# Dynamical Sampling for the Recovery of Spatially Constant Source Terms in
Dynamical Systems

A. Aldroubi, R. Diaz Martin, I, Medri  (Akram Aldroubi) Department of
Mathematics, Vanderbilt University, Nashville, Tennessee 37240-0001 USA
<EMAIL_ADDRESS>(Rocío Martín Díaz) Department of Mathematics,
Vanderbilt University, Nashville, Tennessee 37240-0001 USA
<EMAIL_ADDRESS>(Ivan Medri) Department of Mathematics,
Vanderbilt University, Nashville, Tennessee 37240-0001 USA
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we investigate the problem of source recovery in a dynamical
system utilizing space-time samples. This is a specific issue within the
broader field of dynamical sampling, which involves collecting samples from
solutions to a differential equation across both space and time with the aim
of recovering critical data, such as initial values, the sources, the driving
operator, or other relevant details. Our focus in this study is the recovery
of unknown, stationary sources across both space and time, leveraging space-
time samples. This research may have significant applications; for instance,
it could provide a model for strategically placing devices to measure the
quantity of pollutants emanating from factory smokestacks and dispersing
across a specific area. Space-time samples could be collected using measuring
devices placed at various spatial locations and activated at different times.
We present necessary and sufficient conditions for the positioning of these
measuring devices to successfully resolve this dynamical sampling problem.
This paper provides both a theoretical foundation for the recovery of sources
in dynamical systems and potential practical applications.

###### Key words and phrases:

Dynamical Sampling, Observability, Applications of Linear Algebra

###### 2020 Mathematics Subject Classification:

93B07, 42C15, 15A99

Akram Aldroubi is supported in part by NSF Grant NSF/DMS-2208030.

##  1\. Introduction

In various applications, such as environmental monitoring, it is crucial to
accurately determine the location of pollution sources and assess the
magnitude of their emissions. Typically, this information is obtained by
deploying sensors across different locations. Consequently, these scenarios
can be framed as sampling and reconstruction problems. For instance, in the
case of atmospheric emissions, considered in [34] and the references therein,
the goal is to recover the pollution intensity emitted by $K$ smokestacks by
using $L$ sensors.

In the present work, we examine the case where source emissions are constant
in time, and atmospheric pollution evolves through a linear discrete-time
dynamic. To illustrate, let us consider the following dynamical system:

(1)
$x({n+1})=A^{*}x(n)+\underbrace{\sum\limits_{i=1}^{K}c_{i}\,e_{j_{i}}}_{\text{source
term}},\qquad x(0)=x_{0}.$

Here, $x(n)$ represents the state of contamination at discrete time steps
indexed by $n\in\mathbb{N}_{0}:=\mathbb{N}\cup{0}$. We assume that each state
$x(n)$ is a vector in a finite-dimensional vector space, such as
$\mathbb{R}^{d}$. The system initializes at an unknown vector $x_{0}$, and
then the dynamic is governed by the linear operator $A^{*}$ and by the source
term. For each $i=1,\dots,K$, the canonical vector $e_{j_{i}}$ represents the
location of the $i$-th factory in an industrial zone, which releases plumes at
a constant rate $c_{i}$.

Regardless of the initial state $x_{0}$, the main objective is to estimate the
scalar values $c_{1},\dots,c_{K}$ using space-time samples of the form:

(2) $\langle x(n),e_{k_{\ell}}\rangle$

These samples are collected at specific locations $k_{\ell}$,
$\ell=1,\dots,L$, where $\langle\cdot,\cdot\rangle$ denotes the canonical
inner product in $\mathbb{R}^{d}$. The new set of $L$ canonical vectors
$e_{k_{1}},\dots,e_{k_{L}}$ can be viewed as the sensors in this context. (See
Fig. 1)

The problem above is one of many problems that fall under the umbrella of
dynamical sampling problems. For example, when all $c_{i}$ in system (1) are
set to zero (i.e., $c_{i}=0$ for $i=1,\dots,K$), and assuming the operator
$A^{*}$ is known, the objective is to determine the necessary and sufficient
conditions on the spatial sampling vectors for recovering the initial
condition $x_{0}$. This particular task is referred to as the space-time
trade-off sampling problem (see, for e.g., [2, 4, 5, 6, 7, 9, 13, 18, 20, 23,
29, 36, 37]). If, on the other hand, the operator $A^{*}$ is unknown, it leads
to the system identification problem presented in [11, 12, 35]. Lastly, in [8,
10, 12], the focus is on identifying specific types of source terms that drive
the dynamical system. Dynamical sampling problems are connected to several
areas of mathematics, including frame theory, control theory, functional
analysis, and harmonic analysis [1, 3, 16, 17, 18, 21, 22, 23, 24, 25, 28, 29,
30, 31, 37]. These problems also have numerous applications in science and
engineering [14, 15, 32, 33].

Let us return to our sampling and source reconstruction problem in order to
state it in a more general manner. Instead of using the measurements (2) we
can consider more general samples in the direction of vectors
$b_{\ell}\in\mathbb{R}^{d}$, that is,

(3) $y_{\ell}(n):=\langle x(n),b_{\ell}\rangle,\qquad\text{ for
}\ell=1,\dots,L.$

Moreover, it is worth noting that we can treat the source term (1) as a vector
in the subspace

$W:=span\\{e_{j_{1}},\dots,e_{j_{K}}\\}.$

Then, by expressing any vector in $W$ as a linear combination of
$e_{j_{1}},\dots,e_{j_{K}}$, that is,

(4) $\omega=\sum\limits_{i=1}^{K}c_{i}\,e_{j_{i}},$

we can observe that the problem of recovering $c_{1},\dots,c_{K}$ is
equivalent to that of recovering the vector $\omega$ in the subspace $W$.

Precisely, we state our general problem as follows.

######  Problem 1.

Let $\mathbb{F}$ be either $\mathbb{R}$ or $\mathbb{C}$,
$A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator with adjoint $A^{*}$,
and $W$ be a subspace of $\mathbb{F}^{d}$. Given vectors
$b_{1},\dots,b_{L}\in\mathbb{F}^{d}$, consider the discrete-time dynamic

(5) $\begin{cases}x({n+1})=A^{*}x(n)+\omega,\qquad\text{subject to }\omega\in
W\\\ y_{\ell}(n)=\langle x(n),b_{\ell}\rangle,\qquad\text{ for
}\ell\in\\{1,\dots,L\\},\end{cases}$

where the initial state $x(0)=x_{0}$ is unknown. The source recovery problem
consists of providing necessary and sufficient conditions on such vectors
$b_{1},\dots,b_{L}$ so that the constant source $\omega$ can be recovered from
a finite number of linear operations using a finite number of measurements
$y_{\ell}(n)$.

For convenience and brevity, we adopt the following definitions.

######  Definition 1.1.

  1. (1)

The set of vectors $\\{b_{1},\dots,b_{L}\\}$ used in Problem 1 will be called
spatial observational vectors.

  2. (2)

If the set of spatial observational vectors $\\{b_{1},\dots,b_{L}\\}$ allows
the reconstruction of any $\omega\in W$, we say that the spatial observational
vectors are complete for this source recovery problem.

The spatial observational vectors represent physical measuring devices used to
acquire space-time samples. Specifically, if $b_{j}=e_{l_{j}}$, where
$e_{l_{j}}$ is a standard basis element at spatial position $l_{j}$ for
$\mathbb{F}^{d}$, then the set $\\{b_{1},\dots,b_{L}\\}$ corresponds to
measuring devices positioned at $l_{j}$, $1\leq j\leq l$. These devices can be
activated at various time instances $n$ to collect space-time samples.

We will strive to find answers to the following questions:

  * •

If the subspace $W$ and the sampling vectors $b_{1},\dots,b_{L}$ are given in
advance, is it possible to recover any $\omega\in W$?

  * •

Given the subspace $W$, is there a minimal set $b_{1},\dots,b_{L}$ such that
reconstructing any $\omega\in W$ is possible?

  * •

In case that the source reconstruction is possible, how many time iterations
and spatial measurements are needed?

Figure 1. Left Panel: Image credits [19]. The city of Tandil in Buenos Aires
Province, Argentina is depicted on a map, featuring four factories labeled Fa
1 to Fa 4. In the article [19], the authors investigated the magnetic
properties of the soil’s top layers, exploring the hypothesis of a correlation
with the atmospheric fallout of pollutants generated by metallurgical
factories. This pollution study can be modeled in our scenario by viewing the
wind and precipitation evolution as the dynamical system, and the measurements
taken in the surface layer as the spatial samples. Right panel: As a simple
visualization, the city on the left panel is represented on the right panel as
a $7\times 10$ grid where the pollution sources (factories) are shown as black
pixels and the wind evolution is depicted as dotted vectors from right to
left. Therefore, the city can be viewed as $\mathbb{R}^{70}$. Let
$e_{1}\in\mathbb{R}^{70}$ represent the top left location in the grid, and
then let draw the rest of the canonical vectors in $\mathbb{R}^{70}$
indicating locations from left to right and from top to bottom (so, for e.g.,
$e_{10}\in\mathbb{R}^{70}$ represents the top left location, $e_{61}$ the
bottom left, and $e_{70}$ the top bottom right). Then, the pollution sources
can be symbolized by the canonical vectors $e_{14}$, $e_{19}$, $e_{23}$,
$e_{34}$. Finally, the wind and rain can be modelled as a nilpotent operator
given by a $70\times 70$ matrix with $7$ blocks of size $10\times 10$.

###  1.1. Notation

Throughout this paper, $\mathbb{F}$ will be either the field of real numbers
$\mathbb{R}$ or the field of complex numbers $\mathbb{C}$. Then,
$\mathbb{F}^{d}$ is the natural $d$-dimensional vector space over
$\mathbb{F}$, and $\mathbb{F}[x]$ is the ring of all polynomials with
coefficients in $\mathbb{F}$. As usual, the canonical basis for
$\mathbb{F}^{d}$ will be denoted by $\\{e_{1},\dots,e_{d}\\}$ and, if needed,
a vector $v\in\mathbb{F}^{d}$ is written in canonical coordinates reads as
$v=(v^{1},\dots,v^{d})^{T}$. For a subspace $V$ of $\mathbb{F}^{d}$, its
dimension will be $\mathrm{dim}(V)$, and $V^{\perp}$ will be its orthogonal
complement. For a linear operator $A$, we denote by $A^{*}$ its adjoint
operator and by $\sigma(A)$ its spectrum. Also, $I$ stands for the identity
matrix. We will write ${\bf 0}$ for the zero vector or for matrices with all
entries equal to zero. The set $\mathbb{N}_{0}$ denotes the set of non-
negative integers, defined as $\mathbb{N}_{0}:=\mathbb{N}\cup\\{0\\}$.

A recurring notation used throughout this paper is that of the minimal
$A$-annihilating polynomial of $b$, defined as follows:

######  Definition 1.2.

Given a linear operator $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$. The minimal monic
polynomial that annihilates $A$ is denoted by $m_{A}$. Given a vector
$b\in\mathbb{F}^{d}$, the minimal $A$-annihilating polynomial of $b$ is
defined to be the unique monic polynomial $m_{b}\in\mathbb{F}[x]$ of least
degree such that $m_{b}(A)b=\textbf{0}$ (see, for example, [27]). Note that
$m_{b}$ divides $m_{A}$.

###  1.2. Contributions and Organization

Section 2 contains our main results. We first define the observational
augmented orbits (see (17)) that are fundamental for studying Problem 1.
Theorem 2.1 characterizes the observational augmented orbits. In Subsection
2.1, we determine necessary and sufficient conditions for the spatial
observational vectors $b_{1},\dots,b_{L}$ to be complete (see Definition 1.1
above). Our first result in this regard is presented in Theorem 2.4. This
theorem enables us to derive more concrete necessary and sufficient
conditions. For instance, when $1$ is not an eigenvalue of $A$, we present
Theorem 2.5, which offers easily verifiable necessary and sufficient
conditions for the completeness of observational vectors $b_{1},\dots,b_{L}$
(Sub-subsection 2.1.1). We then provide a general test for verifying the
completeness of the observational vectors $b_{1},\dots,b_{L}$ (Theorem 2.11 in
Sub-subsection 2.1.2). We also establish that the number of observational
vectors $L$ required for completeness must be at least equal to the dimension
$\dim W$ of the source space $W$ (see Corollaries 2.6 and 2.13). This raises
the question of when it is possible to reconstruct the source term using
exactly the same number of spatial observational vectors as the dimension of
$W$. In Subsection 2.2, we investigate this question when $\dim W=1$.
Throughout this article, we include illustrative examples to enhance
understanding, motivate certain ideas, and clarify specific issues. The proofs
of the results are relegated to Section 3.

##  2\. main results

In this section, we aim to establish necessary and/or sufficient conditions
for a set of spatial observational vectors $\\{b_{1},\dots,b_{L}\\}$ for the
dynamic (5) to be complete, as stated in Theorems 2.4, 2.5 and their
corollaries. Before we proceed, we begin by rewriting the states and
measurements of the dynamic in Problem 1.

Denoting the initial state by $x(0)=x_{0}$, the evolution of the states
($x(n)=A^{*}x(n-1)+\omega$ in 1) has the form

(6) $x(n)=(A^{*})^{n}x_{0}+\sum_{j=0}^{n-1}(A^{*})^{j}\omega,\qquad\forall
n\in\mathbb{N}.$

Using (6), the time-space measurements in (5) can be written in the following
form:

(7) $\displaystyle y_{\ell}(n)$ $\displaystyle=\langle
x_{0},A^{n}b_{\ell}\rangle+\langle\omega,\Lambda_{n}b_{\ell}\rangle$ (12)
$\displaystyle=\left[{\begin{array}[]{rrr}A^{n}b_{\ell}\\\
\Lambda_{n}b_{\ell}\end{array}}\right]^{*}\left[{\begin{array}[]{rrr}x_{0}\\\
\omega\end{array}}\right]\qquad$

for $n\in\mathbb{N}_{0}$ and $\ell\in{1,\dots,L}$, where

(13) $\Lambda_{n}:=\sum_{j=0}^{n-1}A^{j}\qquad\text{ for
}n\in\mathbb{N},\qquad\Lambda_{0}:={\bf 0}\text{ (the zero-matrix).}$

The last equality in (7), is written in matrix form, where a $d\times 2d$
matrix is multiplied by a vector in $\mathbb{F}^{2d}$ with the first $d$
components representing $x_{0}$ and the last $d$ components representing
$\omega$.

Equation (7) leads us to introduce two fundamental spaces associated with
$\\{b_{1},\dots,b_{L}\\}$.

The first fundamental space is the observational orbits space
$Z(A;b_{1},\dots,b_{L})$ defined by

(14)
$Z(A;b_{1},\dots,b_{L}):=span\\{A^{n}b_{\ell}:\,n\in\mathbb{N}_{0},\,\,\ell\in\\{1,\dots,L\\}\\}.$

The second space we introduce is referred to as the augmented observational
orbits space defined by

(17) $\displaystyle\mathcal{O}(b_{1},\dots,b_{\ell}):=span$
$\displaystyle\left\\{\left[{\begin{array}[]{rrr}A^{n}b_{\ell}\\\
\Lambda_{n}b_{\ell}\end{array}}\right]\in\mathbb{F}^{2d}:\,n\in\mathbb{N}_{0},\,\ell=1,\dots,L\right\\},$

where $\Lambda_{n}$ is as in (13).

The space $\mathcal{O}(b_{1},\dots,b_{L})$ is a subspace of $\mathbb{F}^{2d}$.
Its dimension plays a crucial role in determining necessary and sufficient
conditions in Theorem 2.4 for the completeness of the spatial observational
vectors ${b_{1},\dots,b_{L}}$ (see Definition 1.2). For the case of $L=1$, it
can be shown that $\dim\mathcal{O}(b_{1})\leq d+1$, and that the dimension $d$
can be achieved by an appropriately chosen vectors $b_{1}$. Intuitively, one
might expect that for any $L\geq 2$, the dimension of
$\mathcal{O}(b_{1},\dots,b_{L})$ could be as large as $2d$. However, the
following theorem states that
$\dim\mathcal{O}(b_{1},\dots,b_{L})\leq\min\\{d+L,2d\\}$.

######  Theorem 2.1.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ and vectors
$b_{1},\dots,b_{L}\in\mathbb{F}^{d}\setminus\\{{\bf 0}\\}$. Then, the subspace
$\mathcal{O}(b_{1},\dots,b_{L})\subset\mathbb{F}^{2d}$ of space-time sampling
vectors defined in (17) coincides with the set of vectors

(20)
$\displaystyle\left\\{\left[{\begin{array}[]{r}(A-I)g+\displaystyle\sum_{\ell=1}^{L}\mu_{\ell}\,b_{\ell}\\\
g\end{array}}\right]\in\mathbb{F}^{2d}:\,g\in
Z(A;b_{1},\dots,b_{L}),\,\mu_{1},\dots,\mu_{L}\in\mathbb{F}\right\\}.$

In particular, $\dim\mathcal{O}(b_{1},\dots,b_{L})\leq\min\\{d+L,2d\\}$.

######  Remark 2.2.

When there is a single observational vector $b_{1}$ (i.e., $L=1$), the space
of orbits $Z(A;b_{1})$ is referred to as the Krylov subspace of the highest
order generated by $A$ and $b_{1}$. It is well-known that the dimension of
$Z(A;b_{1})$ is equal to the degree $r$ of the minimal $A$-annihilating
polynomial $m_{b_{1}}(x)$ of $b_{1}$. In fact, if $r$ denotes the degree of
$m_{b_{1}}$, then ${b_{1},Ab_{1},\dots,A^{r-1}b_{1}}$ forms a basis for
$Z(A;b_{1})$.

######  Corollary 2.3.

Let $d\geq 2$ and $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator.
Consider a non-zero vector $b\in\mathbb{F}^{d}$. Then, the dimension of the
space $\mathcal{O}(b)$ is equal to $r+1$, where $r$ is the degree of the
minimal $A$-annihilating polynomial of $b$.

###  2.1. Necessary and Sufficient conditions for completeness of
observational vectors

Our first result provides the necessary and sufficient conditions for the set
${b_{1},\dots,b_{L}}$ to be a complete observational set for Problem 1.
However, its primary utility lies in deriving additional more useful necessary
and/or sufficient conditions.

######  Theorem 2.4.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator, and let $W$ be a
subspace of $\mathbb{F}^{d}$ with $P_{W}:\mathbb{F}^{d}\to\mathbb{F}^{d}$ the
orthogonal projection onto $W$. A set of spatial observational vectors
$\\{b_{1},\dots,b_{L}\\}\subset\mathbb{F}^{d}$ is complete for Problem 1 (see
Definition 1.1) if and only if there exist vectors $g_{1},\dots,g_{N}\in
Z(A;b_{1},\dots,b_{L})$, for some $N\in\mathbb{N}$, such that

  1. (1)

there exists a solution $M\in\mathbb{F}^{L\times N}$ for the matrix problem

(21) $BM=(A-I)G$

where $B$ is the $d\times L$ matrix having $b_{1},\dots,b_{L}$ as columns, and
$G$ is the $d\times N$ matrix with columns $g_{1},\dots,g_{N}$, and

  2. (2)

$\\{P_{W}(g_{1}),\dots,P_{W}(g_{N})\\}$ spans $W$.

####  2.1.1. Completeness of observational vectors when $1\notin\sigma(A)$.

We now use Theorem 2.4 to derive simple, necessary and sufficient conditions,
enabling us to directly determine whether a set of observational vectors is
complete.

######  Theorem 2.5.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator with spectrum
$\sigma(A)$, and let $W$ be a $K$-dimensional subspace of $\mathbb{F}^{d}$.
Furthermore, assume that $1\notin\sigma(A)$, then, the set of spatial
observational vectors
$\\{b_{1},\dots,b_{L}\\}\subset\mathbb{F}^{d}\setminus\\{{\bf 0}\\}$ is
complete for Problem 1 if and only if

(22) $rank\left(P_{W}(A-I)^{-1}B\right)=K$

where $P_{W}$ denotes the orthogonal projection onto $W$, and $B$ is the
matrix having $b_{1},\dots,b_{L}$ as column vectors.

As a corollary, we conclude that the set of observational vectors must contain
at least $K$ vectors when $\dim W=K$.

######  Corollary 2.6.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator with
$1\notin\sigma(A)$, and let $W$ be a $K$-dimensional subspace of
$\mathbb{F}^{d}$. If the set $\\{b_{1},\dots,b_{L}\\}$ forms a complete
observational set for Problem 1, then $L\geq K$.

It turns out that, when $1\notin\sigma(A)$, a complete set of observational
vectors $\\{b_{1},\dots,b_{K}\\}$ can always be found, where $K=\dim W$.

######  Corollary 2.7.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator, and let $W$ be a
subspace of $\mathbb{F}^{d}$. If $\dim W=K$ and $1\notin\sigma(A)$, then there
exists a complete set of spatial observational vectors
$\\{b_{1},\dots,b_{K}\\}$. In particular, if ${\omega_{1},\dots,\omega_{K}}$
form a basis for $W$, a possible choice is given by $b_{k}=(I-A)\omega_{k}$
for $1\leq k\leq K$.

######  Remark 2.8.

  1. (1)

The significance of Condition (22) in Theorem 2.5 is that it provides a
practical method to verify whether a given set of observational vectors is
complete.

  2. (2)

When the source aligns with (4), the $P_{W}$ in (22) simplifies to a diagonal
matrix with diagonal entries from the set $\\{0,1\\}$.

  3. (3)

The selection of the spatial observational vector in Corollary 2.7 represents
merely one potential choice for a complete set of observational vectors.
However, when each vector $b_{k}$ is required to be a standard basis element
of $\mathbb{F}^{d}$, we must choose a set of vectors
${e_{l_{1}},\dots,e_{l_{K}}}$ and validate their completeness using Theorem
2.5.

Theorem 2.5 above provides a simple test for determining whether a set of
spatial observational vectors
$\\{b_{1},\dots,b_{L}\\}\subset\mathbb{F}^{d}\setminus\\{{\bf 0}\\}$ is
complete for Problem 1 under the assumption that $1\notin\sigma(A)$. Later, in
Theorem 2.11 we will provide a general test without the assumption that
$1\not\in\sigma(A)$. Allowing $1$ to be an eigenvalue of $A$ is more difficult
to analyze. In particular, if $1\in\sigma(A)$, then Corollary 2.7 is false as
the following example shows.

######  Example 1.

In dimension $2$, consider the dynamical system

(23) $x({n+1})=A^{*}x(n)+c_{1}\,\omega_{1}$

where

$A:=\left[{\begin{array}[]{rr}1&0\\\ 1&1\end{array}}\right]\qquad\text{ and
}\qquad\omega_{1}:=\left[{\begin{array}[]{r}1\\\ 0\end{array}}\right].$

The goal is to recover the unknown source intensity $c_{1}\in\mathbb{F}$.

The eigenspace of $A$ corresponding to eigenvalue $1$ is spanned by the vector

$\left[{\begin{array}[]{r}0\\\ 1\end{array}}\right]$

and is orthogonal to $W=span\\{\omega_{1}\\}$. Suppose that there exists a
vector

$b=\left[{\begin{array}[]{r}b^{1}\\\ b^{2}\end{array}}\right]$

such that, by sampling the states $x(n)$, we can recover $c_{1}$ in (23). By
denoting the initial state as

$x_{0}=\left[{\begin{array}[]{r}x_{0}^{1}\\\ x_{0}^{2}\end{array}}\right],$

the measurements will be of the form:

$\begin{cases}y(0)=x_{0}^{1}\,b^{1}+x_{0}^{2}\,b^{2}\\\
y(1)=x_{0}^{1}\,b^{1}+x_{0}^{2}\,(b^{2}+b^{1})+b^{1}\,c_{1}\\\
y(2)=x_{0}^{1}\,b^{1}+x_{0}^{2}\,(b^{2}+2b^{1})+2b^{1}\,c_{1}\\\ \vdots\\\
y(n)=x_{0}^{1}\,b^{1}+x_{0}^{2}\,(b^{2}+nb^{1})+nb^{1}\,c_{1}\\\
\vdots\end{cases}$

The equation above shows that to recover $c_{1}$ from the observations $y(j)$,
then necessarily $b^{1}\neq 0$. However, for any $c_{1}\in\mathbb{F}$, if

$x_{0}=c_{1}\left[{\begin{array}[]{r}b^{2}/b^{1}\\\ -1\end{array}}\right]$

then $y(n)=0$ for all $n\in\mathbb{N}_{0}$. Therefore, for any choice of $b$,
there exists $x_{0}$ such that $y_{n}=0$ for every $c_{1}$, and hence $c_{1}$
cannot be determined. Hence, if $1\in\sigma(A)$ and $\mathrm{dim}(W)=1$,
having only one spatial sampling vector $b_{1}:=b$ might not be enough for
solving Problem 1.

####  2.1.2. Completeness of observational vectors: General case.

In order to state the general test for determining whether or not a set of
spatial observational vectors is complete for Problem 1, we need to introduce
some definitions and notation. This general result is also based on Theorem
2.4.

######  Notation 2.9.

Given $p\in\mathbb{F}[x]$, we denote by $\widehat{p}$ the polynomial

(24) $\widehat{p}(x):=\frac{p(x)-p(1)}{x-1}\in\mathbb{F}[x].$

######  Definition 2.10.

(See [27].) Given $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ a linear operator, a
vector $b\in\mathbb{F}^{d}$, and an $A$-invariant subspace
$V\subseteq\mathbb{F}^{d}$, a polynomial $\kappa\in\mathbb{F}[x]$ is said to
bethe minimal $A$-conductor of $b$ into $V$ if it is the (unique) monic
generator of the polynomial ideal $\mathbb{I}:=\\{p\in\mathbb{F}[x]:\,p(A)b\in
V\\}.$

Given $A$ a linear operator in $\mathbb{F}^{d}$ and non-trivial vectors
$b_{1},\dots,b_{L}\in\mathbb{F}^{d}$, in what follows we will consider, for
$j=1,\dots L,$ the minimal $A$-conductor polynomial
$\kappa_{j}\in\mathbb{F}[x]$ of $b_{j}$ into the observational orbits space

$V_{j}:=Z(A;b_{1},\dots,b_{j-1})=\sum_{i=1}^{j-1}Z(A;b_{i}),$

where $V_{1}:=\\{\textbf{0}\\}$. That is, $\kappa_{j}$ is the generator of the
following principal polynomial ideal

$\mathbb{I}_{j}:=\\{p\in\mathbb{F}[x]:\,p(A)b_{j}\in
Z(A;b_{1})+\dots+Z(A;b_{j-1})\\},$

with the understanding that
$\mathbb{I}_{1}:=\\{p\in\mathbb{F}[x]:\,p(A)b_{1}=\textbf{0}\\}$. For example,
note that $\kappa_{1}$ is the minimal $A$-annihilating polynomial $m_{b_{1}}$
of $b_{1}$. Moreover, if $b_{2}\in Z(A;b_{1})$, then $\kappa_{2}(x)=1$ for all
$x$. If instead $Z(A;b_{1})\cap Z(A;b_{2})=\\{\textbf{0}\\}$, then
$\kappa_{2}=m_{b_{2}}$.

Theorem 2.11, below, extends Theorem 2.5 by eliminating the assumption that
$1\notin\sigma(A)$. It provides a method for determining whether a set of
spatial observational vectors
${b_{1},\dots,b_{L}}\subset\mathbb{F}^{d}\setminus{{\bf 0}}$ is complete for
Problem 1. Before we delve into Theorem 2.11, we need to introduce some
constructs that are essential to its understanding.

Let $\\{b_{1},\dots,b_{L}\\}\subset\mathbb{F}^{d}$ be a set of non-trivial
spatial observational vectors. For each $1\leq j\leq L$, consider the minimal
$A$-conductor polynomial $\kappa_{j}\in\mathbb{F}[x]$ of $b_{j}$ into the
observational orbits space $V_{j}:=\sum_{i=1}^{j-1}Z(A;b_{i})$ (where
$V_{1}:=\\{\textbf{0}\\}$, and $\kappa_{1}:=m_{b_{1}}$). For each $2\leq j\leq
L$, let $q_{j}^{i}\in\mathbb{F}[x]$, for $1\leq i\leq j-1$, be such that

(25) $\kappa_{j}(A)b_{j}=q_{j}^{1}(A)b_{1}+\dots+q_{j}^{j-1}(A)b_{j-1},$

and define the vectors

(26)
$g_{j}:=\widehat{\kappa_{j}}(A)b_{j}-\sum_{i=1}^{j-1}\widehat{q_{j}^{i}}(A)b_{i},\qquad\text{
for }1\leq j\leq L,$

where $\widehat{\kappa_{j}}$ and $\widehat{q_{j}^{i}}$ are defined using
Notation 2.9. Using the constructed vectors above we now state the theorem.

######  Theorem 2.11.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator, let $W$ be a
subspace of $\mathbb{F}^{d}$ with $P_{W}:\mathbb{F}^{d}\to\mathbb{F}^{d}$ the
orthogonal projection onto $W$, and let
$\\{b_{1},\dots,b_{L}\\}\subset\mathbb{F}^{d}$ be a set of non-trivial spatial
observational vectors. Then, the set of spatial observational vectors
$\\{b_{1},\dots,b_{L}\\}$ is complete for Problem 1 if and only if
$L\geq\mathrm{dim}(W)$ and

(27) $span\left\\{P_{W}(g_{j}):\,1\leq j\leq L\right\\}=W,$

where $g_{j}$ are as in (26).

######  Remark 2.12.

Condition (22) in Theorem 2.5 can be rewritten as

$span\\{P_{W}(A-I)^{-1}b_{j}:\,1\leq j\leq L\\}=W$

and it coincides with Condition (27) in Theorem 2.11 for the case
$1\not\in\sigma(A)$. Indeed, if $1\not\in\sigma(A)$, by using (25), we can
rewrite the vectors defined by (26) as

$\displaystyle g_{j}$
$\displaystyle=\widehat{\kappa_{j}}(A)b_{j}-\sum_{i=1}^{j-1}\widehat{q_{j}^{i}}(A)b_{i}$
$\displaystyle=(A-I)^{-1}\left((\kappa_{j}(A)-I)b_{j}-\sum_{i=1}^{j-1}(q_{j}(A)-I)b_{i}\right)$
$\displaystyle=(A-I)^{-1}\left(\left(\kappa_{j}(A)b_{j}-\sum_{i=1}^{j-1}q_{j}(A)b_{i}\right)-b_{j}+\sum_{i=1}^{j-1}b_{i}\right)$
(28) $\displaystyle=(A-I)^{-1}(-b_{j}+\sum_{i=1}^{j-1}b_{i}).$

Since

$span\\{b_{j}:\,1\leq j\leq L\\}=span\\{-b_{j}+\sum_{i=1}^{j-1}b_{i}:\,1\leq
j\leq L\\},$

by using (28), we have that

$\displaystyle span\\{P_{W}(A-I)^{-1}b_{j}:\,1\leq j\leq L\\}$
$\displaystyle=span\\{P_{W}(A-I)^{-1}\left(-b_{j}+\sum_{i=1}^{j-1}b_{i}\right):\,1\leq
j\leq L\\}$ $\displaystyle=span\\{P_{W}(g_{j}):\,1\leq j\leq L\\}.$

Therefore, Theorems 2.5 and 2.11 coincide. Hence, Theorem 2.11 generalizes
Theorem 2.5. However, when $1\notin\sigma(A)$ Condition (22) from Theorem 2.5
is simpler and easier to use than Condition (27).

Theorem 2.11 allows us to generalize Corollary 2.6.

######  Corollary 2.13.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator, and $W$ be a
$K$-dimensional subspace of $\mathbb{F}^{d}$. If $\\{b_{1},\dots,b_{L}\\}$
form a complete set for Problem 1, then $L\geq K$.

When $\dim(W)=K$, if $1\notin\sigma(A)$, one can always obtain a complete set
of observational vectors with cardinality $K$ as in Corollary 2.7. However,
when $1\in\sigma(A)$, we may need $L>K$ observational vectors for
completeness. The following example illustrates the situation where
$1\in\sigma(A)$ when $\dim(W)=1$.

######  Example 2.

Consider the $7\times 7$ matrix in Jordan form

(29) $A=\begin{bmatrix}\begin{matrix}1&1&0\\\ 0&1&1\\\
0&0&1\end{matrix}&\mbox{\Large 0}\\\ \mbox{\Large 0}&\begin{matrix}1&1&0&0\\\
0&1&1&0\\\ 0&0&1&1\\\ 0&0&0&1\end{matrix}\end{bmatrix}$

Let $W:=span\\{\omega_{1}\\}$, where $\omega_{1}$ is any standard vector basis
for $\mathbb{R}^{7}$. For each one of the seven choices for $\omega_{1}$, we
will provide all possible sampling vectors $b_{1}\in\\{e_{1},\dots,e_{7}\\}$
that are complete for Problem 1. Since in this example, $1\in\sigma(A)$ there
are cases where completeness is not achieved using a single observational
vector $b_{1}$. For those situations, we will provide all possible sets of
vectors $\\{b_{1},b_{2}\\}\subset\\{e_{1},\dots,e_{7}\\}$ that are complete
for the source recovery problem 1.

We underline that, in this example, we consider only
$\omega_{1},b_{1},b_{2}\in\\{e_{1},\dots,e_{7}\\}$. This corresponds to
modeling some applications similar to the one depicted in Figure 1.

In general, by fixing an eigenvalue $\lambda$ of a matrix $A$ and a Jordan
block of size $J\times J$ corresponding to $\lambda$, it gives rise to what is
called a Jordan chain: A set of linearly independent vectors $v_{j}$,
$j=1,\ldots,J$. On the one hand, the vector $v_{J}$ satisfying $(A-\lambda
I)^{J}v_{J}=0$ is the so-called lead vector of the chain. On the other hand,
the vector $v_{1}:=(A-\lambda I)^{J-1}v_{J}$ is an eigenvector of $A$. For
$1\leq j\leq J$,

(30) $v_{j-1}:=(A-\lambda I)v_{j}.$

In particular, the lead vector $v_{J}$ generates the chain via multiplication
by $A-\lambda I$.

For this example, the only eigenvalue of $A$ is $\lambda=1$ and $A$ has two
Jordan blocks. For each block, we will work with the following Jordan chains:

  * •

Top-left $3\times 3$ Jordan block: $v_{1}=e_{1}$ (eigenvector), $v_{2}=e_{2}$,
and $v_{3}=e_{3}$ (lead vector).

  * •

Bottom-right $4\times 4$ Jordan block: $v_{1}=e_{1}$ (eigenvector),
$v_{2}=e_{2}$, $v_{3}=e_{3}$, and $v_{4}=e_{4}$ (lead vector).

We start by analyzing the cases where $\omega_{1}$ is an eigenvector of $A$.
That is, $\omega_{1}=e_{1}$ and $\omega_{1}=e_{4}$. In these situations,
Condition (32) from Theorem 2.14 below implies that there is always a complete
vector $b_{1}$ for the source recovery problem since
$\omega_{1}\in\mathop{Ker}(A-I)$. In fact, $\omega_{1}$ is contained in one of
the two Jordan chains, and we will show that any vector $b_{1}$ in that chain
is complete. This situation is summarized in the following table:

$\omega_{1}$ | complete observational vector $b_{1}$
---|---
$e_{1}$ |  $e_{1}$ or $e_{2}$ or $e_{3}$
$e_{4}$ |  $e_{4}$ or $e_{5}$ or $e_{6}$ or $e_{7}$

To show this, consider for example the case $\omega_{1}=e_{1}$ (the case
$\omega_{1}=e_{4}$ is analogous). The fact that $b_{1}=e_{1}$ is complete is
trivial as it samples exactly at the source location $\omega_{1}=e_{1}$. For
the other two choices $b_{1}=e_{2}$, $b_{1}=e_{3}$, their minimal
$A$-annihilator polynomials are $m_{b_{1}}(x)=\kappa_{1}(x)=(x-1)^{2}$ and
$m_{b_{1}}(x)=\kappa_{1}(x)=(x-1)^{3}$, resp. Thus, $\widehat{\kappa_{1}}$ is
given by $(x-1)$ if $b_{1}=e_{2}$ and by $(x-1)^{2}$ if $b_{1}=e_{3}$. We now
show that Condition (27) from Theorem 2.11 is satisfied when $b_{1}=e_{2}$.
For this case, the vector $g_{1}$ in (26) is given by $g_{1}=(A-I)e_{2}$, and
we get

$P_{W}(g_{1})=P_{W}(A-I)e_{2}=P_{W}e_{1}=P_{W}\omega_{1}=\omega_{1}.$

Thus, $b_{1}=e_{2}$ is complete. Similarly, if $b_{1}=e_{3}$, then
$g_{1}=(A-I)^{2}e_{3}$ and we have $P_{W}(g_{1})=\omega_{1}$, and
$b_{1}=e_{3}$ is complete.

Now, if $\omega_{1}\in\\{e_{2},e_{3},e_{5},e_{6},e_{7}\\}$, Theorem 2.14 below
will determine that one spatial observational vector $b_{1}$ is never
complete, that is, it is not enough for solving Problem 1. Indeed, in these
cases $\omega_{1}$ is orthogonal to
$\mathop{Ker}(A-I)=\mathrm{span}\\{e_{1},e_{4}\\}$, and since
$(A^{*}-I)^{7}=\textbf{0}$, it trivially holds that
$\omega_{1}\in\mathop{Ker}((A^{*}-I)^{7})$. Thus, $\omega_{1}$ does not
satisfy Condition (32). Therefore, at least two distinct spatial observational
sampling vectors $b_{1}$ and $b_{2}$ will be needed for a complete
observational set.

We recall that, for this example, we added the constraint
$b_{1},b_{2}\in\\{e_{1},\dots,e_{7}\\}$. Thus, since $A$ is given in Jordan
form, we fall under three possible scenarios:

(31) $Z(A;b_{1})\cap Z(A;b_{2})=\\{\textbf{0}\\}\quad\text{or}\quad
Z(A;b_{2})\subseteq Z(A;b_{1})\quad\text{or}\quad Z(A;b_{1})\subseteq
Z(A;b_{2}).$

Since the last two cases are the same if we switch the roles of $b_{1}$ and
$b_{2}$, we will just focus on the first two.

The first case will never give us a complete set for the source recovery
problem. To see this, note that if $Z(A;b_{1})\cap
Z(A;b_{2})=\\{\textbf{0}\\}$, then $b_{1}\in\\{e_{1},e_{2},e_{3}\\}$ and
$b_{2}\in\\{e_{4},e_{5},e_{6},e_{7}\\}$ (or vice versa). Then, the vectors in
(26) are $g_{1}=\widehat{m}_{b_{1}}(A)b_{1}=e_{1}$ and
$g_{2}=\widehat{m}_{b_{2}}(A)b_{2}=e_{4}$ (or vice versa). Hence,
$P_{W}(g_{1})=0$ and $P_{W}(g_{2})=0$, and $\\{b_{1},b_{2}\\}$ is not
complete.

For the case $Z(A;b_{2})\subseteq Z(A;b_{1})$, let $b_{2}=q_{2}^{1}(A)b_{1}$
for some polynomial $q_{2}^{1}\in\mathbb{F}[x]$. Since $b_{2}\in Z(A;b_{1})$,
the minimal $A$-conductor of $b_{2}$ to $Z(A;b_{1})$ is $\kappa_{2}\equiv 1$.
Thus, $\widehat{\kappa_{2}}\equiv 0$. Then, the vectors defined in (26) are
$g_{1}=\widehat{m}_{b_{1}}(A)b_{1}$, and $g_{2}=\widehat{q_{2}^{1}}(A)b_{1}$.
In particular, $P_{W}(g_{1})=0$. Choosing $b_{2}=(A-I)\omega_{1}$, we have
that $P_{W}(g_{2})\neq\textbf{ 0}$. Therefore, we obtain the following table
summarizing all possible complete spatial observational sets of sampling
vectors $\\{b_{1},b_{2}\\}\subset\\{e_{1},\dots,e_{7}\\}$ for each choice of
the source $\omega_{1}$:

$\omega_{1}$ | $b_{1}$ | $b_{2}$
---|---|---
$e_{2}$ |  $e_{3}$ or $e_{2}$ | $e_{1}$
$e_{3}$ | $e_{3}$ | $e_{2}$
$e_{5}$ |  $e_{7}$ or $e_{6}$ or $e_{5}$ | $e_{4}$
$e_{6}$ |  $e_{7}$ or $e_{6}$ | $e_{5}$
$e_{7}$ | $e_{7}$ | $e_{6}$

###  2.2. Existence of an observational vector for a single source term
problem

Theorem 2.11 provides a test for completeness of a given spatial observations
set of sampling vectors. Given $W$ with $\dim W=K$, when $1\notin\sigma(A)$, a
complete set of $K$ observational vectors always exists. However, when
$1\in\sigma(A)$, Examples 1 and 2 show that we may need more than $K$ vectors.
In order to address the challenges associated with the case where
$1\in\sigma(A)$, we make the assumption for the rest of this section that the
source term belongs to a one-dimensional subspace $W$ of $\mathbb{F}^{d}$
spanned by a known unit vector $\omega_{1}$ and provide the necessary and
sufficient conditions under which there exists a complete set for Problem 1
that consists of a single vector.

######  Theorem 2.14.

Let $A$ be a linear operator in $\mathbb{F}^{d}$, and let $W$ be the one-
dimensional subspace of $\mathbb{F}^{d}$ spanned by a unit vector
$\omega_{1}\in\mathbb{F}^{d}\setminus\\{{\bf 0}\\}$. Then, there exists a
complete observational vector $b$ for Problem 1 if and only if

(32)
$\omega_{1}\not\in\mathop{Ker}(A^{*}-I)^{d}\cap(\mathop{Ker}(A-I))^{\perp}=(Range((A-I)^{d}))^{\perp}\cap(\mathop{Ker}(A-I))^{\perp}.$

##  3\. poofs

###### Proof of Theorem 2.1.

Let $L=1$ and write $b_{1}:=b$. Since $\Lambda_{n}$ in (13) is the sum of
consecutive powers of $A$, it is not difficult to show that

(33)
$span\\{\Lambda_{1}b,\Lambda_{2}b,\dots,\Lambda_{N}b\\}=span\\{b,Ab,\dots,A^{N-1}b\\}.$

From its definition (17), any vector $[f\;\;g]^{T}\in\mathcal{O}(b)$ can be
written as :

$\left[{\begin{array}[]{r}f\\\
g\end{array}}\right]=\sum_{n=0}^{k}a_{n}\left[{\begin{array}[]{r}A^{n}b\\\
\Lambda_{n}b\end{array}}\right],$

for some scalars $a_{0},\dots,a_{k}$. Then, using the simple identity

(34) $(A-I)\Lambda_{k}=(A-I)\sum_{j=0}^{k-1}A^{j}=A^{k}-I,$

we get that

$\displaystyle(A-I)g=\sum_{n=0}^{k}a_{n}(A-I)\Lambda_{n}b=\sum_{n=0}^{k}a_{n}(A^{n}-I)b=f-\sum_{n=0}^{k}a_{n}b.$

From this last expression, it follows that

$f=(A-I)g+\mu b,\qquad\text{where }\mu=-\sum_{n=0}^{k}a_{n},\text{ and
}g=\sum_{n=1}^{k}a_{n}\Lambda_{n}b.$

From (33), and the expression for $g$, we conclude that $g\in Z(A;b)$. Thus,

$span\left\\{\left[{\begin{array}[]{r}A^{n}b\\\
\Lambda_{n}b\end{array}}\right]:\,n\in\mathbb{N}_{0}\right\\}\subseteq\left\\{\left[{\begin{array}[]{rr}(A-I)g+\mu
b\\\ g\end{array}}\right]\in\mathbb{F}^{2d}:\,g\in
Z(A;b),\,\mu\in\mathbb{F}\right\\}.$

Conversely, consider the vector in $\mathbb{F}^{2d}$

$\left[{\begin{array}[]{rr}(A-I)g+\mu b\\\ g\end{array}}\right]$

for an arbitrary vector $g\in Z(A;b)$ and a scalar $\mu\in\mathbb{F}$. By
hypothesis, $g$ is a finite linear combination of vectors of the form
$A^{n}b$. Using (33) again, it follows that

(35) $g=\sum_{n=1}^{k}a_{n}\Lambda_{n}b$

for some scalars $a_{1},\dots,a_{k}$. Now, by utilizing (34), we have

$\displaystyle(A-I)g$
$\displaystyle=(A-I)\sum_{n=1}^{k}a_{n}\Lambda_{n}b=\sum_{n=1}^{k}a_{n}A^{n}b-\sum_{n=1}^{k}a_{n}b$
$\displaystyle=\sum_{n=1}^{k}a_{n}A^{n}b-\sum_{n=1}^{k}a_{n}b\pm
a_{0}b=\sum_{n=0}^{k}a_{n}A^{n}b-\sum_{n=0}^{k}a_{n}b$

where the equality holds for any $a_{0}\in\mathbb{F}$. In particular, if we
choose

$a_{0}:=\mu-\sum_{n=1}^{k}a_{n},$

then

(36) $(A-I)g+\mu
b=\sum_{n=0}^{k}a_{n}A^{n}b-\sum_{n=0}^{k}a_{n}b+\sum_{n=0}^{k}a_{n}b=\sum_{n=0}^{k}a_{n}A^{n}b.$

Since $\Lambda_{0}={\bf 0}$, the expressions (35) and (36) prove that

$\left[{\begin{array}[]{rr}(A-I)g+\mu b\\\
g\end{array}}\right]=\sum_{n=0}^{k}a_{n}\left[{\begin{array}[]{rr}A^{n}b\\\
\Lambda_{n}b\end{array}}\right].$

Hence,

$\left\\{\left[{\begin{array}[]{rr}(A-I)g+\mu b\\\
g\end{array}}\right]\in\mathbb{F}^{2d}:\,g\in
Z(A;b),\,\mu\in\mathbb{F}\right\\}\subseteq
span\left\\{\left[{\begin{array}[]{r}A^{n}b\\\
\Lambda_{n}b\end{array}}\right]:\,n\in\mathbb{N}_{0}\right\\}.$

The case for multiple vectors $b_{1},\dots,b_{L}$ follows by using the same
arguments.

Finally, the dimension of
$\mathcal{O}(b_{1},\dots,b_{L})\subset\mathbb{F}^{2d}$ is at most $d+L$ since

$\left[{\begin{array}[]{rr}(A-I)g+\sum\limits_{i=1}^{L}\mu_{i}b_{i}\\\
g\end{array}}\right]=\left[{\begin{array}[]{cc}(A-I)\\\
I\end{array}}\right]g+\sum\limits_{i=1}^{L}\mu_{i}\left[{\begin{array}[]{cc}b_{i}\\\
\textbf{0}\end{array}}\right],\quad g\in
Z(A;b_{1},\cdots,b_{L})\subseteq\mathbb{F}^{d}.$

∎

###### Proof of Corollary 2.3 .

Consider the operator $F:\mathbb{F}^{d}\to\mathbb{R}^{2d}$ given in matrix
form by

(37) $F:=\begin{bmatrix}A-I\\\ I\end{bmatrix}$

where $I$ denotes the $d\times d$ identity matrix. Since
$\mathrm{dim}(Z(A;b))=r$, where $r$ is the degree of $m_{b}$ (see Remark 2.2),
then $\mathrm{dim}(F(Z(A;b)))\leq r$. On the other hand, using that
$\\{b,Ab,\dots,A^{r-1}b\\}$ is a basis from $Z(A;b)$ (see Remark 2.2), it
follows that vectors

$F(A^{j}b)=\begin{bmatrix}(A-I)A^{j}b\\\ A^{j}b\end{bmatrix}\qquad\text{for
}\quad 0\leq j\leq r-1$

are linearly independent. So, $\mathrm{dim}(F(Z(A;b)))=r$. Since
$b\neq\textbf{0}$, for any non-zero $g\in Z(A;b)$, the vectors

$\begin{bmatrix}(A-I)g\\\ g\end{bmatrix}\quad\text{ and
}\quad\begin{bmatrix}b\\\ \textbf{0}\end{bmatrix}$

are linearly independent. Therefore, from the characterization given by
Theorem 2.1,

(42)
$\displaystyle\mathcal{O}(b):=span\left\\{\left[{\begin{array}[]{r}A^{n}b\\\
\Lambda_{n}b\end{array}}\right]:\,n\in\mathbb{N}_{0}\right\\}$
$\displaystyle=\left\\{\left[{\begin{array}[]{rr}(A-I)g+\mu b\\\
g\end{array}}\right]:\,g\in Z(A;b),\,\mu\in\mathbb{F}\right\\}$

and we have that the dimension of $\mathcal{O}(b)$ is $r+1$. ∎

###  3.1. Proofs of Section 2.1

We start with a lemma that relates linear combinations of measurements as in
(7) with vectors in the augmented observational orbits space
$\mathcal{O}(b_{1},\dots,b_{L})$ defined in (17). Specifically, let
$\\{\alpha_{n,\ell}\\}_{n,\ell}$ be an arbitrary finite set of scalars, and
consider the linear combination of measurements

(43) $\sum_{n,\ell}\alpha_{n,\ell}\,y_{\ell}(n)=\langle
x_{0},\sum_{n,\ell}\alpha_{n,\ell}A^{n}b_{\ell}\rangle+\langle\omega,\sum_{n,\ell}\alpha_{n,\ell}\Lambda_{n}b_{\ell}\rangle.$

We have:

######  Lemma 3.1.

Let $A$ be a linear operator in $\mathbb{F}^{d}$, and $b_{1},\dots,b_{L}$ be
vectors in $\mathbb{F}^{d}$. For each $n\in\mathbb{N}_{0}$ and
$\ell\in\\{1,\dots,L\\}$, let $y_{\ell}(n):=\langle
x_{0},A^{n}b_{\ell}\rangle+\langle\omega,\Lambda_{n}b_{\ell}\rangle$ as in
(7). Then

  1. (a)

For any linear combinations of measurements as in (43), the vectors

(44) $f:=\sum\alpha_{n,\ell}A^{n}b_{\ell}\qquad\text{ and }\qquad
g:=\sum\alpha_{n,\ell}\Lambda_{n}b_{\ell}.$

are such that, $g\in Z(A;b_{1},\dots,b_{L})$ and
$f=(A-I)g+\sum_{\ell=1}^{L}\mu_{\ell}b_{\ell}$ for some
$\mu_{1},...,\mu_{L}\in\mathbb{F}$ (which depend on the scalars
$\alpha_{n,\ell}$). Hence (by Theorem 2.1)
$(f,g)^{T}\in\mathcal{O}(b_{1},\dots,b_{L})$.

  2. (b)

Conversely, for any pair of vectors $f,g$ such that $g\in
Z(A;b_{1},\dots,b_{L})$, $f:=(A-I)g+\sum_{j=1}^{L}\mu_{\ell}b_{\ell},$ where
$\mu_{1},...,\mu_{L}\in\mathbb{F}$, there exists a finite set of scalars
$\\{\alpha_{n,\ell}\\}_{n,\ell}\subset\mathbb{F}$ (depending on
$\mu_{1},\dots,\mu_{L}$ and $g$) such that $f$ and $g$ can be written as in
(44) and Equation (43) is satisfied. In particular

(45) $\sum\alpha_{n,\ell}\,y_{\ell}(n)=\langle
x_{0},f\rangle+\langle\omega,g\rangle.$

###### Proof.

Using linear combinations of the time-space measurements (7), we can generate
any equation of the form

(46) $\underbrace{\sum\alpha_{n,\ell}\,y_{\ell}(n)}_{\widetilde{y}}=\langle
x_{0},\underbrace{\sum\alpha_{n,\ell}A^{n}b_{\ell}}_{f}\rangle+\langle\omega,\underbrace{\sum\alpha_{n,\ell}\Lambda_{n}b_{\ell}}_{g}\rangle.$

From Theorem 2.1 we know that the vectors $f$ and $g$ defined in Equation (46)
are related by $f=(A-I)g+\sum\mu_{\ell}b_{\ell}$ for $g\in
Z(A;b_{1},\dots,b_{\ell})$ and for some scalars $\mu_{\ell}$,
$\ell=1,\dots,L$. Conversely, since Theorem 2.1 gives us the equality

$\displaystyle\mathcal{O}$ $\displaystyle(b_{1},\dots,b_{L})$
$\displaystyle=\left\\{\left[{\begin{array}[]{r}(A-I)g+\displaystyle\sum_{\ell=1}^{L}\mu_{\ell}\,b_{\ell}\\\
g\end{array}}\right]\in\mathbb{F}^{2d}:\,g\in
Z(A;b_{1},\dots,b_{L}),\,\mu_{1},\dots,\mu_{L}\in\mathbb{F}\right\\},$

we have that given any vector $g\in Z(A;b_{1},\dots,b_{\ell})$ and any scalars
$\mu_{1},\dots,\mu_{L}$, the vector $f:=(A-I)g+\sum\mu_{\ell}b_{\ell}$ belongs
to $Z(A;b_{1},\dots,b_{L})$ and thus there exists scalars
$\\{\alpha_{n,\ell}\\}$ that provides us an equation like (46). ∎

###### Proof of Theorem 2.4.

  * •

Suppose there exist vectors $g_{1},\dots,g_{N}\in Z(A;b_{1},\dots,b_{L})$
satisfying Conditions 1 and 2 of Theorem 2.4.

Condition 1 states that there exists a matrix $M\in\mathbb{F}^{L\times N}$
satisfying equation (21) $(A-I)G-BM=0$ (where $B$ and $G$ are the matrices
having the vectors $b_{1},\dots,b_{L}$ and the vectors $g_{1},\dots,g_{N}$,
resp., as columns). Denoting $\mu_{\ell}^{j}:=-M(\ell,j)$, we can rewrite (21)
as $N$ equations of the form
$(A-I)g_{j}+\sum_{\ell=1}^{L}\mu_{\ell}^{j}b_{\ell}={\bf 0}$, for $1\leq j\leq
N$. Defining $f_{j}:=(A-I)g_{j}+\sum\mu_{\ell}^{j}b_{\ell}$, we have that
$f_{j}=\textbf{0}$ for each $1\leq j\leq N$. Then, Part (b) of Lemma 3.1
implies that, for each $1\leq j\leq N$ there exists a finite set of scalars
$\\{\alpha_{n,\ell}^{j}\\}_{n,\ell}\subset\mathbb{F}$ (that depend on
$\mu_{1},\dots,\mu_{L}$ and $g_{j}$) such that we get $N$ equations of the
form

(47)
$\sum_{n,\ell}\alpha_{n,\ell}^{j}\,y_{\ell}(n)=\langle\omega,f_{j}\rangle+\langle\omega,g_{j}\rangle,\qquad
1\leq j\leq N.$

Using the fact that $f_{j}=\textbf{0}$, and that $\omega\in W$, we obtain

(48)
$\sum_{n,\ell}\alpha_{n,\ell}^{j}\,y_{\ell}(n)=\langle\omega,g_{j}\rangle=\langle
P_{W}\omega,g_{j}\rangle=\langle\omega,P_{W}g_{j}\rangle,\qquad 1\leq j\leq
N.$

Since the set of vectors $\\{g_{1},\dots,g_{N}\\}$ also satisfies Condition 2,
$\\{P_{W}g_{1},\dots,P_{W}g_{N}\\}$ is a spanning set for $W$. Thus,
$\\{P_{W}g_{1},\dots,P_{W}g_{N}\\}$ form a frame for $W$ and $\omega$ can be
recovered from the system of equations (47).

  * •

For the converse, assume that there exists a set of spatial observational
vectors $\\{b_{1},\dots,b_{L}\\}$ that are complete for Problem 1. That is,
there exists a finite number $N\geq K$ of equations of the form (43) that
allows us to solve for $\omega\in W$ using linear operations on the
measurements $y_{\ell}(n)$. Due to Lemma 3.1, such a system of $N$ equations
of the form (43) can be written as a system of $N$ equations of the form (45)
and, in matrix notation, we can rewrite the hypothesis that
$\\{b_{1},\dots,b_{L}\\}$ are complete for Problem 1 as follows: There exist
$N$ vectors
$(f_{1},g_{1})^{T},\dots,(f_{N},g_{N})^{T}\in\mathcal{O}(b_{1},\dots,b_{L})$
and a vector
$\widetilde{y}:=(\widetilde{y}_{1},\dots,\widetilde{y}_{N})^{T}\in\mathbb{F}^{N}$
where
$\widetilde{y}_{j}:=\sum\limits_{\ell=1}^{L}\sum_{n}\alpha_{n,\ell}^{j}\,y_{\ell}(n)$
for a finite set of scalars $\\{\alpha_{n,\ell}^{j}\\}$, such that any
$\omega\in W\subseteq\mathbb{F}^{d}$ can be recovered from the matrix system

$\begin{bmatrix}F^{*}&G^{*}\end{bmatrix}\begin{bmatrix}x_{0}\\\
\omega\end{bmatrix}=\widetilde{y},$

where $F^{*}$ and $G^{*}$ are the adjoints of the $d\times N$ matrices $F,G$
having the vectors $f_{1},\dots,f_{N}$ and $g_{1},\dots,g_{N}$ as columns,
respectively. This means that there exists an $N\times d$ matrix $U$ such that

(49)
$\omega=U^{*}\begin{bmatrix}F^{*}&G^{*}\end{bmatrix}\begin{bmatrix}x_{0}\\\
\omega\end{bmatrix}=U^{*}\widetilde{y}.$

Since the map $y\mapsto U^{*}y$ is linear, the right-hand side
$U^{*}\widetilde{y}$ is another linear combination of the samples
$\\{y_{\ell}(n)\\}$ of the dynamical system (5). Hence, (49) is a system of
equations of the form (43). Rewriting the left inequality in (49), we obtain
that

(50) $\omega=(FU)^{*}x_{0}+(GU)^{*}\omega\qquad\text{ for any
}x_{0}\in\mathbb{F}^{d},\text{ and any }\omega\in W.$

Let us denote by $g_{1}^{\prime},\dots,g_{d}^{\prime}$ and
$f_{1}^{\prime},\dots,f_{d}$ the column vectors of the $d\times d$ matrices
$G^{\prime}=GU$ and $F^{\prime}=FU$, respectively. Then, by Lemma 3.1, for
each $1\leq j\leq d$, there exist scalars $\mu_{\ell}^{j}$, ($1\leq\ell\leq
L$) such that

$f_{j}^{\prime}=(A-I)g_{j}^{\prime}+\sum_{\ell=1}^{L}\mu_{\ell}^{j}b_{\ell}^{j}.$

We now show that $\\{P_{W}(g_{1}^{\prime}),\dots,P_{W}(g_{d}^{\prime})\\}$
spans $W$ and $f_{j}^{\prime}\equiv 0$. Taking $x_{0}=\textbf{0}$ in (50), we
get that $\omega=(GU)^{*}\omega$ for any $\omega\in W$ or, equivalently,

$P_{W}(v)=(GU)^{*}P_{W}(v)\qquad\forall v\in\mathbb{F}^{d}.$

This means that $P_{W}GU=PG^{\prime}=P_{W}$. Therefore,
$\\{P_{W}(g_{1}^{\prime}),\dots,P_{W}(g_{d}^{\prime})\\}$ spans $W$. That is,
the vectors $g_{1}^{\prime},\dots,g_{d}^{\prime}$ satisfy Condition 2 for
$N=d$.

Finally, to show that Condition 1 is satisfied, note that since for any
$x_{0}\in\mathbb{F}^{d}$, and any $\omega\in W$

$\omega=(FU)^{*}x_{0}+(GU)^{*}\omega\quad\text{ and
}\quad\omega=(GU)^{*}\omega,$

we have $(FU)^{*}=(F^{\prime})^{*}\equiv\textbf{0}$. Hence,
$f_{j}^{\prime}=(A-I)g_{j}^{\prime}+\sum_{\ell=1}^{L}\mu_{\ell}^{j}b_{\ell}^{j}={\bf
0}$. Thus, letting $M$ be the $L\times d$ matrix given by
$M(\ell,j)=-\mu_{\ell}^{j}$, we and obtain that the vectors
$g_{1}^{\prime},\dots,g_{d}^{\prime}$ also satisfy Condition 1.

∎

###  3.2. Proofs of Section 2.1.1

To prove Theorem 2.5, we need the preliminary results given by Lemma 3.2 and
Proposition 3.3 below.

######  Lemma 3.2.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator, let
$b\in\mathbb{F}^{d}$, and consider the function $f(x)=1/(x-1)$. If the minimal
$A$-annihilating polynomial $m_{b}(x)$ of $b$ is such that $m_{b}(1)\not=0$,
then $f(A)b$ is well-defined and $f(A)b\in Z(A;b)$. In particular, if $A$ does
not have eigenvalue $1$, then, for any $b\in\mathbb{F}^{d}$, $(A-I)^{-1}b\in
Z(A;b).$

###### Proof.

This result follows by applying Theorem 4.1 in the Appendix to the function
$f(x)=1/(1-x)$ and its Hermite interpolating polynomial $q$ on the roots of
$m_{b}(x)$ given by (86). Recall that $p(A)b\in Z(A;b)$ for every polynomial
$p\in\mathbb{F}[x]$. ∎

For the next proposition, we use the same notation as in Theorem 2.4.

######  Proposition 3.3.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator such that
$1\notin\sigma(A)$, and let $\\{b_{1},\dots,b_{L}\\}$ be a set of vectors in
$\mathbb{F}^{d}$. If vectors $g_{1},\dots,g_{T}\in\mathbb{F}^{d}$ satisfy
Condition 1 of Theorem 2.4, then they can be written as the columns of
$G:=(A-I)^{-1}BM$. Moreover, $g_{1},\dots,g_{T}\in Z(A;b_{1},\dots,b_{L})$.

###### Proof.

Since $A$ does not have eigenvalue $1$, $G=(A-I)^{-1}BM$ is well-defined and
from Lemma 3.2 its columns belong to $Z(A;b_{1},\dots,b_{L})$. ∎

We now prove Theorem 2.5 using the two previous results.

###### Proof of Theorem 2.5.

The proof of the necessity of the equality (22) is similar to the proof of
Corollary 2.6 below. To prove sufficiency, assume that (22) is satisfied and
consider the matrix $G:=(A-I)^{-1}B$. Then, clearly, (21) is satisfied, with
$M$ being the $L\times L$ identity matrix. The columns of $G$ are the vectors
$g_{i}:=(A-I)^{-1}b_{i}$ for $i=1,\dots,K$. By Lemma 3.2, for each
$i=1,\dots,K$, we have that $g_{i}\in Z(A;b_{i})$. By equality (22), $P_{W}G$
has rank $K$, so its columns, which are the vectors $P_{W}g_{i}$, span $W$.
Thus, from Theorem 2.4, the observational vectors
$\\{b_{1},\dots,b_{L}\\}\subset\mathbb{F}^{d}\setminus{\\{\bf 0\\}}$ are
complete for Problem 1.

∎

###### Proof of Corollary 2.6.

Let $G$ and $B$ be as in the statement of Theorem 2.4. Condition 2 states that
$rank(P_{W}G)=K$, which implies that ${\rm rank\,}(G)\geq K$. Thus, using
Equation (21), we have that ${\rm rank\,}(G)={\rm
rank\,}\big{(}(A-I)^{-1}BM\big{)}\geq K$. Therefore, ${\rm rank\,}(B)\geq K$.
Consequently, we require at least $K$ linearly independent spatial
observational vectors among ${b_{1},\dots,b_{L}}$, implying $L\geq K$. ∎

###### Proof of Corollary 2.7.

Assume that $A-I$ is a non-singular operator in $\mathbb{F}^{d}$, and that the
source term in (5) is such that $\omega\in W$ where $\mathrm{dim}(W)=K$. One
strategy for positioning the sensors $b_{i}$ is the following: Consider
$\\{\omega_{\ell}\\}_{\ell=1}^{K}$ an orthonormal basis for $W$, and then
define the vectors

(51) $b_{\ell}:=(I-A)\omega_{\ell},\qquad\ell\in\\{1,\dots,K\\}.$

For each $\ell\in\\{1,\dots,K\\}$, consider the scalars
$\\{\alpha_{n,\ell}\\}_{n=0}^{r_{\ell}}$ that are the coefficients of the
minimal $A$-annihilating polynomial
$m_{b_{\ell}}(x)=\sum_{n=0}^{r_{\ell}}\alpha_{n,\ell}x^{n}$ of $b_{\ell}$.
Then, given the measurements

$y_{\ell}(n)=\langle
x_{0},A^{n}b_{\ell}\rangle+\langle\omega,\Lambda_{n}b_{\ell}\rangle$

as in (7), for each $\ell=1,\dots,K$ we use Identity (34) and the fact that
$m_{b_{\ell}}(A)b_{\ell}=\textbf{0}$ to get

$\displaystyle\sum_{n=0}^{r_{\ell}}\alpha_{n,\ell}y_{\ell}(n)$
$\displaystyle=\langle\omega,\sum_{n=0}^{r_{\ell}}\alpha_{n,\ell}\Lambda_{n}b_{\ell}\rangle=m_{b_{\ell}}(1)\langle\omega,(I-A)^{-1}b_{\ell}\rangle=m_{b_{\ell}}(1)\langle\omega,\omega_{\ell}\rangle.$

Hence, since $1\notin\sigma(A)$, $m_{b_{\ell}}(1)\neq 0$, we can solve for
$\langle\omega,\omega_{\ell}\rangle$ above and obtain

$\omega=\left(\frac{1}{m_{b_{1}}(1)}\sum_{n=0}^{r_{1}}\alpha_{n,1}y_{1}(n)\right)\omega_{1}+\dots+\left(\frac{1}{m_{b_{K}}(1)}\sum_{n=0}^{r_{K}}\alpha_{n,K}y_{K}(n)\right)\omega_{K}.$

Moreover, if $r_{j}$ is the degree of the minimal $A$-annihilating polynomial
of $b_{j}$, then the number of time samples that are needed to find $\omega$
is $T=\max\limits_{\ell\in\\{1,\dots,K\\}}\\{r_{\ell}+1\\}$. ∎

###  3.3. Proofs for Section 2.1.2

Now we will prove Theorem 2.11. To do so we need several preliminary results
and their proofs stated in Lemma 3.4, Theorem 3.6, and Corollary 3.8 below.

######  Lemma 3.4.

Using Notation 2.9, we have the following.

  1. (1)

For $u(x),v(x)\in\mathbb{F}[x]$, we have

(52) $\widehat{uv}(x)=\widehat{u}(x)v(x)+u(1)\widehat{v}(x).$

  2. (2)

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator,
$b\in\mathbb{F}^{d}$, and $m_{b}$ be its minimal $A$-annihilating polynomial.
If $p\in\mathbb{F}[x]$ is such that $p(A)b=\textbf{0}$, then there exists
$a\in\mathbb{F}$

$\widehat{p}(A)b=a\,\widehat{m_{b}}(A)b.$

###### Proof.

Identity (52) can be established by a straightforward computation.

To prove Part (2) of the Lemma, we note that, since by hypothesis $p$ is an
$A$-annihilating polynomial of $b$, $m_{b}$ must divide $p$. Thus, there
exists $q\in\mathbb{F}[x]$ such that $p(x)=q(x)m_{b}(x)$. Then, using Part (1)
of the lemma with $q=u$, $m_{b}=v$, using the fact that
$m_{b}(A)b=\textbf{0}$, and setting $a=q(1)$ we get

$\widehat{p}(A)b=\widehat{q}(A)m_{b}(A)b+q(1)\widehat{m_{b}}(A)b=q(1)\widehat{m_{b}}(A)b.$

∎

######  Definition 3.5.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator and let
$b_{1},\dots,b_{L}\in\mathbb{F}^{d}$. We say that $g\in\mathbb{F}^{d}$ is an
$A$-$\\{b_{1},\dots,b_{L}\\}$ characteristic vector if $g\in
Z(A;b_{1},\dots,b_{L})$ and satisfies

(53) $(A-I)g=\sum_{\ell=1}^{L}\mu_{\ell}b_{\ell}$

for some scalars $\mu_{1},\dots,\mu_{L}$. In other words, $g$ is
$A$-$\\{b_{1},\dots,b_{L}\\}$ characteristic if it belongs to the pre-image

$\mathcal{G}:=\\{g\in Z(A;b_{1},\dots,b_{L}):(A-I)g\in
span\\{b_{1},\dots,b_{l}\\}\\}.$

The space $\mathcal{G}$ will be called the $A$-$\\{b_{1},\dots,b_{l}\\}$
characteristic space of the observational space $span\\{b_{1},\dots,b_{l}\\}$.

Inspired by the Cyclic Decomposition Theorem, and using Definition 2.10 of
$A$-conductors, we state and prove the following result.

######  Theorem 3.6.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator and let
$b_{1},\dots,b_{L}\in\mathbb{F}^{d}$ be non-trivial vectors. For each $1\leq
j\leq L$ consider $\kappa_{j}\in\mathbb{F}[x]$ the minimal $A$-conductor
polynomial of $b_{j}$ into the observational orbits space
$V_{j}:=\sum_{i=1}^{j-1}Z(A;b_{i})$ (where $V_{1}:=\\{\textbf{0}\\}$, and
$\kappa_{1}:=m_{b_{1}}$). Also, for each $2\leq j\leq L$, let
$q_{j}^{i}\in\mathbb{F}[x]$, for $1\leq i\leq j-1$, be such that

$\kappa_{j}(A)b_{j}=q_{j}^{1}(A)b_{1}+\dots+q_{j}^{j-1}(A)b_{j-1}.$

Then, $g$ is an $A$-$\\{b_{1},\dots,b_{L}\\}$ characteristic vector, if and
only if it is of the form

(54)
$g=\sum_{j=1}^{L}a_{j}\left(\widehat{\kappa_{j}}(A)b_{j}-\sum_{i=1}^{j-1}\widehat{q_{j}^{i}}(A)b_{i}\right)$

for some scalars $a_{1},\dots,a_{L}\in\mathbb{F}$.

###### Proof.

First, let us prove that the scalars $a_{1},\dots,a_{L}$ in (54) can be chosen
arbitrarily, i.e., that any vector $g$ of the form (54) is an
$A$-$\\{b_{1},\dots,b_{L}\\}$ characteristic vector. Indeed, the expression
(54) defines a vector in $Z(A;b_{1},\dots,b_{L})$ as it is a linear
combination of polynomials on $A$ applied to the vectors $b_{1},\dots,b_{L}$.
Also, by using that $\kappa_{j}(A)b_{j}=\sum_{i=1}^{j-1}q_{j}^{i}(A)b_{i}$,
for $2\leq j\leq L$,

$\displaystyle(A-I)g$
$\displaystyle=(A-I)\left(a_{1}\widehat{m_{b_{1}}}(A)b_{1}+\sum_{j=2}^{L}a_{j}\left(\widehat{\kappa_{j}}(A)b_{j}-\sum_{i=1}^{j-1}\widehat{q_{j}^{i}}(A)b_{i}\right)\right)$
$\displaystyle=a_{1}(m_{b_{1}}(A)-m_{b_{1}}(1))b_{1}+\sum_{j=2}^{L}a_{j}\left((\kappa_{j}(A)-\kappa_{j}(1))b_{j}-\sum_{i=1}^{j-1}(q_{j}^{i}(A)-q_{j}^{i}(1))b_{i}\right)$
$\displaystyle=a_{1}m_{b_{1}}(1)b_{1}+\sum_{j=2}^{L}a_{j}\left(\sum_{i=1}^{j-1}q_{j}^{i}(1)-\kappa_{j}(1)\right)b_{j},$

That is, if $g$ is given by (54) for some scalars $a_{1},\dots,a_{L}$, then
$g$ satisfies (21) with $\mu_{1}:=a_{1}m_{b_{1}}(1)$, and
$\mu_{j}=a_{j}\left(\sum_{i=1}^{j-1}q_{j}^{i}(1)-\kappa_{j}(1)\right)$ for
$2\leq j\leq L$.

Now, given an $A$-$\\{b_{1},\dots,b_{L}\\}$ characteristic vector $g$, we will
prove that $g$ can be written in the form (54). As $g\in
Z(A;b_{1},\dots,b_{L})$, it can be written as

(55) $g=p_{1}(A)b_{1}+p_{2}(A)b_{2}+\dots+p_{L}(A)b_{L}$

for some $p_{1},p_{2},\dots,p_{L}\in\mathbb{F}[x]$. Also, there exist
$\mu_{1},\dots,\mu_{L}\in\mathbb{F}$ such that
$(A-I)g=\sum_{\ell=1}^{L}\mu_{\ell}b_{\ell}$ or, equivalently,

(56)
$\sum_{\ell=1}^{L}\left((A-I)p_{\ell}(A)-\mu_{\ell}\right)b_{\ell}=\textbf{0}.$

To illustrate how to characterize the polynomials $p_{1},\dots,p_{L}$
satisfying (56) for some scalars $\mu_{1},\dots,\mu_{L}$, we will proceed by
cases.

  1. (1)

($L=1$): In this case the expression (55) reads as $g=p_{1}(A)b_{1}$ and the
identity (56) reads as $((A-I)p_{1}(A)-\mu_{1})b_{1}=\textbf{0}$. Thus,
$m_{b_{1}}$ divides the polynomial $(x-1)p_{1}(x)-\mu_{1}$. That is, there
exists a polynomial $\alpha_{1}(x)\in\mathbb{F}[x]$ such that

$\alpha_{1}(x)m_{b_{1}}(x)=(x-1)p_{1}(x)-\mu_{1}.$

Evaluating that expression at $x=1$, we obtain
$\mu_{1}=\alpha_{1}(1)m_{b_{1}}(1)$, and so

$p_{1}(x)=\frac{\alpha_{1}(x)m_{b_{1}}(x)-\alpha_{1}(1)m_{b_{1}}(1))}{x-1}.$

Finally, by using Lemma 3.4, the fact that $m_{b_{1}}(A)b_{1}=0$, and denoting
$a_{1}=\alpha_{1}(1)$, we have that

$g=a_{1}\,\widehat{m_{b_{1}}}(A)b_{1}.$

This concludes the proof for this case since in the statement of the theorem
$\kappa_{1}:=m_{b_{1}}$.

  2. (2)

($L=2$): In this case the expression (55) reads as

(57) $g=p_{1}(A)b_{1}+p_{2}(A)b_{2}$

and the identity (56) reads as

(58) $((A-I)p_{1}(A)-\mu_{1})b_{1}+((A-I)p_{2}(A)-\mu_{2})b_{2}=\textbf{0}.$

From the last expression we can deduce that $((A-I)p_{2}(A)-\mu_{2})b_{2}\in
Z(A;b_{1})\cap Z(A;b_{2})$. Let $\kappa_{2}\in\mathbb{F}[x]$ be the generator
of polynomial ideal $\mathbb{I}_{2}:=\\{k\in\mathbb{F}[x]:\,k(A)b_{2}\in
Z(A;b_{1})\\}$. Then, $\kappa_{2}$ divides the polynomial
$(x-1)p_{2}(x)-\mu_{2}$. That is, there exists a polynomial
$\alpha_{2}(x)\in\mathbb{F}[x]$ such that

(59) $\alpha_{2}(x)\kappa_{2}(x)=(x-1)p_{2}(x)-\mu_{2}$

Evaluating the expression above at $x=1$ and solving for $\mu_{2}$, we obtain

(60)
$p_{2}(x)=\frac{\alpha_{2}(x)\kappa_{2}(x)-\alpha_{2}(1)\kappa_{2}(1)}{x-1}=:\widehat{\alpha_{2}\kappa_{2}}(x).$

Replacing (59) into (58) and taking into account that
$\kappa_{2}(A)b_{2}=q_{2}^{1}(A)b_{1}$, we get

0 $\displaystyle=((A-I)p_{1}(A)-\mu_{1})b_{1}+\alpha_{2}(A)\kappa_{2}(A)b_{2}$
$\displaystyle=\left((A-I)p_{1}(A)-\mu_{1}+\alpha_{2}(A)q_{2}^{1}(A)\right)b_{1}.$

Hence, $m_{b_{1}}$ divides the polynomial
$(x-1)p_{1}(x)-\mu_{1}+\alpha_{2}(x)q_{2}^{1}(x)$, that is, there exists
$\alpha_{1}(x)\in\mathbb{F}[x]$ such that

$\alpha_{1}(x)m_{b_{1}}(x)=(x-1)p_{1}(x)-\mu_{1}+\alpha_{2}(x)q_{2}^{1}(x).$

Evaluating at $x=1$, we can solve for the value of $\mu_{1}$ and obtain

(61)
$p_{1}(x)=\frac{\alpha_{1}(x)m_{b_{1}}(x)-\alpha_{1}(1)m_{b_{1}}(1)}{x-1}-\frac{\alpha_{2}(x)q_{2}^{1}(x)-\alpha_{2}(1)q_{2}^{1}(1)}{x-1}=\widehat{\alpha_{1}m_{b_{1}}}(x)-\widehat{\alpha_{2}q_{2}^{1}}(x).$

Now, we replace (61) and (60) into (57) and get

(62)
$g=\widehat{\alpha_{1}m_{b_{1}}}(A)b_{1}-\widehat{\alpha_{2}q_{2}^{1}}(A)b_{1}+\widehat{\alpha_{2}\kappa_{2}}(A)b_{2}.$

Using Lemma 3.4, the fact that $m_{b_{1}}(A)b_{1}=0$, we have the identity

(63)
$\widehat{\alpha_{1}m_{b_{1}}}(A)b_{1}=\alpha_{1}(1)\widehat{m_{b_{1}}}(A)b_{1}.$

Using Lemma 3.4 again, we rewrite the polynomials
$\widehat{\alpha_{2}q_{2}^{1}}$ and $\widehat{\alpha_{2}\kappa_{2}}$ as

$\displaystyle\widehat{\alpha_{2}q_{2}^{1}}(x)$
$\displaystyle=\widehat{\alpha_{2}}(x)q_{2}^{1}(x)+\alpha_{2}(1)\widehat{q_{2}^{1}}(x)$
$\displaystyle\widehat{\alpha_{2}\kappa_{2}}(x)$
$\displaystyle=\widehat{\alpha_{2}}(x)\kappa_{2}(x)+\alpha_{2}(1)\widehat{\kappa_{2}}(x),$

and use the fact that $\kappa_{2}(A)b_{2}=q_{2}^{1}(A)b_{1}$ to obtain

(64)
$\displaystyle-\widehat{\alpha_{2}q_{2}^{1}}(A)b_{1}+\widehat{\alpha_{2}\kappa_{2}}(A)b_{2}=-\alpha_{2}(1)\widehat{q_{2}^{1}}(A)b_{1}+\alpha_{2}(1)\widehat{\kappa_{2}}(A)b_{2}.$

Finally, by denoting $\kappa_{1}:=m_{b_{1}}$, $a_{1}:=\alpha_{1}(1)$ and
$a_{2}:=\alpha_{2}(1)$, we use (63) and (64) into (62) to obtain

$\displaystyle
g=a_{1}\,\widehat{\kappa_{1}}(A)b_{1}+a_{2}\left(\widehat{\kappa_{2}}(A)b_{2}-\widehat{q_{2}^{1}}(A)b_{1}\right).$

  3. (3)

($L=3$): In this case the expression (55) reads as

(65) $g=p_{1}(A)b_{1}+p_{2}(A)b_{2}+p_{3}(A)b_{3}$

and the identity (56) reads as

(66)
$((A-I)p_{1}(A)-\mu_{1})b_{1}+((A-I)p_{2}(A)-\mu_{2})b_{2}+((A-I)p_{3}(A)-\mu_{3})b_{3}=\textbf{0}.$

From the last expression we can deduce that $((A-I)p_{3}(A)-\mu_{3})b_{3}\in
Z(A;b_{1})+Z(A;b_{2})$. Let $\kappa_{3}\in\mathbb{F}[x]$ be the generator of
polynomial ideal $\mathbb{I}_{3}:=\\{k\in\mathbb{F}[x]:\,k(A)b_{3}\in
Z(A;b_{1})+Z(A;b_{2})\\}$. Then, $\kappa_{3}$ divides the polynomial
$(x-1)p_{3}(x)-\mu_{3}$. That is, there exists a polynomial
$\alpha_{3}(x)\in\mathbb{F}[x]$ such that

(67) $\alpha_{3}(x)\kappa_{3}(x)=(x-1)p_{3}(x)-\mu_{3}$

Similarly to the previous cases, we can obtain

(68)
$p_{3}(x)=\frac{\alpha_{3}(x)\kappa_{3}(x)-\alpha_{3}(1)\kappa_{3}(1)}{x-1}=:\widehat{\alpha_{3}\kappa_{3}}(x).$

Replacing (67) into (66), and using that
$\kappa_{3}(A)b_{3}=q_{3}^{1}(A)b_{1}+q_{3}^{2}(A)b_{2}$, we get

(69)
$\displaystyle((A-I)p_{1}(A)-\mu_{1}+\alpha_{3}(A)q_{3}^{1}(A))b_{1}+((A-I)p_{1}(A)-\mu_{1}+\alpha_{3}(A)q_{3}^{2}(A))b_{2}=\textbf{0}.$

From the last expression we can deduce that
$((A-I)p_{2}(A)-\mu_{2}+\alpha_{3}(A)q_{3}^{2}(A))b_{2}\in Z(A;b_{1})$. Let
$\kappa_{2}\in\mathbb{F}[x]$ be the generator of polynomial ideal
$\mathbb{I}_{2}:=\\{k\in\mathbb{F}[x]:\,k(A)b_{2}\in Z(A;b_{1})\\}$. Then,
$\kappa_{2}$ divides the polynomial
$(x-1)p_{2}(x)-\mu_{2}+\alpha_{3}(x)q_{3}^{2}(x)$. That is, there exists a
polynomial $\alpha_{2}(x)\in\mathbb{F}[x]$ such that

(70)
$\alpha_{2}(x)\kappa_{2}(x)=(x-1)p_{2}(x)-\mu_{2}+\alpha_{3}(x)q_{3}^{2}(x).$

As before, we can deduce

(71)
$p_{2}(x)=\frac{\alpha_{2}(x)\kappa_{2}(x)-\alpha_{2}(1)\kappa_{2}(1)}{x-1}-\frac{\alpha_{3}(x)q_{3}^{2}(x)-\alpha_{3}(1)q_{3}^{2}(1)}{x-1}=\widehat{\alpha_{2}\kappa_{2}}(x)-\widehat{\alpha_{3}q_{3}^{2}}(x).$

Replacing (67) and (70) into (69), and using that
$\kappa_{2}(A)b_{2}=q_{2}^{1}(A)b_{1}$, we get

$\displaystyle((A-I)p_{1}(A)-\mu_{1}+\alpha_{2}(A)q_{2}^{1}(A)+\alpha_{3}(A)q_{3}^{1}(A))b_{1}=\textbf{0}.$

Hence, $m_{b_{1}}$ divides the polynomial
$(x-1)p_{1}(x)-\mu_{1}+\alpha_{2}(x)q_{2}^{1}(x)+\alpha_{3}(x)q_{3}^{1}(x)$,
that is, there exists $\alpha_{1}(x)\in\mathbb{F}[x]$ such that

(72)
$\alpha_{1}(x)m_{b_{1}}(x)=(x-1)p_{1}(x)-\mu_{1}+\alpha_{2}(x)q_{2}^{1}(x)+\alpha_{3}(x)q_{3}^{1}(x).$

Once again, evaluating at $x=1$, we can solve for the value of $\mu_{1}$ and
then obtain

$\displaystyle p_{1}(x)$
$\displaystyle=\frac{\alpha_{1}(x)m_{b_{1}}(x)-\alpha_{1}(1)m_{b_{1}}(1)}{x-1}-\frac{\alpha_{2}(x)q_{2}^{1}(x)-\alpha_{2}(1)q_{2}^{1}(1)}{x-1}$
$\displaystyle\quad-\frac{\alpha_{3}(x)q_{3}^{1}(x)-\alpha_{3}(1)q_{3}^{1}(1)}{x-1}$
(73)
$\displaystyle=\widehat{\alpha_{1}m_{b_{1}}}(x)-\widehat{\alpha_{2}q_{2}^{1}}(x)-\widehat{\alpha_{3}q_{3}^{1}}(x).$

Now, we replace (3) and (71) into (65) and get

$\displaystyle g$
$\displaystyle=\left(\widehat{\alpha_{1}m_{b_{1}}}(A)-\widehat{\alpha_{2}q_{2}^{1}}(A)-\widehat{\alpha_{3}q_{3}^{1}}(A)\right)b_{1}+\left(\widehat{\alpha_{2}\kappa_{2}}(A)-\widehat{\alpha_{3}q_{3}^{2}}(A)\right)b_{2}+\widehat{\alpha_{3}\kappa_{3}}(A)b_{3}$
(74)
$\displaystyle=\underbrace{\widehat{\alpha_{1}m_{b_{1}}}(A)b_{1}}_{(I)}+\underbrace{\left(\widehat{\alpha_{2}\kappa_{2}}(A)b_{2}-\widehat{\alpha_{2}q_{2}^{1}}(A)b_{1}\right)}_{(II)}+\underbrace{\left(\widehat{\alpha_{3}\kappa_{3}}(A)b_{3}-\widehat{\alpha_{3}q_{3}^{1}}(A)b_{1}-\widehat{\alpha_{3}q_{3}^{2}}(A)b_{2}\right)}_{(III)}.$

By applying Lemma 3.4 on the polynomials that appear in the terms $(I)$,
$(II)$ and $(III)$ of (3), together with the identities
$\kappa_{2}(A)b_{2}=q_{2}^{1}(A)b_{1}$ and
$\kappa_{3}(A)b_{3}=q_{3}^{1}(A)b_{1}+q_{3}^{2}(A)b_{2}$, we can reach,

$\displaystyle(I)=\alpha_{1}(1)\widehat{m_{b_{1}}}(A)b_{1};$
$\displaystyle(II)=\cancel{\widehat{\alpha_{2}}(A)\kappa_{2}(A)b_{2}}+\alpha_{2}(1)\widehat{\kappa_{2}(A)}b_{2}-\cancel{\widehat{\alpha_{2}}(A)q_{2}^{1}(A)b_{1}}-\alpha_{2}(1)\widehat{q_{2}^{1}}(A)b_{1};$
$\displaystyle(III)=\cancel{\widehat{\alpha_{3}}(A)\kappa_{3}(A)b_{3}}+\alpha_{3}(1)\widehat{\kappa_{3}}(A)b_{3}$
$\displaystyle\qquad-\left(\cancel{\widehat{\alpha_{3}}(A)q_{3}^{1}(A)b_{1}}+\alpha_{3}(1)\widehat{q_{3}^{1}}(A)b_{1}+\cancel{\widehat{\alpha_{3}}(A)q_{3}^{2}(A)b_{2}}+\alpha_{3}(1)\widehat{q_{3}^{2}}(A)b_{2}\right).$

Finally, using the notation $\kappa_{1}:=m_{b_{1}}$, $a_{1}:=\alpha_{1}(1)$
$a_{2}:=\alpha_{2}(1)$, and $a_{3}:=\alpha_{3}(1)$, we obtain

$\displaystyle
g=a_{1}\widehat{\kappa_{1}}(A)b_{1}+a_{2}\left(\widehat{\kappa_{2}}(A)b_{2}-\widehat{q_{2}^{1}}(A)b_{1}\right)+a_{3}\left(\widehat{\kappa_{3}}(A)b_{3}-\widehat{q_{3}^{2}}(A)b_{2}-\widehat{q_{3}^{1}}(A)b_{1}\right).$

  4. (4)

($L\geq 3$): Follows recursively mimicking the previous cases.

∎

######  Remark 3.7.

In Theorem 3.6, once we fix the minimal $A$-conductor polynomials
$\kappa_{j}$, the polynomials $q_{j}^{i}$ satisfying (25) are not necessarily
unique. Indeed, for any choice of $q_{j}^{i}\in\mathbb{F}[x]$ ($2\leq j\leq
L$, $1\leq i\leq j-1$) satisfying (25), we can build the vectors in (26),
i.e.,

$g_{j}=\widehat{\kappa_{j}}(A)b_{j}-\sum_{i=1}^{j-1}\widehat{q_{j}^{i}}(A)b_{i}\qquad\text{
for }1\leq j\leq L.$

Using formula (54), the $A$-$\\{b_{1},\dots,b_{L}\\}$ characteristic space
$\mathcal{G}$ (see Definition 3.5) is spanned by the set
$\\{g_{1},\dots,g_{L}\\}$. Hence, $\mathrm{dim}(\mathcal{G})\leq L$. However,
if the spatial observational vectors $b_{1},\dots,b_{L}$ are linearly
independent, then $\mathrm{dim}(\mathcal{G})=L$, as stated in the next
corollary.

######  Corollary 3.8.

Let $A:\mathbb{F}^{d}\to\mathbb{F}^{d}$ be a linear operator and let
$b_{1},\dots,b_{L}$ be linearly independent vectors in $\mathbb{F}^{d}$. Then
the $A$-$\\{b_{1},\dots,b_{L}\\}$ characteristic space $\mathcal{G}$ (see
Definition 3.5) has dimension $L$.

###### Proof.

From Remark 3.7, we already know that $\mathrm{dim}(\mathcal{G})\leq L$.

The polynomials $\kappa_{j}$, for $1\leq j\leq L$, are unique and completely
determined (see the statement of Theorem 3.6). However, the polynomials
$q_{j}^{i}$ satisfying (25) are not necessarily unique.

We will show that once the polynomials $q_{j}^{i}$ satisfying (25) are fixed,
the vectors of the form (26) are linearly independent as long as
$b_{1},\dots,b_{L}$ are linearly independent. As a result, we will obtain
$\mathrm{dim}(\mathcal{G})=L$.

We will proceed by induction.

  * •

First case. We recall that $\kappa_{1}=m_{b_{1}}$, the minimal
$A$-annihilating polynomial of $b_{1}$. Then,

(75) $g_{1}:=\widehat{\kappa_{1}}(A)b_{1}\not=\textbf{0}$

since $\mathrm{deg}(\widehat{\kappa_{1}})<\mathrm{deg}(\kappa_{1})$, and
$\kappa_{1}$ is the unique monic polynomial with the smallest degree such that
$\kappa_{1}(A)b_{1}=\textbf{0}$.

  * •

Let $1<\ell\leq L$. Assuming that that $g_{1},\dots,g_{\ell-1}$ given by (26)
are linearly independent, we will prove that
$g_{1},\dots,g_{\ell-1},g_{\ell}$, are linearly independent by separating into
two cases.

If $\kappa_{\ell}\not\equiv 1$, then
$\mathrm{deg}(\widehat{\kappa_{\ell}})<\mathrm{deg}(\kappa_{\ell})$. Thus, the
vector $\widehat{\kappa_{\ell}}(A)b_{\ell}$ does not belong to
$Z(A;b_{1},\dots,b_{\ell-1})=\sum_{i=1}^{\ell-1}Z(A;b_{i})$ since by
definition, $\kappa_{\ell}$ is the unique monic polynomial of the least degree
such that $\kappa_{\ell}(A)b_{\ell}\in Z(A;b_{1},\dots,b_{\ell-1})$. In
particular, $\widehat{\kappa_{\ell}}(A)b_{\ell}$ is non-trivial. Consequently,
$g_{\ell}=\widehat{\kappa_{\ell}}(A)b_{\ell}-\sum_{i=1}^{\ell-1}\widehat{q_{\ell}^{i}}(A)b_{i}\not\in
Z(A;b_{1},\dots,b_{\ell-1})$. Hence, it canxxnot be written as a linear
combination of the vectors $g_{1}=\widehat{\kappa_{1}}(A)b_{1}$,…,
$g_{\ell-1}=\widehat{\kappa_{\ell-1}}(A)b_{\ell-1}-\sum_{i=1}^{\ell-2}\widehat{q_{\ell-1}^{i}}(A)b_{i}$
that lie in $\sum_{i=1}^{\ell-1}Z(A;b_{i})$.

If $\kappa_{3}\equiv 1$, then
$g_{\ell}=\widehat{\kappa_{\ell}}(A)b_{\ell}-\sum_{i=1}^{\ell-1}\widehat{q_{\ell}^{i}}(A)b_{i}=-\sum_{i=1}^{\ell-1}\widehat{q_{\ell}^{i}}(A)b_{i}$.
By way of contradiction, suppose that the vectors $g_{1}$,…, $g_{\ell-1}$,
$g_{\ell}$ are linearly dependent, then there exist scalars
$\beta_{1},\dots,\beta_{\ell}$, not all zero, such that

(76)
$\displaystyle\textbf{0}=\beta_{1}g_{1}+\dots+\beta_{\ell-1}g_{\ell-1}-\beta_{\ell}\left(\sum_{i=1}^{\ell-1}\widehat{q_{\ell}^{i}}(A)b_{i}\right).$

If we apply the operator $A-I$ to the equality above, and note that

$\displaystyle(A-I)g_{j}$
$\displaystyle=\cancel{\kappa_{j}(A)b_{j}}-\kappa_{j}(1)b_{j}-\sum_{i=1}^{j-1}\cancel{q_{j}^{i}(A)b_{i}}+\sum_{i=1}^{j-1}q_{j}^{i}(1)b_{i}=\sum_{i=1}^{j-1}q_{j}^{i}(1)b_{i}-\kappa_{j}(1)b_{j},$

we obtain

(77) 0
$\displaystyle=-\beta_{1}\kappa_{1}(1)b_{1}+\beta_{2}\left(q_{2}^{1}(A)b_{1}-\kappa_{2}(A)b_{2}\right)+\dots$
$\displaystyle\qquad\dots+\beta_{\ell-1}\left(\sum_{i=1}^{\ell-2}q_{\ell-1}^{i}(1)b_{i}-\kappa_{\ell-1}(1)b_{\ell-1}\right)+\beta_{\ell}\left(\sum_{i=1}^{\ell-1}q_{\ell}^{i}(1)b_{i}-b_{\ell}\right).$

Since by hypothesis $b_{1},\dots,b_{\ell-1},b_{\ell}$ are linearly
independent, and the last term in (77) is the only one where the vector
$b_{\ell}$ appears, we conclude that $\beta_{\ell}=0$. Replacing
$\beta_{\ell}=0$ into (76) we get
$\textbf{0}=\beta_{1}g_{1}+\dots+\beta_{\ell-1}g_{\ell-1}$ with
$\beta_{1},\dots,\beta_{\ell-1}$ not all zero (since $\beta_{\ell}$ is already
zero), which is a contradiction because of the inductive hypothesis.

∎

###### Proof of Theorem 2.11.

The proof of Theorem 2.11 is a direct consequence of Theorem 3.6 and Theorem
2.4. ∎

###### Proof of Corollary 2.13.

Let $G$ and $B$ as in the statement of Theorem 2.4. Without assuming any
condition on the spectrum of $A$, the result is a direct consequence of
Corollary 3.8 as the dimension of the $A$-$\\{b_{1},\dots,b_{L}\\}$
characteristic space $\mathcal{G}$ equals the rank of $B$. Then, as we need
the projected subspace $P_{W}(\mathcal{G})$ to have dimension $K$, we need at
least $L=K$ spatial observational vectors $b_{j}$. ∎

######  Remark 3.9.

As we mention before, the polynomials $\kappa_{j}$, for $1\leq j\leq L$, are
unique and completely determined even though the polynomials $q_{j}^{i}$
satisfying (25) are not necessarily unique. In what follows we will make
particular choices for those polynomials $q_{j}^{i}$. These elections will
correspond to particular choices of bases for the observational orbits spaces
$Z(A;b_{1})$, $Z(A;b_{1},b_{2})$, …, $Z(A;b_{1},\dots,b_{L})$.

For each $1\leq j\leq L$, let
$r_{j}:=\mathrm{deg}(m_{b_{j}})=\mathrm{dim}(Z(A;b_{j}))$ and
$s_{j}:=\mathrm{deg}(\kappa_{j}).$

Notice that for each $2\leq j\leq L$,

$\widehat{\kappa_{j}}=0\Longleftrightarrow\kappa_{j}\equiv
1\Longleftrightarrow b_{j}\in Z(A;b_{1})+\dots+Z(A;b_{j-1}).$

Since $\mathrm{deg}(m_{b_{1}})=r_{1}=s_{1}$, we can consider the basis
$\mathcal{B}_{1}:=\\{b_{1},Ab_{1},\dots,A^{s_{1}-1}b_{1}\\}$ for $Z(A;b_{1})$.
Since $\kappa_{2}(A)b_{2}\in Z(A;b_{1})$, it can be written as a unique linear
combination of the vectors in $\mathcal{B}_{1}$. Thus, we pick $q_{2}^{1}$ as
the unique polynomial of degree at most $s_{1}-1$ such that
$\kappa_{2}(A)b_{2}=q_{2}^{1}(A)b_{1}$.

Having chosen $q_{2}^{1}(x)$ and a basis $\mathcal{B}_{1}$ for $Z(A;b_{1})$,
let us consider a basis $\mathcal{B}_{2}$ for
$Z(A;b_{1},b_{2})=Z(A;b_{1})+Z(A;b_{2})$ by extending
$\mathcal{B}_{1}=\\{b_{1},Ab_{1},\dots,A^{s_{1}-1}b_{1}\\}$ with some vectors
from $\\{b_{2},Ab_{2},\dots,A^{r_{2}-1}b_{2}\\}$ (which is a basis for
$Z(A;b_{2})$). As $\mathrm{deg}(\kappa_{2})=s_{2}$, we will show that we can
extend $\mathcal{B}_{1}$ to a basis $\mathcal{B}_{2}$ for $Z(A;b_{1},b_{2})$
by choosing some vectors of the form $A^{n}b_{2}$ for $n\leq s_{2}-1$. Indeed,
let $\kappa_{2}(x)=\sum_{j=0}^{s_{2}}u_{j}x^{j}$, with $u_{s_{2}}=1$. We know
that $\kappa_{2}(A)b_{2}\in Z(A;b_{1})$, thus there exits scalars
$v_{1},\dots,v_{s_{1}-1}$ such that

$\sum_{j=0}^{s_{2}}u_{j}A^{j}b_{2}=\sum_{i=0}^{s_{1}-1}v_{i}A^{i}b_{1}.$

In particular,

(78)
$A^{s_{2}}b_{2}=\sum_{i=0}^{s_{1}-1}{v_{i}}A^{i}b_{1}-\sum_{j=0}^{s_{2}-1}{u_{j}}A^{j}b_{2}.$

That is, $A^{s_{2}}b_{2}$ can be written in terms of $\mathcal{B}_{1}$ and
some vectors of the form $A^{n}b_{2}$ with $n<s_{2}$. Now, given the vector
$A^{s_{2}+1}b_{2}$, by applying $A$ to (78) we have

(79)
$A^{s_{2}+1}b_{2}=\sum_{i=0}^{s_{1}-1}{v_{i}}A^{i+1}b_{1}-\sum_{j=1}^{s_{2}-1}{u_{j-1}}A^{j}b_{2}+u_{s_{2}-1}A^{s_{2}}b_{2}.$

Since the first $s_{1}$ vectors on the right-hand side of (79) can be written
in terms of $\mathcal{B}_{1}$ (including $A^{s_{1}+1}b_{1}$) and since we have
already shown that $A^{s_{2}}b_{2}$ is a linear combination of vectors of the
form $A^{n}b_{2}$ with $n<s_{2}$, we conclude that $A^{s_{2}+1}b_{2}$ can also
be written as a linear combination of $\mathcal{B}_{1}$ and some vectors of
the form $A^{n}b_{2}$ with $n<s_{2}$. By repeating this argument recursively
we obtain that all vectors $A^{m}b_{2}$ with $m\geq s_{2}$ are linear
combinations of vectors in $\mathcal{B}_{1}$ and vectors of the form
$A^{n}b_{2}$ with $n<s_{2}$ (in other words, every vector in $Z(A;b_{2})$ is a
linear combination of vectors in $\mathcal{B}_{1}$ and vectors $A^{n}b_{2}$
with $n<s_{2}$). Finally, notice that the vector $A^{s_{2}-1}b_{2}\not\in
Z(A;b_{1})$ (i.e., it cannot be written in terms of $\mathcal{B}_{1}$ since
$\kappa_{2}(x)$ has degree $s_{2}$ and it is the polynomial with the least
degree satisfying $\kappa_{2}(A)b_{2}$ that can be written in terms of
$\mathcal{B}_{1}$).

Once we have chosen $\mathcal{B}_{2}$ in that way, one can pick accordingly
unique polynomials $q_{3}^{1}$ and $q_{3}^{2}$ satisfying (25) of degree less
than $s_{1}-1$ and $s_{2}-1$ respectively.

These choices can be repeated recursively.

###  3.4. Proofs for Section 2.2

In Theorem 2.14, we assume that the source term belongs to one-dimensional
subspace $W$ of $\mathbb{F}^{d}$ spanned by a known unit vector $\omega_{1}$.
That is, for an arbitrary linear operator $A$ in $\mathbb{F}^{d}$ with adjoint
$A^{*}$, we consider the non-homogeneous discrete-time dynamical system

(80) $x({n+1})=A^{*}x(n)\quad+\underbrace{c_{1}\,\omega_{1}}_{\text{source
term }\omega},$

and spatial measurements at only one location $b$:

(81) $y(n)=\langle x(n),b\rangle.$

To prove Theorem 2.14, we will need several auxiliary results.

######  Remark 3.10.

We recall that if $1$ does not belong to the spectrum of $A$, one strategy to
pick the spatial sampling vector is to consider $b:=(I-A)\omega_{1}$ as in
Corollary 2.7. Then, we consider scalars $\\{\alpha_{n}\\}_{n=0}^{r}$ such
that $m_{b}(x)=\sum_{n=0}^{r}\alpha_{n}x^{n}$ is the minimal $A$-annihilating
of $b$. As $m_{b}$ divides the minimal polynomial of $A$, it holds that
$m_{b}(1)\not=0$. Therefore, by considering $r+1$ space-time measurements
$y(0),\dots,y(r)$ and performing the following linear combinations

$\displaystyle\sum_{n=0}^{r}\alpha_{n}y(n)$ $\displaystyle=\langle
c_{1}\,\omega_{1},\sum_{n=0}^{r}\alpha_{n}\Lambda_{n}b\rangle=\langle
c_{1}\,\omega_{1},(I-A)^{-1}m_{b}(1)b\rangle=c_{1}\,m_{b}(1)\,\langle\omega_{1},\omega_{1}\rangle,$

we can recover the intensity $c_{1}$ of the source term
$\omega:=c_{1}\,\omega_{1}$ as

$c_{1}=\frac{1}{m_{b}(1)}\sum_{n=0}^{r}\alpha_{n}y(n).$

The following result provides a simple strategy to pick a space sampling
vector $b$ in order to recover the scalar $c_{1}$ in (80) in a wide range of
scenarios.

######  Proposition 3.11.

Let $A$ be an operator in $\mathbb{F}^{d}$, and $W$ be the one-dimensional
subspace of $\mathbb{F}^{d}$ spanned by a unit vector
$\omega_{1}\in\mathbb{F}^{d}$. If there exists an eigenvector $b$ of $A$ such
that $\omega_{1}$ is not orthogonal to $b$, then $b$ is, in fact, complete for
Problem 1. In particular, if $A$ is diagonalizable, it is always possible to
recover $c_{1}$ by using only one sampling vector $b$.

###### Proof.

Consider an eigenvector $b$ of $A$ such that is not orthogonal to $\omega_{1}$
(this can be achieved, for example, if $A$ is diagonalizable). If we know two
consecutive observations, for instance, $y({0})$ and $y(1)$, then

$c_{1}=\frac{y({1})-\lambda y(0)}{\langle\omega_{1},b\rangle},$

where $\lambda$ is such that $Ab=\overline{\lambda}b$. (Notice that in this
case, since $b$ is an eigenvector for $A$, its minimal $A$-annihilating
polynomial has degree 1.) ∎

As a particular case of Theorem 3.6 we have the following corollary.

######  Corollary 3.12.

Let $A$ be a linear operator in $\mathbb{F}^{d}$ and let $b\in\mathbb{F}^{d}$.
Let $m_{b}(x)$ be the minimal $A$-annihilating polynomial of $b$. Then, every
= $A$-$\\{b\\}$ characteristic vector $g\in\mathbb{F}^{d}$ (see Definition
3.5) is a multiple of $\widehat{m_{b}}(A)b$, where
$\widehat{m_{b}}\in\mathbb{F}[x]$ is given by

(82) $\widehat{m_{b}}(x):=\frac{m_{b}(x)-m_{b}(1)}{x-1}\in\mathbb{F}[x].$

In particular, if $m_{b}(1)=0$, then every non-zero $A$-$\\{b\\}$
characteristic vector $g$ is an eigenvector of $A$ corresponding to eigenvalue
$1$.

As an application of Corollary 3.12, the following result gives us a general
test for discriminating whether a vector $b$ is an appropriate or
inappropriate choice as a spatial sampling vector for recovering $c$ in (80).

######  Lemma 3.13.

Given a linear operator $A$ in $\mathbb{F}^{d}$, and $W$ a one-dimensional
subspace of $\mathbb{F}^{d}$ spanned by a unit vector
$\omega_{1}\in\mathbb{F}^{d}$. A spatial sampling vector $b\in\mathbb{F}^{d}$
is complete for the Problem 1 if and only if

$\langle\omega_{1},\widehat{m_{b}}(A)b\rangle\not=0,$

where $m_{b}$ is the minimal $A$-annihilating polynomial of $b$.

###### Proof.

The proof follows from checking Condition 2 in Theorem 2.4 in the case $L=1$,
$b_{1}:=b$, and using Corollary 3.12. ∎

###### Proof of Theorem 2.14.

Let $\sigma(A)$ (resp. $\sigma(A^{*})$) be the spectrum of $A$ (resp. of
$A^{*}$). First simply notice that $1\in\sigma(A)$ if and only if
$1\in\sigma(A^{*})$ (we recall that $\sigma(A^{*})=\overline{\sigma(A)}$).
Therefore, if $1$ is not an eigenvalue for $A$, the space
$\mathop{Ker}(I-A^{*})^{d}$ is the trivial subspace $\\{{\bf 0}\\}$. Thus,
Condition (32) is trivially satisfied as the source location $\omega_{1}$ is
assumed different from the zero vector, and there is always a complete set of
vectors for Problem 1 by considering only one spatial sampling vector as
pointed out in Remark 3.10.

Now, assume $1\in\sigma(A)$. In the notation of Remark 4.2 given in the
Appendix, let us assume $\lambda_{1}=1$ and so $E_{1}:=E_{\lambda_{1}}$ is the
projector ($(E_{1})^{2}=E_{1}$) onto the generalized eigenspace of $A$
corresponding to the eigenvalue $1$. Then, from (94),
$\omega_{1}\in\mathop{Ker}(A^{*}-I)^{d}$ if and only if
$\omega_{1}\in\mathop{Ker}(I-E_{1}^{*})$. Now, we separate the rest of the
proof into two cases.

  1. (1)

Assume $\omega_{1}\in\mathop{Ker}(I-E_{1}^{*})$, that is,
$E_{1}^{*}\omega_{1}=\omega_{1}$. We claim that there exists a complete
$b\in\mathbb{F}^{d}$ for Problem 1 if and only if
$\omega_{1}\not\in(\mathop{Ker}(A-I))^{\perp}$. Indeed, since $E_{1}$ is a
polynomial on $A$ (see Remark 4.2), we have by Lemma 3.13 that
$b\in\mathbb{F}^{d}$ is complete if and only if

$0\not=\langle\omega_{1},\widehat{m_{b}}(A)b\rangle=\langle
E_{1}^{*}\omega_{1},\widehat{m_{b}}(A)b\rangle=\langle\omega_{1},\widehat{m_{b}}(A)E_{1}b\rangle,$

where we recall that $\widehat{m_{b}}(x)\in\mathbb{F}[x]$ is as in (82).

Thus, if $b$ is a complete for Problem 1, it is necessary that
$E_{1}b\not=\textbf{0}$. This implies that $m_{b}(1)=0$ (see Lemma 4.3 in the
Appendix). Then, by the particular case in Corollary 3.12, we have that the
vector $\widehat{m_{b}}(A)b$ is an eigenvector of $A$ corresponding to the
eigenvalue $1$. Moreover, notice that

$\mathop{Ker}(A-I)=\\{\widehat{m}_{b}(A)b:\,b\in\mathbb{F}^{d}\text{ with
}m_{b}(1)=0\\}$

(see Lemma 4.3.) As a result, we can find a complete $b$ as long as
$\omega_{1}$ is not orthogonal to the eigenspace $\mathop{Ker}(A-I)$. We
recall that, if $\omega_{1}\not\in(\mathop{Ker}(A-I))^{\perp}$, we are in the
setting of Proposition 3.11 which provides a strategy for picking an accurate
sampling vector $b$.

  2. (2)

If $\omega_{1}\not\in\mathop{Ker}(I-E_{1}^{*})$, we claim that it is always
possible to find a complete observational vector $b$ for Problem 1 given by

(83) $b:=(A-I)(I-E_{1})(I-E_{1}^{*})\omega_{1}.$

First, we will see that $b\not=\textbf{0}$. By assumption,
$(I-E_{1}^{*})\omega_{1}\not=\textbf{0}$. Then, we have

(84)
$0\not=\langle(I-E_{1}^{*})\omega_{1},(I-E_{1}^{*})\omega_{1}\rangle=\langle\omega_{1},(I-E_{1})(I-E_{1}^{*})\omega_{1}\rangle,$

and so $(I-E_{1})(I-E_{1}^{*})\omega_{1}\not=\textbf{0}$. Finally, we know
that $(I-E_{1})$ projects onto
$\bigoplus_{\lambda\in\sigma(A),\lambda\not=1}\mathop{Ker}(A-\lambda I)^{d}$
and has kernel $\mathop{Ker}(A-I)^{d}$ (see Remark 4.2). Since
$\mathop{Ker}(A-I)\subset\mathop{Ker}(A-I)^{d}$, we get that
$b=(A-I)(I-E_{1})(I-E_{1}^{*})\omega_{1}\not=\textbf{0}$. Moreover, as

$(I-E_{1})(I-E_{1}^{*})\omega_{1}\in\bigoplus_{\lambda\in\sigma(A),\lambda\not=1}\mathop{Ker}(A-\lambda
I)^{d},$

and that the direct sum of generalized eigenspaces of $A$ is invariant under
$A$ (and hence under any polynomial on $A$), we have that
$b\in\bigoplus_{\lambda\in\sigma(A),\lambda\not=1}\mathop{Ker}(A-\lambda
I)^{d}$. Hence, $m_{b}(1)\not=0$.

Now, consider
$\widehat{m_{b}}(x)=\frac{m_{b}(x)-m_{b}(1)}{x-1}\in\mathbb{F}[x].$ Then, we
have that

$\displaystyle\widehat{m_{b}}(A)b$
$\displaystyle=\widehat{m_{b}}(A)(A-I)(I-E_{1})(I-E_{1}^{*})\omega_{1}$
$\displaystyle=\big{(}m_{b}(A)-m_{b}(1)\big{)}(I-E_{1})(I-E_{1}^{*})\omega_{1}.$

We will show that

$m_{b}(A)(I-E_{1})(I-E_{1}^{*})\omega_{1}={\bf 0}$

by using the fact that $I-E_{1}=(I-E_{1})^{2}$ and that $I-E_{1}=p(A)$, where
$p(x):=\widetilde{h_{1}}(x)(x-1)^{n_{1}}$ is the polynomial given in (91) (for
eigenvalue $\lambda_{i}=1$). Notice that $(x-1)$ divides $p(x)$, and so we can
factor $p(x)=\frac{p(x)}{(x-1)}(x-1)=\widehat{p}(x)(x-1)$. Then, when
evaluating at $A$ we have

$\displaystyle m_{b}(A)(I-E_{1})(I-E_{1}^{*})\omega_{1}$
$\displaystyle=m_{b}(A)(I-E_{1})(I-E_{1})(I-E_{1}^{*})\omega_{1}$
$\displaystyle=m_{b}(A)\widehat{p}(A)(A-I)(I-E_{1})(I-E_{1}^{*})\omega_{1}$
$\displaystyle=\widehat{p}(A)m_{b}(A)b=\textbf{0}.$

Therefore,

(85) $\widehat{m_{b}}(A)b=-m_{b}(1)(I-E_{1})(I-E_{1}^{*})\omega_{1}.$

Using (85) and (84) we obtain

$\displaystyle\langle\omega_{1},\widehat{m_{b}}(A)b\rangle$
$\displaystyle=-m_{b}(1)\langle\omega_{1},(I-E_{1})(I-E_{1}^{*})\omega_{1}\rangle$
$\displaystyle=|m_{b}(1)|\|(I-E_{1}^{*})\omega_{1}\|^{2}\not=0.$

Thus, by Lemma 3.13 the vector $b$ defined by (83) is complete for Problem 1.

∎

##  4\. Appendix: Matrix Analysis

######  Theorem 4.1.

[26, Theorem 13.2] Let $f$ be an analytic function defined on a domain
containing the $s$ roots of the polynomial
$m_{b}(x)=\prod_{i=1}^{s}(x-\lambda_{i})^{n_{i}}$ which is the minimal
$A$-annihilating of $b$ (each root $\lambda_{i}$ has multiplicity $n_{i}$).
Then

$f(A)b=q(A)b$

for $q$ the (unique) Hermite interpolating polynomial of degree less than or
equal to $\sum_{i=1}^{s}n_{i}=deg(m_{b})$ that satisfies the interpolation
conditions

(86)
$\frac{d^{j}}{dx^{j}}q(\lambda_{i})=\frac{d^{j}}{dx^{j}}f(\lambda_{i})\qquad\text{
for all }j=0,\dots,n_{i}-1,\,i=1,\dots,s.$

In particular, if $f$ is an analytic function on the spectrum of $A$ and
$m_{b}$ is replaced by the minimal polynomial of $A$, then $f(A)=q(A)$.

######  Remark 4.2.

Let $A$ be a linear operator in $\mathbb{F}^{d}$, and let

$m_{A}(x):=\prod_{j=1}^{s}(x-\lambda_{j})^{n_{j}}$

be its minimal polynomial (i.e., the unique monic polynomial with the least
degree satisfying $m_{A}(A)={\bf 0}$). (Notice that,
$\lambda_{1},\ldots,\lambda_{s}\in\mathbb{C}$ are the distinct eigenvalues of
$A$ and, when writing $A$ in Jordan form, $n_{i}$ determines the size of the
largest Jordan block in which $\lambda_{i}$ appears.) By the Lagrange-Hermite
formula, the Hermite interpolating polynomial satisfying (86) for the case
$f(x)=1$ is exactly the constant polynomial $1$ and it can be written as

(87)
$1=\sum_{i=1}^{s}\left[\left(\sum_{k=0}^{n_{i}-1}\frac{1}{k!}\phi_{i}^{(k)}\left(\lambda_{i}\right)\left(x-\lambda_{i}\right)^{k}\right)\prod_{j\neq
i}\left(x-\lambda_{j}\right)^{n_{j}}\right]$

where $\phi_{i}(x):=1/\prod_{j\neq i}\left(x-\lambda_{j}\right)^{n_{j}}$ (see
[26]). Let

(88) $p_{\lambda_{i}}(x):=h_{\lambda_{i}}(x)\prod_{j\neq
i}\left(x-\lambda_{j}\right)^{n_{j}},\quad\text{ where }\quad
h_{\lambda_{i}}(x):=\sum_{k=0}^{n_{i}-1}\frac{1}{k!}\phi_{i}^{(k)}\left(\lambda_{i}\right)\left(x-\lambda_{i}\right)^{k}\,$

and consider the operators

(89) $E_{\lambda_{i}}:=p_{\lambda_{i}}(A).$

Therefore,

(90) $I=E_{\lambda_{1}}+\dots+E_{\lambda_{s}}.$

It is easy to check that

$E_{\lambda_{i}}E_{\lambda_{k}}=\begin{cases}{\bf 0}&\text{ if }i\not=k\\\
E_{\lambda_{i}}&\text{ if }i=k\end{cases}.$

Indeed, if $i\not=k$, $p_{\lambda_{i}}(x)p_{\lambda_{k}}(x)$ has the same
roots as $m_{A}(x)$, and so $m_{A}|p_{\lambda_{i}}p_{\lambda_{k}}$ having that
$E_{\lambda_{i}}E_{\lambda_{k}}=p_{\lambda_{i}}(A)p_{\lambda_{k}}(A)={\bf 0}$.
Also, notice that when we multiply by $E_{\lambda_{i}}$ on both sides of
Identity (90) we get that $E_{\lambda_{i}}=(E_{\lambda_{i}})^{2}$ (i.e.,
$E_{\lambda_{i}}$ is an idempotent operator). Moreover, $E_{\lambda_{i}}$ and
$I-E_{\lambda_{i}}$ (where $I-E_{\lambda_{i}}=\sum_{j\not=i}E_{\lambda_{j}}$)
are idempotent projectors such that $I=E_{\lambda_{i}}+(I-E_{\lambda_{i}})$
and $E_{\lambda_{i}}(I-E_{\lambda_{i}})={\bf 0}$. In addition, we can write

(91) $E_{\lambda_{i}}=h_{\lambda_{i}}(A)\prod_{j\neq
i}\left(A-\lambda_{j}I\right)^{n_{j}},\qquad\text{ and }\qquad
I-E_{\lambda_{i}}=\widetilde{h_{\lambda_{i}}}(A)(A-\lambda_{i}I)^{n_{i}}$

where the polynomial $h_{\lambda_{i}}(x)$ is given in (88), and the explicit
formulas for $\widetilde{h_{\lambda_{i}}}(x)$ can be deduced from (87)
obtaining

(92)
$\widetilde{h_{\lambda_{i}}}(x):=\sum_{k=1,k\not=j}^{s}h_{\lambda_{k}}(x)\prod_{j\neq
k,j\not=i}\left(x-\lambda_{j}I\right)^{n_{j}}.$

Therefore, using the above characterization and the decomposition of
$\mathbb{F}^{d}$ as a direct sum (not necessarily orthogonal) of the
generalized eigenspaces of $A$

$\mathbb{F}^{d}={\bigoplus_{j=1}^{s}}\mathop{Ker}(A-\lambda_{j}I)^{d},$

we obtain

(93)
$\mathop{Ker}(E_{\lambda_{i}})={\bigoplus_{j\not=i}}\mathop{Ker}(A-\lambda_{j}I)^{d},\qquad\mathop{Ker}(I-E_{\lambda_{i}})=\mathop{Ker}(A-\lambda_{i}I)^{d}.$

In sum, $E_{\lambda_{i}}$ is the projection onto the generalized eigenspace
corresponding to eigenvalue $\lambda_{i}$ whose kernel is the direct sum of
the rest of the generalized eigenspaces of $A$. Finally, taking the adjoint in
the resolution of the identity (90) we get

$\displaystyle I$
$\displaystyle=\left(\sum_{i=1}^{s}\left[\left(\sum_{j=0}^{n_{i}-1}\frac{1}{j!}\phi_{i}^{(j)}\left(\lambda_{i}\right)\left(A-\lambda_{i}I\right)^{j}\right)\prod_{j\neq
i}\left(A-\lambda_{j}I\right)^{n_{j}}\right]\right)^{*}$
$\displaystyle=\sum_{i=1}^{s}\left[\left(\sum_{j=0}^{n_{i}-1}\frac{1}{j!}\overline{\phi_{i}^{(j)}\left(\lambda_{i}\right)}\left(A^{*}-\overline{\lambda_{i}}I\right)^{j}\right)\prod_{j\neq
i}\left(A^{*}-\overline{\lambda_{j}}I\right)^{n_{j}}\right]$
$\displaystyle=\sum_{i=1}^{s}\underbrace{\left[\left(\sum_{j=0}^{n_{i}-1}\frac{1}{j!}(\phi_{i}^{*})^{(j)}\left(\overline{\lambda_{i}}\right)\left(A^{*}-\overline{\lambda_{i}}I\right)^{j}\right)\prod_{j\neq
i}\left(A^{*}-\overline{\lambda_{j}}I\right)^{n_{j}}\right]}_{E_{\lambda_{i}}^{*}}$

where $\phi_{i}^{*}(x):=1/\prod_{j\neq
i}\left(x-\overline{\lambda_{j}}\right)^{n_{j}}$, which means that the
operators $\\{E_{\lambda_{i}}^{*}\\}_{i=1}^{s}$ are the corresponding
idempotent projectors onto the generalized eigenspaces associated to $A^{*}$.
Using the notation given in (88), we can write
$E_{\lambda_{i}}^{*}=p_{\overline{\lambda}_{i}}(A^{*})$. Therefore, in analogy
to (93)

(94)
$\begin{array}[]{cc}\mathop{Ker}(E_{\lambda_{i}}^{*})=\displaystyle{\oplus_{j\not=i}}\mathop{Ker}(A^{*}-\overline{\lambda_{j}}I)^{d},&\qquad\mathop{Ker}(I-E_{\lambda_{i}}^{*})=\mathop{Ker}(A^{*}-\overline{\lambda_{i}}I)^{d}.\end{array}$

We remark that it is not necessarily true that the projectors
$E_{\lambda_{i}}$ are self-adjoint operators, in fact, they are not orthogonal
projections in general. In the special case when $A$ is in Jordan form (i.e.,
the matrix representation of the operator $A$ with respect to the canonical
basis gives rise to its Jordan matrix representation), we have that the
generalized eigenspaces are mutually orthogonal, and so the projectors
$E_{\lambda_{i}}$ are orthogonal projections. In this case,
$\mathop{Ker}(A-\lambda_{i}I)=\mathop{Ker}(A^{*}-\overline{\lambda_{i}}I)$.

######  Lemma 4.3.

Let $A$ be a linear operator in $\mathbb{F}^{d}$, and $b$ be a vector in
$\mathbb{F}^{d}$. Let $\lambda_{i}$ be an eigenvalue of $A$ and
$E_{\lambda_{i}}$ be the projection operator onto the generalized eigenspace
of $A$ corresponding to $\lambda_{i}$. Then, $E_{\lambda_{i}}b\not=\textbf{0}$
if and only if $m_{b}(\lambda_{i})=0$ (where $m_{b}$ is the unique minimal
$A$-annihilating polynomial of $b$.)

Moreover, the eigenspace corresponding to $\lambda_{i}$ can be characterized
by

$\mathop{Ker}(A-\lambda_{i}I)=\left\\{p_{b,\lambda_{i}}(A)b:\,b\in\mathbb{F}^{d}\text{
with }m_{b}(\lambda_{i})=0,\text{ and
}p_{b,\lambda_{i}}(x):=\frac{m_{b}(x)}{x-\lambda_{i}}\right\\}.$

(Notice that when $\lambda_{i}=1$, then $p_{b,1}(x)=\widehat{m_{b}}(x)$.)

###### Proof.

First, observe that $E_{\lambda_{i}}b\neq\textbf{0}$ if and only if
$p_{\lambda_{i}}(A)b\not=\textbf{0}$ where $p_{\lambda_{i}}(x)$ is given by
(88). From its expression $p_{\lambda_{i}}(\lambda_{i})\neq 0$, and
$p_{\lambda_{i}}(\lambda_{j})=0$ for all $j\neq i$. Thus, if $(x-\lambda_{i})$
is not a factor of the minimal $A$-annihilator $m_{b}$ of $b$, then,
$p_{\lambda_{i}}(A)b=\textbf{0}$ since for this case $m_{b}$ divides
$p_{\lambda_{i}}$. Conversely, if $m_{b}(\lambda_{i})=0$ then
$(x-\lambda_{i})$ is a factor of all $A$-annihilators of $b$. Since
$p_{\lambda_{i}}(\lambda_{i})\neq 0$, then
$p_{\lambda_{i}}(A)b\not=\textbf{0}$.

For the last assertion, it is easy to check that

$\left\\{p_{b,\lambda_{i}}(A)b:\,b\in\mathbb{F}^{d}\text{ with
}m_{b}(\lambda_{i})=0,\text{ and
}p_{b,\lambda_{i}}(x):=\frac{m_{b}(x)}{x-\lambda_{i}}\right\\}\subseteq\mathop{Ker}(A-\lambda_{i}I).$

Indeed,

$(A-\lambda_{i}I)p_{b,\lambda_{i}}(A)b=m_{b}(A)b=\textbf{0}.$

The other inclusion also follows easily by noticing that, in general, a vector
$v$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda_{i}$ if and
only if its minimal $A$-annihilating polynomial is $m_{v}(x)=(x-\lambda_{i})$.
Therefore, if $v\in\mathbb{F}^{d}$ is an eigenvector of $A$ corresponding to
eigenvalue $\lambda_{i}$, we have that

$p_{b,\lambda_{i}}(x)=\frac{m_{v}(x)}{x-\lambda_{i}}=\frac{x-\lambda_{i}}{x-\lambda_{i}}\equiv
1.$

So, we can trivially write $v=p_{v,\lambda_{i}}(A)v$ which belongs to

$\left\\{p_{b,\lambda_{i}}(A)b:\,b\in\mathbb{F}^{d}\text{ with
}m_{b}(\lambda_{i})=0,\text{ and
}p_{b,\lambda_{i}}(x):=\frac{m_{b}(x)}{x-\lambda_{i}}\right\\}.$

∎

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