#  $\mathcal{P}$-canonical forms and Drazin inverses

M. MOUÇOUF

###### Abstract.

We show that the $\mathcal{P}$-canonical form of $(A^{k})_{k}$ (or simply of
the matrix $A$) can be written as a sum of two parts, the geometric and the
non-geometric parts of $A$. We also show that if we plug $-k$ for $k$ in the
geometric part of $A$, we get the $\mathcal{P}$-canonical form of the Drazin
inverse $A_{d}$ of $A$.

###### Key words and phrases:

Matrix, $\mathcal{P}$-Canonical form, Powers, Linear recurrence sequences,
Binomial coefficients, Drazin inverses

###### 2010 Mathematics Subject Classification:

15AXX, 05A10, 11B37, 15A09

Department of Mathematics, Faculty of Science, Chouaib Doukkali University,
Morocco
Email<EMAIL_ADDRESS>
##  1\. Introduction

Let $A$ be a nonsingular matrix over a field $F$. It is a well known fact
that, for many numerical examples of matrices $A$, replacing $k$ with $-k$ in
certain forms of $A^{k}$, one can obtain the kth power of the inverse $A^{-1}$
of $A$. However, this fact is not proven in general. The problem here is that
the kth power of a matrix can be represented in several forms. In fact, It can
happen that in certain forms of $A^{k}$ it is not even possible to substitute
$k=-1$ and, in other forms we can substitute $k=-1$ but we does not obtain
$A^{-1}$. For example, Let $A$ be an $r$-circulant matrix. The expressions of
$A^{k}$ given in Theorem 4.1 of [7] and Theorem 3.1 of [8] do not provide
$A^{-1}$ when we plug into them $k=-1$.
In this paper, we consider an arbitrary matrix $A$, singular or nonsingular,
and we prove that if we plug $-k$ for $k$ into the geometric part of $A^{k}$
we get the kth power of the Drazin inverse of $A$. In order to avoid any
confusion that may arise by using the plugging-in operation, we begin by
showing that the representations of $A^{k}$ into them we plug $-k$ for $k$,
are canonical.
Throughout this paper, we use the following notations:

  * •

$F$ denotes a field and $\mathcal{C}_{F}$ denotes the set of all sequences
$\boldsymbol{s}=(s_{k})_{k\geqslant 0}$ over $F$. It is well known that
$\mathcal{C}_{F}$ is an $F$-algebra under componentwise addition,
multiplication and scalar multiplication.

  * •

$\mathcal{S}^{\ast}=\\{\boldsymbol{\lambda}=(\lambda^{k})_{k\geqslant
0}/\lambda\in F,\lambda\neq 0\\}$ be the set of all nonzero geometric
sequences.

  * •

$\mathcal{S}^{\circ}=\\{\boldsymbol{0}_{i}\in\mathcal{C}_{F}/i\in\mathbb{N},\boldsymbol{0}_{i}(k)=\delta_{ik}\\}$.

  * •

$\mathcal{S}=\mathcal{S}^{\ast}\cup\mathcal{S}^{\circ}$.

  * •

$F_{\mathcal{S}}$ denotes the $F$-vector spaces spanned by $\mathcal{S}$. It
is well known that $F_{\mathcal{S}}$ is the set of all linear recurrence
sequences in $F$ whose carateristic polynomials are of the form $X^{m}P(X)$
where $P(X)$ is square-free and split with nonzero constant terms.

  * •

$F_{\mathcal{S}^{\ast}}$ denotes the $F$-vector spaces spanned by
$\mathcal{S}^{\ast}$. It is well known that $F_{\mathcal{S}^{\ast}}$ is the
set of all linear recurrence sequences in $F$ whose carateristic polynomials
are square-free and split with nonzero constant terms.

  * •

$F_{\mathcal{S}^{\circ}}$ denotes the $F$-vector spaces spanned by
$\mathcal{S}^{\circ}$. It is well known that $F_{\mathcal{S}^{\circ}}$ is the
set of all linear recurrence sequences in $F$ whose carateristic polynomials
split and have zero as its only root.

It is well known from the general theory of linear recurrence sequences that

  * •

$\mathcal{S}$, $\mathcal{S}^{\ast}$ and $\mathcal{S}^{\circ}$ are basis of
$F_{\mathcal{S}}$, $F_{\mathcal{S}^{\ast}}$ and $F_{\mathcal{S}^{\circ}}$,
respectively.

  * •

$F_{\mathcal{S}}$, $F_{\mathcal{S}^{\ast}}$ and $F_{\mathcal{S}^{\circ}}$ are
subalgebras of $\mathcal{C}_{F}$. More precisely, the set $\mathcal{S}^{\ast}$
is a group, and hence $F_{\mathcal{S}^{\ast}}$ is exactly the group algebra of
$\mathcal{S}^{\ast}$ over $F$.

  * •

$F_{\mathcal{S}^{\circ}}$ and $F_{\mathcal{S}^{\ast}}$ are supplementary
vector spaces relative to $F_{\mathcal{S}}$, i.e.,
$F_{\mathcal{S}}=F_{\mathcal{S}^{\circ}}\oplus F_{\mathcal{S}^{\ast}}$.

##  2\. Canonical forms for linear recurrence sequences

Let $\mathcal{T}=\\{\Lambda_{i}/i\geqslant 0\\}$ be the set of all sequences
$\Lambda_{i}=(\binom{k}{i})_{k\geq 0}$ and consider the $F$-vector space
$F\langle\mathcal{T}\rangle$ spanned by $\mathcal{T}$. Then we have the
following result

######  Proposition 2.1.

Let $F$ be a field. Let $\widetilde{\mathcal{T}}$ be the set of all sequences
$\widetilde{\Lambda}_{i}=(\binom{-k}{i})_{k\geq 0}$ and consider the sequence
$\Gamma=(0,1,2,\ldots)$. Then we have
1\. $F\langle\mathcal{T}\rangle$ is a subalgebra of $\mathcal{C}_{F}$ and
$\mathcal{T}$ is a basis of $F\langle\mathcal{T}\rangle$.
2\. $\widetilde{\mathcal{T}}$ is a basis of $F\langle\mathcal{T}\rangle$, and
the linear automorphism $\chi$ of $F\langle\mathcal{T}\rangle$ which maps
$\Lambda_{i}$ to $\widetilde{\Lambda}_{i}$ is an involution of the algebra
$F\langle\mathcal{T}\rangle$.
3\. $F\langle\mathcal{T}\rangle$ and $F_{\mathcal{S}^{\ast}}$ are $F$-linearly
disjoint.
4\. If $F$ is of characteristic $0$, then the sequence $\Gamma$ is
transcendental over the ring $F_{\mathcal{S}^{\ast}}$.
5\. $F_{\mathcal{S}}[\mathcal{T}]=F_{\mathcal{S}^{\circ}}\oplus
F_{\mathcal{S}^{\ast}}[\mathcal{T}]$, where $F_{\mathcal{S}}[\mathcal{T}]$ is
the subalgebras of $\mathcal{C}_{F}$ obtained by adjoining to
$F_{\mathcal{S}}$ the set $\mathcal{T}$.

###### Proof.

1\. Using the following formula (due to Riordan [5])

$\displaystyle\binom{k}{i}\binom{k}{j}$
$\displaystyle=\sum_{m=i}^{i+j}\binom{m}{i}\binom{i}{m-j}\binom{k}{m}$
$\displaystyle=\sum_{m=0}^{i}\binom{i+j-m}{m,i-m,j-m}\binom{k}{i+j-m}$

we can easily deduce that $\Lambda_{i}\Lambda_{j}$ is a linear combinaison of
elements of $\mathcal{T}$ for all $i,j$. Then $F\langle\mathcal{T}\rangle$ is
a subalgebra of $\mathcal{C}_{F}$. More precisely,
$F\langle\mathcal{T}\rangle$ is the set of all linear recurrence sequences in
$F$ with charateristic polynomials split over $F$ and have one as its only
root. It is well known that $\mathcal{T}$ is a basis of
$F\langle\mathcal{T}\rangle$.
2\. Let $i\in\mathbb{N}$. It is clear that
$\widetilde{\Lambda}_{0}=\Lambda_{0}$ and then $\widetilde{\Lambda}_{0}\in
F\langle\mathcal{T}\rangle$. Suppose that $i\geqslant 1$. Then we have
$\widetilde{\Lambda}_{i}=(-1)^{i}(\binom{k+i-1}{i})$. An easy application of
the Vandermonde convolution shows that

$\widetilde{\Lambda}_{i}=\sum_{j=1}^{i}(-1)^{i}\binom{i-1}{i-j}\Lambda_{j}.$

Then $\widetilde{\Lambda}_{i}\in F\langle\mathcal{T}\rangle$ for all
$i\in\mathbb{N}$. Since the coordinate matrix of the family
$(\widetilde{\Lambda}_{0},\ldots,\widetilde{\Lambda}_{i})$ relative to
$(\Lambda_{0},\ldots,\Lambda_{i})$ is the involutory lower triangular Pascal
matrix of order $i+1$, it follows that $\widetilde{\mathcal{T}}$ is a basis of
$F\langle\mathcal{T}\rangle$ and that $\chi$ is an involution.
We note that $\chi$ is an $F$-algebra automorphism is due to the fact that
Riordan’s formula remains true for any negative integer $k$.
3\. Let $\boldsymbol{\lambda}_{1},\ldots,\boldsymbol{\lambda}_{n}$ be any
family of pairwise distinct elements of $\mathcal{S}^{\ast}$ and $m$ be a
positive integer, and set
$P(X)=\displaystyle\prod_{j=1}^{n}(X-\lambda_{j})^{m+1}$. From Theorem 1 and
Theorem 2 of [4] assure that
$\\{\Lambda_{i}\boldsymbol{\lambda}_{j}/1\leqslant j\leqslant
n\,\,\text{and}\,\,0\leqslant i\leqslant m\\}$ is a basis of the $F$-vector
space of all linear recurrence sequences with characteristic polynomial $P$.
It follows that the set
$\\{\Lambda_{i}\boldsymbol{\lambda}/i\in\mathbb{N},\boldsymbol{\lambda}\in\mathcal{S}^{\ast}\\}$
is linearly independent over the field $F$. The result now follows from
Proposition 11.6.1. of [2] and the fact that $\mathcal{S}^{\ast}$ and
$\mathcal{T}$ are bases of the $F$-vector spaces $F_{\mathcal{S}^{\ast}}$ and
$F\langle\mathcal{T}\rangle$, respectively.
4\. It is well known that when $F$ is of characteristic $0$, the family
$\\{\Gamma^{i}/i\in\mathbb{N}\\}$) plays the same algebraic role as
$\mathcal{T}$. By the same argument as in the proof of 3., it turns that the
$F$-vector spaces $F_{\mathcal{S}^{\ast}}$ and $F[\Gamma]$ are linearly
disjoint. It then follows that $\Gamma$ is transcendental over the ring
$F_{\mathcal{S}^{\ast}}$.
5\. First note that
$F_{\mathcal{S}^{\circ}}[\mathcal{T}]=F_{\mathcal{S}^{\circ}}$ since
$\boldsymbol{0}_{i}\Lambda_{j}=\binom{i}{j}\boldsymbol{0}_{i}$ for all
integers $i$ and $j$. Now, since

$\\{\boldsymbol{0}_{0},\ldots,\boldsymbol{0}_{n^{\prime}}\\}\cup\\{\Lambda_{i}\boldsymbol{\lambda}_{j}/\boldsymbol{\lambda}_{0},\ldots,\boldsymbol{\lambda}_{n}\in\mathcal{S}^{\ast}\,\,\text{and}\,\,0\leqslant
i\leqslant m\\}$

is linearly independent over the field $F$, because it is a basis of the
$F$-vector space of all linear recurrence sequence with characteristic
polynomial $X^{n^{\prime}}\displaystyle\prod_{j=1}^{n}(X-\lambda_{j})^{m+1}$,
the result $F_{\mathcal{S}}[\mathcal{T}]=F_{\mathcal{S}^{\circ}}\oplus
F_{\mathcal{S}^{\ast}}[\mathcal{T}]$ follows. ∎

######  Remark 2.2.

To be more precise, Proposition 11.6.1. of [2] shows that if $A$ and $B$ are
$F$-subalgebras of $\Omega$, then the following statement are equivalent:
1\. $A$ and $B$ are $F$-linearly disjoint.
2\. $\\{u_{i}v_{j}\\}_{i,j}$ is linearly independent over $F$ whenever
$\\{u_{i}\\}_{i}$ is a $F$-basis of $A$ and $\\{v_{j}\\}_{j}$ is a $F$-basis
of $B$.
However, using this result it is easy to prove that the two following
statements are also equivalent:
1\. $A$ and $B$ are $F$-linearly disjoint.
2\. There exists a $F$-basis $\\{u_{i}\\}_{i}$ of $A$ and a $F$-basis
$\\{v_{j}\\}_{j}$ of $B$ such that $\\{u_{i}v_{j}\\}_{i,j}$ is linearly
independent over $F$.

From Proposition 2.1, it follows that each $\boldsymbol{u}\in
F_{\mathcal{S}}[\mathcal{T}]$ can be written in exactly one way as
$\boldsymbol{u}=\boldsymbol{v}+\boldsymbol{u}_{0}\Lambda_{0}+\cdots+\boldsymbol{u}_{m}\Lambda_{m}$,
in which $\boldsymbol{v}\in F_{\mathcal{S}^{\circ}}$ and
$\boldsymbol{u}_{0},\ldots,\boldsymbol{u}_{m}\in F_{\mathcal{S}^{\ast}}$ and
that this representation is canonical. For ease of reference, let us call

  * •

$\boldsymbol{v}+\boldsymbol{u}_{0}\Lambda_{0}+\cdots+\boldsymbol{u}_{m}\Lambda_{m}$
the canonical form of $\boldsymbol{u}$ relative to
$(\mathcal{S}^{\ast},\mathcal{T})$.

  * •

$\boldsymbol{u}_{0}\Lambda_{0}+\cdots+\boldsymbol{u}_{m}\Lambda_{m}$ the
geometric part of $\boldsymbol{u}$ relative to
$(\mathcal{S}^{\ast},\mathcal{T})$.

  * •

$\boldsymbol{v}$ the non-geometric part of $\boldsymbol{u}$.

Before going further let us state the following definition.

######  Definition 2.3.

We say a sequence of $F_{\mathcal{S}}[\mathcal{T}]$ to be purely geometric if
its non-geometric part is zero.

######  Lemma 2.4.

Let $\boldsymbol{u}$ a sequence over $F$. Suppose that there exists a sequence
$\boldsymbol{w}=\boldsymbol{w}_{0}\Lambda_{0}+\cdots+\boldsymbol{w}_{m}\Lambda_{m}\in
F_{\mathcal{S}^{\ast}}[\mathcal{T}]$ and a nonnegative integer $\tau$ such
that $\boldsymbol{u}_{k}=\boldsymbol{w}_{k}$ for all $k\geq\tau$. Then
$\boldsymbol{u}\in F_{\mathcal{S}}[\mathcal{T}]$; in this case,
$\boldsymbol{w}_{0}\Lambda_{0}+\cdots+\boldsymbol{w}_{m}\Lambda_{m}$ and
$(\boldsymbol{u}_{0}-\boldsymbol{w}_{0})\boldsymbol{0}_{0}+\cdots+(\boldsymbol{u}_{\tau-1}-\boldsymbol{w}_{\tau-1})\boldsymbol{0}_{\tau-1}$
are the geometric part and the non-geometric part of $\boldsymbol{u}$,
respectively.

###### Proof.

Follows immediately from the fact that

$\boldsymbol{u}=(\boldsymbol{u}_{0}-\boldsymbol{w}_{0})\boldsymbol{0}_{0}+\cdots+(\boldsymbol{u}_{\tau-1}-\boldsymbol{w}_{\tau-1})\boldsymbol{0}_{\tau-1}+\boldsymbol{w}.$

∎

######  Remark 2.5.

It is easily seen that two sequences $\boldsymbol{u},\boldsymbol{v}\in
F_{\mathcal{S}}[\mathcal{T}]$ are shift equivalent if and only if they have
the same geometric parts.

##  3\. $\mathcal{P}$-canonical forms of matrices

Let $\mathbf{M_{q}}(\mathcal{C}_{F})$ be the set of all matrices of order $q$
over $\mathcal{C}_{F}$ and let

$\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}=\mathbf{M_{q}}(F_{\mathcal{S}})\langle\mathcal{T}\rangle$

be the subalgebra of $\mathbf{M_{q}}(\mathcal{C}_{F})$ obtained by adjoining
to $\mathbf{M_{q}}(F_{\mathcal{S}})$ the set $\mathcal{T}$. It is
straightforward to check that $\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}$
is the set of all sequence matrices of order $q$ over $\mathcal{C}_{F}$ that
are linear recurrence sequences with characteristic polynomials split over
$F$, and that $\mathbf{M_{q}}(F_{\mathcal{S}^{\circ}})$ is the subset of
$\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}$ consisting of all sequence
matrices whose terms vanish from some point onwards. Since

$\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}=\mathbf{M_{q}}(F_{\mathcal{S}^{\circ}})\oplus\mathbf{M_{q}}(F_{\mathcal{S}^{\ast}})_{\mathcal{T}},$

it follows that each
$\boldsymbol{U}\in\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}$ can be
written in exactly one way as $\boldsymbol{U}=\boldsymbol{V}+\boldsymbol{W}$,
in which $\boldsymbol{V}\in\mathbf{M_{q}}(F_{\mathcal{S}^{\circ}})$ and
$\boldsymbol{W}\in\mathbf{M_{q}}(F_{\mathcal{S}^{\ast}})_{\mathcal{T}}$. Let
$\boldsymbol{U_{ij}}$ be the $(i,j)$-th entrie of $\boldsymbol{U}$, then
$\boldsymbol{V_{ij}}$ and $\boldsymbol{W_{ij}}$ are the non-geometric part and
the geometric part of $\boldsymbol{U_{ij}}$, respectively. We say then that
$\boldsymbol{V}$ and $\boldsymbol{W}$ are the non-geometric part and the
geometric part of $\boldsymbol{U}$, respectively. We note that there exist
matrices $\mathcal{V}_{0},\ldots,\mathcal{V}_{n}$ with coefficients in $F$ and
$\mathcal{W}_{0},\ldots,\mathcal{W}_{m}$ with coefficients in
$F_{\mathcal{S}^{\ast}}$ such that

$\boldsymbol{V}=\mathcal{V}_{0}\boldsymbol{0}_{0}+\cdots+\mathcal{V}_{n}\boldsymbol{0}_{n}\quad\text{is
the non-geometric part of}\quad\boldsymbol{U}$

and

$\boldsymbol{W}=\mathcal{W}_{0}\Lambda_{0}+\cdots+\mathcal{W}_{m}\Lambda_{m}\quad\text{is
the geometric part of}\quad\boldsymbol{U}.$

We note that, in view of Proposition 2.1, the matrices
$\mathcal{V}_{0},\ldots,\mathcal{V}_{n}$ and
$\mathcal{W}_{0},\ldots,\mathcal{W}_{m}$ are uniquely determined by
$\boldsymbol{U}$. We conclude that every matrix $\boldsymbol{U}$ of
$\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}$ can be written in the form

$\boldsymbol{U}=\mathcal{V}_{0}\boldsymbol{0}_{0}+\cdots+\mathcal{V}_{n}\boldsymbol{0}_{n}+\mathcal{W}_{0}\Lambda_{0}+\cdots+\mathcal{W}_{m}\Lambda_{m},$

where $\mathcal{V}_{0},\ldots,\mathcal{V}_{n}\in\mathbf{M_{q}}(F)$ and
$\mathcal{W}_{0},\ldots,\mathcal{W}_{m}\in\mathbf{M_{q}}(F_{\mathcal{S}^{\ast}})$,
and that these matrices are uniquely determined by $\boldsymbol{U}$. Let us
call this representation the $\mathcal{P}$-canonical form of $\boldsymbol{U}$,
and let us abbreviate this with “$\mathcal{P}$-cf”.
Let now $A$ be a matrix over $F$ and let $P_{A}$ be its characteristic
polynomial. It is well known that the entries of
$\boldsymbol{A}=(A^{k})_{k\geq 0}$ are linear recurrence sequences with
characteristic polynomial $P_{A}$. From this it follows that if $P_{A}$ splits
over $F$, then

$\boldsymbol{A}\in\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}=\mathbf{M_{q}}(F_{\mathcal{S}^{\circ}})\oplus\mathbf{M_{q}}(F_{\mathcal{S}^{\ast}})_{\mathcal{T}},$

and hence from the discussion above , there exist matrices
$\mathcal{A}_{0},\ldots,\mathcal{A}_{l}$ with coefficients in
$F_{\mathcal{S}^{\ast}}$ and a matrix $N(A)$ with coefficients in
$F_{\mathcal{S}^{\circ}}$ such that

$\boldsymbol{A}=N(A)+\mathcal{A}_{0}\Lambda_{0}+\cdots+\mathcal{A}_{l}\Lambda_{l},$

and this representation is the “$\mathcal{P}$-cf” of $\boldsymbol{A}$, or
simply of $A$. In the following proposition we give some properties of the
matrices $N(A),\mathcal{A}_{0},\ldots,\mathcal{A}_{l}$.

######  Proposition 3.1.

Let $A$ be a square matrix of order $q$ over $F$, and suppose that its minimal
polynomial is of the form
$m_{A}=X^{t_{0}}\prod_{j=1}^{p}\displaystyle(X-\lambda_{j})^{t_{j}}$ with the
$\lambda_{j}$ distinct and belong to $F$(possibly $t_{0}=0$). Let
$\pi_{0},\ldots,\pi_{p}$ be the spectral projections of $A$ at
$0,\lambda_{1},\ldots,\lambda_{p}$, respectively. Then we have
1\.
$\mathcal{A}_{i}=\displaystyle\sum_{j=1}^{p}\boldsymbol{\lambda}_{j}\lambda_{j}^{-i}(A-\lambda_{j}I_{q})^{i}\pi_{j}$,
$i\geqslant 1$,
$\mathcal{A}_{0}=\boldsymbol{\lambda}_{1}\pi_{1}+\cdots+\boldsymbol{\lambda}_{p}\pi_{p}$.
2\. $N(A)=\displaystyle\sum_{i=0}^{t_{0}-1}\boldsymbol{0}_{i}A^{i}\pi_{0}$
($N(A)=0$ if and only if $t_{0}=0$).
3\. $A$ is nilpotent if and only if each $\mathcal{A}_{i}$ is zero, i.e.,
$\boldsymbol{A}=N(A)$.
4\. $\mathcal{A}_{i}=0$ if and only if $i>m=\max\\{t_{1}-1,\ldots,t_{p}-1\\}$.
5\. If $A$ is not nilpotent, then no coefficient $\mathcal{A}_{i}$, $1\leq
i\leq m$, is zero.
6\.
$\mathcal{A}_{i}=\displaystyle\sum_{j=1}^{p}\boldsymbol{\lambda}_{j}\lambda_{j}^{-i}(A-\lambda_{j}I_{q})^{i}\pi_{j}$
is a nilpotent matrix, $i\geqslant 1$.
7\. $N(A)=\boldsymbol{0}_{0}\pi_{0}+N$ where $N$ is a nilpotent matrix.
8\.
$\boldsymbol{A}=(N+\mathcal{A}_{1}\Lambda_{1}+\cdots+\mathcal{A}_{m}\Lambda_{m})+(\boldsymbol{0}_{0}\pi_{0}+\mathcal{A}_{0}\Lambda_{0})$
is the unique decomposition of $\boldsymbol{A}$ as a sum of commuting
nilpotent and diagonalizable matrices.

###### Proof.

1\. and 2. We know that

$I_{q}=\pi_{0}+\pi_{1}+\cdots+\pi_{p},$

then

$A=(A\pi_{0})\pi_{0}+(A\pi_{1})\pi_{1}+\cdots+(A\pi_{p})\pi_{p}$

and hence

$A^{k}=(A\pi_{0})^{k}\pi_{0}+(A\pi_{1})^{k}\pi_{1}+\cdots+(A\pi_{p})^{k}\pi_{p}$

for all $k\geqslant 0$. But since
$A\pi_{i}=(A-\lambda_{i}I_{q})\pi_{i}+\lambda_{i}\pi_{i}$ and
$(A-\lambda_{i}I_{q})^{t_{i}}\pi_{i}=0$, we see that

$A^{k}=\sum_{i=0}^{t_{0}-1}0^{k-i}\binom{k}{i}A^{i}\pi_{0}+\sum_{i=1}^{p}(\sum_{j=0}^{s}\lambda_{i}^{k-j}(A-\lambda_{i}I_{q})^{j}\pi_{i}\binom{k}{j})$

for all $k\geqslant 0$ and $s\geqslant\max\\{t_{1}-1,\ldots,t_{p}-1\\}$.
Therefore,

$\boldsymbol{A}=N(A)+\mathcal{A}_{0}\Lambda_{0}+\cdots+\mathcal{A}_{s}\Lambda_{s},$

where

$\mathcal{A}_{i}=\displaystyle\sum_{j=1}^{p}\boldsymbol{\lambda}_{j}\lambda_{j}^{-i}(A-\lambda_{j}I_{q})^{i}\pi_{j},1\leqslant
i\leqslant s,$
$\mathcal{A}_{0}=\boldsymbol{\lambda}_{1}\pi_{1}+\cdots+\boldsymbol{\lambda}_{p}\pi_{p}$

and

$N(A)=\displaystyle\sum_{i=0}^{t_{0}-1}\boldsymbol{0}_{i}A^{i}\pi_{0}.$

3\. Is evident.
4\. Let $E^{1},\ldots,E^{p}$ be the generalized eigenspaces corresponding to
$\lambda_{1},\ldots,\lambda_{p}$, respectively. Let $v\in E^{j}$ and suppose
$\mathcal{A}_{i}=0$. Then we have
$\displaystyle\sum_{j=1}^{p}\boldsymbol{\lambda}_{j}\lambda_{j}^{-i}(A-\lambda_{j}I_{q})^{i}\pi_{j}(v)=0$,
hence $\boldsymbol{\lambda}_{j}\lambda_{j}^{-i}(A-\lambda_{j}I_{q})^{i}(v)=0$.
So $(A-\lambda_{j}I_{q})^{i}(v)=0$. Therefore, $E^{j}\subseteq
ker(A-\lambda_{j}I_{q})^{i}$ and then $i\geqslant t_{j}$. Consequently, we
have $i>m=\max\\{t_{1}-1,\ldots,t_{p}-1\\}$.
The converse follows immediately from the fact that

$A^{t_{0}}\pi_{0}=(A-\lambda_{j}I_{q})^{t_{j}}\pi_{j}=0,1\leqslant j\leqslant
p.$

5\. Is evident.
6\. Follows from the fact that if $i\geqslant 1$ then $\mathcal{A}_{i}$ is a
sum of commuting nilpotent matrices.
7\. Is evident.
8\. Let $A=A_{n}+A_{s}$ be the Jordan-Chevalley decomposition of $A$. Then
$A^{k}=(A_{n}+A_{s})^{k}=B_{k}+A^{k}_{s}$ where $B_{k}$ is a matrix which
commutes with $A_{n}$ and $A_{s}$. More precisely, $A^{k}=B_{k}+A^{k}_{s}$ is
the Jordan-Chevalley decomposition of $A^{k}$. Since
$(A^{k}_{s})=\boldsymbol{0}_{0}\pi_{0}+\mathcal{A}_{0}\Lambda_{0}$ it follows
that $(B_{k})=N+\mathcal{A}_{1}\Lambda_{1}+\cdots+\mathcal{A}_{m}\Lambda_{m}$.
The fact that the matrices $A_{s}^{k},k\geqslant 0$ are simultaneously
diagonalizable guarantees that the sequence matrix $(A_{s}^{k})$ is
diagonalizable. ∎

######  Remark 3.2.

The elements
$\boldsymbol{0}_{0},\boldsymbol{\lambda}_{1},\ldots,\boldsymbol{\lambda}_{p}$
are the only geometric sequences that are eigenvalues of the sequence matrix
$\boldsymbol{A}$.

######  Remark 3.3.

$A$ is invertible if and only if $N(A)=0$.

######  Remark 3.4.

If we plug $k=0$ into
$(\mathcal{A}_{0}\Lambda_{0}+\cdots+\mathcal{A}_{m}\Lambda_{m})_{k}$, we get
$\pi_{1}+\cdots+\pi_{p}=I_{q}-\pi_{0}$. Thus, if we have already determined
the geometric part of a matrix $A$, then we can easily check whether or not
$A$ is nonsingular. For this, it is sufficient to plug $k=0$ into the
geometric part of $A$ and see if we get the identity matrix or not.

We have the following corollaries

######  Corollary 3.5.

Let $A$ be a square matrix over $F$ and let $t_{0}$ be its index. Suppose
there exists a purely geometric sequence $A(k)_{k\geq 0}$ and a nonnegative
integer $\tau$ such that $A^{k}=A(k)$ for all $k\geq\tau$. Then we have
1\. $t_{0}\leq\tau$.
2\. $t_{0}$ is the smallest nonnegative integer for which
$A^{t_{0}}=A(t_{0})$. In this case we have $A^{k}=A(k)$ for all $k\geq t_{0}$.

###### Proof.

On the one hand, according to Lemma 2.4,

$N(A)=(I-A(0))\boldsymbol{0}_{0}+(A-A(1))\boldsymbol{0}_{1}+\cdots+(A^{\tau-1}-A(\tau-1))\boldsymbol{0}_{\tau-1}.$

On the other hand, according to Proposition 3.1,

$N(A)=\displaystyle\sum_{i=0}^{t_{0}-1}\boldsymbol{0}_{i}A^{i}\pi_{0}.$

Therefore, it follows that $t_{0}\leq\tau$ Since $t_{0}$ is the smallest
nonnegative integer for which $A^{t_{0}}\pi_{0}=0$. Furthermore, the unicity
of the representation of $N(A)$ ensures that $A^{i}-A(i)=A^{i}\pi_{0}\neq 0$,
$0\leq i\leq t_{0}-1$ and $A^{i}-A(i)=0$, $t_{0}\leq i\leq\tau$. Therefore,
$t_{0}$ is the smallest nonnegative integer for which $A^{t_{0}}=A(t_{0})$.
This completes the proof of the corollary. ∎

######  Corollary 3.6.

Let $A$ be a square matrix over $F$. Then the minimal polynomial of $A$ is
$m_{A}(X)=X^{t_{0}}\prod_{j=1}^{p}(X-\lambda_{j})^{t_{j}}$, where $t_{0}$ and
$t_{j},j\neq 0$, are respectively the greatest integers such that
$\boldsymbol{0}_{t_{0}-1}$ and $\boldsymbol{\lambda}_{j}\Lambda_{t_{j}-1}$
appear in the $\mathcal{P}$-cf of $A$.

###### Proof.

The proof of the Corollary follows immediately from the fact that $t_{0}$ and
$t_{j}$, $1\leq j\leq p$ are the the smallest nonnegative integer for which
$A^{t_{0}}\pi_{0}=0$ and $(A-\lambda_{j}I_{q})^{t_{j}}\pi_{j}=0$. ∎

######  Corollary 3.7.

Let $A$ be a square matrix over $F$ such that its characteristic polynomial
splits over $F$. Then $A$ is diagonalizable if and only if the geometric part
of $A$ is an element of $\mathbf{M_{q}}(F_{\mathcal{S}^{\ast}})$ and the non-
geometric part of $A$ is an element of $\boldsymbol{0}_{0}\mathbf{M_{q}}(F)$.

###### Proof.

A direct consequence of Proposition 3.1. ∎

The following example is from [6].

######  Example 3.8.

Let

$A=\begin{pmatrix}1&1&1&0\\\ 1&1&1&-1\\\ 0&0&-1&1\\\ 0&0&1&-1\end{pmatrix}$

and let $A(k)$ be the sequence matrix

$A(k)=\begin{pmatrix}2^{-1+k}&2^{-1+k}&\frac{1}{16}2^{k}((-1)^{1+k}+5)&\frac{1}{16}2^{k}((-1)^{k}-1)\vspace*{0.1pc}\\\
2^{-1+k}&2^{-1+k}&\frac{5}{16}2^{k}((-1)^{1+k}+1)&\frac{1}{16}2^{k}(5(-1)^{k}-1)\vspace*{0.1pc}\\\
0&0&(-1)^{k}2^{-1+k}&(-1)^{1+k}2^{-1+k}\vspace*{0.1pc}\\\
0&0&(-1)^{1+k}2^{-1+k}&(-1)^{k}2^{-1+k}\end{pmatrix}.$

We can verify that $A^{k}=A(k)$ for all $k\geq 4$. We have $A(0)\neq I_{4}$,
then $A$ is singular.
$A(1)\neq A$, $A(2)=A^{2}$. Then $A^{k}=A(k)$ for all $k\geq 2$.
Hence we have

  * •

The non-geometric part of $A$ is
$(I_{4}-A(0))\boldsymbol{0}_{0}+(A-A(1))\boldsymbol{0}_{1}$.

  * •

The geometric part of $A$ is

$\begin{pmatrix}2^{-1}(2^{k})&2^{-1}(2^{k})&\frac{5}{16}(2^{k})-\frac{1}{16}((-2)^{k})&\frac{-1}{16}(2^{k})+\frac{1}{16}((-2)^{k})\vspace*{0.1pc}\\\
2^{-1}(2^{k})&2^{-1}(2^{k})&\frac{5}{16}(2^{k})-\frac{5}{16}((-2)^{k})&\frac{-1}{16}(2^{k})+\frac{5}{16}((-2)^{k})\vspace*{0.1pc}\\\
0&0&2^{-1}((-2)^{k})&-2^{-1}((-2)^{k})\vspace*{0.1pc}\\\
0&0&-2^{-1}((-2)^{k})&2^{-1}((-2)^{k})\end{pmatrix}$

  * •

The minimal polynomial of $A$ is $X^{2}(X-2)(X+2)$.

  * •

$\pi_{0}=I_{4}-A(0)=$,
$\pi_{2}=\begin{pmatrix}2^{-1}&2^{-1}&\frac{5}{16}&\frac{-1}{16}\vspace*{0.1pc}\\\
2^{-1}&2^{-1}&\frac{5}{16}&\frac{-1}{16}\vspace*{0.1pc}\\\
0&0&0&0\vspace*{0.1pc}\\\ 0&0&0&0\end{pmatrix}$ and
$\pi_{-2}=\begin{pmatrix}0&0&\frac{-1}{16}&\frac{1}{16}\vspace*{0.1pc}\\\
0&0&\frac{-5}{16}&\frac{5}{16}\vspace*{0.1pc}\\\
0&0&2^{-1}&-2^{-1}\vspace*{0.1pc}\\\ 0&0&-2^{-1}&2^{-1}\end{pmatrix}$

Consider now the following linear map

$\displaystyle\theta:F_{\mathcal{S}^{\ast}}$ $\displaystyle\longrightarrow
F_{\mathcal{S}^{\ast}}$ $\displaystyle\boldsymbol{\lambda}$
$\displaystyle\longmapsto\boldsymbol{\lambda}^{-1}$

Obviously, $\theta(\mathcal{S}^{\ast})=\mathcal{S}^{\ast}$ and
$\theta^{-1}=\theta$. Since
$\theta(\boldsymbol{\lambda}\boldsymbol{\mu})=\theta(\boldsymbol{\lambda})\theta(\boldsymbol{\mu})$
for all $\boldsymbol{\lambda},\boldsymbol{\mu}\in\mathcal{S}^{\ast}$, it
follows that $\theta$ is an $F$-algebra automorphism of
$F_{\mathcal{S}^{\ast}}$.
Let

$\displaystyle\Psi:F_{\mathcal{S}^{\circ}}\longrightarrow
F_{\mathcal{S}^{\circ}}$

be any $F$-endomorphism of $F_{\mathcal{S}^{\circ}}$ and consider

$\displaystyle\Psi\oplus\theta:F_{\mathcal{S}}\longrightarrow F_{\mathcal{S}}$

The direct sum of $\theta$ and $\Psi$. It is obvious that $\theta_{\Psi}$ is
an $F$-endomorphism of $F_{\mathcal{S}}$, involutory if $\Psi$ is, but it is
not an algebra homomorphism; in fact, we have
$\boldsymbol{\lambda}\boldsymbol{0}_{i}=\lambda^{i}\boldsymbol{0}_{i}$, then
$\theta_{\Psi}(\boldsymbol{\lambda}\boldsymbol{0}_{i})=\lambda^{i}\Psi(\boldsymbol{0}_{i})$
which is not always equal to
$\theta_{\Psi}(\boldsymbol{\lambda})\theta_{\Psi}(\boldsymbol{0}_{i})=\boldsymbol{\lambda}^{-1}\Psi(\boldsymbol{0}_{i})$.
Let us denote by $\widetilde{\theta}$ the $F$-algebra automorphism

$\displaystyle\widetilde{\theta}:\mathbf{M_{q}}(F_{\mathcal{S}^{\ast}})$
$\displaystyle\longrightarrow\mathbf{M_{q}}(F_{\mathcal{S}^{\ast}})$
$\displaystyle(A_{ij})$ $\displaystyle\longmapsto(\theta(A_{ij}))$

Note that $\widetilde{\theta}$ is an involution of
$\mathbf{M_{q}}(F_{\mathcal{S}^{\ast}})$.
Let $\widetilde{\theta}_{\Psi}$ denotes the linear endomorphism induced by
$\theta_{\Psi}$ on
$\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}=\mathbf{M_{q}}(F_{\mathcal{S}^{\circ}})\oplus\mathbf{M_{q}}(F_{\mathcal{S}^{\ast}})_{\mathcal{T}}$
defined by

$\displaystyle\widetilde{\theta}_{\Psi}:\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}$
$\displaystyle\longrightarrow\mathbf{M_{q}}(F_{\mathcal{S}})_{\mathcal{T}}$
$\displaystyle\boldsymbol{U}$
$\displaystyle\longmapsto\mathcal{V}_{0}\Psi(\boldsymbol{0}_{0})+\cdots+\mathcal{V}_{n}\Psi(\boldsymbol{0}_{n})+\widetilde{\theta}(\mathcal{W}_{0})\widetilde{\Lambda_{0}}+\cdots+\widetilde{\theta}(\mathcal{W}_{m})\widetilde{\Lambda_{m}},$

where

$\mathcal{V}_{0}\boldsymbol{0}_{0}+\cdots+\mathcal{V}_{n}\boldsymbol{0}_{n}+\mathcal{W}_{0}\Lambda_{0}+\cdots+\mathcal{W}_{m}\Lambda_{m}$

is the $\mathcal{P}$-cf of $\boldsymbol{U}$. The map
$\widetilde{\theta}_{\Psi}$ is well-defined, since $\widetilde{\Lambda_{i}}\in
F\langle\mathcal{T}\rangle$ for all non-negative integer $i$ (see Proposition
2.1).
In the remainder of this section, we will only be interested in the special
case when $\Psi$ is the zero map. Let us denote the map

$0\oplus\theta:F_{\mathcal{S}}\longrightarrow F_{\mathcal{S}}$

simply by $\theta_{0}$. It should be noted that the image of a sequence
$\boldsymbol{u}$ of $F_{\mathcal{S}}$ under $\theta_{0}$ can be obtained by
simply plugging in $-k$ for $k$ in the geometric part of $\boldsymbol{u}$ and
neglecting its non-geometric part.
Let $A$ be a square matrix of index $t_{0}$ with characteristic polynomial
splits over $F$. We may assume that the Jordan canonical form of $A$ has the
form as follows

$A=P\begin{pmatrix}D&0\\\ 0&N\end{pmatrix}P^{-1}$

where $P$ is a nonsingular matrix, $D$ is a nonsingular matrix of order
$r=\text{rank}(A^{t_{0}})$, and $N$ is a nilpotent matrix such that
$N^{t_{0}}=0$. Then we can write the matrix $\boldsymbol{A}$ in the form

$\boldsymbol{A}=P\begin{pmatrix}0&0\\\
0&N^{k}\end{pmatrix}_{k}P^{-1}+P\begin{pmatrix}D^{k}&0\\\
0&0\end{pmatrix}_{k}P^{-1}.$

It is clear that these two matrices are the non-geometric and the geometric
parts of $\boldsymbol{A}$ written in a form other than the $\mathcal{P}$-cf.
If we plug $-k$ for $k$ into this form of the geometric part of
$\boldsymbol{A}$, we get the matrix
$\boldsymbol{A}_{d}-\pi_{0}\boldsymbol{0}_{0}$. In the theorem below, we show
that the same remains true for the $\mathcal{P}$-cf of $\boldsymbol{A}$.

######  Theorem 3.9.

Let $A$ be a square matrix of index $t_{0}$ with coefficients in $F$ such that
its characteristic polynomial splits over $F$. Then we have
$\boldsymbol{A}_{d}=\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0}$,
where $\boldsymbol{A}_{d}=(A_{d}^{k})_{k}$.

###### Proof.

We must proof that

$\begin{array}[]{rrlc}\boldsymbol{A}^{t_{0}+1}(\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0})&=&\boldsymbol{A}^{t_{0}}&(1^{t_{0}})\\\
(\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0})\boldsymbol{A}(\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0})&=&\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0}&(3)\\\
\boldsymbol{A}(\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0})&=&(\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0})\boldsymbol{A}&(5)\end{array}$

From Proposition 3.1 it follows that

$\boldsymbol{A}=\displaystyle\sum_{i=0}^{t_{0}-1}\boldsymbol{0}_{i}A^{i}\pi_{0}+\sum_{i=0}^{m}(\sum_{j=1}^{p}\boldsymbol{\lambda}_{j}\lambda_{j}^{-i}(A-\lambda_{j}I_{q})^{i}\pi_{j})\Lambda_{i}$

and then

$\widetilde{\theta}_{0}(\boldsymbol{A})=\displaystyle\sum_{i=0}^{m}(\sum_{j=1}^{p}\boldsymbol{\lambda}^{-1}_{j}\lambda_{j}^{-i}(A-\lambda_{j}I_{q})^{i}\pi_{j})\widetilde{\Lambda}_{i},$

where $m=\max\\{t_{1}-1,\ldots,t_{p}-1\\}$.
Clearly that
$\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0}$ satisfies
$(5)$.
In order to prove that
$\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0}$ satisfies
$(3)$, observe first that

$\mathcal{A}_{i}\Lambda_{i}\widetilde{\theta}_{0}(\mathcal{A}_{l}\Lambda_{l})=\displaystyle\sum_{j=1}^{p}\lambda_{j}^{-(i+l)}(A-\lambda_{j}I_{q})^{i+l}\pi_{j}\Lambda_{i}\widetilde{\Lambda}_{l}.$

As $(A-\lambda_{j}I_{q})^{m+1}\pi_{j}=0$, it follows that

$\boldsymbol{A}\widetilde{\theta}_{0}(\boldsymbol{A})=\displaystyle\sum_{j=1}^{p}\sum_{i=0}^{m}\lambda_{j}^{-i}(A-\lambda_{j}I_{q})^{i}\pi_{j})(\sum_{l=0}^{i}\Lambda_{l}\widetilde{\Lambda}_{i-l}).$

Furthermore, since $\displaystyle\sum_{l=0}^{i}\binom{k}{l}\binom{-k}{i-l}=0$
for all positive integer $i$, we also have

$\displaystyle\sum_{l=0}^{i}\Lambda_{l}\widetilde{\Lambda}_{i-l}=\boldsymbol{0}.$

Therefore,

$\boldsymbol{A}\widetilde{\theta}_{0}(\boldsymbol{A})=\displaystyle\sum_{j=1}^{p}\pi_{j}=I_{q}-\pi_{0}.$

Hence

$\widetilde{\theta}_{0}(\boldsymbol{A})\boldsymbol{A}\widetilde{\theta}_{0}(\boldsymbol{A})=\widetilde{\theta}_{0}(\boldsymbol{A})-\pi_{0}\widetilde{\theta}_{0}(\boldsymbol{A})=\widetilde{\theta}_{0}(\boldsymbol{A})$

and because of
$\boldsymbol{A}\boldsymbol{0}_{0}^{2}\pi_{0}^{2}=\boldsymbol{0}_{0}\pi_{0}$,
we conclude that

$(\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0})\boldsymbol{A}(\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0})=\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0}.$

To complete the proof, it remains to show that
$\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0}$ satisfies
$(1^{t_{0}})$. Since
$\boldsymbol{A}\widetilde{\theta}_{0}(\boldsymbol{A})=I_{q}-\pi_{0}$ and
$\boldsymbol{A}^{t_{0}}\pi_{0}=\boldsymbol{0}_{0}\pi_{0}$, it follows that

$\boldsymbol{A}^{t_{0}+1}\widetilde{\theta}_{0}(\boldsymbol{A})=\boldsymbol{A}^{t_{0}}-\boldsymbol{A}^{t_{0}}\pi_{0}=\boldsymbol{A}^{t_{0}}-\boldsymbol{0}_{0}\pi_{0}.$

But since
$\boldsymbol{A}^{t_{0}+1}\boldsymbol{0}_{0}\pi_{0}=\boldsymbol{0}_{0}\pi_{0}$,
it follows that

$\boldsymbol{A}^{t_{0}+1}(\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0})=\boldsymbol{A}^{t_{0}}.$

The uniqueness of a $\\{1^{t_{0}},3,5\\}$-inverse $(B_{k})_{k}$ of
$\boldsymbol{A}$ follows from the uniqueness of each $B_{k}$ which is the
unique $\\{1^{t_{0}},3,5\\}$-inverse of $A^{k}$, for all positive integer $k$.
We conclude that
$\widetilde{\theta}_{0}(\boldsymbol{A})+\boldsymbol{0}_{0}\pi_{0}$ is the
$\\{1^{t_{0}},3,5\\}$-inverse of $\boldsymbol{A}$. ∎

######  Remark 3.10.

Although we have specified that the set $F$ is a field and that the
characteristic polynomial of $A$ splits over $F$, the theorem 3.9 remains true
in any integral domain $R$ and for any square matrix. In this case, $A_{d}$ is
a matrix over the algebraic closure of the quotient field of $R$.

######  Remark 3.11.

If $F=\mathbb{C}$ is the complex field, then

$\displaystyle A_{d}$ $\displaystyle=$
$\displaystyle(\boldsymbol{A}_{d})_{1}=\widetilde{\theta}_{0}(\boldsymbol{A})_{1}$
$\displaystyle=$
$\displaystyle\displaystyle\sum_{j=1}^{p}\pi_{j}\sum_{i=0}^{m}\lambda_{j}^{-i-1}(A-\lambda_{j}I_{q})^{i})(\widetilde{\Lambda}_{i})_{1}$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{p}\pi_{j}\sum_{i=0}^{m}(-1)^{i}\lambda_{j}^{-i-1}(A-\lambda_{j}I_{q})^{i}).$

We find then the well-known result that $A_{d}=\frac{1}{z}(A)$ is the matrix
function corresponding to the reciprocal $f(z)=\frac{1}{z}$, defined on
nonzero eigenvalues (see e.g. Corollary 1. p. 165 of [1]).

Let now $\mathcal{H}$ be a basis of the free $F_{\mathcal{S}^{\ast}}$-module
$F_{\mathcal{S}^{\ast}}\langle\mathcal{T}\rangle$. Since $\mathcal{H}$ has the
same cardinality as $\mathcal{T}$, it can be indexed by $\mathbb{N}$, i.e.,
$\mathcal{H}=\\{H_{i}\\}_{i\in\mathbb{N}}$.

######  Proposition 3.12.

Let $F$ be a field of characteristic $0$ and
$\mathcal{H}=\\{H_{i}\\}_{i\in\mathbb{N}}$ be a basis of the
$F_{\mathcal{S}^{\ast}}$-module
$F_{\mathcal{S}^{\ast}}\langle\mathcal{T}\rangle$. Assume that $\mathcal{H}$
is contained in the $F$-vector space spanned by $\mathcal{T}$. Then we have
the following:
1\. For all nonnegative integer $i$, there exists a polynomial $P_{i}(X)\in
F[X]$ such that $H_{i}=(P_{i}(k))_{k\geq 0}$.
2\. $\widetilde{\theta}(H_{i})=(P_{i}(-k))_{k\geq 0}$ For all nonnegative
integer $i$.
3\. Let $A$ be a square matrix over $F$ whose characteristic polynomial splits
over $F$. Let

$\boldsymbol{A}=N(A)+\mathcal{B}_{0}(P_{0}(k))_{k}+\cdots+\mathcal{B}_{m}(P_{m}(k))_{k}$

be the representation of $A$ with respect to
$(\mathcal{S}^{\ast},\mathcal{H})$, which we call the $\mathcal{P}$-cf of $A$
relative to $(\mathcal{S}^{\ast},\mathcal{H})$. Then the sequence matrix
obtained by plugging $-k$ for $k$ in the geometric part

$\mathcal{B}_{0}(P_{0}(k))_{k}+\cdots+\mathcal{B}_{m}(P_{m}(k))_{k}$

of $A$ is $\boldsymbol{A}_{d}-\pi_{0}\boldsymbol{0}_{0}$.

###### Proof.

1\. It is well known that when the characteristic of $F$ is $0$,
$\binom{k}{j}$ is a polynomial in $k$, i.e., there exists $Q_{j}(X)\in F[X]$
such that $\binom{k}{j}=Q_{j}(k)$ and then $\Lambda_{j}=(Q_{j}(k))_{k\geq 0}$.
Since $\mathcal{H}$ is contained in the $F$-vector space spanned by
$\mathcal{T}$, there exist elements $aij$ in $F$, such that

$H_{i}=\sum_{finite}a_{ij}\Lambda_{j}.$

In other words, $H_{i}=(P_{i}(k))_{k\geq 0}$, where

$P_{i}(k)=\sum_{finite}a_{ij}Q_{j}(k).$

2\. It is well known that the equality $\binom{k}{j}=Q_{j}(k)$, which is
between two different forms of the same sequence, still valid when we plug in
$-k$ for $k$. Then

$\displaystyle\widetilde{\theta}(H_{i})$ $\displaystyle=$
$\displaystyle\sum_{finite}a_{ij}\widetilde{\theta}(\Lambda_{j})$
$\displaystyle=$ $\displaystyle\sum_{finite}a_{ij}(\binom{-k}{j})_{k\geq 0}$
$\displaystyle=$ $\displaystyle\sum_{finite}a_{ij}(Q_{j}(-k))_{k\geq 0}$
$\displaystyle=$ $\displaystyle(P_{i}(-k))_{k\geq 0}.$

3\. The result follows immediately from $2.$ above together with Theorem 3.9.
∎

As a particular consequence of the above proposition, we obtain the following
corollary.

######  Corollary 3.13.

Let $F$ be a field of characteristic $0$ and
$\mathcal{H}=(\Gamma^{i})_{i\geqslant 0}$, where $\Gamma$ is the sequence
$\Gamma=(0,1,2,\ldots)$. Let $A$ be a square matrix over $F$ whose
characteristic polynomial splits over $F$. Then the sequence matrix obtained
by plugging $-k$ for $k$ in the geometric part of $A$ relative to
$(\mathcal{S}^{\ast},\mathcal{H})$ is
$\boldsymbol{A}_{d}-\pi_{0}\boldsymbol{0}_{0}$.

###### Proof.

It is well known that

$(\Gamma^{i})_{i\geqslant 0}=T(\Lambda_{i})_{i\geqslant 0}$

where $T$ is the infinite invertible lower triangular matrix
$(m!S(n,m))_{n,m\geqslant 0}$ and $S(n,m)$ is the Stirling number of the
second kind (see, e.g. Riordan [5]).
From this result, we deduce on the one hand that $\mathcal{H}$ is contained in
the $F$-vector space spanned by $\mathcal{T}$ and on the other hand that
$\mathcal{H}$ is, as is well known, a basis of the
$F_{\mathcal{S}^{\ast}}$-module
$F_{\mathcal{S}^{\ast}}\langle\mathcal{T}\rangle$. The corollary is then a
consequence of Proposition 3.12. ∎

Now let us consider the case where $F=\mathbb{C}$ the complex field. Consider
the following sets

  * •

$\mathcal{S}^{\ast}=\\{\boldsymbol{\lambda}=(\lambda^{k})_{k\geqslant
0}/\lambda\in\mathbb{C},\lambda\neq 0\\}$

  * •

$\widetilde{S}=\\{\boldsymbol{\lambda}/\lambda\in\mathbb{R},\lambda\neq 0\\}$

  * •

$\mathcal{S}^{+}=\\{\boldsymbol{\lambda}\in\mathcal{S}^{\ast}/Im(\lambda)>0\\}$

  * •

$\mathcal{S}^{-}=\\{\boldsymbol{\lambda}\in\mathcal{S}^{\ast}/Im(\lambda)<0\\}=\\{\boldsymbol{\overline{\lambda}}/\boldsymbol{\lambda}\in\mathcal{S}^{+}\\}$

  * •

$\mathcal{S}^{+}_{1}=\\{\frac{\boldsymbol{\lambda}+\boldsymbol{\overline{\lambda}}}{2}/\boldsymbol{\lambda}\in\mathcal{S}^{+}\\}=\\{[r,\cos(\theta)]/0\neq
r\in\mathbb{R}^{+},\theta\in]0,\pi[\\}$, where
$[r,\cos(\theta)]=(r^{k}\cos(k\theta))_{k\geq 0}$

  * •

$\mathcal{S}^{+}_{2}=\\{\frac{\boldsymbol{\lambda}-\boldsymbol{\overline{\lambda}}}{2\mathrm{i}}/\boldsymbol{\lambda}\in\mathcal{S}^{+}\\}=\\{[r,\sin(\theta)]/0\neq
r\in\mathbb{R}^{+},\theta\in]0,\pi[\\}$, where
$[r,\sin(\theta)]=(r^{k}\sin(k\theta))_{k\geq 0}$

Clearly $\widetilde{S},\mathcal{S}^{+},\mathcal{S}^{-}$ constitute a partition
of $\mathcal{S}^{\ast}$, and then the set
$\mathbb{S}=\widetilde{S}\bigcup\mathcal{S}^{+}_{1}\bigcup\mathcal{S}^{+}_{2}$
is $\mathbb{C}$-linearly independent and then is $\mathbb{R}$-linearly
independent.
From the the general theory of linear recurrence sequences we have the
following:

  * •

$\mathbb{R}_{\mathcal{S}^{\circ}}\bigoplus\mathbb{R}_{\mathbb{S}}$ is the
$\mathbb{R}$-vector space of real linear recurrence sequences whose
carateristic polynomials are of the form $X^{m}P(X)$ where
$P(X)\in\mathbb{R}[X]$ is square-free with nonzero constant terms.

  * •

$\mathbb{R}_{\mathcal{S}^{\circ}}\bigoplus\mathbb{R}_{\mathbb{S}}[\Gamma]$ is
the $\mathbb{R}$-vector space of real linear recurrence sequences whose
carateristic polynomials $P(X)\in\mathbb{R}[X]$.

  * •

If
$u\in\mathbb{R}_{\mathcal{S}^{\circ}}\bigoplus\mathbb{R}_{\mathbb{S}}[\Gamma]$
then there exist
$\boldsymbol{\rho}_{1},\ldots,\boldsymbol{\rho}_{l}\in\widetilde{S}$,
$\boldsymbol{\lambda}_{1}=(r_{1}^{k}e^{\mathrm{i}\theta_{1}}),\ldots,\boldsymbol{\lambda}_{m}=(r_{m}^{k}e^{\mathrm{i}\theta_{m}})\in\mathcal{S}^{+}$,
$\mu_{1},\ldots,\mu_{1}\in\mathbb{R}$, $P_{1},\ldots P_{l}\in\mathbb{R}[X]$
and $Q_{1},\ldots Q_{m}\in\mathbb{C}[X]$ such that, for all $k\geq 0$,

$\displaystyle u_{k}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{q}\mu_{i}\boldsymbol{0}_{i}+\sum_{i=0}^{l}P_{i}(k)\rho_{i}^{k}+\sum_{j=0}^{m}(Q_{j}(k)\lambda_{j}^{k}+\overline{Q}_{j}(k)\overline{\lambda}_{j}^{k})$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{q}\mu_{i}\boldsymbol{0}_{i}+\sum_{i=0}^{l}P_{i}(k)\rho_{i}^{k}+\sum_{i=0}^{m}(Q_{i}(k)\lambda_{i}^{k}+\overline{Q_{i}(k)\lambda_{i}^{k}})$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{q}\mu_{i}\boldsymbol{0}_{i}+\sum_{i=0}^{l}P_{i}(k)\rho_{i}^{k}+\sum_{i=0}^{m}2\Re(Q_{i}(k))[r_{i},\cos(\theta_{i})]_{k}$
$\displaystyle-\sum_{i=0}^{m}2\Im(Q_{i}(k))[r_{i},\sin(\theta_{i})]_{k},$

this means that
$\mathcal{S}^{\circ}\bigcup\\{\boldsymbol{\lambda}\Gamma^{i}/0\neq\lambda\in\mathbb{R},i\in\mathbb{N}\\}\bigcup\\{[r,\cos(\theta)]\Gamma^{i},[r,\sin(\theta)]\Gamma^{i}/i\in\mathbb{N},\,\,0\neq
r\in\mathbb{R}^{+},\,\,\theta\in]0,\pi[\\}$ spans the $\mathbb{R}$-vector
space
$\mathbb{R}_{\mathcal{S}^{\circ}}\bigoplus\mathbb{R}_{\mathbb{S}}[\Gamma]$,
and hence is a basis for it.

  * •

$\Gamma$ is transcendental over $\mathbb{C}_{\mathcal{S}^{\ast}}$ and it is so
over $\mathbb{R}_{\mathbb{S}}$

######  Proposition 3.14.

Let $A\in M_{q}(\mathbb{R})$ be a real square matrix. Let
$\mathcal{H}=(\Gamma^{i})_{i\geq 0}$ and

$\mathbb{S}=\\{\boldsymbol{\lambda}\Gamma^{i}/0\neq\lambda\in\mathbb{R},i\in\mathbb{N}\\}\bigcup\\{[r,\cos(\theta)]\Gamma^{i},[r,\sin(\theta)]\Gamma^{i}/i\in\mathbb{N},\,\,0\neq
r\in\mathbb{R}^{+},\,\,\theta\in]0,\pi[\\}.$

Then the sequence matrix obtained by plugging $-k$ for $k$ in the geometric
part of $A$ relative to $(\mathbb{S},\mathcal{H})$ is
$\boldsymbol{A}_{d}-\pi_{0}\boldsymbol{0}_{0}$.

###### Proof.

The proof follows simply from the fact that the equalities

$\boldsymbol{\lambda}+\boldsymbol{\overline{\lambda}}=2[r,\cos(\theta)]$

and

$\boldsymbol{\lambda}-\boldsymbol{\overline{\lambda}}=2\mathrm{i}[r,\sin(\theta)]$

still valid when we plug in $-k$ for $k$. ∎

######  Proposition 3.15.

Let $A\in M_{q}(\mathbb{R})$ be a real square matrix. Let

$\mathbb{S}=\\{\boldsymbol{\lambda}\Gamma^{i}/0\neq\lambda\in\mathbb{R},i\in\mathbb{N}\\}\bigcup\\{[r,\cos(\theta)]\Gamma^{i},[r,\sin(\theta)]\Gamma^{i}/i\in\mathbb{N},\,\,0\neq
r\in\mathbb{R}^{+},\,\,\theta\in]0,\pi[\\}.$

Then the sequence matrix obtained by plugging $-k$ for $k$ in the geometric
part of $A$ relative to $(\mathbb{S},\mathcal{T})$ is
$\boldsymbol{A}_{d}-\pi_{0}\boldsymbol{0}_{0}$.

###### Proof.

The proof follows immediately from the fact that $\mathcal{T}$ is contained in
the $\mathbb{R}$-vector space spanned by $\mathcal{H}=(\Gamma^{i})_{i\geq 0}$,
since $(\Lambda_{i})_{i\geq 0}=D(\Gamma^{i})_{i\geq 0}$ where

$D=(\frac{s(n,k)}{n!})_{n,k\geqslant 0}$

is the infinite lower triangular matrices with coefficients in $\mathbb{R}$
and $s(n,k)$ is the Stirling number of the first kind (see, e.g. Riordan [5]).
∎

For the purpose of illustration, let us consider the following examples.

######  Example 3.16.

Let

$B=\begin{pmatrix}1&1&0&0\\\ -2&0&1&0\\\ 2&0&0&1\\\ -2&-1&-1&-1\end{pmatrix}.$

We have for all $k\geq 1$, $B^{k}=B(k)$ where
$B(k)=$

$\begin{pmatrix}\cos(\frac{k\pi}{2})+\sin(\frac{k\pi}{2})&\frac{2\sin(\frac{k\pi}{2})-k\cos(\frac{k\pi}{2})}{2}&\frac{(1-k)\sin(\frac{k\pi}{2})-k\cos(\frac{k\pi}{2})}{2}&\frac{(1-k)\sin(\frac{k\pi}{2})}{2}\\\
-2\sin(\frac{k\pi}{2})&\frac{(k+2)\cos(\frac{k\pi}{2})+(k-1)\sin(\frac{k\pi}{2})}{2}&k\sin(\frac{k\pi}{2})&\frac{-k\cos(\frac{k\pi}{2})+(k-1)\sin(\frac{k\pi}{2})}{2}\\\
2\sin(\frac{k\pi}{2})&\frac{-k\cos(\frac{k\pi}{2})+(1-k)\sin(\frac{k\pi}{2})}{2}&(1-k)\sin(\frac{k\pi}{2})+\cos(\frac{k\pi}{2})&\frac{k\cos(\frac{k\pi}{2})+(3-k)\sin(\frac{k\pi}{2})}{2}\\\
-2\sin(\frac{k\pi}{2})&\frac{k\cos(\frac{k\pi}{2})+(k-3)\sin(\frac{k\pi}{2})}{2}&(k-2)\sin(\frac{k\pi}{2})&\frac{(2-k)\cos(\frac{k\pi}{2})+(k-3)\sin(\frac{k\pi}{2})}{2}\end{pmatrix}$

Since $B(0)=I_{4}$, $B$ is nonsingular and

$B^{-1}=B(-1)=\begin{pmatrix}-1&-1&-1&-1\\\ 2&1&1&1\\\ -2&-1&-2&-2\\\
2&2&3&2\end{pmatrix}.$

Also we have $B^{-k}=B(-k)$.

######  Example 3.17.

Let us take example 3.8. We have

$A_{d}=A(-1)=\begin{pmatrix}\frac{1}{4}&\frac{1}{4}&\frac{3}{16}&\frac{-1}{16}\vspace*{0.1pc}\\\
\frac{1}{4}&\frac{1}{4}&\frac{5}{16}&\frac{-3}{16}\vspace*{0.1pc}\\\
0&0&\frac{-1}{4}&\frac{1}{4}\vspace*{0.1pc}\\\
0&0&\frac{1}{4}&\frac{-1}{4}\end{pmatrix}.$

For all $k\geq 1$,

$A_{d}^{k}=A(-k)=\begin{pmatrix}2^{-1-k}&2^{-1-k}&\frac{1}{16}2^{-k}((-1)^{1+k}+5)&\frac{1}{16}2^{-k}((-1)^{k}-1)\vspace*{0.1pc}\\\
2^{-1-k}&2^{-1-k}&\frac{5}{16}2^{-k}((-1)^{1+k}+1)&\frac{1}{16}2^{-k}(5(-1)^{k}-1)\vspace*{0.1pc}\\\
0&0&(-1)^{k}2^{-1-k}&(-1)^{1+k}2^{-1-k}\vspace*{0.1pc}\\\
0&0&(-1)^{1+k}2^{-1-k}&(-1)^{k}2^{-1-k}\end{pmatrix}.$

######  Example 3.18.

Let $x\in\mathbb{C}$ and let

$E=\begin{pmatrix}2\sqrt{3}-x-10&2\sqrt{3}-2x-23&\sqrt{3}-x-5\\\
4&\sqrt{3}+9&2\\\ -2\sqrt{3}+2x+2&-4\sqrt{3}+4x+5&-\sqrt{3}+2x+1\end{pmatrix}$

Let

$\displaystyle\delta_{s}(x)=\begin{cases}0&\text{if}\quad x=0\\\
x^{s}&\text{if}\quad x\neq 0\end{cases}$

and put

$E(k)=\begin{pmatrix}e_{11}(k)&e_{12}(k)&e_{13}(k)\\\
e_{21}(k)&e_{22}(k)&e_{23}(k)\\\ e_{31}(k)&e_{32}(k)&e_{33}(k)\end{pmatrix}$

where

$\displaystyle e_{11}(k)$ $\displaystyle=$ $\displaystyle
2^{k+1}(\cos(\frac{k\pi}{6})-5\sin(\frac{k\pi}{6}))-\delta_{k}(x)$
$\displaystyle e_{12}(k)$ $\displaystyle=$ $\displaystyle
2^{k+1}(\cos(\frac{k\pi}{6})-\frac{23}{2}\sin(\frac{k\pi}{6}))-2\delta_{k}(x)$
$\displaystyle e_{13}(k)$ $\displaystyle=$ $\displaystyle
2^{k}(\cos(\frac{k\pi}{6})-5\sin(\frac{k\pi}{6}))-\delta_{k}(x)$
$\displaystyle e_{21}(k)$ $\displaystyle=$ $\displaystyle
2^{k+2}\sin(\frac{k\pi}{6})$ $\displaystyle e_{22}(k)$ $\displaystyle=$
$\displaystyle 2^{k}(\cos(\frac{k\pi}{6})+9\sin(\frac{k\pi}{6}))$
$\displaystyle e_{23}(k)$ $\displaystyle=$ $\displaystyle
2^{k+1}\sin(\frac{k\pi}{6})$ $\displaystyle e_{31}(k)$ $\displaystyle=$
$\displaystyle-2^{k+1}(\cos(\frac{k\pi}{6})-\sin(\frac{k\pi}{6}))+2\delta_{k}(x)$
$\displaystyle e_{32}(k)$ $\displaystyle=$
$\displaystyle-2^{k+2}(\cos(\frac{k\pi}{6})-\frac{5}{4}\sin(\frac{k\pi}{6}))+4\delta_{k}(x)$
$\displaystyle e_{33}(k)$ $\displaystyle=$
$\displaystyle-2^{k}(\cos(\frac{k\pi}{6})-\sin(\frac{k\pi}{6}))+2\delta_{k}(x)$

We have, for all $k\geq 1$, $E^{k}=E(k)$ and

$E(0)=\begin{pmatrix}2-\delta_{0}(x)&2-2\delta_{0}(x)&1-\delta_{0}(x)\\\
0&1&0\\\ -2+2\delta_{0}(x)&-4+4\delta_{0}(x)&-1+2\delta_{0}(x)\end{pmatrix}$

Then $E$ is nonsingular if and only if $x\neq 0$.

$E_{d}=\begin{pmatrix}\frac{\sqrt{3}+5}{2}-\delta_{-1}(x)&\frac{2\sqrt{3}+23}{4}-2\delta_{-1}(x)&\frac{\sqrt{3}+5}{4}-\delta_{-1}(x)\vspace*{0.1pc}\\\
-1&\frac{\sqrt{3}-9}{4}&\frac{-1}{2}\vspace*{0.1pc}\\\
\frac{-\sqrt{3}-1}{2}+2\delta_{-1}(x)&\frac{-4\sqrt{3}-5}{4}+4\delta_{-1}(x)&\frac{-\sqrt{3}-1}{4}+2\delta_{-1}(x)\end{pmatrix}$

and for all $k\geq 1$

$E_{d}^{k}=\begin{pmatrix}a_{11}(k)&a_{12}(k)&a_{13}(k)\\\
a_{21}(k)&a_{22}(k)&a_{23}(k)\\\ a_{31}(k)&a_{32}(k)&a_{33}(k)\end{pmatrix}$

where

$\displaystyle a_{11}(k)$ $\displaystyle=$ $\displaystyle
2^{-k+1}(\cos(\frac{k\pi}{6})+5\sin(\frac{k\pi}{6}))-\delta_{-k}(x)$
$\displaystyle a_{12}(k)$ $\displaystyle=$ $\displaystyle
2^{-k+1}(\cos(\frac{k\pi}{6})+\frac{23}{2}\sin(\frac{k\pi}{6}))-2\delta_{-k}(x)$
$\displaystyle a_{13}(k)$ $\displaystyle=$ $\displaystyle
2^{-k}(\cos(\frac{k\pi}{6})+5\sin(\frac{k\pi}{6}))-\delta_{-k}(x)$
$\displaystyle a_{21}(k)$ $\displaystyle=$
$\displaystyle-2^{-k+2}\sin(\frac{k\pi}{6})$ $\displaystyle a_{22}(k)$
$\displaystyle=$ $\displaystyle 2^{-k}\cos(\frac{k\pi}{6})-9\times
2^{-k}\sin(\frac{k\pi}{6})$ $\displaystyle a_{23}(k)$ $\displaystyle=$
$\displaystyle-2^{-k+1}\sin(\frac{k\pi}{6})$ $\displaystyle a_{31}(k)$
$\displaystyle=$
$\displaystyle-2^{-k+1}(\cos(\frac{k\pi}{6})+\sin(\frac{k\pi}{6}))+2\delta_{-k}(x)$
$\displaystyle a_{32}(k)$ $\displaystyle=$
$\displaystyle-2^{-k+2}(\cos(\frac{k\pi}{6})-\frac{5}{4}\sin(\frac{k\pi}{6}))+4\delta_{-k}(x)$
$\displaystyle a_{33}(k)$ $\displaystyle=$
$\displaystyle-2^{-k}(\cos(\frac{k\pi}{6})+\sin(\frac{k\pi}{6}))+2\delta_{-k}(x)$

######  Example 3.19.

Let $p\in\mathbb{N},p\geqslant 2$ and consider the following circulant
matrices over $\mathbb{C}$

$A=\begin{pmatrix}1&\cdots&1\\\ \vdots&&\vdots\\\
1&\cdots&1\end{pmatrix},\quad B=\frac{1}{p-1}\begin{pmatrix}0&1&\cdots&1\\\
\vdots&&&\vdots\\\ 1&\cdots&1&0\end{pmatrix}\quad
C=\frac{1}{p}\begin{pmatrix}p-1&1&\cdots&1\\\ \vdots&&&\vdots\\\
1&\cdots&1&p-1\end{pmatrix}$

where $p\in\mathbb{N},p\geqslant 2$.
It is clear that $A^{k}=p^{k-1}A$ for all $k\geqslant 1$. Since
$B=\frac{1}{p-1}(A-I_{p})$, it follows that
$B^{k}=a_{k}B+b_{k}I_{p},k\geqslant 1$, where

$a_{k}=\frac{p-1}{p}(1)^{k}-\frac{p-1}{p}\left(\frac{-1}{p-1}\right)^{k}\quad\text{and}\quad
b_{k}=\frac{1}{p}(1)^{k}+\frac{p-1}{p}\left(\frac{-1}{p-1}\right)^{k},\quad
k\geqslant 1,$

From this last result we deduce the following:

  * •

From Lemma 2.4 it follows that The geometric part of $\boldsymbol{B}$ is
$\boldsymbol{a}B+\boldsymbol{b}I_{p}$, where $a_{0}=0$ and $b_{0}=1$ are
obtained by plugging $0$ for $k$ in $a_{k}$ and $b_{k}$, and that The non-
geometric part of $\boldsymbol{B}$ is
$(B^{0}-(a_{0}B+b_{0}I_{p}))\boldsymbol{0}_{0}=\boldsymbol{0}$.

  * •

The eigenvalues of $B$ are $1$ and $\displaystyle\frac{-1}{p-1}$ and the
nonzero eigenvalue of $A$ is $p$.

  * •

The geometric part of $\boldsymbol{A}$ is
$\displaystyle\frac{1}{p}\boldsymbol{p}A$ and the non-geometric part of
$\boldsymbol{A}$ is
$\displaystyle(A^{0}-\frac{1}{p}A)\boldsymbol{0}_{0}=C\boldsymbol{0}_{0}$.

  * •

The spectral projections of $A$ and $B$ are
$\pi(A)_{0}=C,\pi(A)_{p}=\displaystyle\frac{1}{p}A$ and
$\pi(B)_{1}=\displaystyle\frac{p-1}{p}B+\frac{1}{p}I_{p},\pi(B)_{\frac{1}{1-p}}=\displaystyle\frac{p-1}{p}(I_{p}-B)$.

  * •

From Corollary 3.6 it follows that the minimal polynomials of $A$ and $B$ are
$m_{A}(X)=X(X-p)$ and $m_{B}(X)=\displaystyle(X-1)(X+\frac{1}{p-1})$.

  * •

$B^{k}=\pi(B)_{1}+(1-p)^{-k}\pi(B)_{\frac{1}{1-p}}$, for all $k\in\mathbb{Z}$.

  * •

$A_{d}^{k}=p^{-k-1}A$ for all $k\geqslant 1$.

## References

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