# On prescribed characteristic polynomials

Peter Danchev  Institute of Mathematics and Informatics, Bulgarian Academy of
Sciences, 1113 Sofia, Bulgaria<EMAIL_ADDRESS>,  Esther García
Departamento de Matemática Aplicada, Ciencia e Ingeniería de los Materiales y
Tecnología Electrónica, Universidad Rey Juan Carlos, 28933 Móstoles (Madrid),
Spain<EMAIL_ADDRESS>and  Miguel Gómez Lozano  Departamento de
Álgebra, Geometría y Topología, Universidad de Málaga, 29071 Málaga, Spain
<EMAIL_ADDRESS>
###### Abstract.

Let $\mathbb{F}$ be a field. We show that given any $n$th degree monic
polynomial $q(x)\in\mathbb{F}[x]$ and any matrix
$A\in\mathbb{M}_{n}(\mathbb{F})$ whose trace coincides with the trace of
$q(x)$ and consisting in its main diagonal of $k$ 0-blocks of order one, with
$k<n-k$, and an invertible non-derogatory block of order $n-k$, we can
construct a square-zero matrix $N$ such that the characteristic polynomial of
$A+N$ is exactly $q(x)$. We also show that the restriction $k<n-k$ is
necessary in the sense that, when the equality $k=n-k$ holds, not every
characteristic polynomial having the same trace as $A$ can be obtained by
adding a square-zero matrix. Finally, we apply our main result to decompose
matrices into the sum of a square-zero matrix and some other matrix which is
either diagonalizable, invertible, potent or torsion.

The first author was partially supported by the BIDEB 2221 of TÜBÍTAK

The second author and third authors were partially supported by Ayuda Puente
2023, URJC, and MTM2017-84194-P (AEI/FEDER, UE)

The three authors were partially supported by the Junta de Andalucía FQM264

Key words: characteristic polynomial, square-zero matrix

2010 Mathematics Subject Classification: 15A15, 15A21, 15A83

##  1\. Introduction and Basic Facts

Throughout the current article, we denote by $\mathbb{M}_{n}(\mathbb{F})$ the
matrix ring consisting of all square matrices of size $n\times n$ over an
arbitrary field $\mathbb{F}$.

The problem of completing a matrix, when some of its entries are fixed, such
that the resulting matrix satisfies a prescribed characteristic polynomial is
classical and has been studied by several authors; see, for more detailed
information, the survey by G. Cravo [2]. Back in 1958, H.K. Farahat and W.
Ledermann showed that any matrix of the form

$A=\left(\begin{array}[]{c|ccc}0&0&\dots&0\\\ \hline\cr 0&&&\\\ \vdots&&C&\\\
0&&&\\\ \end{array}\right)\in\mathbb{M}_{n}(\mathbb{F}),$

where $C\in\mathbb{M}_{n-1}(\mathbb{F})$ is non-derogatory, can be modified by
adding a matrix of the form

$N=\left(\begin{array}[]{c|ccc}a&u_{1}&\dots&u_{n-1}\\\ \hline\cr v_{1}&&&\\\
\vdots&&{\bf 0}_{n-1,n-1}&\\\ v_{n-1}&&&\\\
\end{array}\right)\in\mathbb{M}_{n}(\mathbb{F})$

such that the characteristic polynomial of $A+N$ is any prescribed monic
polynomial of degree $n$ (cf. [7, Theorem 3.4]). Furthermore, this problem was
generalized by G.N. Oliveira who raised the question of when a block matrix of
the form

$\left(\begin{array}[]{c|c}A_{11}&A_{12}\\\ \hline\cr A_{21}&A_{22}\\\
\end{array}\right),$

where part of its blocks are known, can be completed to achieve any prescribed
characteristic polynomial, and addressed it in several of his works (see [8],
[9], [10], [11]).

In particular, in his work [8], Oliveira proved that any matrix of the form

$A=\left(\begin{array}[]{c|c }{\bf 0}_{k,k}&{\bf 0}_{k,n-k}\\\ \hline\cr{\bf
0}_{n-k,k}&A_{22}\\\ \end{array}\right)\in\mathbb{M}_{n}(\mathbb{F}),$

where $A_{22}\in\mathbb{M}_{n-k}(\mathbb{F})$ is fixed, can be completed by
adding a matrix of the form

$N=\left(\begin{array}[]{c|c }X_{11}&X_{12}\\\ \hline\cr X_{21}&{\bf
0}_{n-k,n-k}\\\ \end{array}\right)\in\mathbb{M}_{n}(\mathbb{F})$

such that the characteristic polynomial of $A+N$ is any prescribed monic
polynomial $q(x)$ of degree $n$ if, and only if, the product of certain
elementary divisors of the matrix $A_{22}$ divides $q(x)$.

The prescription of several entries of a matrix together with a fixed
characteristic polynomial has been recently used by S. Breaz and G.
Călugăreanu when showing that every square matrix with nilpotent trace over a
general ring can be expressed as the sum of three nilpotent matrices; see [1].

With the aim of addressing decomposition problems involving square-zero
matrices, in this work we study the problem of modifying a matrix of the form

$A=\left(\begin{array}[]{c|c }{\bf 0}_{k,k}&{\bf 0}_{k,n-k}\\\ \hline\cr{\bf
0}_{n-k,k}&A_{22}\\\ \end{array}\right)\in\mathbb{M}_{n}(\mathbb{F}),$

with fixed $A_{22}\in\mathbb{M}_{n-k}(\mathbb{F})$, by adding to it a square-
zero matrix $N$ such that the characteristic polynomial of $A+N$ is any
prescribed polynomial. Since square-zero matrices always have zero trace, the
trace of $A+N$ equals the trace of $A$, and by this modification of $A$ we can
only get characteristic polynomials having the same trace as $A$. We will show
in the sequel that, when $k<n-k$ and $A_{22}$ is invertible and non-
derogatory, given any monic degree $n$ polynomial $q(x)$ with the same trace
as $A$ there exists a square-zero matrix $N$ such that the characteristic
polynomial of $A+N$ is precisely $q(x)$. The first and last $k$ rows of such
matrix $N$ will be non-zero in general, so this is not exactly a completion
problem, because, in order to get the desired characteristic polynomial by
adding a square-zero matrix to $A$, we modify the last $k$ rows of $A_{22}$.

We also illustrate that the restriction $k<n-k$ cannot be generally ignored in
the sense that, when the equality $k=n-k$ holds, not every characteristic
polynomial having the same trace as $A$ can be obtained by adding a square-
zero matrix.

Further, in the last section of this work, we will apply this result to the
problem of decomposing matrices into the sum of a square-zero matrix and a
matrix satisfying some “good” condition, where “good” stands for the
properties of being either diagonalizable, invertible, $n$-torsion or
$n$-potent. This fits in our general project of decomposing square matrices
into the sum of a square-zero matrix and another matrix satisfying some
properties such as been diagonalizable, potent, invertible, etc. (see, for
instance, [3], [4], [5] and [6]).

##  2\. Main Theorem

Recall that the trace of a monic polynomial

$q(x)=x^{n}+a_{n-1}x^{n-1}+\dots+a_{0}\in\mathbb{F}[x]$

is $-a_{n-1}\in\mathbb{F}$. If such polynomial is the characteristic
polynomial of a certain matrix, the trace coincides with the trace (i.e., the
sum of the diagonal elements) of such matrix. Recall that a matrix $N$ is
nilpotent if there exists $k\in\mathbb{N}$ with $N^{k}=0$, and it is called a
square-zero matrix when $N^{2}=0$.

######  Remark 2.1.

Given a non-derogatory matrix $A$, there always exists a square-zero matrix
$N$ such that the characteristic polynomial of $A+N$ is any given monic
polynomial whose trace coincides with the trace of $A$. In fact, we can
suppose that $A$ is the companion matrix of a certain polynomial

$p(x)=x^{n}+u_{n-1}x^{n-1}+u_{n-2}x^{n-2}+\dots+u_{0}$

thus:

$A=\left(\begin{array}[]{ccc c}0&\dots&0&-u_{0}\\\ 1&&0&\vdots\\\
&\ddots&&-u_{n-2}\\\ 0&&1&-u_{n-1}\\\ \end{array}\right).$

Given any polynomial

$q(x)=x^{n}+u_{n-1}x^{n-1}+v_{n-2}x^{n-2}+\dots+v_{0}$

(where the trace of $q(x)$ coincides with the trace of $A$) it is enough to
consider the square-zero matrix

$N=\left(\begin{array}[]{ccc c}0&&0&u_{0}-v_{0}\\\ 0&&0&\vdots\\\
\vdots&&\vdots&u_{n-2}-v_{n-2}\\\ 0&&0&0\\\ \end{array}\right)$

to get that

$A+N=\left(\begin{array}[]{ccc c}0&&0&-v_{0}\\\ 1&&0&\vdots\\\
&\ddots&&-v_{n-2}\\\ 0&&1&-u_{n-1}\\\ \end{array}\right)$

has characteristic polynomial equal to $q(x)$.

The goal of this section is to generalize the previous result to any
invertible non-derogative matrix completed with $k$ rows/columns of zeros. In
order to do that, we begin with a technical lemma.

######  Lemma 2.2.

Let $\mathbb{F}$ be a field and let
$a_{1},\dots,a_{n},u_{2},\dots,u_{n}\in\mathbb{F}$. Then the determinant of
the matrix

$A=\left(\begin{array}[]{ccccccc}a_{1}+x&a_{2}&a_{3}&\cdots&a_{n-2}&a_{n-1}&a_{n}\\\
0&x&0&\cdots&0&0&u_{2}\\\ 0&0&x&\cdots&0&0&u_{3}\\\ 0&0&0&\ddots&0&0&u_{4}\\\
\vdots&\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&0&x&u_{n-1}\\\ x&0&0&\cdots&0&0&x+u_{n}\\\ \end{array}\right)$

is
$|A|=x^{n}+u_{n}x^{n-1}+(a_{1}-a_{n})x^{n-1}+a_{1}u_{n}x^{n-2}+\sum_{i=2}^{n-1}a_{i}u_{i}x^{n-2}$.

###### Proof.

By $R(i)$ and $C(i)$, we respectively denote the $i^{\rm th}$-row and the
$i^{\rm th}$-column of the involved matrix. To perform the calculations, we
will assume that our matrix is a matrix with coefficients in the field of
fractions of $\mathbb{F}[x]$.

For $s=2,\dots,n-1$, we add $-\frac{a_{s}}{x}R(s)$ to row $R(1)$ as follows:

$|A|=\left|\begin{array}[]{ccccccc}a_{1}+x&0&0&\cdots&0&0&a_{n}-\sum_{s=2}^{n-1}\frac{a_{s}u_{s}}{x}\\\
0&x&0&\cdots&0&0&u_{2}\\\ 0&0&x&\cdots&0&0&u_{3}\\\ 0&0&0&\ddots&0&0&u_{4}\\\
\vdots&\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&0&x&u_{n-1}\\\ x&0&0&\cdots&0&0&x+u_{n}\\\ \end{array}\right|.$

By the expansion of the determinant using $C(2)$, …, $C(n-2)$, we get:

$\displaystyle|A|$
$\displaystyle=x^{n-2}\left|\begin{array}[]{cc}a_{1}+x&a_{n}-\sum_{s=2}^{n-1}\frac{a_{s}u_{s}}{x}\\\
x&x+u_{n}\\\ \end{array}\right|$
$\displaystyle=x^{n-2}(a_{1}+x)(x+u_{n})-x^{n-1}(a_{n}-\sum_{s=2}^{n-1}\frac{a_{s}u_{s}}{x})$
$\displaystyle=x^{n}+u_{n}x^{n-1}+(a_{1}-a_{n})x^{n-1}+a_{1}u_{n}x^{n-2}+\sum_{i=2}^{n-1}a_{i}u_{i}x^{n-2}.$

∎

We are now prepared to establish the following chief result.

######  Theorem 2.3.

Let $\mathbb{F}$ be a field, let $n,k\in\mathbb{N}$ with $k<n-k$, and consider
the block matrix

$A=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&{\bf 0}_{k,n-k}\\\ \hline\cr{\bf
0}_{n-k,k}&A_{22}\\\ \end{array}\right)\in\mathbb{M}_{n}(\mathbb{F})$

consisting of $k$ rows and columns of zeros and an invertible non-derogatory
matrix $A_{22}$. Then, for any monic polynomial $q(x)$ of degree $n$ whose
trace coincides with the trace of $A$, there exists a square-zero matrix $N$
such that the characteristic polynomial of $A+N$ coincides with $q(x)$.

###### Proof.

The case $k=0$ was treated in Remark 2.1, so we can suppose that $k>0$. Under
an appropriate change of the existing basis, we can also suppose that $A_{22}$
is a companion matrix $C(p(x))$ for a polynomial of the form
$p(x)=x^{n-k}+u_{n-k-1}x^{n-k-1}+\dots+u_{0}$.

Take arbitrary $a_{ij}\in\mathbb{F}$, $i=1,\dots,k$, $j=0,\dots,n-2k$, and
define the matrix $N:=(\alpha_{ij})\in\mathbb{M}_{n}(\mathbb{F})$ like this:

$\alpha_{i,j}:=\begin{cases}-a_{i,n-k+1}&\ \text{ if }\ i=1,\dots,k\text{ and
}j=k,\\\ -a_{i,j}&\ \text{ if }\ i=1,\dots,k\text{ and }j=k+1,\dots,n-k+1,\\\
a_{n-i+1,n-k+1}&\ \text{ if }\ i=n-k+1,\dots,n\text{ and }j=k,\\\
a_{n-i+1,j}&\ \text{ if }\ i=n-k+1,\dots,n\text{ and }j=k+1,\dots,n-k+1,\\\
-1&\ \text{ if }\ i=1,\dots,k-1\text{ and }j=i,n-i+1,\\\ 1&\ \text{ if }\
i=n-k+1,\dots,n\text{ and }j=i,n-i+1,\\\ 0&\ \text{ otherwise. }\end{cases}$

Let us first show that $N^{2}=0$: in fact, if we denote by
$\\{e_{1},\dots,e_{n}\\}$ the vectors of the canonical basis of
$\mathbb{F}^{n}$, and by $n:\mathbb{F}^{n}\to\mathbb{F}^{n}$ the endomorphism
whose matrix with respect to the canonical basis is $N$, one can check that

$n(e_{j})\in{\rm
span}\\{e_{1}-e_{n},e_{2}-e_{n-1},\dots,e_{k}-e_{n-k+1}\\},\qquad j=1,\dots,n$

and $n(e_{i}-e_{n-i+1})=0$, $i=1,\dots,k$, so $n^{2}=0$, i.e., $N$ is really a
square-zero matrix, as claimed.

Let us calculate the characteristic polynomial of the matrix $A+N$, i.e., let
us calculate the determinant of the matrix $M=x\operatorname{Id}_{n}-(A+N)$.
We will perform several elementary transformations on the rows and columns of
$M$ in order to obtain a matrix of the form of Lemma 2.2, and then we will use
the formula for the determinant of such matrix proved in Lemma 2.2. To this
purpose, we denote by $R(i)$ and $C(i)$ the respective $i^{\rm th}$-row and
the $i^{\rm th}$-column of the involved matrices. For our calculations, we
will assume that our matrix is a matrix with coefficients in the field of
fractions of the polynomial ring possessing coefficients in $\mathbb{F}$.

Since it is not possible to explicitly write the generic matrix $M$, we will
represent the transformations on the example $n=10$ and $k=4$. In this
concrete case, we have:

$A=\left(\begin{array}[]{cccc|cccccc}0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0&0\\\ \hline\cr
0&0&0&0&0&0&0&0&0&-u_{0}\\\ 0&0&0&0&1&0&0&0&0&-u_{1}\\\
0&0&0&0&0&1&0&0&0&-u_{2}\\\ 0&0&0&0&0&0&1&0&0&-u_{3}\\\
0&0&0&0&0&0&0&1&0&-u_{4}\\\ 0&0&0&0&0&0&0&0&1&-u_{5}\\\ \end{array}\right),$
$N=\left(\begin{array}[]{cccccccccc}-1&0&0&-a_{1,7}&-a_{1,5}&-a_{1,6}&-a_{1,7}&0&0&-1\\\
0&-1&0&-a_{2,7}&-a_{2,5}&-a_{2,6}&-a_{2,7}&0&-1&0\\\
0&0&-1&-a_{3,7}&-a_{3,5}&-a_{3,6}&-a_{3,7}&-1&0&0\\\
0&0&0&-a_{4,7}&-a_{4,5}&-a_{4,6}&-a_{4,7}&0&0&0\\\ 0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0\\\ 0&0&0&a_{4,7}&a_{4,5}&a_{4,6}&a_{4,7}&0&0&0\\\
0&0&1&a_{3,7}&a_{3,5}&a_{3,6}&a_{3,7}&1&0&0\\\
0&1&0&a_{2,7}&a_{2,5}&a_{2,6}&a_{2,7}&0&1&0\\\
1&0&0&a_{1,7}&a_{1,5}&a_{1,6}&a_{1,7}&0&0&1\\\ \end{array}\right),\hbox{ and}$
$M=\left(\begin{array}[]{cccccccccc}x+1&0&0&a_{1,7}&a_{1,5}&a_{1,6}&a_{1,7}&0&0&1\\\
0&x+1&0&a_{2,7}&a_{2,5}&a_{2,6}&a_{2,7}&0&1&0\\\
0&0&x+1&a_{3,7}&a_{3,5}&a_{3,6}&a_{3,7}&1&0&0\\\
0&0&0&x+a_{4,7}&a_{4,5}&a_{4,6}&a_{4,7}&0&0&0\\\ 0&0&0&0&x&0&0&0&0&u_{0}\\\
0&0&0&0&-1&x&0&0&0&u_{1}\\\
0&0&0&-a_{4,7}&-a_{4,5}&-a_{4,6}&x-a_{4,7}&0&0&u_{2}\\\
0&0&-1&-a_{3,7}&-a_{3,5}&-a_{3,6}&-a_{3,7}-1&x-1&0&u_{3}\\\
0&-1&0&-a_{2,7}&-a_{2,5}&-a_{2,6}&-a_{2,7}&-1&x-1&u_{4}\\\
-1&0&0&-a_{1,7}&-a_{1,5}&-a_{1,6}&-a_{1,7}&0&-1&x-1+u_{5}\\\
\end{array}\right).$

(1) For $s=1,2,....k$, we add row $R(s)$ to row $R(n-s+1)$:

$\left(\begin{array}[]{cccccccccc}x+1&0&0&a_{1,7}&a_{1,5}&a_{1,6}&a_{1,7}&0&0&1\\\
0&x+1&0&a_{2,7}&a_{2,5}&a_{2,6}&a_{2,7}&0&1&0\\\
0&0&x+1&a_{3,7}&a_{3,5}&a_{3,6}&a_{3,7}&1&0&0\\\
0&0&0&x+a_{4,7}&a_{4,5}&a_{4,6}&a_{4,7}&0&0&0\\\ 0&0&0&0&x&0&0&0&0&u_{0}\\\
0&0&0&0&-1&x&0&0&0&u_{1}\\\ 0&0&0&x&0&-1&x&0&0&u_{2}\\\
0&0&x&0&0&0&-1&x&0&u_{3}\\\ 0&x&0&0&0&0&0&-1&x&u_{4}\\\
x&0&0&0&0&0&0&0&-1&x+u_{5}\\\ \end{array}\right)$

(2) For $s=2,....k$, we add $-\sum_{i=2}^{s}\frac{1}{x^{i-2}}C(n-s+i-1)$ to
column $C(s)$:

$\left(\begin{array}[]{cccccccccc}x+1&0&0&0&a_{1,5}&a_{1,6}&a_{1,7}&0&0&1\\\
0&x&-1/x&-1/x^{2}&a_{2,5}&a_{2,6}&a_{2,7}&0&1&0\\\
0&0&x&-1/x&a_{3,5}&a_{3,6}&a_{3,7}&1&0&0\\\
0&0&0&x&a_{4,5}&a_{4,6}&a_{4,7}&0&0&0\\\ 0&0&0&0&x&0&0&0&0&u_{0}\\\
0&0&0&0&-1&x&0&0&0&u_{1}\\\ 0&0&0&0&0&-1&x&0&0&u_{2}\\\
0&0&0&0&0&0&-1&x&0&u_{3}\\\ 0&0&0&0&0&0&0&-1&x&u_{4}\\\
x&1&1/x&1/x^{2}&0&0&0&0&-1&x+u_{5}\\\ \end{array}\right)$

(3) For $s=2,....k$, we add $-\frac{1}{x^{s-1}}C(1)$ to column $C(s)$:

$\left(\begin{array}[]{cccccccccc}x+1&-(x+1)/x&-(x+1)/x^{2}&-(x+1)/x^{3}&a_{1,5}&a_{1,6}&a_{1,7}&0&0&1\\\
0&x&-1/x&-1/x^{2}&a_{2,5}&a_{2,6}&a_{2,7}&0&1&0\\\
0&0&x&-1/x&a_{3,5}&a_{3,6}&a_{3,7}&1&0&0\\\
0&0&0&x&a_{4,5}&a_{4,6}&a_{4,7}&0&0&0\\\ 0&0&0&0&x&0&0&0&0&u_{0}\\\
0&0&0&0&-1&x&0&0&0&u_{1}\\\ 0&0&0&0&0&-1&x&0&0&u_{2}\\\
0&0&0&0&0&0&-1&x&0&u_{3}\\\ 0&0&0&0&0&0&0&-1&x&u_{4}\\\
x&0&0&0&0&0&0&0&-1&x+u_{5}\\\ \end{array}\right)$

(4) For $s=k+1,\dots,n-1$, we add $\frac{1}{x}R(s)$ to row $R(s+1)$:

$\left(\begin{array}[]{cccccccccc}x+1&-(x+1)/x&-(x+1)/x^{2}&-(x+1)/x^{3}&a_{1,5}&a_{1,6}&a_{1,7}&0&0&1\\\
0&x&-1/x&-1/x^{2}&a_{2,5}&a_{2,6}&a_{2,7}&0&1&0\\\
0&0&x&-1/x&a_{3,5}&a_{3,6}&a_{3,7}&1&0&0\\\
0&0&0&x&a_{4,5}&a_{4,6}&a_{4,7}&0&0&0\\\ 0&0&0&0&x&0&0&0&0&u_{0}\\\
0&0&0&0&0&x&0&0&0&\sum_{i=0}^{1}u_{i}/x^{1-i}\\\
0&0&0&0&0&0&x&0&0&\sum_{i=0}^{2}u_{i}/x^{2-i}\\\
0&0&0&0&0&0&0&x&0&\sum_{i=0}^{3}u_{i}/x^{3-i}\\\
0&0&0&0&0&0&0&0&x&\sum_{i=0}^{4}u_{i}/x^{4-i}\\\
x&0&0&0&0&0&0&0&0&x+\sum_{i=0}^{5}u_{i}/x^{5-i}\\\ \end{array}\right)$

Let us denote by $v_{s}:=\sum_{i=0}^{s}\frac{u_{i}}{x^{s-i}}$,
$s=0,\dots,n-k-1$.

(5) For $s=3,\dots,k$, we add $\sum_{i=2}^{s-1}\frac{C(i)}{x^{s-i+1}}$ to
column $C(s)$:

$\left(\begin{array}[]{cccccccccc}x+1&-(x+1)/x&-(x+1)(1/x^{2}+1/x^{3})&-(x+1)(1/x^{3}+2/x^{4}+1/x^{5})&a_{1,5}&a_{1,6}&a_{1,7}&0&0&1\\\
0&x&0&0&a_{2,5}&a_{2,6}&a_{2,7}&0&1&0\\\
0&0&x&0&a_{3,5}&a_{3,6}&a_{3,7}&1&0&0\\\
0&0&0&x&a_{4,5}&a_{4,6}&a_{4,7}&0&0&0\\\ 0&0&0&0&x&0&0&0&0&v_{0}\\\
0&0&0&0&0&x&0&0&0&v_{1}\\\ 0&0&0&0&0&0&x&0&0&v_{2}\\\
0&0&0&0&0&0&0&x&0&v_{3}\\\ 0&0&0&0&0&0&0&0&x&v_{4}\\\
x&0&0&0&0&0&0&0&0&x+v_{5}\\\ \end{array}\right)$

Let us now prove by induction on $k$ that, for any $t\leq k$, in the above
matrix the coefficient in position $(1,t)$ is $-z_{t}$, where

$z_{t}:=(x+1)\left(\sum_{r=0}^{t-2}{\binom{t-2}{r}\over x^{t-1+r}}\right).$

Indeed, for $k=1$ there is nothing to prove. Let us suppose that

$z_{i}:=(x+1)\left(\sum_{r=0}^{i-2}{\binom{i-2}{r}\over x^{i-1+r}}\right)$

for $i=1,\dots,t-1$, then one computes that

$\displaystyle z_{t}$
$\displaystyle=(x+1)/x^{t-1}+z_{2}/x^{t-1}+\cdots+z_{t-2}/x^{3}+z_{t-1}/x^{2}$
$\displaystyle=(x+1)\left(1/x^{t-1}+\sum_{i=0}^{t-3}\sum_{j=0}^{i}{\binom{i}{j}\over
x^{t+j}}\right)=(x+1)\left(1/x^{t-1}+\sum_{i=0}^{t-3}\sum_{j=i}^{t-3-i}{\binom{j}{i}\over
x^{t+i}}\right)$
$\displaystyle=(x+1)\left(1/x^{t-1}+\sum_{i=0}^{t-3}{\sum_{j=i}^{t-3-i}\binom{j}{i}\over
x^{t+i}}\right)=(x+1)\left(1/x^{t-1}+\sum_{i=0}^{t-3}{\binom{t-2}{i+1}\over
x^{t+i}}\right)$
$\displaystyle=(x+1)\left(1/x^{t-1}+\sum_{i=1}^{t-2}{\binom{t-2}{i}\over
x^{t-1+i}}\right)=(x+1)\left(\sum_{r=0}^{t-2}{\binom{t-2}{r}\over
x^{t-1+r}}\right).$

(6) For $s=k+1,\dots,n-k+1$, we add $-\sum_{i=2}^{s-1}\frac{a_{i,s}}{x}C(i)$
to column $C(s)$:

$\left(\begin{array}[]{cccccccccc}x+1&-z_{2}&-z_{3}&-z_{4}&a_{1,5}+\sum_{i=2}^{k}a_{i,5}z_{i}/x&a_{1,6}+\sum_{i=2}^{k}a_{i,6}z_{i}/x&a_{1,7}+\sum_{i=2}^{k}a_{i,7}z_{i}/x&0&0&1\\\
0&x&0&0&0&0&0&0&1&0\\\ 0&0&x&0&0&0&0&1&0&0\\\ 0&0&0&x&0&0&0&0&0&0\\\
0&0&0&0&x&0&0&0&0&v_{0}\\\ 0&0&0&0&0&x&0&0&0&v_{1}\\\
0&0&0&0&0&0&x&0&0&v_{2}\\\ 0&0&0&0&0&0&0&x&0&v_{3}\\\
0&0&0&0&0&0&0&0&x&v_{4}\\\ x&0&0&0&0&0&0&0&0&x+v_{5}\\\ \end{array}\right)$

Let us denote by

$w_{s}:=a_{1,s}+\sum_{i=2}^{k}a_{i,s}z_{i}/x,$

$s=k+1,\dots,n-k+1$.

(7) For $s=2,\dots,k-1$, we add $-\frac{1}{x}C(s)$ to column $C(n-s+1)$:

$\left(\begin{array}[]{cccccccccc}x+1&-z_{2}&-z_{3}&-z_{4}&w_{5}&w_{6}&w_{7}&z_{3}/x&z_{2}/x&1\\\
0&x&0&0&0&0&0&0&0&0\\\ 0&0&x&0&0&0&0&0&0&0\\\ 0&0&0&x&0&0&0&0&0&0\\\
0&0&0&0&x&0&0&0&0&v_{0}\\\ 0&0&0&0&0&x&0&0&0&v_{1}\\\
0&0&0&0&0&0&x&0&0&v_{2}\\\ 0&0&0&0&0&0&0&x&0&v_{3}\\\
0&0&0&0&0&0&0&0&x&v_{4}\\\ x&0&0&0&0&0&0&0&0&x+v_{5}\\\ \end{array}\right)$

Thereby, we have obtained a matrix equivalent to $M$ which is of the form
required in Lemma 2.2. Therefore, we get:

$\displaystyle|M|$
$\displaystyle=x^{n}+v_{n-k-1}x^{n-1}+v_{n-k-1}x^{n-2}+\sum_{s=k+1}^{n-k+1}w_{s}v_{s-k-1}x^{n-2}$
$\displaystyle+\sum_{i=1}^{k-2}{z_{k-i}\over x}v_{n-2k+i}x^{n-2}.$

Next, substituting the $v^{\prime}s$, $w^{\prime}s$ and $z^{\prime}s$ in the
above expression, we conclude that

$\displaystyle|M|=x^{n}+\sum_{i=0}^{n-k-1}u_{i}x^{k+i}+\sum_{i=0}^{n-k-1}u_{i}x^{k+i-1}$
$\displaystyle+\sum_{s=k+1}^{n-k+1}\left(a_{1,s}+\sum_{i=2}^{k}\sum_{r=0}^{i-2}\binom{i-2}{r}a_{i,s}{1\over
x^{i+r-2}}+\sum_{i=2}^{k}\sum_{r=0}^{i-2}\binom{i-2}{r}a_{i,s}{1\over
x^{i+r-1}}\right)\left(\sum_{j=0}^{s-k-1}u_{j}/x^{s-k-1-j}\right)x^{n-2}$
$\displaystyle+\sum_{i=1}^{k-2}{(1+x)\over
x}\left(\sum_{r=0}^{k-i-2}{\binom{k-i-2}{r}\over
x^{k-i-1+r}}\right)\left(\sum_{j=0}^{n-2k+i}u_{j}/x^{n-2k+i-j}\right)x^{n-2}$
$\displaystyle=x^{n}+\sum_{i=0}^{n-k-1}u_{i}x^{k+i}+\sum_{i=0}^{n-k-1}u_{i}x^{k+i-1}\hskip
216.2411pt(1)$
$\displaystyle+\sum_{s=k+1}^{n-k+1}\sum_{j=0}^{s-k+1}a_{1,s}u_{j}x^{n+k+j-s-1}\hskip
239.00298pt(2)$
$\displaystyle+\sum_{s=k+1}^{n-k+1}\sum_{i=2}^{k}\sum_{r=0}^{i-2}\sum_{j=0}^{s-k+1}\binom{i-2}{r}a_{i,s}u_{j}x^{n+k+j-i-
r-s-1}\hskip 169.29381pt(3)$
$\displaystyle+\sum_{s=k+1}^{n-k+1}\sum_{i=2}^{k}\sum_{r=0}^{i-2}\sum_{j=0}^{s-k+1}\binom{i-2}{r}a_{i,s}u_{j}x^{n+k+j-i-
r-s}\hskip 177.82971pt(4)$
$\displaystyle+\sum_{i=1}^{k-2}\sum_{r=0}^{k-i-2}\sum_{j=0}^{n-2k+i}\binom{k-i-2}{r}u_{j}x^{k+j-r-2}\hskip
204.85974pt(5)$
$\displaystyle+\sum_{i=1}^{k-2}\sum_{r=0}^{k-i-2}\sum_{j=0}^{n-2k+i}\binom{k-i-2}{r}u_{j}x^{k+j-r-1}.\hskip
204.85974pt(6)$

From now on, we are going to focus on the elements $a_{i,s}$ and we are going
to determine the monomial of smallest degree, among the monomials appearing in
the determinant of $M$, where each $a_{i,s}$ appears. Recall that $u_{0}\neq
0$.

$\bullet$ $i=1$. The element $a_{1,s}$ appears only in the summands labelled
by (2)

$None$ $\sum_{s=k+1}^{n-k+1}\sum_{j=0}^{s-k+1}a_{1,s}u_{j}x^{n+k+j-s-1}.$

To get the monomial of smallest degree, the index $j$ has to be equal to $0$:

$\sum_{s=k+1}^{n-k+1}a_{1,s}u_{0}x^{n+k-s-1},$

hence each $a_{1,s}$ appears in the monomial $a_{1,s}u_{0}x^{n+k-s-1}$,
$s=k+1,\dots,n-k+1$.

Let us we represent the degrees of such monomials in the following table:

degree | $s=k+1$ | … | … | $s=n-k+1$
---|---|---|---|---
$i=1$ | $n-2$ | … | … | $2k-2$

$\bullet$ $i>1$. Set $i>1$ and $s\in\\{k+1,\dots,n-k+1\\}$ and let us
determine the monomial of smallest degree, where $a_{i,s}$ appears: it only
appears in summands labelled by (3) and (4) in the expansion of the
determinant of $M$ and, consequently, it is in the summand labelled by $(3)$
where it appears with the smallest degree (notice that the summand labelled by
$(4)$ coincides with the summand $(3)$ multiplied by $x$):

$None$
$\sum_{s=k+1}^{n-k+1}\sum_{i=2}^{k}\sum_{r=0}^{i-2}\sum_{j=0}^{s-k+1}\binom{i-2}{r}a_{i,s}u_{j}x^{n+k+j-i-
r-s-1}.$

There, the index $j$ has to be as small as possible (whence $j=0$) and $r$ has
to be as large as possible (so $r=i-2$):

$\sum_{s=k+1}^{n-k+1}\sum_{i=2}^{k}\binom{i-2}{i-2}a_{i,s}u_{0}x^{n+k-2i-s+1},$

hence each $a_{i,s}$ appears in the monomial $u_{0}x^{n+k-2i-s+1}$.
Representing the degrees of such monomials in a table we get

degree | $s=k+1$ | … | … | $s=n-k+1$
---|---|---|---|---
$i>1$ | $n-2i$ | … | … | $2k-2i$

Let us now collect all the information about the degrees in a single table:

degree | $s=k+1$ | $s=k+2$ | … | $s=n-k-1$ | $s=n-k$ | $s=n-k+1$
---|---|---|---|---|---|---
$i=1$ | $n-2$ | $n-3$ | … | $2k$ | $2k-1$ | $2k-2$
$i=2$ | $n-4$ | $n-5$ | … | $2k-2$ | $2k-3$ | $2k-4$
$i=3$ | $n-6$ | $n-7$ | … | $2k-4$ | $2k-5$ | $2k-6$
$\vdots$ | $\vdots$ | $\vdots$ |  | $\vdots$ | $\vdots$ | $\vdots$
$i=k-2$ | $n-2k+4$ | $n-2k+3$ | … | 6 | 5 | 4
$i=k-1$ | $n-2k+2$ | $n-2k+1$ | … | 4 | 3 | 2
$i=k$ | $n-2k$ | $n-2k-1$ | … | 2 | 1 | 0

If we concentrate on the last row and the first two columns of this table, we
obtain all the possible degrees from 0 to $n-2$.

For $i=1,\dots,k-1$ and $s=k+1,k+2$, as well as for $i=k$ and
$s=k+1,\dots,n-k+1$, let us define the polynomial $f_{i,s}$ consisting of the
sum of all the monomials of the determinant of $M$ which contain the variable
$a_{i,s}$. Thus, $\\{f_{i,s}\\}$ is a family of $n-2$ polynomials of smallest
degree from 0 to $n-2$, which implies that the set $\\{f_{i,s}\\}$ forms a
basis of the vector space of polynomials of degree less than or equal to $n-2$
over $\mathbb{F}$.

Therefore, for any monic polynomial $q(x)$ of degree $n$ whose trace coincides
with the trace of $p(x)$, we can choose appropriate elements $a_{i,s}$ and use
such basis to get $q(x)$ as characteristic polynomial of $A+N$, which
completes the proof. ∎

We will show now that the restriction $k<n-k$ in the previous theorem is
unremovable. Before giving a counter-example when $n=2k$, let us study how
square-zero matrices $N\in\mathbb{M}_{2k}(\mathbb{F})$ must be in order to get
an invertible matrix $A+N$ for

$A=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&{\bf 0}_{k,k}\\\ \hline\cr{\bf
0}_{k,k}&C\\\ \end{array}\right)\in\mathbb{M}_{2k}(\mathbb{F}),$

where $C$ is invertible and non-derogatory.

The following assertion is pivotal for our purpose mentioned above.

######  Proposition 2.4.

For any matrix of the form

$A=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&{\bf 0}_{k,k}\\\ \hline\cr{\bf
0}_{k,k}&C(p(x))\\\ \end{array}\right)\in\mathbb{M}_{2k}(\mathbb{F}),$

where $C(p(x))$ is the companion matrix of a degree $k$ polynomial whose zero-
coefficient is non-zero and any square-zero matrix
$N\in\mathbb{M}_{2k}(\mathbb{F})$ such that $A+N$ is invertible, there exists
a basis of $\mathbb{F}^{n}$ with respect to which the matrix $A$ remains the
same and $N$ has the form

$N=\left(\begin{array}[]{c|c}X&-X^{2}\\\ \hline\cr{\rm Id}_{k}&-X\\\
\end{array}\right)$

for some $X\in\mathbb{M}_{k}(\mathbb{F})$.

###### Proof.

Suppose that

$C(p(x))=\left(\begin{array}[]{ccc c}0&\dots&0&-u_{0}\\\ 1&&0&\vdots\\\
&\ddots&&-u_{n-2}\\\ 0&&1&-u_{n-1}\\\
\end{array}\right)\in\mathbb{M}_{k}(\mathbb{F})$

with $u_{0}\neq 0$ and

$A=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&{\bf 0}_{k,k}\\\ \hline\cr{\bf
0}_{k,k}&C(p(x))\\\ \end{array}\right)\in\mathbb{M}_{2k}(\mathbb{F}).$

Let $N=(n_{i,j})$ be any square-zero matrix such that $A+N$ is invertible.
Then, the matrix $(A+N)^{2}=A^{2}+AN+NA$ is also invertible and has the form

$(A+N)^{2}=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&*\\\ \hline\cr M&*\\\
\end{array}\right),$

where $M$ is given by:

$M=\left(\begin{array}[]{ccccc}-u_{0}n_{2k,1}&-u_{0}n_{2k,2}&\dots&\dots&-u_{0}n_{2k,k}\\\
n_{k+1,1}-u_{1}n_{2k,1}&n_{k+1,2}-u_{1}n_{2k,2}&\dots&\dots&n_{k+1,k}-u_{1}n_{2k,k}\\\
n_{k+2,1}-u_{2}n_{2k,1}&n_{k+2,2}-u_{2}n_{2k,2}&\dots&\dots&n_{k+2,k}-u_{2}n_{2k,k})\\\
\vdots&\vdots&&&\vdots\\\
n_{2k-1,1}-u_{k-1}n_{2k,1}&n_{2k-1,2}-u_{k-1}n_{2k,2}&\dots&\dots&n_{2k-1,2}-u_{k-1}n_{2k,2}\\\
\end{array}\right).$

Since $u_{0}\neq 0$, from the form of $M$ we obtain that the block matrix

$N_{21}=\left(\begin{array}[]{cccc}n_{k+1,1}&n_{k+1,2}&\dots&n_{k+1,k}\\\
n_{k+2,1}&n_{k+2,2}&\dots&n_{k+2,k}\\\ \vdots&&&\vdots\\\
n_{2k,1}&n_{2k,2}&&n_{2k,k}\\\ \end{array}\right)\hbox{ in
}N=\left(\begin{array}[]{c|c}{N}_{11}&N_{12}\\\ \hline\cr N_{21}&N_{22}\\\
\end{array}\right)$

is invertible too.

Let $\\{e_{1},\dots,e_{n}\\}$ be the canonical basis of $\mathbb{F}^{n}$, and
consider the subspace $S$ of $\mathbb{F}^{n}$ generated by
$e_{1},\dots,e_{k}$; since the block $N_{21}$ is invertible, there exists
linear combinations of the first $k$-columns of $N$ that give rise to
$e_{k+1},e_{k+2},\dots,e_{2k}$ modulo $S$, i.e., there exists scalars
$\alpha_{ij}\in\mathbb{F}$ such that

$(\alpha_{1i}N^{(1)}+\dots+\alpha_{ki}N^{(k)})+S=e_{k+i}+S,\quad i=1,\dots,k,$

where $N^{(1)},\dots,N^{(k)}$ denote the first $k$-columns of $N$. Define the
basis

$\\{e_{1}^{\prime},\dots,e_{k}^{\prime},e_{k+1},\dots,e_{2k}\\},$

where

$e^{\prime}_{i}=\alpha_{1i}e_{1}+\dots+\alpha_{ki}e_{k},\quad i=1,\dots,k.$

With respect to this basis, the matrix $N$ is of the form

$N=\left(\begin{array}[]{c|c}*&*\\\ \hline\cr{\rm Id}_{k}&*\\\
\end{array}\right)$

and the matrix $A$ remains as

$A=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&{\bf 0}_{k,k}\\\ \hline\cr{\bf
0}_{k,k}&C(p(x))\\\ \end{array}\right)\in\mathbb{M}_{2k}(\mathbb{F}).$

Suppose that

$N=\left(\begin{array}[]{c|c}X&Y\\\ \hline\cr{\rm Id}_{k}&Z\\\
\end{array}\right).$

From the condition $N^{2}=0$, we easily infer that $Y=-X^{2}$ and $Z=-X$,
i.e.,

$N=\left(\begin{array}[]{c|c}X&-X^{2}\\\ \hline\cr{\rm Id}_{k}&-X\\\
\end{array}\right)$

for some $X\in\mathbb{M}_{k}(\mathbb{F})$. ∎

We are now ready to give the promised counter-example.

######  Remark 2.5.

Let $\mathbb{F}=\mathbb{R}$, the field of real numbers. Consider the matrix

$A=\left(\begin{array}[]{cc|cc}0&0&0&0\\\ 0&0&0&0\\\ \hline\cr 0&0&0&-1\\\
0&0&1&0\\\ \end{array}\right)\in\mathbb{M}_{4}(\mathbb{F}).$

We claim that there does not exist any square-zero matrix $N$ such that the
characteristic polynomial of $A+N$ is $x^{4}+1$. Otherwise, according to
Proposition 2.4, we can suppose that $N$ is of the form

$N=\left(\begin{array}[]{c|c}X&-X^{2}\\\ \hline\cr{\rm Id}_{2}&-X\\\
\end{array}\right)\hbox{ for some }X=\left(\begin{array}[]{cc}n_{11}&n_{12}\\\
n_{21}&n_{22}\\\ \end{array}\right).$

It is readily checked that the characteristic polynomial of $A+N$ is

$p_{A+N}(x)=x^{4}+(n_{12}-n_{21}+1)x^{2}-(n_{11}+n_{22})x+n_{11}n_{22}-n_{12}n_{21}$

and, if this polynomial equals $x^{4}+1$, then

$\left\\{\begin{array}[]{ll}n_{12}-n_{21}+1=0\\\ n_{11}+n_{22}=0\\\
n_{11}n_{22}-n_{12}n_{21}=1\end{array}\right.$

so that $n_{22}=-n_{11}$ and $n_{21}=n_{12}+1$ would imply
$1=-n_{11}^{2}-n_{12}(n_{12}+1)=-n_{11}^{2}-n_{12}^{2}-n_{12}$; therefore,

$n_{11}^{2}=-n_{12}^{2}-n_{12}-1,$

but, for any real number, the left hand side of this equality is positive,
while the right hand side of this equality is less than or equal to $-3/4$,
which is an absurd.

Besides, one also observes that:

######  Remark 2.6.

In Remark 2.1, we showed how to modify an invertible non-derogatory matrix $A$
by adding a square-zero matrix $N$ to obtain a matrix $A+N$ with any
prescribed characteristic polynomial with the same trace as $A$. In addition,
by construction, the resulting matrix $A+N$ was again non-derogatory. With
that idea in mind, we close this section by posing the following problem:

Question. Given a matrix of the form

$A=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&{\bf 0}_{k,n-k}\\\ \hline\cr{\bf
0}_{n-k,k}&A_{22}\\\ \end{array}\right)\in\mathbb{M}_{n}(\mathbb{F})$

consisting of $k$ rows and columns of zeros and an invertible non-derogatory
matrix $A_{22}$, and a prescribed polynomial $q(x)$ of degree $n$ whose trace
coincides with the trace of $A$, is it possible to find a square-zero matrix
such that $A+N$ is non-derogatory and the characteristic polynomial of $A+N$
coincides with $q(x)$?

##  3\. Some consequences

In the present section, some results related to our project of decomposing a
square matrix $A$ into the sum of a square-zero matrix $N$ and another matrix
satisfying some fundamental properties such as been diagonalizable,
invertible, torsion, etc. (see, for example, [3], [4], [5] and [6]) can be
obtained by forcing that $A-N$ verifies a certain prescribed characteristic
polynomial. Specifically, the key point in [5] when showing that any nilpotent
matrix of rank at least $\frac{n}{2}$ can be expressed as the sum of a torsion
matrix and a square-zero matrix was the fact that the original nilpotent
matrix could be written as the direct sum of blocks, each of them expressed as
the sum of a square-zero matrix and a matrix satisfying a certain
characteristic polynomial that forced this last matrix to be torsion.

So, with the aid of Theorem 2.3, we can elementarily prove the following three
corollaries.

######  Corollary 3.1.

Let $\mathbb{F}$ be a field, let $n,k\in\mathbb{N}$ with $k<n-k$, and consider
the block matrix

$A=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&{\bf 0}_{k,n-k}\\\ \hline\cr{\bf
0}_{n-k,k}&A_{22}\\\ \end{array}\right)\in\mathbb{M}_{n}(\mathbb{F})$

consisting of $k$ rows and columns of zeros and an invertible non-derogatory
matrix $A_{22}$. Then, if the field $\mathbb{F}$ has enough elements, there
exists a square-zero matrix $N$ such that $A+N$ is diagonalizable (with $n$
different eigenvalues).

###### Proof.

If $\mathbb{F}$ has enough elements, we can take different elements
$\alpha_{1},\dots,\alpha_{n}\in\mathbb{F}$ such that
$-(\alpha_{1}+\dots+\alpha_{n})$ equals the trace of $A$. Then fix the
polynomial $q(x)=(x-\alpha_{1})\cdots(x-\alpha_{n})$ and use Theorem 2.3 to
find a square-zero matrix $N$ such that the characteristic polynomial of $A+N$
is $q(x)$. Since $A+N$ has $n$ different eigenvalues, $A+N$ is diagonalizable,
as asserted. ∎

######  Corollary 3.2.

Let $\mathbb{F}$ be a field, let $n,k\in\mathbb{N}$ with $k<n-k$, and consider
the block matrix

$A=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&{\bf 0}_{k,n-k}\\\ \hline\cr{\bf
0}_{n-k,k}&A_{22}\\\ \end{array}\right)\in\mathbb{M}_{n}(\mathbb{F})$

consisting of $k$ rows and columns of zeros and an invertible matrix $A_{22}$.
Then, there exists an invertible matrix $T$ and a square-zero matrix $N$ such
that $A=T+N$.

###### Proof.

Without loss of generality, assume that $A_{22}$ is the direct sum of the
invertible and non-derogatory companion matrices of its elementary divisors.
Combine the $0$’s in the main diagonal of $A$ with each of these invertible
and non-derogatory companion matrices and utilize Theorem 2.3 to get
characteristic polynomials with nonzero zero-degree coefficients by adding
appropriate square-zero matrices, as desired. ∎

Recall that a matrix $P\in\mathbb{M}_{n}(\mathbb{F})$ is $m$-potent if
$P^{m}=P$ for some $1<m\in\mathbb{N}$; a matrix
$T\in\mathbb{M}_{n}(\mathbb{F})$ is $k$-torsion (or just a $k$-root of the
unity) if $T^{k}=\operatorname{Id}$ for some $k\in\mathbb{N}$.

######  Corollary 3.3.

Let $\mathbb{F}$ be a field, let $n,k\in\mathbb{N}$ with $k<n-k$, and consider
the block matrix

$A=\left(\begin{array}[]{c|c}{\bf 0}_{k,k}&{\bf 0}_{k,n-k}\\\ \hline\cr{\bf
0}_{n-k,k}&A_{22}\\\ \end{array}\right)\in\mathbb{M}_{n}(\mathbb{F})$

consisting of $k$ rows and columns of zeros and an invertible non-derogatory
matrix $A_{22}$. If the trace of $A$ is zero, then the following two
statements are true:

  * (i)

if $n\geq 3$, there exists an $n$-potent matrix $P$ and a square-zero matrix
$N_{1}$ such that $A=P+N_{1}$;

  * (ii)

if $n\geq 2$, there exists an $n$-torsion matrix $T$ and a square-zero matrix
$N_{2}$ such that $A=T+N_{2}$;

###### Proof.

One observes that it suffices to prescribe the zero-trace polynomials
$p_{1}(x)=x^{n}-x$ and $p_{2}(x)=x^{n}-1$ and employ Theorem 2.3 to get
square-zero matrices $N_{1},N_{2}$ such that $A+N_{i}$ has characteristic
polynomial equal to $p_{i}(x)$, $i=1,2$. Then, decompose $A=(A+N_{1})-N_{1}$
to get (i) and $A=(A+N_{2})-N_{2}$ to get (ii), as wanted. ∎

Funding: The first-named author (Peter Danchev) was supported in part by the
BIDEB 2221 of TÜBÍTAK; the second and third-named authors (Esther García and
Miguel Gómez Lozano) were partially supported by Ayuda Puente 2023, URJC, and
MTM2017-84194-P (AEI/FEDER, UE). The all three authors were partially
supported by the Junta de Andalucía FQM264.

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