On Powers, Roots and Moore–Penrose Inverses of Matrices Via Generalized Fibonacci Numbers

Sinan Karakaya; Halim Özdemir; Ahmet Yaşar Özban

doi:10.2478/amsil-2026-0002

Full Article

1.

Introduction and preliminaries

The sequence {F_n} defined by the recurrence relation F_n = F_n−1 + F_n−2 for all integers n ≥ 2, where F₀ = 0 and F₁ = 1 is the well-known Fibonacci sequence. Similarly, the sequence {L_n} defined by L_n = L_n−1 + L_n−2, n ≥ 2, where L₀ = 2 and L₁ = 1 is the Lucas sequence and the sequence {M_n} defined by M_n = 3M_n−1 − 2M_n−2, n ≥ 2 with the initial conditions M₀ = 0 and M₁ = 1 is the Mersenne sequence; see [2, 3, 9, 16].

Throughout the study ℕ, ℤ, ℝ and ℂ will denote, respectively, the set of natural numbers, the set of integers, the field of real numbers and the field of complex numbers. Moreover ℳ_k×l (𝔽), where 𝔽 = ℝ or 𝔽 = ℂ, will denote the set of all k × l matrices over the field 𝔽, and I will denote the appropriately sized unit matrix.

The generalized Fibonacci sequence {U_n} and the generalized Lucas sequence {V_n} are, respectively, defined by U_n = pU_n−₁ + qU_n−₂ and V_n = pV_n−₁ + qV_n−₂, n ∈ ℕ, n ≥ 2, where U₀ = 0, U₁ = 1, V₀ = 2 and V₁ = p with p, q ∈ ℝ\{0}. Based on these, negatively indexed generalized Fibonacci and generalized Lucas numbers are defined by $U_{- n} = - \frac{U_{n}}{{(- q)}^{n}}$ {U_{- n}} = - {{{U_n}} \over {{{(- q)}^n}}} and $V_{- n} = \frac{y_{n}}{{(- q)}^{n}}$ {V_{- n}} = {{{y_n}} \over {{{(- q)}^n}}} , respectively, for all n ∈ ℕ. Meanwhile, under the condition that p² + 4q > 0 we have $U_{n} = \frac{α^{n} - β^{n}}{α - β}$ {U_n} = {{{\alpha^n} - {\beta^n}} \over {\alpha - \beta}} and V_n = αⁿ + βⁿ, known as Binet formulas for all n ∈ ℤ, where $α = (p + \sqrt{p^{2} + 4 q}) / 2$ \alpha = \left({p + \sqrt {{p^2} + 4q}} \right)/2 and $β = (p - \sqrt{p^{2} + 4 q}) / 2$ \beta = \left({p - \sqrt {{p^2} + 4q}} \right)/2 are the solutions of the characteristic equation x² − px − q = 0. The Fibonacci sequence {F_n} and the Lucas sequence {L_n} are, respectively, the particular cases of the sequences {U_n} and {V_n} for p = q = 1. On the other hand, the Mersenne sequence {M_n} is the particular case of the sequence {U_n} for p = 3 and q = −2. Also, it can be easily seen that M_n = 2ⁿ − 1 for all n ∈ ℤ by using Binet formula; see [2, 3, 8, 13, 14].

There are many identities in the related literature about generalized Fibonacci and Lucas numbers; see [8, 13, 14]. Some important identities that we will use in this work are given below. Under the conditions that p, q ∈ ℝ\{0} and p² + 4q > 0, the following identities hold for all m, n, r ∈ ℤ [14]. (1.1) $U_{m} V_{n} = U_{m + n} + {(- q)}^{n} U_{m - n},$ {U_m}{V_n} = {U_{m + n}} + {(- q)^n}{U_{m - n}}, (1.2) $V_{n} = U_{n + 1} + q U_{n - 1},$ {V_n} = {U_{n + 1}} + q{U_{n - 1}}, (1.3) $U_{n + r} U_{n - r} - U_{n}^{2} = - {(- q)}^{n - r} U_{r}^{2},$ {U_{n + r}}{U_{n - r}} - U_n^2 = - {(- q)^{n - r}}U_r^2, (1.3) $U_{m + n} = U_{m} U_{n + 1} + q U_{m - 1} U_{n} .$ {U_{m + n}} = {U_m}{U_{n + 1}} + q{U_{m - 1}}{U_n}. A matrix A ∈ ℳ_m×m (ℝ) that satisfies the equality Aⁿ = B is called a nth root of the matrix B ∈ ℳ_m×m (ℝ), where n is a positive integer [11].

The roots of matrices need to be calculated in solving some problems in different fields such as statistics, economics, and healthcare. For example, the Markov chain model is used in finance and healthcare. The transitions of a Markov chain can be described by a stochastic matrix, and determining the stochastic nth root of a stochastic matrix is an important problem for Markov chain models. For this reason, matrix roots are in the field of interest of those working in the areas mentioned above and different methods of obtaining them have been offered in literature; see [5,6,7, 15, 17].

It is known that the problems that arise in applied sciences basically include the solutions or the functions of the solutions of the system of linear equations Ax = g, where A ∈ ℳ_m×l (ℝ), x ∈ ℳ_l×1 (ℝ) is a vector of unknowns, and g ∈ ℳ_m×1 (ℝ) is a vector of known values. On the other hand, if A is an n × n nonsingular matrix, then the system has the unique solution x = A⁻¹b. However, there are cases where A is singular or not a square matrix. In these cases, the system may have a unique solution or infinitely many solutions. That’s why a general theory that lays out all these is desired. One such theory involves the use of generalized and Moore–Penrose inverses of matrices [4]. Moreover, note that there are, of course, different methods to calculate the Moore–Penrose inverse of a matrix; see [1, 4, 12].

A matrix G ∈ ℳ_l×m (ℝ) is called a generalized inverse of a matrix A ∈ ℳ_m×l (ℝ) if AGA = A [1]. For any A ∈ ℳ_m×l (ℝ), there is a unique matrix A^† ∈ ℳ_l×m (ℝ) satisfying the conditions (AA^†)^T = AA^†, (A^†A)^T = A^†A, AA^†A = A, and A^†AA^† = A^†, where the matrix A^T is the transpose of the matrix A and the matrix A^† is known as the Moore–Penrose inverse of the matrix A. Moreover, the property (1.5) $A^{†} = {(A^{T} A)}^{†} A^{T} = A^{T} {(A A^{T})}^{†}$ {A^\dagger} = {({A^T}A)^\dagger}{A^T} = {A^T}{(A{A^T})^\dagger} is a well-known fact [1, 4].

The aim of the study is to give results which can be used to calculate positive integer powers, roots and Moore–Penrose inverses of some class of matrices, more specifically, the class of matrices X ∈ ℳ_m×m (ℝ) satisfying the matrix equation X³ − V_nX² + (−q)ⁿ X = 0, via generalized Fibonacci numbers and to support these results with some numerical examples.

Before giving the main results, it will be beneficial to introduce some terminologies which will be used in the subsequent lines. For an X ∈ ℳ_m×m, p_X (λ) = det (λI − X) is the characteristic polynomial of X. The eigenvalues of X are the roots of the characteristic equation p_X (λ) = 0 or, equivalently, are the zeros of the characteristic polynomial p_X (λ). Well-known Cayley-Hamilton theorem states that p_X (X) = 0. The minimal polynomial of X is the monic polynomial m_X (λ) of least degree such that m_X (X) = 0. If P (λ) is a polynomial such that P (X) = 0 then the minimal polynomial m_X (λ) divides P (λ). Moreover, m_X (λ) divides p_X (λ) and, m_X (λ) and p_X (λ) have the same zeros. All the eigenvalues of a real symmetric matrix X are real.

Now consider the polynomial P (λ) = λ³ − V_nλ² + (−q)ⁿ λ, n ∈ ℕ∪{0} and let X ∈ ℳ_m×m (ℝ) satisfies P (X) = 0 for some n ∈ ℕ∪ {0}. The polynomial P can be written as P (λ) = (λ − ϕ₁) (λ − ϕ₂) λ, where $ϕ_{1} = (V_{n} + \sqrt{V_{n}^{2} - 4 {(- q)}^{n}}) / 2$ {\phi_1} = ({V_n} + \sqrt {V_n^2 - 4{{(- q)}^n}})/2 and $ϕ_{2} = (V_{n} - \sqrt{V_{n}^{2} - 4 {(- q)}^{n}}) / 2$ {\phi_2} = ({V_n} - \sqrt {V_n^2 - 4{{(- q)}^n}})/2 . So, σ_X ⊆ {ϕ₁, ϕ₂, 0}, where σ_X is the spectrum of the matrix X.

As we have stated earlier $α = (p + \sqrt{p^{2} + 4 q}) / 2$ \alpha = \left({p + \sqrt {{p^2} + 4q}} \right)/2 , $β = (p - \sqrt{p^{2} + 4 q}) / 2 \in ℝ$ \beta = \left( {p - \sqrt {{p^2} + 4q} } \right)/2 \in {\mathbb R} , where p, q ∈ R, pq ≠ 0 and p² + 4q > 0, are the roots of the characteristic equation x² − px − q = 0. So, α + β = p and αβ = −q. Then using Binet formula V_n = αⁿ + βⁿ for all n ∈ ℤ, we obtain $\begin{array}{l} V_{n}^{2} - 4 {(- q)}^{n} & = {(α^{n} + β^{n})}^{2} - 4 {(- q)}^{n} \\ = α^{2 n} + β^{2 n} + 2 {(α β)}^{n} - 4 {(- q)}^{n} \\ = α^{2 n} + β^{2 n} - 2 {(α β)}^{n} \\ = {(α^{n} - β^{n})}^{2} \geq 0. \end{array}$ \matrix{{V_n^2 - 4{{(- q)}^n}} \hfill & {= {{({\alpha^n} + {\beta^n})}^2} - 4{{(- q)}^n}} \hfill \cr {} \hfill & {= {\alpha^{2n}} + {\beta^{2n}} + 2{{(\alpha \beta)}^n} - 4{{(- q)}^n}} \hfill \cr {} \hfill & {= {\alpha^{2n}} + {\beta^{2n}} - 2{{(\alpha \beta)}^n}} \hfill \cr {} \hfill & {= {{({\alpha^n} - {\beta^n})}^2} \ge 0.} \hfill \cr} Since α ≠ ±β, $V_{n}^{2} - 4 {(- q)}^{n} = 0$ V_n^2 - 4{(- q)^n} = 0 only if n = 0 and n = 0 leads to V₀ = 2, it follows that $ϕ_{1} = ϕ_{2} = \frac{V_{0}}{2} = 1 \in ℝ$ {\phi _1} = {\phi _2} = {{{V_0}} \over 2} = 1 \in {\mathbb R} and hence, P (λ) = λ (λ − 1)². So, if $V_{n}^{2} - 4 {(- q)}^{n} = 0$ V_n^2 - 4{(- q)^n} = 0 , then n = 0 and m_X (λ) ∈ 𝒮₁, where $S_{1} = {λ, (λ - 1), λ (λ - 1), λ {(λ - 1)}^{2}} .$ {S_1} = \{\lambda,\;\left({\lambda - 1} \right),\;\lambda \left({\lambda - 1} \right),\;\lambda {(\lambda - 1)^2}\}. On the other hand, for n ∈ ℕ we have $V_{n}^{2} - 4 {(- q)}^{n} > 0$ V_n^2 - 4{(- q)^n} > 0 . Then ϕ₁, ϕ₂ ∈ ℝ\{0} and ϕ₁ ≠ ϕ₂. So, P (λ) = λ (λ − ϕ₁) (λ − ϕ₂). Therefore, m_X (λ) ∈ 𝒮₂, where $\begin{array}{l} S_{2} = {λ, (λ - ϕ_{i}), λ (λ - ϕ_{i}), (λ - ϕ_{i}) (λ - ϕ_{j}), λ (λ - ϕ_{i}) (λ - ϕ_{j}) : \\ i, j = 1, 2, i \neq j} . \end{array}$ \eqalign{& {S_2} = \{\lambda,\;(\lambda - {\phi_i}),\;\lambda (\lambda - {\phi_i}),\;(\lambda - {\phi_i})(\lambda - {\phi_j}),\;\lambda (\lambda - {\phi_i})(\lambda - {\phi_j}):\, \cr & \,\,\,\,\,\,\,\,\,\,\,\,i,\;j = 1,2,\;i \ne j\}. \cr} So, for any given X ∈ ℳ_m×m, if X³ − V_nX² + (−q)ⁿ X = 0 for some n ∈ ℕ∪{0}, then m_X (λ) ∈ 𝒮₁ or m_X (λ) ∈ 𝒮₂. Conversely, if m_X (λ) ∈ 𝒮₁ or m_X (λ) ∈ 𝒮₂, then m_X (λ) divides P (λ) therefore X³ − V_nX² + (−q)ⁿ X = 0.

Having characterized the matrices satisfying the matrix equation X³ − V_nX² + (−q)ⁿ X = 0, we can now begin to give our main results.

The relationships between some sequences and matrices have been investigated in different studies. In [14], it has been shown that the integer powers of the matrices X ∈ ℳ_m×m(ℝ) satisfying the matrix equation X² − pX − qI = 0 can be found using the generalized Fibonacci sequence {U_n} and in this way algebraic properties of the sequence {U_n} are obtained. Based on these, in [10] it has been shown that the integer powers of the matrices satisfying X² − V_nX + (−q)ⁿ I = 0 can be obtained using the sequence {U_n} and some applications have been given.

In this study we will deal with powers of matrices X ∈ ℳ_m×m(ℝ) satisfying a higher degree matrix equation, namely, X³ − V_nX² + (−q)ⁿ X = 0. First, we will give a relation between generalized Fibonacci numbers and positive integer powers of the matrices satisfying the matrix equation. Then, we will develop a formula for the nth roots of such matrices. It is clear that the matrices X ∈ ℳ_m×m(ℝ) satisfying the matrix equation X³ − V_nX² + (−q)ⁿ X = 0 may be singular or not. For that reason, in the last part of the study we will give some results which can be used for finding the Moore–Penrose inverses of such matrices. Note that the Moore–Penrose inverse applies not only to square matrices, but also to rectangular matrices. In the rest of the work it will be assumed that p, q ∈ ℝ\{0} and p² + 4q > 0.

2.

Results

The main result given below shows the relationship between generalized Fibonacci numbers and positive integer powers of matrices X ∈ ℳ_m×m (ℝ) satisfying the third-degree matrix equation given above and is key to the rest of the work.

Theorem 2.1

If a matrix X ∈ ℳ_m×m (ℝ) satisfies the equation X³ − V_nX² + (−q)ⁿ X = 0 for some n ∈ ℕ, then $\begin{matrix} X^{k} = \frac{1}{U_{n}} (U_{nk - n} X - {(- q)}^{n} U_{nk - 2 n} I) X & for all k \in ℕ . \end{matrix}$ \matrix{{{X^k} = {1 \over {{U_n}}}({U_{nk - n}}X - {{( - q)}^n}{U_{nk - 2n}}I)X} & {for\,all\,\,k \in {\mathbb N}.}}

Proof

Proof is obtained by using induction method with equality (1.1) for m = nk − n.

It is clear that for n = 1, we have V₁ = p and Theorem 2.1 provides the following formula for the positive integer powers of the matrix X.

Corollary 2.2

If X ∈ ℳ_m×m (ℝ) satisfies the equation X³ − pX² − qX = 0, then X^k = U_k−1X² + qU_k−2X for all k ∈ ℕ.

Example 2.3

Let $X = (\begin{matrix} 4 & 2 & 2 & 4 \\ 2 & - 1 & - 3 & - 1 \\ 3 & - 3 & - 1 & - 1 \\ - 4 & 1 & 1 & - 1 \end{matrix}) .$ X = \left({\matrix{4 \hfill & 2 \hfill & 2 \hfill & 4 \hfill \cr 2 \hfill & {- 1} \hfill & {- 3} \hfill & {- 1} \hfill \cr 3 \hfill & {- 3} \hfill & {- 1} \hfill & {- 1} \hfill \cr {- 4} \hfill & 1 \hfill & 1 \hfill & {- 1} \hfill \cr}} \right). Then, since p_X (λ) = λ(λ + 3)(λ − 2)² and hence m_X (λ) = λ (λ + 3) (λ − 2) = λ³ + λ² − 6λ, it is easily seen that X³ + X² − 6X = 0. Therefore, X³ − V_nX² + (−q)ⁿX = 0 if V_n = −1 and (−q)ⁿ = −6 for some n ∈ ℕ. Now, if we let, for example, n = 1, then V₁ = p = −1 and q = 6. Then, U_k = − U_k−1 + 6U_k−2 and hence, by Corollary 2.2, $\begin{array}{l} X^{k} = U_{k - 1} X^{2} + q U_{k - 2} X = \\ (\begin{matrix} 8 U_{k - 1} + 24 U_{k - 2} & 4 U_{k - 1} + 12 U_{k - 2} & 4 U_{k - 1} + 12 U_{k - 2} & 8 U_{k - 1} + 24 U_{k - 2} \\ 4 U_{k - 1} + 12 U_{k - 2} & 13 U_{k - 1} - 6 U_{k - 2} & 9 U_{k - 1} - 18 U_{k - 2} & 13 U_{k - 1} - 6 U_{k - 2} \\ 4 U_{k - 1} + 12 U_{k - 2} & 9 U_{k - 1} - 18 U_{k - 2} & 13 U_{k - 1} - 6 U_{k - 2} & 13 U_{k - 1} - 6 U_{k - 2} \\ - 8 U_{k - 1} - 24 U_{k - 2} & 6 U_{k - 2} - 13 U_{k - 1} & 6 U_{k - 2} - 13 U_{k - 1} & - 17 U_{k - 1} - 6 U_{k - 2} \end{matrix}) \end{array}$ \eqalign{& {X^k} = {U_{k - 1}}{X^2} + q{U_{k - 2}}X = \cr & \left({\matrix{{8{U_{k - 1}} + 24{U_{k - 2}}} \hfill & {4{U_{k - 1}} + 12{U_{k - 2}}} \hfill & {4{U_{k - 1}} + 12{U_{k - 2}}} \hfill & {8{U_{k - 1}} + 24{U_{k - 2}}} \hfill \cr {4{U_{k - 1}} + 12{U_{k - 2}}} \hfill & {13{U_{k - 1}} - 6{U_{k - 2}}} \hfill & {9{U_{k - 1}} - 18{U_{k - 2}}} \hfill & {13{U_{k - 1}} - 6{U_{k - 2}}} \hfill \cr {4{U_{k - 1}} + 12{U_{k - 2}}} \hfill & {9{U_{k - 1}} - 18{U_{k - 2}}} \hfill & {13{U_{k - 1}} - 6{U_{k - 2}}} \hfill & {13{U_{k - 1}} - 6{U_{k - 2}}} \hfill \cr {- 8{U_{k - 1}} - 24{U_{k - 2}}} \hfill & {6{U_{k - 2}} - 13{U_{k - 1}}} \hfill & {6{U_{k - 2}} - 13{U_{k - 1}}} \hfill & {- 17{U_{k - 1}} - 6{U_{k - 2}}} \hfill \cr}} \right) \cr} for all k ∈ ℕ.

Having stated the result, which enables us to find the positive integer powers of the matrices under consideration, now we can give the following that gives the nth roots of the matrices satisfying the matrix equation.

Theorem 2.4

If X ∈ ℳ_m×m (ℝ) satisfies X³ − V_nX² + (−q)ⁿ X = 0 for some n ∈ ℕ then Cⁿ = X, where $C = \frac{1}{{(- q)}^{n - 1} U_{n}} (- U_{n - 1} X^{2} + U_{2 n - 1} X)$ C = {1 \over {{{(- q)}^{n - 1}}{U_n}}}(- {U_{n - 1}}{X^2} + {U_{2n - 1}}X) .

Proof

If we use (1.1) for m = n − 1, and necessary arrangements are made, (2.1) $CX = \frac{1}{U_{n}} (X^{2} - q U_{n - 1} X)$ CX = {1 \over {{U_n}}}({X^2} - q{U_{n - 1}}X) is obtained. Now, let $A = \frac{1}{{(- q)}^{n - 1} U_{n}} (- U_{n - 1} X + U_{2 n - 1} I)$ A = {1 \over {{{(- q)}^{n - 1}}{U_n}}}(- {U_{n - 1}}X + {U_{2n - 1}}I) . It is obvious that the matrices C, X and A are mutually commutative. If we use (2.1), then we easily obtain (2.2) $C^{2} = CXA = \frac{1}{U_{n}} (X - q U_{n - 1} I) C$ {C^2} = CXA = {1 \over {{U_n}}}(X - q{U_{n - 1}}I)C and (2.3) $C^{3} = \frac{1}{U_{n}^{2}} {(X - q U_{n - 1} I)}^{2} C,$ {C^3} = {1 \over {U_n^2}}{(X - q{U_{n - 1}}I)^2}C, since C = XA. So, using (1.2), (2.2) and (2.3) we get (2.4) $C^{3} - p C^{2} = \frac{1}{U_{n}^{2}} (q (U_{n + 1} U_{n - 1} + {(- q)}^{n - 1})) C .$ {C^3} - p{C^2} = {1 \over {U_n^2}}(q({U_{n + 1}}{U_{n - 1}} + {(- q)^{n - 1}}))C. Now, to use (1.3) for r = 1 in (2.4) leads to C³ − pC² − qC = 0. So, we can use Corollary 2.2 for the matrix C, and hence we obtain that Cⁿ = U_n−1C² + qU_n−2C for all n ∈ ℕ. So, if n − 1 and 1 are taken instead of n and r in (1.3), respectively, and performed some necessary operations, we get (2.5) $C^{n} = \frac{1}{U_{n}} ({(- q)}^{n - 1} C + U_{n - 1} CX) .$ {C^n} = {1 \over {{U_n}}}\left({{{(- q)}^{n - 1}}C + {U_{n - 1}}CX} \right). Using (2.1) in (2.5) then gives (2.6) $C^{n} = \frac{1}{U_{n}^{2}} (U_{2 n - 1} - q U_{n - 1}^{2}) X .$ {C^n} = {1 \over {U_n^2}}({U_{2n - 1}} - qU_{n - 1}^2)X. If the equation (1.4) used for n = m − 1, $U_{2 m - 1} = U_{m}^{2} + q U_{m - 1}^{2}$ {U_{2m - 1}} = U_m^2 + qU_{m - 1}^2 or, equivalently, (2.7) $U_{2 n - 1} - q U_{n - 1}^{2} = U_{n}^{2}$ {U_{2n - 1}} - qU_{n - 1}^2 = U_n^2 is obtained. Combining the equations (2.6) and (2.7), we get Cⁿ = X, and thus the proof is completed.

Example 2.5

Let $A = (\begin{matrix} 29 & 41 & 29 & 29 \\ 70 & 99 & 70 & 70 \\ 41 & 58 & 41 & 41 \\ 29 & 41 & 29 & 29 \end{matrix}) .$ A = \left({\matrix{{29} & {41} & {29} & {29} \cr {70} & {99} & {70} & {70} \cr {41} & {58} & {41} & {41} \cr {29} & {41} & {29} & {29} \cr}} \right). Then, it can be easily seen that A³ − 198A² + A = 0. If we use the sequences {U_n} and {V_n} established with p = 14, q = 1, then, we get A³ − V₂A² + (−q)² A = 0.

If we use Theorem 2.4 for n = 2, $C = \frac{1}{- U_{2}} (- U_{1} A^{2} + U_{3} A) = \frac{1}{14} (A^{2} - 197 A) = (\begin{matrix} 2 & 3 & 2 & 2 \\ 5 & 7 & 5 & 5 \\ 3 & 4 & 3 & 3 \\ 2 & 3 & 2 & 2 \end{matrix})$ C = {1 \over {- {U_2}}}(- {U_1}{A^2} + {U_3}A) = {1 \over {14}}({A^2} - 197A) = \left({\matrix{2 & 3 & 2 & 2 \cr 5 & 7 & 5 & 5 \cr 3 & 4 & 3 & 3 \cr 2 & 3 & 2 & 2 \cr}} \right) is obtained, and C² = A. Again, using {U_n} and {V_n} for p = 6, q = −1, we get A³ − V₃A² + (−q)³ A = 0. Hence, if we use Theorem 2.4 for n = 3, then, $D = \frac{1}{U_{3}} (- U_{2} A^{2} + U_{5} A) = \frac{1}{35} (- 6 A^{2} + 1189 A) = (\begin{matrix} 1 & 1 & 1 & 1 \\ 2 & 3 & 2 & 2 \\ 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{matrix})$ D = {1 \over {{U_3}}}\left({- {U_2}{A^2} + {U_5}A} \right) = {1 \over {35}}\left({- 6{A^2} + 1189A} \right) = \left({\matrix{1 & 1 & 1 & 1 \cr 2 & 3 & 2 & 2 \cr 1 & 2 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr}} \right) is obtained, and D³ = A. That is, C and D are square root and cube root of the matrix A, respectively.

In the literature, there are many studies in which some identities related to some sequences of numbers are obtained by using matrices. Similarly, some identities related to generalized Fibonacci numbers can be obtained by using Theorem 2.1 for some class of matrices. But, we will consider different results of Theorem 2.1. Recall that, for a matrix X ∈ ℳ_m×m (ℝ), if P (X) = 0 for some polynomial P (λ), then m_X (λ), the minimal polynomial of X, divides P (λ). So, the matrices X ∈ ℳ_m×m (ℝ) satisfying the equation X³ − V_nX² + (−q)ⁿ X = 0 may be singular or not. But of course one can consider the Moore–Penrose inverse of such matrices. The next result is about the Moore–Penrose inverses of the matrices satisfying the equation X³ − V_nX² + (−q)ⁿ X = 0.

Theorem 2.6

If X ∈ ℳ_m×m (ℝ) satisfies the matrix equation X³ − V_nX² + (−q)ⁿ X = 0 for some n ∈ ℕ and X² − V_nX is symmetric, then $G = \frac{1}{{(- q)}^{2 n} U_{n}} (- U_{2 n} X^{2} + U_{3 n} X)$ G = {1 \over {{{(- q)}^{2n}}{U_n}}}(- {U_{2n}}{X^2} + {U_{3n}}X) is the Moore–Penrose inverse of the matrix X.

Proof

First of all, from (1.1) for m = 2n and m = n, we get (2.8) $U_{3 n} - U_{2 n} V_{n} = - {(- q)}^{n} U_{n}$ {U_{3n}} - {U_{2n}}{V_n} = - {(- q)^n}{U_n} and (2.9) $U_{2 n} = U_{n} V_{n},$ {U_{2n}} = {U_n}{V_n}, respectively. Now, we have to show that four conditions in the definition of the Moore–Penrose inverse are satisfied for the matrices X and G.

If we use (2.8) and (2.9), (2.10) $XG = - \frac{1}{{(- q)}^{n}} (X^{2} - V_{n} X)$ XG = - {1 \over {{{(- q)}^n}}}({X^2} - {V_n}X) is obtained. Since X² − V_nX is symmetric, XG is also symmetric.
Since GX = XG, from (1) GX is also symmetric.
We have to show that XGX = X. Using (2.10) we get $XGX = - \frac{1}{{(- q)}^{n}} (X^{3} - V_{n} X^{2}) .$ XGX = - {1 \over {{{(- q)}^n}}}({X^3} - {V_n}{X^2}). Since X³ − V_nX² = − (−q)ⁿ X it follows that XGX = X.
Finally, we must show that GXG = G. $GX = - \frac{1}{{(- q)}^{n}} (X^{2} - V_{n} X)$ GX = - {1 \over {{{(- q)}^n}}}({X^2} - {V_n}X) , since GX = XG. If X³ − V_nX² + (−q)ⁿ X = 0 is taken into consideration and necessary arrangements are made, then GXG = G is easily obtained and the proof is completed.

Corollary 2.7

If X ∈ ℳ_m×m (ℝ) is a symmetric matrix satisfying X³ − V_nX² + (−q)ⁿ X = 0 for some n ∈ ℕ then the matrix $G = \frac{1}{{(- q)}^{2 n} U_{n}} (- U_{2 n} X^{2} + U_{3 n} X)$ G = {1 \over {{{(- q)}^{2n}}{U_n}}}(- {U_{2n}}{X^2} + {U_{3n}}X) is the Moore–Penrose inverse of the matrix X.

Proof

Since X is a symmetric matrix, X² − V_nX is also symmetric. So, the proof directly follows from Theorem 2.6.

As is seen, Theorem 2.6 provides a new way to find the Moore–Penrose inverses of matrices satisfying the matrix equation X³ − V_nX² + (−q)ⁿ X = 0 using generalized Fibonacci numbers. It is clear that, Corollary 2.7 can be used easily for symmetric matrix X satisfying the matrix equation. Now, we will find the Moore–Penrose inverses of some rectangular matrices X ∈ ℳ_m×l (ℝ) or non-symmetric matrices X ∈ ℳ_m×m (ℝ) for which X^T X or XX^T satisfy the cubic matrix equation. Let X ∈ ℳ_m×l (ℝ). According to (1.5), it is sufficient to find the matrices (X^T X)^† or (XX^T)^† to find the Moore–Penrose inverse of X. Since X^T X and XX^T are symmetric, they satisfy the first condition of Corollary 2.7. Now, if the matrix X^T X or XX^T satisfies an equation of the form x³ − ax² + bx = 0, where a, b ∈ ℝ, ab ≠ 0 and a² − 4b > 0, then we can establish a sequence {V_n} that meets the conditions V_n = a and (−q)ⁿ = b. Although we can choose any number from the set ℕ as n, notice that the formula that requires the least number of calculations to find the Moore–Penrose inverse of a matrix X is $G = \frac{1}{{(- q)}^{2} U_{1}} (- U_{2} X^{2} + U_{3} X)$ G = {1 \over {{{(- q)}^2}{U_1}}}(- {U_2}{X^2} + {U_3}X) , which is obtained for n = 1. So taking n = 1 and using the conditions V₀ = 2 and V₁ = p ∈ ℝ\{0}, we can establish the sequences {V_n} and {U_n} for p = a and q = −b. Then, we can get the Moore–Penrose inverse of the matrix X.

Example 2.8

Let $A = (\begin{matrix} 1 & - 1 & 2 \\ 2 & - 1 & 1 \\ 4 & - 3 & 5 \end{matrix})$ A = \left({\matrix{1 & {- 1} & 2 \cr 2 & {- 1} & 1 \cr 4 & {- 3} & 5 \cr}} \right) . Then, ${AA}^{T} = (\begin{matrix} 6 & 5 & 17 \\ 5 & 6 & 16 \\ 17 & 16 & 50 \end{matrix})$ A{A^T} = \left({\matrix{6 & 5 & {17} \cr 5 & 6 & {16} \cr {17} & {16} & {50} \cr}} \right) and the characteristic polynomial is p_A (λ) = λ³ − 62λ² + 66λ. From Cayley-Hamilton Theorem, we get (AA^T³) − 62 (AA^T)² + 66 (AA^T) = 0. To use Corollary 2.7, we need to construct a sequence {V_n} so that V_n = 62 and (−q)ⁿ = 66. Of course, such a construction is not unique as we stated earlier. For example, we can establish a sequence {V_n} satisfying the conditions V₂ = 62 and (−q)² = 66 or satisfying the conditions V₁ = 62 and (−q)¹ = 66. Let {V_n} be the generalized Lucas sequence established by using p = 62, q = −66. So, V₁ = p = 62. So, we can say that ${({AA}^{T})}^{3} - V_{1} {({AA}^{T})}^{2} + {(- q)}^{1} ({AA}^{T}) = 0.$ {(A{A^T})^3} - {V_1}{(A{A^T})^2} + {(- q)^1}\left({A{A^T}} \right) = 0. Thus, the matrix AA^T satisfies the condition of Corollary 2.7 for n = 1. According to Corollary 2.7, ${({AA}^{T})}^{†} = \frac{1}{{(- q)}^{2} U_{1}} (- U_{2} {({AA}^{T})}^{2} + U_{3} ({AA}^{T})),$ {(A{A^T})^\dagger} = {1 \over {{{(- q)}^2}{U_1}}}(- {U_2}{(A{A^T})^2} + {U_3}(A{A^T})), where U₁, U₂ and U₃ are elements of sequences {U_n} which established by using p = 62, q = −66. So, ${({AA}^{T})}^{†} = \frac{1}{198} (\begin{matrix} 44 & - 77 & 11 \\ - 77 & 137 & - 17 \\ 11 & - 17 & 5 \end{matrix})$ {(A{A^T})^\dagger} = {1 \over {198}}\left({\matrix{{44} & {- 77} & {11} \cr {- 77} & {137} & {- 17} \cr {11} & {- 17} & 5 \cr}} \right) is obtained. Hence, from (1.5), the Moore–Penrose inverse of A is given by $A^{†} = A^{T} {({AA}^{T})}^{†} = \frac{1}{66} (\begin{matrix} - 22 & 43 & - 1 \\ 0 & - 3 & - 3 \\ 22 & - 34 & 10 \end{matrix}) .$ {A^\dagger} = {A^T}{(A{A^T})^\dagger} = {1 \over {66}}\left({\matrix{{- 22} & {43} & {- 1} \cr 0 & {- 3} & {- 3} \cr {22} & {- 34} & {10} \cr}} \right).

Noteworthy that Theorem 2.6 can be used not only for square matrices but also for rectangular matrices as the following two examples show.

Example 2.9

Let $C = (\begin{matrix} 1 & 3 & 5 & 7 \\ 2 & 4 & 6 & 8 \end{matrix})$ C = \left({\matrix{1 & 3 & 5 & 7 \cr 2 & 4 & 6 & 8 \cr}} \right) . Then, $X = C^{T} C = (\begin{array}{l} 5 & 11 & 17 & 23 \\ 11 & 25 & 39 & 53 \\ 17 & 39 & 61 & 83 \\ 23 & 53 & 83 & 113 \end{array}) .$ X = {C^T}C = \left({\matrix{5 \hfill & {11} \hfill & {17} \hfill & {23} \hfill \cr {11} \hfill & {25} \hfill & {39} \hfill & {53} \hfill \cr {17} \hfill & {39} \hfill & {61} \hfill & {83} \hfill \cr {23} \hfill & {53} \hfill & {83} \hfill & {113} \hfill \cr}} \right). So, p_X (λ) = λ⁴ − 204λ³ + 80λ² and m_X (λ) = λ³ − 204λ² + 80λ. It is clear that X³ − 204X² + 80X = 0. Therefore, X³ − V_nX² + (−q)ⁿ X = 0 if V_n = 204 and (−q)ⁿ = 80 for some n ∈ ℕ. Letting n = 1, which requires the least computation yields V₁ = p = 204 and q = −80. Then, according to Corollary 2.7, we get $X^{†} = {(C^{T} C)}^{†} = \frac{1}{{(- q)}^{2} U_{1}} (- U_{2} X^{2} + U_{3} X) .$ {X^\dagger} = {({C^T}C)^\dagger} = {1 \over {{{(- q)}^2}{U_1}}}\left({- {U_2}{X^2} + {U_3}X} \right). Here, U₁, U₂ and U₃ are the elements of the sequence {U_n} where U_n = 204U_n−1 − 80U_n−2. So, $X^{†} = {(C^{T} C)}^{†} = \frac{1}{400} (\begin{matrix} 689 & 353 & 17 & - 319 \\ 353 & 181 & 9 & - 163 \\ 17 & 9 & 1 & - 7 \\ - 319 & - 163 & - 7 & 149 \end{matrix})$ {X^\dagger} = {({C^T}C)^\dagger} = {1 \over {400}}\left({\matrix{{689} & {353} & {17} & {- 319} \cr {353} & {181} & 9 & {- 163} \cr {17} & 9 & 1 & {- 7} \cr {- 319} & {- 163} & {- 7} & {149} \cr}} \right) is obtained. According to (1.5), $C^{†} = \frac{1}{20} (\begin{matrix} - 20 & 17 \\ - 10 & 9 \\ 0 & 1 \\ 10 & - 7 \end{matrix}) .$ {C^\dagger} = {1 \over {20}}\left({\matrix{{- 20} & {17} \cr {- 10} & 9 \cr 0 & 1 \cr {10} & {- 7} \cr}} \right).

Example 2.10

Let $D = \frac{1}{2} (\begin{matrix} 2 & 2 & 1 \\ - 2 & - 2 & 1 \\ 2 & - 2 & 1 \\ 2 & - 2 & - 1 \end{matrix}) .$ D = {1 \over 2}\left({\matrix{2 & 2 & 1 \cr {- 2} & {- 2} & 1 \cr 2 & {- 2} & 1 \cr 2 & {- 2} & {- 1} \cr}} \right). Then, ${DD}^{T} = \frac{1}{4} (\begin{array}{l} 9 & - 7 & 1 & - 1 \\ - 7 & 9 & 1 & - 1 \\ 1 & 1 & 9 & 7 \\ - 1 & - 1 & 7 & 9 \end{array})$ D{D^T} = {1 \over 4}\left({\matrix{9 \hfill & {- 7} \hfill & 1 \hfill & {- 1} \hfill \cr {- 7} \hfill & 9 \hfill & 1 \hfill & {- 1} \hfill \cr 1 \hfill & 1 \hfill & 9 \hfill & 7 \hfill \cr {- 1} \hfill & {- 1} \hfill & 7 \hfill & 9 \hfill \cr}} \right) is obtained. Also, it can be easily seen that (DD^T)³ − 5(DD^T)² + 4(DD^T) = 0. Now, we can establish {V_n} with p = 3 and q = −2. So, we get (DD^T)³ − V₂(DD^T)² + (−q)²(DD^T) = 0. So, matrix DD^T satisfies the condition of Corollary 2.7 for n = 2. Now, we can establish {U_n} with p = 3 and q = −2. In this case, {U_n} be Mersenne sequence {M_n}. If Corollary 2.7 is used, then $\begin{array}{l} {({DD}^{T})}^{†} & = \frac{1}{{(2)}^{4} M_{2}} (- M_{4} {({DD}^{T})}^{2} + M_{6} ({DD}^{T}) \\ = \frac{1}{{(2)}^{4} 3} (- 15 {({DD}^{T})}^{2} + 63 ({DD}^{T})) \\ = \frac{1}{8} (\begin{matrix} 3 & 1 & 2 & - 2 \\ 1 & 3 & 2 & - 2 \\ 2 & 2 & 3 & - 1 \\ - 2 & - 2 & - 1 & 3 \end{matrix}) \end{array}$ \matrix{{{{(D{D^T})}^\dagger}} \hfill & {= {1 \over {{{(2)}^4}{M_2}}}(- {M_4}{{(D{D^T})}^2} + {M_6}(D{D^T})} \hfill \cr {} \hfill & {= {1 \over {{{(2)}^4}3}}(- 15{{(D{D^T})}^2} + 63(D{D^T}))} \hfill \cr {} \hfill & {= {1 \over 8}\left({\matrix{3 & 1 & 2 & {- 2} \cr 1 & 3 & 2 & {- 2} \cr 2 & 2 & 3 & {- 1} \cr {- 2} & {- 2} & {- 1} & 3 \cr}} \right)} \hfill \cr} is obtained. According to (1.5), $D^{†} = \frac{1}{4} (\begin{matrix} 1 & - 1 & 1 & 1 \\ 1 & - 1 & - 1 & - 1 \\ 2 & 2 & 2 & - 2 \end{matrix})$ {D^\dagger} = {1 \over 4}\left({\matrix{1 & {- 1} & 1 & 1 \cr 1 & {- 1} & {- 1} & {- 1} \cr 2 & 2 & 2 & {- 2} \cr}} \right) is found.

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