3. H-matrix generated by Fibonacci numbers

Let X and Y be two subsets of ω. Let A ¼ ð Þ ank be an infinite matrix of real or complex numbers ank, where n, k ∈ N. Then, the matrix A defines the A-transformation from X into Y, if for every sequence x ¼ ð Þ xk ∈ X the sequence Ax ¼ ð Þ Ax <sup>n</sup> � �, the A-transform of x exists and is in Y where

$$(A\mathfrak{x})\_n = \sum\_k a\_{nk}\mathfrak{x}\_k.$$

For simplicity in notation, here and in what follows, the summation without limits runs from 0 to ∞. By ð Þ X; Y , we denote the class of all such matrices. A sequence x is said to be A-summable to l if Ax converges to l which is called as the A-limit of x.

For a sequence space X, the matrix domain XA of an infinite matrix A is defined as

$$X\_A = \{ \mathbf{x} = (\mathbf{x}\_k) \in \boldsymbol{\omega} : A \mathbf{x} \in \mathbf{X} \}, \tag{1}$$

which is a sequence space.

An infinite matrix A ¼ ð Þ ank is said to be regular if and only if the following conditions (or Toplitz conditions) hold [17–19]:

i. lim<sup>n</sup>!<sup>∞</sup> X∞ k¼0 ank ¼ 1,

phyllotaxis objects, it is used usually the number ratios of the left-hand and right-hand spirals observed on the surface of the phyllotaxis objects. Botanists proved that these ratios are equal

By using hyperbolic Fibonacci functions, he had developed an original geometric theory of phyllotaxis and explained why Fibonacci spirals arise on the surface of the phyllotaxis objects namely, pine cones, cacti, pine apple, heads of sunflower, and so on, in process of their growths. Bodnar's geometry [15] confirms that these functions are 'natural' functions of the nature, which show their value in the botanic phenomenon of phyllotaxis. This fact allows us to assert that these functions can be attributed to the class of fundamental mathematical discoveries of contemporary science because they reflect natural phenomena, in particular,

From above discussion, it gave us motivation to see the behavior of the infinite matrices

Fibonacci numbers f <sup>n</sup> and we call it as H-matrix generated by Fibonacci numbers f <sup>n</sup> and introduce some new sequence spaces related to matrix domain of H in the sequence spaces

Sheikh and Ganie [16] introduced the Riesz sequence space <sup>r</sup><sup>q</sup>ð Þ <sup>u</sup>; <sup>p</sup> and studied its various topological properties where u ¼ ð Þ uk is a sequence such that uk 6¼ 0 for all k ∈ N and qk

qk, ∀n∈ N

0, if k > n:

� � is given by

if 0 ≤ k ≤ n,

Qn <sup>¼</sup> <sup>X</sup><sup>n</sup> k¼0

> 8 < :

� � is regular if and only if Qn ! <sup>∞</sup> as <sup>n</sup> ! <sup>∞</sup>.

ukqk Qn

Let X and Y be two subsets of ω. Let A ¼ ð Þ ank be an infinite matrix of real or complex numbers ank, where n, k ∈ N. Then, the matrix A defines the A-transformation from X into Y, if for every

� � of the Riesz mean Ru; qn

r q nk ¼

3. H-matrix generated by Fibonacci numbers

In the present chapter, we have introduced a new type of matrix <sup>H</sup> <sup>¼</sup> hu

<sup>8</sup> , … <sup>¼</sup> <sup>1</sup> <sup>þ</sup> ffiffiffi

5 p 2

nk

� � n, k∈ N by using

� � the

to the ratios of the adjacent Fibonacci numbers, that is,

phyllotaxis phenomenon.

80 Matrix Theory-Applications and Theorems

generated by Fibonacci numbers.

lp, l∞, c and c0, where 1 ≤ p < ∞.

sequence of positive numbers and

<sup>u</sup> ¼ r q nk

2.8. The space rqð Þ <sup>u</sup>; <sup>p</sup>

Then, the matrix R<sup>q</sup>

The Riesz mean Ru; qn

<sup>f</sup> <sup>i</sup>þ<sup>1</sup> f i : 2 1 , 3 2 , 5 3 , 8 5 , 13


In the present paper, we introduce <sup>H</sup>-matrix with <sup>H</sup> <sup>¼</sup> hu nk � � n, k<sup>∈</sup> <sup>N</sup> as follows:

$$h\_{nk}^u = \begin{cases} \frac{\mu\_k f\_k^{-2}}{f\_n f\_{n+1}} & \text{if } \ 0 \le k \le n, \\ 0, & \text{if } \ k > n. \end{cases}$$

Thus, for uk ¼ 1 and for all k∈ N, we have

$$H = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & \cdots \\ 1/2 & 1/2 & 0 & 0 & 0 & \cdots \\ 1/6 & 1/6 & 4/6 & 0 & 0 & \cdots \\ 1/15 & 1/15 & 4/15 & 9/15 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \end{pmatrix}.$$

It is obvious that the matrix H is a triangle, that is, hu nn 6¼ 0 and hu nk ¼ 0 for k > n and for all n∈ N. Also, since it satisfies the conditions of Toeplitz matrix and hence it is regular matrix.

Note that if we take qk ¼ f 2 <sup>k</sup> , then the matrix <sup>H</sup> is special case of the matrix Rq <sup>u</sup>, where

$$Q\_n = \sum\_{k=0}^n f\_k^2 = f\_n f\_{n+1'}$$

introduced by Sheikh and Ganie [16].

The approach of constructing a new sequence space by means of matrix domain of a particular limitation method has been studied by several authors [17–26].

Throughout the text of the chapter, X denotes any of the spaces l∞, c, c<sup>0</sup> and lp ð Þ 1 ≤ p < ∞ . Then, the Fibonacci sequence space X Hð Þ is defined by

$$X(H) = \{ \mathfrak{x} = (\mathfrak{x}\_k) \in \omega : y = (y\_k) \in X \}\_{\prime \prime}$$

where the sequence y ¼ yk � � is the <sup>H</sup>-transform of the sequence <sup>x</sup> <sup>¼</sup> ð Þ xk and is given by

$$y\_k = H\_k(\mathbf{x}) = \frac{1}{f\_k f\_{k+1}} \sum\_{i=0}^k f\_i^2 u\_i \mathbf{x}\_i \text{ for all } k \in \mathbb{N}.\tag{2}$$

This shows that H xð Þ¼ y and since y ∈ X, we conclude that H xð Þ∈ X. Thus, we deduce that

which shows that T is norm preserving. Hence, T is isometry. Consequently, the spaces X Hð Þ

1 <sup>f</sup> <sup>j</sup> <sup>f</sup> <sup>j</sup>þ<sup>1</sup> 1 A < ∞:

> ¼ 1 f n

<sup>f</sup> <sup>k</sup>þ<sup>1</sup> � <sup>f</sup> <sup>k</sup> <sup>f</sup> <sup>k</sup> <sup>f</sup> <sup>k</sup>þ<sup>1</sup> � �

<sup>f</sup> <sup>k</sup>�<sup>1</sup> <sup>f</sup> <sup>k</sup> <sup>f</sup> <sup>k</sup>þ<sup>1</sup> � �

1 <sup>f</sup> <sup>k</sup> <sup>f</sup> <sup>k</sup>þ<sup>1</sup> � �

> 5 <sup>p</sup> <sup>þ</sup><sup>1</sup> <sup>2</sup> .◊

Nature of Phyllotaxy and Topology of H-matrix http://dx.doi.org/10.5772/intechopen.74676 83

<sup>∣</sup>uj<sup>∣</sup> for all j∈ N. Thus, we have for every

∥T xð Þ∥ ¼ ∥y∥ ¼ ∥H xð Þ∥<sup>X</sup> ¼ ∥x∥<sup>X</sup>

and X are isometrically isomorphic. Hence, the proof of the Theorem is complete.◊

f 2 i X∞ j¼i

0 @

X∞ k¼n

1 ¼ f <sup>n</sup>

¼ f 2 n 1 f n X∞ k¼n

¼ f 2 n 1 f n X∞ k¼n

≥ f 2 n <sup>f</sup> <sup>n</sup>�<sup>1</sup> f n

and the conclusion follows because <sup>f</sup> <sup>n</sup> <sup>f</sup> <sup>n</sup>�<sup>1</sup> is bounded since it converges to ffiffi

∣Hið Þx ∣ ≤

which shows that H xð Þ ∈l∞. Therefore, we deduce that x∈ l<sup>∞</sup> implies x∈ l∞ð Þ H .

≤ K <sup>f</sup> <sup>i</sup> <sup>f</sup> <sup>i</sup>þ<sup>1</sup>

x∈ l∞. Then, there is a constant K > 0 such that ∣xj∣ < <sup>K</sup>

1 f k � <sup>1</sup> <sup>f</sup> <sup>k</sup>þ<sup>1</sup> � �

X∞ k¼n

1 f k � <sup>1</sup> <sup>f</sup> <sup>k</sup>þ<sup>1</sup> � �

X∞ k¼n

Proof: It is obvious that c<sup>0</sup> ⊂ c0ð Þ H and c⊂ c Hð Þ, since the matrix H is regular matrix. Now, let

1 <sup>f</sup> <sup>i</sup> <sup>f</sup> <sup>i</sup>þ<sup>1</sup> X i

j¼0 f 2 <sup>j</sup> ∣ujxj∣

X i

j¼0 f 2 <sup>j</sup> ¼ K

sup i

n o be Fibonacci number sequences. Then, we have

x∈ X Hð Þ and Tx ¼ y. Hence, T is surjective.

Theorem 3: Let f <sup>j</sup>

Proof: We have,

This gives,

i∈ N that

Theorem 4: X ⊂ X Hð Þ holds.

Furthermore, for any x∈ X Hð Þ, we have by (3) that

With the definition of matrix domain given by Eq. (1), we can redefine the space X Hð Þ as the matrix domain of the triangle H in the space X, that is,

$$\mathcal{X}(H) = \mathcal{X}\_H.$$

Theorem 1: The space X Hð Þ is a BK-space with the norm given by

$$\|\mathbf{x}\| = |H(\mathbf{x})|\_{X} = \|y\|\_{X} = \begin{cases} \left[\sum\_{k=0}^{\infty} |y\_k|^p\right]^{\frac{1}{p}} & \text{for } \text{ for } X \in \{l\_p\}. \\\sup\_k y\_k & \text{for } X \in \{l\_\ast, \mathbf{c}, \mathbf{c}\_0\}. \end{cases} \tag{3}$$

Proof: Since the matrix <sup>H</sup> <sup>¼</sup> <sup>h</sup><sup>u</sup> nk � � is a triangle, that is, hu nn 6¼ 0 and hu nk ¼ 0 for k > n for all n. We have the result by Eq. (3) and Theorem 4.3.2 of Wilansky [6] gives the fact that X Hð Þ is a BK-space.◊

Theorem 2: The space X Hð Þ is isometrically isomorphic to the space X.

Proof: To prove the result, we should show the linear bijection between the spaces X Hð Þ and X. For that, consider the transformation T from X Hð Þ to X by x ! y ¼ Tx. Then, the linearity of T follows from Eq. (2). Further, we see that x ¼ 0 whenever Tx ¼ 0 and consequently T is injective.

Moreover, let y ¼ yk � �<sup>∈</sup> <sup>X</sup> be given and define the sequence <sup>x</sup> <sup>¼</sup> ð Þ xk by

$$\mathbf{x}\_{k} = \frac{f\_{k+1}}{\mu\_{k} f\_{k}} y\_{k} - \frac{f\_{k-1}}{\mu\_{k} f\_{k}} y\_{k-1}; k \in \mathbf{N}. \tag{4}$$

Then, by using (2) and (4), we have for every k∈ N that

$$\begin{aligned} H(\mathbf{x}) &= \frac{1}{f\_k f\_{k+1}} \sum\_{i=0}^k f\_i^2 u\_i \mathbf{x}\_i \\ &= \frac{1}{f\_k f\_{k+1}} \sum\_{i=0}^k f\_i (f\_{i+1} y\_i - f\_{i-1} y\_{i-1}) \\ &= y\_k. \end{aligned}$$

This shows that H xð Þ¼ y and since y ∈ X, we conclude that H xð Þ∈ X. Thus, we deduce that x∈ X Hð Þ and Tx ¼ y. Hence, T is surjective.

Furthermore, for any x∈ X Hð Þ, we have by (3) that

$$\|\|T(\mathfrak{x})\|\| = \|\mathfrak{y}\| = \|H(\mathfrak{x})\|\_{X} = \|\mathfrak{x}\|\_{X}$$

which shows that T is norm preserving. Hence, T is isometry. Consequently, the spaces X Hð Þ and X are isometrically isomorphic. Hence, the proof of the Theorem is complete.◊

Theorem 3: Let f <sup>j</sup> n o be Fibonacci number sequences. Then, we have

$$\sup\_{i} \left( f\_i^2 \sum\_{j=i}^{\infty} \frac{1}{f\_j f\_{j+1}} \right) < \infty$$

Proof: We have,

The approach of constructing a new sequence space by means of matrix domain of a particular

Throughout the text of the chapter, X denotes any of the spaces l∞, c, c<sup>0</sup> and lp ð Þ 1 ≤ p < ∞ . Then,

X k

i¼0 f 2

With the definition of matrix domain given by Eq. (1), we can redefine the space X Hð Þ as the

X Hð Þ¼ XH:

P<sup>∞</sup> <sup>k</sup>¼<sup>0</sup> yk � � � � <sup>p</sup> � �<sup>1</sup>

have the result by Eq. (3) and Theorem 4.3.2 of Wilansky [6] gives the fact that X Hð Þ is a

Proof: To prove the result, we should show the linear bijection between the spaces X Hð Þ and X. For that, consider the transformation T from X Hð Þ to X by x ! y ¼ Tx. Then, the linearity of T follows from Eq. (2). Further, we see that x ¼ 0 whenever Tx ¼ 0 and consequently T is injective.

� �<sup>∈</sup> <sup>X</sup> be given and define the sequence <sup>x</sup> <sup>¼</sup> ð Þ xk by

X k

i¼0 f 2 <sup>i</sup> uixi

X k

<sup>ð</sup><sup>f</sup> <sup>i</sup>þ<sup>1</sup>yi � <sup>f</sup> <sup>i</sup>�<sup>1</sup>yi�<sup>1</sup><sup>Þ</sup>

i¼0 f i

yk � <sup>f</sup> <sup>k</sup>�<sup>1</sup> uk f <sup>k</sup>

sup k

8 < :

� � is a triangle, that is, hu

� �∈ X � �,

� � is the <sup>H</sup>-transform of the sequence <sup>x</sup> <sup>¼</sup> ð Þ xk and is given by

<sup>i</sup> uixi for all k ∈ N: (2)

� �:

yk�<sup>1</sup>; k <sup>∈</sup> <sup>N</sup>: (4)

nk ¼ 0 for k > n for all n. We

(3)

<sup>p</sup> for for X ∈ lp

yk for X ∈f g l∞; c; c<sup>0</sup> :

nn 6¼ 0 and hu

X Hð Þ¼ x ¼ ð Þ xk ∈ ω : y ¼ yk

1 <sup>f</sup> <sup>k</sup> <sup>f</sup> <sup>k</sup>þ<sup>1</sup>

limitation method has been studied by several authors [17–26].

yk ¼ Hkð Þ¼ x

Theorem 1: The space X Hð Þ is a BK-space with the norm given by

kxk k ¼ jH xð Þ <sup>X</sup> ¼ k ky <sup>X</sup> ¼

nk

Theorem 2: The space X Hð Þ is isometrically isomorphic to the space X.

xk <sup>¼</sup> <sup>f</sup> <sup>k</sup>þ<sup>1</sup> uk f <sup>k</sup>

<sup>H</sup>ðxÞ ¼ <sup>1</sup>

<sup>f</sup> <sup>k</sup> <sup>f</sup> <sup>k</sup>þ<sup>1</sup>

<sup>¼</sup> <sup>1</sup> <sup>f</sup> <sup>k</sup> <sup>f</sup> <sup>k</sup>þ<sup>1</sup>

¼ yk:

Then, by using (2) and (4), we have for every k∈ N that

matrix domain of the triangle H in the space X, that is,

the Fibonacci sequence space X Hð Þ is defined by

where the sequence y ¼ yk

82 Matrix Theory-Applications and Theorems

Proof: Since the matrix <sup>H</sup> <sup>¼</sup> <sup>h</sup><sup>u</sup>

BK-space.◊

Moreover, let y ¼ yk

$$\sum\_{k=n}^{n} \left( \frac{1}{f\_k} - \frac{1}{f\_{k+1}} \right) = \frac{1}{f\_n}$$

This gives,

$$1 = f\_n \sum\_{k=n}^{\infty} \left(\frac{1}{f\_k} - \frac{1}{f\_{k+1}}\right)$$

$$= f\_n^2 \frac{1}{f\_n} \sum\_{k=n}^{\infty} \left(\frac{f\_{k+1} - f\_k}{f\_k f\_{k+1}}\right)$$

$$= f\_n^2 \frac{1}{f\_n} \sum\_{k=n}^{\infty} \left(\frac{f\_{k-1}}{f\_k f\_{k+1}}\right)$$

$$\ge f\_n^2 \frac{f\_{n-1}}{f\_n} \sum\_{k=n}^{\infty} \left(\frac{1}{f\_k f\_{k+1}}\right)$$

and the conclusion follows because <sup>f</sup> <sup>n</sup> <sup>f</sup> <sup>n</sup>�<sup>1</sup> is bounded since it converges to ffiffi 5 <sup>p</sup> <sup>þ</sup><sup>1</sup> <sup>2</sup> .◊ Theorem 4: X ⊂ X Hð Þ holds.

Proof: It is obvious that c<sup>0</sup> ⊂ c0ð Þ H and c⊂ c Hð Þ, since the matrix H is regular matrix. Now, let x∈ l∞. Then, there is a constant K > 0 such that ∣xj∣ < <sup>K</sup> <sup>∣</sup>uj<sup>∣</sup> for all j∈ N. Thus, we have for every i∈ N that

$$\begin{aligned} |H\_i(\mathbf{x})| &\le \frac{1}{f\_i f\_{i+1}} \sum\_{j=0}^i f\_j^2 |\mu\_j \mathbf{x}\_j| \\ &\le \frac{K}{f\_i f\_{i+1}} \sum\_{j=0}^i f\_j^2 = K \end{aligned}$$

which shows that H xð Þ ∈l∞. Therefore, we deduce that x∈ l<sup>∞</sup> implies x∈ l∞ð Þ H .

We now consider the case 1 ≤ p < ∞. We only consider the case 1 < p < ∞ and by similar argument will follow for p ¼ 1. So, let x∈ lp. Then, for every i∈ N and by Holder's inequality, we have

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$$\begin{aligned} \|H\_i(\mathbf{x})\|^p &\le \left(\sum\_{j=0}^i \frac{f\_j^2}{f\_i f\_{i+1}} |u\_j \mathbf{x}\_j|\right)^p \\ &\le \left(\sum\_{j=0}^i \frac{f\_j^2}{f\_i f\_{i+1}} |u\_j \mathbf{x}\_j|\right)^p \left(\sum\_{j=0}^i \frac{f\_j^2}{f\_i f\_{i+1}}\right)^{p-1} \\ &= \frac{1}{f\_i f\_{i+1}} \sum\_{j=0}^i f\_j^2 |u\_j \mathbf{x}\_j|^p. \end{aligned}$$

Hence, we have

$$\sum\_{i=0}^{\infty} |H\_i(\mathbf{x})|^p \le \sum\_{i=0}^{\infty} \frac{1}{f\_i f\_{i+1}} \sum\_{j=0}^i f\_j^2 \left| u\_j \mathbf{x}\_j \right|^p$$

$$= \sum\_{i=0}^{\infty} |\mathbf{x}\_j|^p \left| u\_j \right|^p f\_j^2 \sum\_{i=j}^{\infty} \frac{1}{f\_i f\_{i+1}}.$$

Hence, the right-hand side of above inequality can be made arbitrary small, since, sup<sup>j</sup> f 2 j P<sup>∞</sup> i¼j 1 <sup>f</sup> <sup>i</sup> <sup>f</sup> <sup>i</sup>þ<sup>1</sup> � � <sup>&</sup>lt; <sup>∞</sup> by Theorem 3 (above) and <sup>x</sup><sup>∈</sup> lp. This shows that <sup>x</sup>∈lpð Þ <sup>H</sup> . This completes the proof of the theorem.◊
