

j

)2jFn−j,

5mLn =∑j

(2m+ 1

j

)2jFn−j.

18

3.3 Multiplication of the Two matrices

This is the generalization of [10], [8] and [2]. While [10], [8] are cases when m = 1and reduces to case of one-matrix. [2] generalizes [10]. Here we try to generalize [8].

For a 2-by-2 matrix R to be commute with P , R must be in the form:

R :=

[a bb a− b

]We define Tn,m as following:

Definition. [Tn+1,m Tn,mTn,m Tn−1,m

]=

[1 11 0

]n·[a bb a− b

]m.

for some constant a and b.

Corollary 3.3.1.Tn,m+1 = bTn+1,m + (a− b)Tn,m.

Proof. This follows by comparing the lower left entries of[Tn+1,m+1 Tn,m+1

Tn,m+1 Tn−1,m+1

]=

[a bb a− b

]·[Tn+1,m Tn,mTn,m Tn−1,m

]

3.3.1 Some Identities!

Proposition 3.3.2 (Entries as a sum of two multiplications).

Tn+s,m+t = Tn+1,mTs,t + Tn,mTs−1,t = Tn,mTs+1,t + Tn−1,mTs,t.

Proof. Notice the following matrix multiplication then compare the entries on bothsides.[

Tn+s+1,m+t Tn+s,m+t

Tn+s,m+t Tn+s−1,m+t

]=

[1 11 0

]n+s

·[a bb a− b

]m+t

=

[1 11 0

]n·[a bb a− b

]m·[

1 11 0

]s·[a bb a− b

]t=

[Tn+1,m Tn,mTn,m Tn−1,m

]·[Ts+1,t Ts,tTs,t Ts−1,t

].

Proposition 3.3.3 (Generalize Cassini’s identity).

Tn+1,mTn−1,m − T 2n,m = (−1)n(a2 − ab− b2)m.

19

Proof. By taking the determinants on both sides of our definition.

Theorem 3.3.4 (Generalize Vajda’s identity).

Tn+i,m+kTn+j,m+l − Tn,mTn+i+j,m+k+l = (−1)n(a2 − ab− b2)mTi,kTj,l.

Proof. Take the determinants on both sides of the following equation:

[Tn+i,m+k Tn,m

Tn+i+j,m+k+l Tn+j,m+l

]=

[Tn+1,0 Tn,0Tn+j+1,l Tn+j,l

]·[

Ti,m+k T0,mTi−1,m+k T−1,m

]

=

([1 0

Tj+1,l Tj,l

]·[Tn+1,0 Tn,0Tn,0 Tn−1,0

])·([

T1,m T0,mT0,m T−1,m

]·[

Ti,k 0Ti−1,k 1

]).

=

([1 0

Tj+1,l Tj,l

]·[

1 11 0

]n)·([

a bb a− b

]m·[

Ti,k 0Ti−1,k 1

]).

3.4 Fibonacci as Sum of Binomial

It is possible to find a complex number b and integers A,B so that PARB is a diagonalmatrix cI. I list their values in the table below:

A B b c Note0 1 0 1 Identity0 2 2 5 Lucas1 2 ±i 1± 2i Complex Fibo2 1 −1 1

2 2−1

2

5

42 4 2±

√5 25(9± 4

√5)

3 1 −2 −13 2 −1± i −1± 2i

4 1 −3

2

1

24 2 −3 5

4 4−1±

√5i

2−25

4

5 1 −5

3−1

3

5 2−3± i

2−1± 1

2i

...

20

Theorem 3.4.1. If there is an integers A,B and a complex number b satisfy thefollowing: [

1 11 0

]A·[

1 bb 1− b

]B=

[c 00 c

]then

cmFn =∑j

(Bm

j

)bjFn+Am−j, for any n ∈ Z and m ∈ N.

Proof. Consider the entries of P n ·RBm

On one hand:

P n+Am ·RBm = P n ·(PARB

)m= cmP n.

On the other hand:

P n+Am ·RBm = P n+Am−Bm · (PR)Bm

= P n+Am−Bm · (P + b)Bm

= P n+Am−Bm ·

(∑j

(Bm

j

)bjPBm−j

)

=∑j

(Bm

j

)bjP n+Am−j.

Then compare the right hand side of both equations to get the result.

3.4.1 Sum Involving Binomial

We let P :=

[1 11 0

]and Rb :=

[1 bb 1− b

]We consider the identities arise from the matrix multiplication of the form[

Tn+1,m Tn,mTn,m Tn−1,m

]:=

[1 11 0

]n·[

1 bb 1− b

]m.

The idea of two-matrix multiply is quite powerful. We start by using this method toshow theorem 2 of Keskin and Demirturk, [4].

Proposition 3.4.2.

5nF2mn+k =∑j

(2n

j

)LjmL

2n−jm−1Fj+k,

5nL2mn+k =∑j

(2n

j

)LjmL

2n−jm−1Lj+k,

21

5n+1F(2n+1)m+k =∑j

(2n+ 1

j

)LjmL

2n+1−jm−1 Lj+k,

and

5nL(2n+1)m+k =∑j

(2n+ 1

j

)LjmL

2n+1−jm−1 Fj+k,

Proof. First note that R22 = 5I.

For the first two equations,

R2n2 P

2mn+kRb = (PmR2)2nP kRb

= (LmP + Lm−1I)2nP kRb

=∑j

(2n

j

)LjmL

2n−jm−1P

j+kRb.

Then set b = 0 for the first identity and b = 2 for the second identity .

The last two equations can be done similarly.

The following theorem is the generalization of theorem 3.2.5, theorem 3.4.2 andtheorem 5.3.2 .

Theorem 3.4.3. Assume PARBb = cI then

cnsTBmn,0 =∑j

(Bn

j

)T jm,sT

Bn−jm−1,sTAns+j,0.

Proof.

(PARBb )nsPBmn = (PmRs

b)BnPAns

= (Tm,sP + Tm−1,sI)BnPAns

=∑j

(Bn

j

)T jm,sT

Bn−jm−1,sP

Ans+j.

The following proposition is similar to the above theorem. We rewrite it in a moreconvenience form. Lemma 3 in Keskin and Demirturk, [4], is also a special case ofthis proposition.

Proposition 3.4.4. Assume PARBb = cI then

cnG(mB−A)n+k =∑j

(Bn

j

)T jm,1T

Bn−jm−1,1Gj+k,

where

[Gn+1 Gn

Gn Gn−1

]=

[1 11 0

]n· T.

22

Proof. Consider

(RBb P

A)nP (mB−A)n+kT = (RbPm)BnP kT = (Tm,1P + Tm−1,1I)BnP kT

=∑j

(Bn

j

)T jm,1T

Bn−jm−1,1P

j+kT.

Corollary 3.4.5 (Lemma 3 of Keskin and Demirturk, [4] ).

Let m, k ∈ Z with m 6= 1 and m 6= 0. Then

Fmn+k =n∑

j=0

(n

j

)F jmF

n−jm−1Fj+k,

and

Lmn+k =n∑

j=0

(n

j

)F jmF

n−jm−1Lj+k,

for all n ∈ N.

Proof. Notice that

P 2R−1 =

[1 11 0

]2 [1 −1−1 2

]=

[1 00 1

]Therefore PmR−1 = Pm−2 and Tm,1 = Fm−2.Now by the previous proposition, we have

G(m−2)n+k =∑j

(n

j

)T jm,1T

n−jm−1,1Gj+k

After substitute m by m+ 2, we have

Gmn+k =∑j

(n

j

)T jm+2,1T

n−jm+1,1Gj+k =

∑j

(n

j

)F jmF

n−jm+1Gj+k.

Then let T = I to obtain the first identity and let T =

[1 22 −1

]to obtain the

second identity.

Another consequence of the proposition:

Corollary 3.4.6. Assume PARBb = cI for constants A,B, b, c, then

TmB+k,nB = cnF(mB−A)n+k

Proof. Notice that PmB+kRnBb = (PARB

b )nP (mB−A)n+k.

23

Next are the identities among Fibonacci terms. (Although it might not be so spec-tacular.)

Proposition 3.4.7. For any n,m, s ∈ Z and b ∈ R,

Tn,m = Fn+1Gm + FnGm−1 = Gm+1Fn +GmFn−1 =∑j

(m

j

)Gm−j

s−1 GjsFn−sm+j

where Gs is defined by Gs = Gs−1 +Gs−2 where G0 = b,G1 = 1.

Proof.

P nRm = (P sR)mP n−sm

= (GsP +Gs−1I)mP n−sm

=

(∑j

(m

j

)Gm−j

s−1 GjsP

j

)P n−sm

=∑j

(m

j

)Gm−j

s−1 GjsP

n−sm+j.

Some special cases:

For s = 0, Tn,m =∑j

(m

j

)(1− b)m−jbjFn+j

For s = 1, Tn,m =∑j

(m

j

)bm−jFn−m+j

For s = 2, Tn,m =∑j

(m

j

)(b+ 1)jFn−2m+j

For s = 3, Tn,m =∑j

(m

j

)(b+ 1)m−j(b+ 2)jFn−3m+j.

4 Multiplication of Three (or more) Matrices

4.1 Multiplication of Three Matrices

Definition. We define Fn,m,p as following[Fn+1,m,p Fn,m,p

Fn,m,p Fn−1,m,p

]:=

[1 11 0

]n [a bb a− b

]m [A BB A−B

]p.

Proposition 4.1.1. Fn,m,p = aFn,m−1,p + bFn−1,m−1,p = A · Fn,m,p−1 +B · Fn−1,m,p−1

24

Proof. Consider from the definition:

[Fn+1,m,p Fn,m,p

Fn,m,p Fn−1,m,p

]=

[Fn+1,m−1,p Fn,m−1,pFn,m−1,p Fn−1,m−1,p

] [a bb a− b

]=

[Fn+1,m,p−1 Fn,m,p−1Fn,m,p−1 Fn−1,m,p−1

] [A BA A−B

].

Proposition 4.1.2.

Fn,m,p = Fn+1,m,0F0,0,p + Fn,m,0F−1,0,p = Fn,m,0F1,0,p + Fn−1,m,0F0,0,p

= Fn+1,0,pF0,m,0 + Fn,0,pF−1,m,0 = Fn,0,pF1,m,0 + Fn−1,0,pF0,m,0.

Proof. Consider from the definition:

[Fn+1,m,p Fn,m,p

Fn,m,p Fn−1,m,p

]=

[Fn+1,m,0 Fn,m,0

Fn,m,0 Fn−1,m,0

] [F1,0,p F0,0,p

F0,0,p F−1,0,p

]=

[Fn+1,0,p Fn,0,p

Fn,0,p Fn−1,0,p

] [F1,m,0 F0,m,0

F0,m,0 F−1,m,0

].

Proposition 4.1.3 (Generalize Cassini’s identity).

Fn+1,m,pFn−1,m,p − F 2n,m,p = (−1)n(a2 − ab− b2)m(A2 − AB −B2)p.

Proof. By taking determinant on both sides of the definition.

Theorem 4.1.4 (Generalize Vajda’s identity).

Fn+i1,m+i2,p+i3Fn+j1,m+j2,p+j3 − Fn,m,pFn+i1+j1,m+i2+j2,p+i3+j3

= (−1)n(a2 − ab− b2)m(A2 − AB −B2)pFi1,i2,i3Fj1,j2,j3 .

Proof.[Fn+i1,m+i2,p+i3 Fn,m,p

Fn+i1+j1,m+i2+j2,p+i3+j3 Fn+j1,m+j2,p+j3

]=

[Fn+1,0,0 Fn,0,0

Fn+j1+1,j2,j3 Fn+j1,j2,j3

]·[

Fi1,m+i2,p+i3 F0,m,p

Fi1−1,m+i2,p+i3 F−1,m,p

]=

([1 0

Fj1+1,j2,j3 Fj1,j2,j3

]·[Fn+1,0,0 Fn,0,0

Fn,0,0 Fn−1,0,0

])·([

F1,m,p F0,m,p

F0,m,p F−1,m,p

]·[

Fi1,i2,i3 0Fi1−1,i2,i3 1

])=

([1 0

Fj1+1,j2,j3 Fj1,j2,j3

]·[

1 11 0

]n)·([

a bb a− b

]m·[A BB A−B

]p·[

Fi1,i2,i3 0Fi1−1,i2,i3 1

]),

Then take determinant on both sides.

25

4.2 Quaternions

Next, we generalize Berzsenyi results, [2], with Quaternions. This one is harder sincethe element in the matrix does not commute.

Definition. We define Qn,m,p,q as following[Qn+1,m,p,q Qn,m,p,q

Qn,m,p,q Qn−1,m,p,q

]:=

[1 11 0

]n [1 ii 1− i

]m [1 jj 1− j

]p [1 kk 1− k

]q.

Corollary 4.2.1.Qn,m,p,q = Qn−1,m,p,q +Qn−2,m,p,q,

Qn,m,p,q = iQn+1,m−1,p,q + (1− i)Qn,m−1,p,q,

Qn,m,p,q = Qn,m−1,p,q + iQn−1,m−1,p,q,

Qn,m,p,q = Fn+1Q0,m,p,q + FnQ−1,m,p,q.

Qn,m,p,q = Qn+1,m,0,0Q0,0,p,q +Qn,m,0,0Q−1,0,p,q.

Proof. From matrix multiplication and compare the entry on both sides.

Corollary 4.2.2 (Generalize of the first Corollary).

Qn1+n2,m1+m2,p1+p2,q1+q2 = Qn1+1,m1,p1,q1Qn2,m2,p2,q2 +Qn1,m1,p1,q1Qn2−1,m2,p2,q2 .

Proof. Notice the following matrix multiplication then compare the entry on bothsides. [

Qn1+1,m1,p1,q1 Qn1,m1,p1,q1

Qn1,m1,p1,q1 Qn1−1,m1,p1,q1

]·[Qn2+1,m2,p2,q2 Qn2,m2,p2,q2

Qn2,m2,p2,q2 Qn2−1,m2,p2,q2

]

=

[Qn1+n2+1,m1+m2,p1+p2,q1+q2 Qn1+n2,m1+m2,p1+p2,q1+q2

Qn1+n2,m1+m2,p1+p2,q1+q2 Qn1+n2−1,m1+m2,p1+p2,q1+q2

].

The following identities can be shown in a straight forward way using the matrix.

Theorem 4.2.3 (Generalize the results of Berzsenyi, [2]).

1. Qn,m+1,p,q −Qn,m,p,q = i(Qn+1,m,p,q −Qn,m,p,q), (monodiffric)Qn,m,p,q+1 −Qn,m,p,q = (Qn+1,m,p,q −Qn,m,p,q)k,By (1) and (2), we get:i[Qn,m,p,q+1 −Qn,m,p,q] = [Qn,m+1,p,q −Qn,m,p,q]k.

2. Qn,m,p,q =∑

s1,s2,s3

(ms1

)(ps2

)(qs3

)is1js2ks3Fn−s1−s2−s3 ,

3. Qm,2m,0,0 = Qp,0,2p,0 = Qq,0,0,2q = 0,

4. Qm±1,2m,0,0 = (1 + 2i)m, Qp±1,0,2p,0 = (1 + 2j)p, Qq±1,0,0,2q = (1 + 2k)q.

26

5. Qn,2m,p,q = Qm+1,2m,0,0Qn−m,0,p,q = (1 + 2i)mQn−m,0,p,q,

6. Qn,2m+1,p,q = (1 + 2i)m [Qn−m,0,p,q + iQn−m−1,0,p,q] .

Proof. Note that N2 = N + I and N2M2 = (N + iI)2 = (1 + 2i)Nwhich implies NM2 = 1 + 2i. Also NM = N + i.

2) NnMmP pQq

= Nn−m−p−q · [NM ]m · [NP ]p · [NQ]q

= Nn−m−p−q · [N + i]m · [N + j]p · [N + k]q

= Nn−m−p−q ·

(m∑

s1=0

(m

s1

)is1Nm−s1

)·

(p∑

s2=0

(p

s2

)js2Np−s2

)·

(q∑

s3=0

(q

s3

)ks3N q−s3

)

=∑

s1,s2,s3

(m

s1

)(p

s2

)(q

s3

)is1js2ks3Nn−s1−s2−s3 .

3) and 4) follows directly from (NM2)m = (1 + 2i)m.5) follows from NnM2mP pQq = (NM2)m ·Nn−mP pQq.[Qn+1,2m,p,q Qn,2m,p,q

Qn,2m,p,q Qn−1,2m,p,q

]=

[Qm+1,2m,0,0 Qm,2m,0,0

Qm,2m,0,0 Qm−1,2m,0,0

] [Qn−m+1,0,p,q Qn−m,0,p,q

Qn−m,0,p,q Qn−m−1,0,p,q

].

6) follows from NnM2m+1P pQq = M · (NM2)m · (Nn−mP pQq).

[Qn+1,2m+1,p,q Qn,2m+1,p,q

Qn,2m+1,p,q Qn−1,2m+1,p,q

]=

[1 ii 1− i

] [Qm+1,2m,0,0 Qm,2m,0,0

Qm,2m,0,0 Qm−1,2m,0,0

] [Qn−m+1,0,p,q Qn−m,0,p,q

Qn−m,0,p,q Qn−m−1,0,p,q

]=

[1 ii 1− i

] [Qm+1,2m,0,0Qn−m+1,0,p,q Qm+1,2m,0,0Qn−m,0,p,q

Qm−1,2m,0,0Qn−m,0,p,q Qm−1,2m,0,0Qn−m−1,0,p,q

].

and compare the entry on the bottom left entry.

Corollary 4.2.4 (Generalize Cassini’s identity).

Qn+1,m,0,0Qn−1,m,0,0 −Q2n,m,0,0 = (−1)n(2− i)m.

Proof. By taking determinant on both sides of our definition.

Remark 1. No more general version of Cassini’s identity as the entry in the matrixdoes not commute.

27

5 Numbers related to 3-by-3 Matrix

5.1 Tribonacci Numbers

By letting M :=

1 1 11 0 00 1 0

, we also have M−1 =

0 1 00 0 11 −1 −1

.Then Tribonacci numbers,

tn := tn−1 + tn−2 + tn−3 where t0 = t1 = 0 and t2 = 1

can be defined using this matrix by

Let An :=

tn+2 tn+1 + tn tn+1

tn+1 tn + tn−1 tntn tn−1 + tn−2 tn−1

.Then An = Mn and we also have the relation An+1 = MAn = AnM .

5.1.1 Properties

1. t−n =

∣∣∣∣ tn+1 tn+2

tn tn+1

∣∣∣∣ .From M−n = (Mn)−1:

t−n =1

|M |n

∣∣∣∣ tn+1 tn + tn−1tn tn−1 + tn−2

∣∣∣∣ =

∣∣∣∣ tn+1 tn+1 + tn + tn−1tn tn + tn−1 + tn−2

∣∣∣∣ =

∣∣∣∣ tn+1 tn+2

tn tn+1

∣∣∣∣ .2.

∣∣∣∣∣∣tn+1 tn+2 tn+3

tn tn+1 tn+2

tn−1 tn tn+1

∣∣∣∣∣∣ = 1.

1 = |Mn| =

∣∣∣∣∣∣tn+2 tn+1 + tn tn+1

tn+1 tn + tn−1 tntn tn−1 + tn−2 tn−1

∣∣∣∣∣∣ =

∣∣∣∣∣∣tn+2 tn+2 + tn+1 + tn tn+1

tn+1 tn+1 + tn + tn−1 tntn tn + tn−1 + tn−2 tn−1

∣∣∣∣∣∣=

∣∣∣∣∣∣tn+2 tn+3 tn+1

tn+1 tn+2 tntn tn+1 tn−1

∣∣∣∣∣∣ =

∣∣∣∣∣∣tn+1 tn+2 tn+3

tn tn+1 tn+2

tn−1 tn tn+1

∣∣∣∣∣∣ .3. tn+m = tntm+2 + (tn−1 + tn−2)tm+1 + tn−1tm.

Consider the (3,1) entry of Mn+m,

Mn+m =

tn+2 tn+1 + tn tn+1

tn+1 tn + tn−1 tntn tn−1 + tn−2 tn−1

· tm+2 tm+1 + tm tm+1

tm+1 tm + tm−1 tmtm tm−1 + tm−2 tm−1

.28

5.2 Trucas Numbers

un := un−1 + un−2 + un−3 where u0 = 3, u1 = 1 and u2 = 3

can be defined using the matrix.

Define Un :=

un+2 un+1 + un un+1

un+1 un + un−1 unun un−1 + un−2 un−1

.5.2.1 Properties

1. Un = Mn ·

3 4 11 2 33 −2 −1

.2. un = tn+1 + 2tn + 3tn−1.

Noticed that

3 4 11 2 33 −2 −1

= M + 2I + 3M−1.

3.

∣∣∣∣∣∣un+1 un+2 un+3

un un+1 un+2

un−1 un un+1

∣∣∣∣∣∣ = 44 for all integer n.

4. tn = 122

(2un + un−1 + 5un−2) = 122

(5un+1 − 3un − 4un−1).

Notice that

3 4 11 2 33 −2 −1

−1 =1

22

2 1 55 −3 −4−4 9 1

.Therefore

tn+2 tn+1 + tn tn+1

tn+1 tn + tn−1 tntn tn−1 + tn−2 tn−1

=1

22

un+2 un+1 + un un+1

un+1 un + un−1 unun un−1 + un−2 un−1

· 2 1 5

5 −3 −4−4 9 1

=

1

22

2 1 55 −3 −4−4 9 1

· un+2 un+1 + un un+1

un+1 un + un−1 unun un−1 + un−2 un−1

.5.3 Multiplication of Two Matrices

We found the new relations by generalizing the matrix of the form MARB.

29

Lemma 5.3.1. Let the 3-by-3 matrix M be

1 1 11 0 00 1 0

and let the 3-by-3 matrix R be

1 a+ b bb 1− b aa b− a 1− b− a

.

Then MR = RM = M + (a+ b) · I + b ·M−1.

Proof. This can be verified directly.

Using the same notation, below is the table of MA ·RB = cI.

A B a b c Note0 1 0 0 1 Identity3 1 0 −1 14 1 −2 0 −16 1 1

2−32

12

7 1 −43

−13

−13

8 1 −8 4 19 1 5

4−2 1

4

10 1 −78

−58

−18

1 3 −1 1 4

2 3 E −13E 1

16(121± 143

√33

9) E =

3±√

33

4

3 3 −23E − 1

3E 121± 187

√33

9E =

5±√

33

24 3 1 −1 45 3 −1 0 26 37 3 −1 −1 −48 310 3 −3 1 411 3 1 −2 −213 3 3 −3 4-2 3 1 1 4 Lucas...

The application of the table above is as below:

Theorem 5.3.2. If there is an integers A,B and a complex number b satisfy thefollowing: 1 1 1

1 0 00 1 0

A

·

1 a+ b bb 1− b aa b− a 1− b− a

B

=

c 0 00 c 00 0 c

.30

then

cmtn =∑i,j

(Bm

i, j, Bm− i− j

)(a+ b)ibjtn+Am−i−2j

, for any n ∈ Z and m ∈ N.

Proof. Consider the entries of Mn+Am ·RBm

On one hand:

Mn+Am ·RBm = Mn ·(MARB

)m= cmMn.

On the other hand:

Mn+Am ·RBm = Mn+Am−Bm · (MR)Bm

= Mn+Am−Bm · (M + (a+ b)I + bM−1)Bm

= Mn+Am−Bm ·

(∑i,j

(Bm

i, j, Bm− i− j

)(a+ b)ibjMBm−i−2j

)

=∑i,j

(Bm

i, j, Bm− i− j

)(a+ b)ibjMn+Am−i−2j.

Then compare the right hand side of both equations to get the result.

6 Another Form of Complex Fibonacci Numbers

We find another matrix representation of the results presented by Harman, [7]. Wecan define G(n,m) as an entry in the matrix below:

A(n,m) :=

[G(n+ 1,m+ 1) G(n,m+ 1)G(n+ 1,m) G(n,m)

]=

[1 11 0

]m·[

1 + i i1 0

]·[

1 11 0

]n.

Remark 2. The matrix is well defined. (which can be shown by 1. and 2. below)

6.1 Identities

1. With n = 0,m = 0, we have that G(0, 0) = 0, G(1, 0) = 1, G(0, 1) = i andG(1, 1) = 1 + i.

2. G(n+1,m) = G(n,m)+G(n−1,m) and G(n,m+1) = G(n,m)+G(n,m−1).

Consider (2,1) entry:[G(n+ 1,m+ 1) G(n,m+ 1)G(n+ 1,m) G(n,m)

]=

[G(n,m+ 1) G(n− 1,m+ 1)G(n,m) G(n− 1,m)

]·[

1 11 0

].

31

Consider (1,2) entry:[G(n+ 1,m+ 1) G(n,m+ 1)G(n+ 1,m) G(n,m)

]=

[1 11 0

]·[

G(n+ 1,m) G(n,m)G(n+ 1,m− 1) G(n,m− 1)

].

3. G(n+ 1,m+ 1) = G(n,m) +G(n,m− 1) +G(n− 1,m) +G(n− 1,m− 1).

Follows from the previous.

4. G(n, 0) = Fn and G(0,m) = iFm.

This can be shown using G(n+ 2, 0) = G(n+ 1, 0) +G(n, 0) and G(0,m+ 2) =G(0,m+ 1) +G(0,m) along the initial conditions.

5. G(n,m) = FmG(n, 1) + Fm−1G(n, 0) = FnG(1,m) + Fn−1G(0,m).

Consider the (2,2) entry in[G(n+ 1,m+ 1) G(n,m+ 1)G(n+ 1,m) G(n,m)

]=

[Fm+1 Fm

Fm Fm−1

]·[G(n+ 1, 1) G(n, 1)G(n+ 1, 0) G(n, 0)

]=

[G(1,m+ 1) G(0,m+ 1)G(1,m) G(0,m)

]·[Fn+1 Fn

Fn Fn−1

].

6. G(n,m) = FnFm+1 + iFn+1Fm.

Consider the (2,2) entry in[G(n+ 1,m+ 1) G(n,m+ 1)G(n+ 1,m) G(n,m)

]=

[Fm+1 Fm

Fm Fm−1

]·[

1 + i i1 0

]·[Fn+1 Fn

Fn Fn−1

]=

[Fm+2 + iFm+1 iFm+1

Fm+1 + iFm iFm

]·[Fn+1 Fn

Fn Fn−1

]7. G(n,m) +G(m,n) = (1 + i)(FmFn+1 + Fm+1Fn).

The result follows immediately from the previous.

8. G(n,m) +G(n− 1,m− 1) = (1 + i)Fm+n.

Consider the (2,2) entry in

A(n,m) +A(n− 1,m− 1) =

[1 11 0

]m·[

1 + i 1 + i1 + i 0

]·[

1 11 0

]n=

[1 + i 0

0 1 + i

]·[

1 11 0

]m+n+1

.

32

9. G(n,m) = (1 + i)FnFm +G(m− 1, n− 1)

By combining the previous 2 results.

10. G(n+ 2k,m+ 2k) = (1 + i)2k∑j=1

Fn+jFm+j +G(n,m)

Telescoping version of 9.

11. G(n+ 2k + 1,m+ 2k + 1) = (1 + i)2k+1∑j=1

Fn+jFm+j +G(m,n)

Telescoping version of 9.

6.2 Constant matrix representation of the symmetric form

We also show that the matrix can generated by a constant matrix.[G(n, 1) G(n, 0)G(n, 0) G(n,−1)

]=

[Fn Fn

Fn 0

]+

[iFn+1 0

0 iFn+1

]

=

[Fn+1 Fn

Fn Fn−1

]−[Fn−1 0

0 Fn−1

]+

[iFn+1 0

0 iFn+1

]

=

[1 11 0

]n−D(Sn−1 − T n−1) +

[i 00 i

]·D(Sn+1 − T n+1)

=

[1 11 0

]n− Sn−1D

(I −

[i 00 i

]· S2

)+ T n−1D

(I −

[i 00 i

]· T 2

)

=

[1 11 0

]n+

[a 00 a

] [α 00 α

]n+

[b 00 b

] [β 00 β

]n.

where α =1 +√

5

2, β =

1−√

5

2, a =

−(5−√

5) + (5 +√

5)i

10and b =

−(5 +√

5) + (5−√

5)i

10.

Second we write the matrix product for the general matrix.

[G(n,m+ 1) G(n,m)G(n,m) G(n,m− 1)

]=

[G(n, 1) G(n, 0)G(n, 0) G(n,−1)

]·[Fm+1 Fm

Fm Fm−1

]

=

([1 11 0

]n−[Fn−1 0

0 Fn−1

]+

[iFn+1 0

0 iFn+1

])·[

1 11 0

]m.

33

6.3 Third Order Recurrence

In this section we define H(n,m) as an entry in the matrix of the following:

B(n,m) :=

H(n+ 2,m+ 2) H(n+ 1,m+ 2) H(n,m+ 2)H(n+ 2,m+ 1) H(n+ 1,m+ 1) H(n,m+ 1)H(n+ 2,m) H(n+ 1,m) H(n,m)

=

1 1 11 0 00 1 0

m

·

2 + 2i 2 + i i1 + 2i 1 + i i

1 1 0

· 1 1 0

1 0 11 0 0

n

.

Remark 3. The matrix is well defined. (which can be shown by 1. and 2. below)

6.3.1 Identities

1. With n = 0,m = 0, we have the following as initial conditionsH(0, 0) = 0, H(1, 0) = 1, H(2, 0) = 1, H(0, 1) = i,H(1, 1) = 1 + i, H(2, 1) = 1 + 2i, H(0, 2) = i, H(1, 2) = 2 + i andH(2, 2) = 2 + 2i.

2. H(n+ 2,m) = H(n+ 1,m) +H(n,m) +H(n− 1,m) andH(n,m+ 2) = H(n,m+ 1) +H(n,m) +H(n,m− 1).

Consider (1,1), (2,1) and (3,1) entries: H(n+ 2,m+ 2) H(n+ 1,m+ 2) H(n,m+ 2)H(n+ 2,m+ 1) H(n+ 1,m+ 1) H(n,m+ 1)H(n+ 2,m) H(n+ 1,m) H(n,m)

=

H(n+ 1,m+ 2) H(n,m+ 2) H(n− 1,m+ 2)H(n+ 1,m+ 1) H(n,m+ 1) H(n− 1,m+ 1)H(n+ 1,m) H(n,m) H(n− 1,m)

· 1 1 0

1 0 11 0 0

.Consider (1,1), (1,2) and (1,3) entries: H(n+ 2,m+ 2) H(n+ 1,m+ 2) H(n,m+ 2)

H(n+ 2,m+ 1) H(n+ 1,m+ 1) H(n,m+ 1)H(n+ 2,m) H(n+ 1,m) H(n,m)

=

1 1 11 0 00 1 0

· H(n+ 2,m+ 1) H(n+ 1,m+ 1) H(n,m+ 1)

H(n+ 2,m) H(n+ 1,m) H(n,m)H(n+ 2,m− 1) H(n+ 1,m− 1) H(n,m− 1)

.3.

H(n+ 1,m+ 1) =H(n+ 1,m+ 1) +H(n+ 1,m) +H(n+ 1,m− 1)+

H(n,m+ 1) +H(n,m) +H(n,m− 1)+

H(n− 1,m+ 1) +H(n− 1,m) +H(n− 1,m− 1).

34

Follows from the previous identity.

4. H(n, 0) = tn+1 and H(0,m) = itm+1.

This can be shown using H(n+ 2, 0) = H(n+ 1, 0) +H(n, 0) +H(n− 1, 0) andH(0,m+2) = H(0,m+1)+H(0,m)+H(0,m−1) along the initial conditions.

5. H(n,m) = tm+1H(n, 1)+(tm+tm−1)H(n, 0) = tn+1H(1,m)+(tn+tn−1)H(0,m).

Consider the (3,3) entry in H(n+ 2,m+ 2) H(n+ 1,m+ 2) H(n,m+ 2)H(n+ 2,m+ 1) H(n+ 1,m+ 1) H(n,m+ 1)H(n+ 2,m) H(n+ 1,m) H(n,m)

=

tm+3 tm+2 + tm+1 tm+2

tm+2 tm+1 + tm tm+1

tm+1 tm + tm−1 tm

· H(n+ 2, 1) H(n+ 1, 1) H(n, 1)

H(n+ 2, 0) H(n+ 1, 0) H(n, 0)H(n+ 2,−1) H(n+ 1,−1) H(n,−1)

=

H(1,m+ 2) H(0,m+ 2) H(−1,m+ 2)H(1,m+ 1) H(0,m+ 1) H(−1,m+ 1)H(1,m) H(0,m) H(−1,m)

· tn+3 tn+2 tn+1

tn+2 + tn+1 tn+1 + tn tn + tn−1tn+2 tn+1 tn

.And apply the fact that H(n,−1) = H(−1,m) = 0.

6. H(n,m) = tn+1tm+2 + itn+2tm+1.

Consider the (3,3) entry in H(n+ 2,m+ 2) H(n+ 1,m+ 2) H(n,m+ 2)H(n+ 2,m+ 1) H(n+ 1,m+ 1) H(n,m+ 1)H(n+ 2,m) H(n+ 1,m) H(n,m)

=

tm+2 tm+1 + tm tm+1

tm+1 tm + tm−1 tmtm tm−1 + tm−2 tm−1

· 2 + 2i 2 + i i

1 + 2i 1 + i i1 1 0

· tn+2 tn+1 tntn+1 + tn tn + tn−1 tn−1 + tn−2tn+1 tn tn−1

=

tm+4 tm+3 + tm+2 tm+3

tm+3 tm+2 + tm+1 tm+2

tm+2 tm+1 + tm tm+1

· 0 0 1

0 0 0i 0 0

· tn+4 tn+3 tn+2

tn+3 + tn+2 tn+2 + tn+1 tn+1 + tntn+3 tn+2 tn+1

=

itm+3 0 tm+4

itm+2 0 tm+3

itm+1 0 tm+2

· tn+4 tn+3 tn+2

tn+3 + tn+2 tn+2 + tn+1 tn+1 + tntn+3 tn+2 tn+1

=

X X XX X XX X tm+2tn+1 + itm+1tn+2

.It could be more identities for this section. Also we can generalize H(n,m) toany linear recurrence with constant coefficient of order 3 with some specific initialcondition that mentioned in page 2 (page 145 in the book) of Pethe, [12].

35

6.4 Higher Dimensions

Consider 3 dimension analog of Harman from the same paper, [7].

Define G(l,m, n) by

G(l + 1,m, n) = G(l,m, n) +G(l − 1,m, n),

G(l,m+ 1, n) = G(l,m, n) +G(l,m− 1, n),

G(l,m, n+ 1) = G(l,m, n) +G(l,m, n− 1)

where G(a, b, c) = [a, b, c] for a, b, c ∈ {0, 1}.

Theorem 6.4.1 (This is an analog of (6) in previous section).

G(l,m, n) = [FlFm+1Fn+1, Fl+1FmFn+1, Fl+1Fm+1Fn].

Proof. We could define a 2− 2− 2 matrix which contains values of G(l,m, n). Thenthe calculation is similar to (6) of previous section.

References

[1] A. Benjamin and J. Quinn, Proofs that really count, Mathematical Associationof America, 2003

[2] George Berzsenyi, Gaussian Fibonacci Numbers, The Fibonacci Quarterly15(1977): 233-236.

[3] Bloom, D.M., 1996, Problem 55, Math Horizons (Sept.), 32.

[4] Bahar Demirturk and Refik Keskin, Some new Fibonacci and Lucas identitiesby matrix methods , International Journal of Mathematical Education in Scienceand Technology, 41:3, 379-387, October 2009.

[5] Bahar Demirturk, Fibonacci and Lucas Sums by matrix methods , InternationalMathematical Forum, 5, 2010, no.3, 99-107.

[6] S. Halici, On Fibonacci Quaternions, Advances in Applied Clifford Algebras, June2012

[7] C. J. Harman, Complex Fibonacci Numbers, The Fibonacci Quarterly 19(1981):82-86.

[8] Muthulakshmi R. Iyer, Some Results on Fibonacci Quaternions, The FibonacciQuarterly, Volume 7, Number 2, April, 1969: 201-210.

[9] Robert C Johnson ,Fibonacci numbers and matrices, available atmaths.dur.ac.uk/∼dma0rcj/PED/fib.pdf.

36

[10] J.H. Jordan, Gaussian Fibonacci and Lucas Numbers, The Fibonacci Quarterly,Volume 3, Number 4, December, 1965: 315-318.

[11] T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley 2001

[12] S. Pethe, Some identities For Tribonacci Sequences, The Fibonacci Quarterly,Volume 26, Number 2, May, 1988: 144-151.

37