# Decompositions of Matrices Into a Sum of
Torsion Matrices and Matrices of Fixed Nilpotence

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.

For $n\geq 2$ and fixed $k\geq 1$, we study when a square matrix $A$ over an
arbitrary field $\mathbb{F}$ can be decomposed as $T+N$ where $T$ is a torsion
matrix and $N$ is a nilpotent matrix with $N^{k}=0$. For fields of prime
characteristic, we show that this decomposition holds as soon as the
characteristic polynomial of $A\in\mathbb{M}_{n}(\mathbb{F})$ is algebraic
over its base field and the rank of $A$ is at least $\frac{n}{k}$, and we
present several examples that show that the decomposition does not hold in
general. Furthermore, we completely solve this decomposition problem for $k=2$
and nilpotent matrices over arbitrary fields (even over division rings). This
somewhat continues our recent publications in Lin. & Multilin. Algebra (2023)
and Internat. J. Algebra & Computat. (2022) as well as it strengthens results
due to Calugareanu-Lam in J. Algebra & Appl. (2016).

The first author was partially supported by the Bulgarian National Science
Fund under Grant KP-06 No. 32/1 of December 07, 2019.

The second author was partially supported by Ayuda Puente 2022, URJC

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

Key words: (torsion) matrices, nilpotents, zero-square elements, polynomials

2010 Mathematics Subject Classification: 15A21, 15A24, 15B99, 16U60

##  1\. Introduction and Fundamentals

The decomposition of matrices over an arbitrary field into the sum of some
special elements, like nilpotent elements, idempotent elements, potent
elements, units, etc., was in the focus of so many researchers for a long time
(see, e.g., [1], [2], [3], [4], [5], [6], [20], [22] and [23] and the
bibliography cited therewith). Specifically, concerning our own work on the
subject, in [14] we found some necessary (and sufficient) conditions when any
square matrix over a field (finite or infinite) is expressible as a sum of a
diagonalizable matrix and a nilpotent matrix of index less than or equal to
two. In particular, we also obtained some results on the expression of square
matrices into the sum of a potent matrix and a square-zero matrix over finite
fields. Nevertheless, such a decomposition does not hold for fields of zero
characteristic (see [14, Example 4.3]). Further insight in that matter over
some special finite rings was achieved by us in [13]. We also refer the
interested reader to [11] for some other aspects of the realization of
matrices into the sum of specific elements over certain fields.

By combining the notions of invertibility and nilpotence, Cǎlugǎreanu and Lam
introduced in 2016 the notion of fine rings [7]: those in which every nonzero
element can be written as the sum of an invertible element and a nilpotent
one, proving in that work that every nonzero square matrix over a division
ring is the sum of an invertible matrix and a nilpotent matrix. The rings
whose nonzero idempotents are fine turned out to be an interesting class of
indecomposable rings and were studied in [8] by Cǎlugǎreanu and Zhou. In 2021,
the same authors focused on rings in which every nonzero nilpotent element is
fine, which they called $NF$ rings, and showed that for a commutative ring $R$
and $n\geq 2$, the matrix ring $\mathbb{M}_{n}(R)$ is $NF$ if and only if $R$
is a field; see [9]. A slightly more general class of rings than fine rings
was defined in [12] under the name nil-good rings (every element $a$ can be
expressed the sum $a=n+u$ where $n$ is nilpotent and $u$ is either zero or a
unit); in [18] it is shown that the matrix ring $\mathbb{M}_{n}(D)$ over a
division ring $D$ is nil-good. In general, no restriction in the index of
nilpotence is required in these decompositions.

In our work [15] we considered the ring of matrices
$\mathbb{M}_{n}(\mathbb{F})$ over an arbitary field $\mathbb{F}$, we fixed a
bound $k$ for the index of nilpotence, and studied when a matrix $A$ in
$\mathbb{M}_{n}(\mathbb{F})$ could be expressed as the sum of an invertible
matrix $U$ and a nilpotent matrix $N$ with $N^{k}=0$. Here we will continue
our study in this branch by replacing the invertibility condition of $U$ by
being a torsion matrix. Recall that a torsion matrix $T$ is the one for which
there is a positive integer $s$ such that $T^{s}$ is the identity matrix. One
elementarily sees that such a matrix is necessarily invertible as well as that
it is $s+1$-potent, i.e., $T^{s+1}=T$. Canonical forms of torsion matrices
were studied by D. Sjerve and a full classification over the rational numbers
is presented in his paper [21].

The paper is organized as follows: in the first section we will show that the
desired decomposition holds as soon as the characteristic polynomial of $A$ is
algebraic over its base field and its rank satisfies a certain bound, and we
present several examples that show that the decomposition does not hold in
general. In the second section, we focus on nilpotent matrices and deal with
the problem of finding a necessary and sufficient condition to decompose such
matrices as the sum of a torsion matrix and a zero-square matrix (fixed
nilpotence $k\leq 2$). Since we solve this problem by dealing with the Jordan
canonical blocks of the considered nilpotent matrix, our result also holds for
nilpotent matrices over division rings.

##  2\. Decomposing Matrices Into a Sum of Torsion Matrices and Matrices of
Fixed Nilpotence

As usual, for convenience of the presentation, let us specify that the letter
$\mathbb{F}$ will stand an arbitrary field unless it is not specified
something else, and the symbol $\mathbb{M}_{n}(\mathbb{F})$ is reserved for
the matrix ring over $\mathbb{F}$. All other unexplained explicitly notations
are standard and will be in an agreement with the book [19].

In our work [15] we showed:

######  Theorem 2.1.

[15, Theorem 2.7] Let $n\geq 2$. Let $\mathbb{F}$ be a field, consider the
ring $\mathbb{M}_{n}(\mathbb{F})$ and let us fix $k\geq 1$. Given a nonzero
matrix $A\in\mathbb{M}_{n}(\mathbb{F})$, there exists an invertible matrix
$U\in\mathbb{M}_{n}(\mathbb{F})$ and a nilpotent matrix
$N\in\mathbb{M}_{n}(\mathbb{F})$ with $N^{k}=0$ such that $A=U+N$ if and only
if the rank of $A$ is greater than or equal to $\frac{n}{k}$

In this section, we will address the following query:

Problem: Given a fixed $k\geq 1$, find necessary and sufficient conditions to
decompose any non-zero square matrix $A$ over a field $\mathbb{F}$ as a sum of
a torsion matrix $T$ and a nilpotent matrix $N$ with $N^{k}=0$.

Notice that the proposed Problem is already solved for matrices over finite
fields by using Theorem 2.1 and the obvious fact that the unit group of finite
rings is always torsion. Nevertheless, the rank condition is not enough to
guarantee this decomposition when working over infinite fields. In the rest of
this section, we will show some cases when this decomposition holds, and some
counterexamples showing that the decomposition does not hold in general.

######  Remark 2.2.

Let $A\in\mathbb{M}_{n}(\mathbb{F})$. If there exists a nilpotent matrix
$N\in\mathbb{M}_{n}(\mathbb{F})$ ($N^{k}=0$) and $T=A-N$ satisfies
$T^{s}=\operatorname{Id}$ for some $s\in\mathbb{N}$, then the following three
points are fulfilled:

  * •

the trace of $A$ coincides with the trace of $T$;

  * •

the minimal polynomial of $T$ divides $X^{s}-1$ and therefore the eigenvalues
of $T$ (in some extension of $\mathbb{F}$) are $s$-roots of the unity.
Moreover, if $X^{s}-1$ is separable, $T$ is diagonalizable;

  * •

the trace of $T$ coincides with the sum of its eigenvalues, so it is an
algebraic number over the base field of $\mathbb{F}$.

For example, a matrix $A\in\mathbb{M}_{n}(\mathbb{F})$, even of full rank, and
whose trace is transcendent over its base field can never be decomposed into
the sum $T+N$, where $T$ is a torsion matrix and $N$ is a nilpotent matrix.

Let $n\geq 2$. Recall that the trace of a polynomial
$p(x)=x^{n}+b_{n-1}x^{n-1}+\dots+b_{0}\in\mathbb{F}[x]$ is the scalar
$-b_{n-1}$ and coincides with the trace of the companion matrix
$C(p(x))\in\mathbb{M}_{n}(\mathbb{F})$. Notice that the rank of a companion
matrix is always greater than or equal to $n-1$.

We can now give a partial solution to the proposed above Problem. Concretely,
the following statements hold.

######  Proposition 2.3.

Let $n\geq 2$, let $p(x)\in\mathbb{F}[x]$ be a polynomial of degree $n$ and
let $C(p(x))\in\mathbb{M}_{n}(\mathbb{F})$ be its companion matrix. If the
trace of $p(x)$ can be expressed as the sum of $n$ different roots of the
unity in some extension of $\mathbb{F}$, then $A$ can be decomposed (in some
extension of $\mathbb{F}$) into $T+N$, where $T$ is a torsion matrix and
$N^{2}=0$. In particular, this always holds if the trace of $p(x)$ is either
$1$, or $-1$, or $0$.

###### Proof.

By hypothesis, the trace of $p(x)=x^{n}+b_{n-1}x^{n-1}+\dots+b_{0}$ can be
expressed as $\alpha_{1}+\dots+\alpha_{n}$ for some different roots of unity
$\alpha_{1},\dots,\alpha_{n}$ in some extension of $\mathbb{F}$. Let us
consider the polynomial
$q(x)=(x-\alpha_{1})\cdots(x-\alpha_{n})=x^{n}+a_{n-1}x^{n-1}+\dots+a_{0}$.
Thus, we have

$\displaystyle C$
$\displaystyle(p(x))=\left(\begin{array}[]{cccc}0&0&\dots&-b_{0}\\\
1&0&&\vdots\\\ &\ddots&\ddots&\\\ 0&&1&-b_{n-1}\\\ \end{array}\right)$
$\displaystyle=\underbrace{\left(\begin{array}[]{cccc}0&0&\dots&-a_{0}\\\
1&0&&\vdots\\\ &\ddots&\ddots&\\\ 0&&1&-a_{n-1}\\\
\end{array}\right)}_{C(q(x))}+\underbrace{\left(\begin{array}[]{cccc}0&0&\dots&a_{0}-b_{0}\\\
0&0&&\vdots\\\ &\ddots&\ddots&\\\ 0&&0&a_{n-1}-b_{n-1}\\\
\end{array}\right)}_{N},$

where $T:=C(q(x))$ is a diagonalizable torsion matrix and $N^{2}=0$, because
$a_{n-1}-b_{n-1}=0$, as required. In view of these arguments, the last claim
follows now at once. ∎

As an immediate consequence, we obtain:

######  Corollary 2.4.

Let $A\in\mathbb{M}_{n}(\mathbb{F})$ and let $p_{1}(x),\dots,p_{k}(x)$ be the
elementary divisors of $A$ (respectively, the invariant factors of $A$). If
each $p_{i}(x)$ has degree $n_{i}$ and its trace is a sum of $n_{i}^{\prime}s$
different roots of the unity in some extension of $\mathbb{F}$, then $A$ can
be decomposed (in some extension of $\mathbb{F}$) into $T+N$, where $T$ is a
torsion matrix and $N^{2}=0$.

###### Proof.

If $p_{1}(x),\dots,p_{k}(x)$ are the elementary divisors (respectively, the
invariant factors of $A$), then the matrix $A$ is similar to a direct sum of
companion matrices of each $p_{i}(x)$. However, utilizing Proposition 2.3, we
can express every companion matrix of $p_{i}(x)$ into the sum $T_{i}+N_{i}$,
where $T_{i}$ is a torsion matrix and $N_{i}^{2}=0$, as needed. ∎

The next two curious comments are worthwhile.

######  Remark 2.5.

Not every matrix $A\in\mathbb{M}_{n}(\mathbb{F})$ whose trace is a sum of $n$
roots of unity can be written as $T+N$, where $T$ is a torsion matrix and
$N^{2}=0$ (even if those roots are different and $A$ satisfies the rank
condition). Indeed, let $n\geq 2$ and let us consider the element
$a\in\mathbb{F}$ such that $na$ is a sum of roots of unity, but $a$ itself is
not a root of unity (notice that such an element $a$ always exists and is easy
to be constructed, so we leave out the details). Then the matrix
$A=a\operatorname{Id}\in\mathbb{M}_{n}(\mathbb{F})$ cannot be written as a sum
$T+N$; otherwise there would exist $m\in\mathbb{N}$ such that
$T^{m}=\operatorname{Id}$; but then
$\operatorname{Id}=T^{m}=(A-N)^{m}=a^{m}\operatorname{Id}-ma^{m-1}N$, so that
$(a^{m}-1)^{2}\operatorname{Id}=((a^{m}-1)\operatorname{Id})^{2}=(ma^{m-1}N)^{2}=0$,
a contradiction.
For example, the matrix $A=\left(\begin{array}[]{cc}\frac{1}{2}&0\\\
0&\frac{1}{2}\\\ \end{array}\right)$ cannot be expressed as $T+N$ even if its
trace is the sum of two (different) $6^{\rm th}$-roots of unity:
$\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{1}{2}-\frac{\sqrt{3}}{2}i$.

######  Remark 2.6.

Let $\mathbb{F}$ be a field of characteristic 0. When $n\geq 3$, if $p(x)$ is
a polynomial of degree $n$ and the trace of $p(x)$ is the sum of $n$ equal
roots of unity, then the matrix $C(p(x))$ can never be written as $T+N$.
Indeed, let $\alpha$ be a root of unity, and consider a degree $n$ polynomial
$p(x)$ whose trace is $n\alpha$ and its companion matrix $C(p(x))$. Suppose
now that $C(p(x))=T+N$ where $T$ satisfies $T^{m}=\operatorname{Id}$ for some
$m\in\mathbb{N}$ and $N^{2}=0$. Since the minimal polynomial of $T$ divides
$X^{m}-1$ and this polynomial has no multiple roots, then $T$ is
diagonalizable and its eigenvalues are all roots of unity whose sum coincides
with the trace of $T$ (which, on the other side, coincides with the trace of
$p(x)$), so it is exactly $n\alpha$. The only solution to
$n\alpha=\alpha_{1}+\dots+\alpha_{n}$, $\alpha$, $\alpha_{1}$, $\dots$,
$\alpha_{n}$ being roots of unity, is $\alpha=\alpha_{1}=\dots=\alpha_{n}$.
Therefore, the eigenvalues of $T$ are all equal to $\alpha$ and thus
$T=\alpha\operatorname{Id}$. But then $N=C(p(x))-\alpha\operatorname{Id}$
should have zero square, which is manifestly untrue (the rank of
$C(p(x))-\alpha\operatorname{Id}$ is at least $n-1$).

For example, the matrix

$C((x-1)^{3})=\left(\begin{array}[]{ccc}0&0&1\\\ 1&0&-3\\\ 0&1&3\\\
\end{array}\right)$

cannot be expressed as $T+N$, where $T$ is a torsion matrix and $N^{2}=0$.

Nevertheless, when we focus on fields of prime characteristic, we can
partially solve the Problem. Let $\mathbb{F}$ be a field of prime
characteristic $p$ and let us denote by $\mathbb{F}_{p}$ its base field.

######  Lemma 2.7.

Let $n\geq 2$. If $\mathbb{F}$ is a field of characteristic $p$ and we fix
$k\geq 1$, then for every matrix $A\in\mathbb{M}_{n}(\mathbb{F})$ of rank
greater than or equal to $\frac{n}{k}$ and whose entries are algebraic over
$\mathbb{F}_{p}$ there exists a matrix $N$ with $N^{k}=0$ such that $A-N$ is a
torsion matrix.

###### Proof.

The subfield $\mathbb{K}$ of $\mathbb{F}$ generated by the base field
$\mathbb{F}_{p}$ and by the entries of matrix $A$ is a finite field, and
$A\in\mathbb{M}_{n}(\mathbb{K})$. Apply Theorem 2.1 to decompose $A$ as $U+N$,
where $U\in\mathbb{M}_{n}(\mathbb{K})$ is invertible and
$N\in\mathbb{M}_{n}(\mathbb{K})$ satisfies $N^{k}=0$. Since $U$ is an
invertible matrix over a finite field, being invertible is equivalent to being
a torsion matrix, as wanted. ∎

The conditions on the entries of the matrix can be translated to the
coefficients of the characteristic polynomial of the matrix $A$. We will say
that a polynomial is algebraic over $\mathbb{F}_{p}$ is all its coefficients
are algebraic over $\mathbb{F}_{p}$.

######  Theorem 2.8.

Let $\mathbb{F}$ be a field of characteristic $p$, let us fix an index of
nilpotence $k\geq 1$ and let $A\in\mathbb{M}_{n}(\mathbb{F})$ of rank greater
than or equal to $\frac{n}{k}$. If the characteristic polynomial of $A$ is
algebraic over $\mathbb{F}_{p}$, then $A$ can be written as $T+N$, where $T$
is a torsion matrix and $N^{k}=0$. In particular, this decomposition always
holds for nilpotent matrices of rank greater than or equal to $\frac{n}{k}$.

###### Proof.

Let us consider the primary rational canonical form $C$ of $A$, whose
characteristic polynomial is algebraic over $\mathbb{F}_{p}$. The eigenvalues
of $C$ (roots in some extension of $\mathbb{F}$ of the characteristic
polynomial) are algebraic over $\mathbb{F}_{p}$ and, therefore, all the
elementary divisors of $A$ are algebraic polynomials over $\mathbb{F}_{p}$.
Consequently, the entries of $C$ are all algebraic over $\mathbb{F}_{p}$ and
we can apply Lemma 2.7 to get the proof. ∎

Open Question 1: Given a fixed index of nilpotence $k\geq 1$, find a suitable
criterion for the decomposition of an arbitrary matrix over a field of zero
characteristic into the sum of a torsion matrix and a nilpotent matrix of
index of nilpotence $\leq k$.

In the following section we will answer this question for $k\leq 2$ and
nilpotent matrices of rank at least $\frac{n}{2}$. Since our arguments are
quite technical, we leave open the question of decomposing nilpotent matrices
of rank at least $\frac{n}{k}$ into torsion and nilpotent matrices of index
less than or equal to $k$.

##  3\. Decomposing Nilpotent Matrices Into a Sum of Torsion and Square-Zero
Matrices

The goal of this section is to show that, for any field $\mathbb{F}$, every
nilpotent matrix in $\mathbb{M}_{n}(\mathbb{F})$ whose rank is at least
$\frac{n}{2}$ can always be decomposed as the sum of a torsion matrix and a
square-zero matrix. Recall that every nilpotent matrix is similar to a direct
sum of its Jordan blocks – all of them associated to the eigenvalue 0.

As in the previous section, the main difficulties arise when dealing with
Jordan blocks of size 1. The condition on the rank guarantees that they can
always be combined with Jordan blocks of bigger size. Let us consider the
following two examples.

######  Example 3.1.

Let $A$ be the nilpotent matrix

$A=\left(\begin{array}[]{ccc|c}0&0&0&0\\\ 1&0&0&0\\\ 0&1&0&0\\\ \hline\cr
0&0&0&0\\\ \end{array}\right),$

which is the smallest nilpotent matrix decomposed into Jordan blocks with one
Jordan block of size 1 and rank of $A$ greater than or equal to $\frac{n}{2}$
($n=4$). Let us define

$N=\left(\begin{array}[]{cccc}0&1&0&-1\\\ 0&0&-1&0\\\ 0&0&0&0\\\ 0&0&-1&0\\\
\end{array}\right)$

and let $T=A-N$. Then $N^{2}=0$ and the characteristic polynomial of $T$ is
$x^{4}-1$, so $A=T+N$ with $T^{4}=\operatorname{Id}$ and $N^{2}=0$.

######  Example 3.2.

In some situations, more than one Jordan block of size 1 has to be combined
with a Jordan block of bigger size. Consider for example

$A=\left(\begin{array}[]{cccc|cc}0&0&0&0&0&0\\\ 1&0&0&0&0&0\\\ 0&1&0&0&0&0\\\
0&0&1&0&0&0\\\ \hline\cr 0&0&0&0&0&0\\\ 0&0&0&0&0&0\\\ \end{array}\right),$

which again is the smallest nilpotent matrix expressed in Jordan blocks having
two blocks of size 1 and satisfying ${\rm rank}(A)\geq\frac{n}{2}$ ($n=6$
here). Let us consider the zero-square matrix

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

Then $T=A-N$ has characteristic polynomial equal to $x^{6}-1$, so $A=T+N$ with
$T^{6}=\operatorname{Id}$ and $N^{2}=0$.

The construction of the zero-square matrices above depend on how many Jordan
blocks of size 1 we have to combine with a Jordan block of bigger size. Let
$s$ be the number of Jordan blocks of size 1 (recall that the condition on the
rank requires that $2(s+1)\leq n$). The construction of $N$ also depends on a
certain parameter, called $r$, that must satisfy $1+2s+r<n$ and that it is
zero in the above examples. With such ingredients, we are going to define a
family of zero-square matrices called $N_{s,r}$ and an auxiliary family of
zero-square matrices denoted $N^{\prime}_{s,r}$.

######  Definition 3.3.

Let $s,r\in\mathbb{N}\cup\\{0\\}$ such that $2(s+1)\leq n$ and $1+2s+r<n$. Let
$A\in\mathbb{M}_{n}(\mathbb{F})$ be a nilpotent matrix consisting on a single
Jordan block and $s$ Jordan blocks of size $1$ (notice that the condition
$2(s+1)\leq n$ is equivalent to $A$ having rank greater or equal to
$\frac{n}{2}$). Let us define the matrices

$\displaystyle N_{0,r}:$ $\displaystyle=-e_{1,n},$ $\displaystyle N_{s,r}:$
$\displaystyle=\sum_{i=n-2s-r}^{n-s-r-2}(e_{i+1,i}+e_{i+1+s+r+1,i})+e_{1,n-s-r-1}$
$\displaystyle-\sum_{i=n-s}^{n-1}(e_{i+1,i}+e_{i+1-s-r-1,i})-e_{1,n},\hbox{ if
$s\geq 1$, }$ $\displaystyle N^{\prime}_{0,r}:$ $\displaystyle=-e_{1,n},$
$\displaystyle N^{\prime}_{s,r}:$
$\displaystyle=Q_{n-2s-r,n-s-r-1}N_{s,r}Q_{n-2s-r,n-s-r-1}=$
$\displaystyle=\sum_{i=n-2s-r}^{n-s-r-2}(e_{i+1,i}-e_{i+1+s+r+1,i})-e_{1,n-s-r-1}$
$\displaystyle-\sum_{i=n-s}^{n-1}(e_{i+1,i}-e_{i+1-s-r-1,i})-e_{1,n},\hbox{ if
$s\geq 1$, }$ $\displaystyle T_{s,r}:$ $\displaystyle=A-N_{s,r},\hbox{ for
$s\geq 0$,}$ $\displaystyle T^{\prime}_{s,r}:$
$\displaystyle=A-N^{\prime}_{s,r}\hbox{ for $s\geq 0$.}$

Now we are going to show that the matrices $N_{s,r}$ and $N^{\prime}_{s,r}$
are zero-square matrices. Let us recall the elementary matrices of
$\mathbb{M}_{n}(\mathbb{F})$:

  * $P_{i,j}(t):=\operatorname{Id}+te_{i,j}$, $t\in\mathbb{F}$.

  * $P_{i,j}:=\operatorname{Id}+e_{i,j}+e_{j,i}-e_{i,i}-e_{j,j}$

  * $Q_{i}(s)=\operatorname{Id}+(s-1)e_{i,i}$, $0\neq s\in\mathbb{F}$.

All of them are invertible with inverses $(P_{i,j}(t))^{-1}=P_{i,j}(-t)$,
$(P_{i,j})^{-1}=P_{i,j}$ and $(Q_{i}(s))^{-1}=Q_{i}(1/s)$. Let us define
$Q_{i,j}:=\sum_{r=i}^{j}Q_{r}(-1)$.

######  Proposition 3.4.

Let $\mathbb{F}$ be a field and take $n\in\mathbb{N}$ and
$s,r\in\mathbb{N}\cup\\{0\\}$ such that $2(s+1)\leq n$ and $1+2s+r<n$. Then,
the matrices $N_{s,r}$ and $N^{\prime}_{s,r}$ in $\mathbb{M}_{n}(\mathbb{F})$
are nilpotent of index $2$.

###### Proof.

Along this proof, let us identify matrices with the endomorphisms acting on
the canonical basis $\\{e_{1},\dots,e_{n}\\}$ of $\mathbb{F}^{n}$. Clearly
$N_{0,r}^{2}=(N^{\prime}_{0,r})^{2}=0$. Suppose that $s\geq 1$ and let us
denoted by $N$ the matrix $N_{s,r}$. For every $i=1\dots,n$ we have:

  * (1)

if $i=1,\dots,n-2s-r-1$, $N(e_{i})=0$,

  * (2)

if $i=n-2s-r,\dots,n-s-r-2$, $N(e_{i})=e_{i+1}+e_{i+1+s+r+1}$,

  * (3)

$N(e_{n-s-r-1})=e_{1}$,

  * (4)

if $i=n-s-r,\dots,n-s-1$, $N(e_{i})=0$,

  * (5)

if $i=n-s,\dots,n-1$, $N(e_{i})=-e_{i+1}-e_{i+1-s-r-1}$,

  * (6)

$N(e_{n})=-e_{1}$

Therefore, the image of $N$ is spanned by

$\\{e_{1}\\}\cup\\{e_{n-2s-r+1}+e_{n-s+2},\dots,e_{n-s-r-1}+e_{n}\\}\cup\\{e_{n-s+1}+e_{n-2s-r},\dots,e_{n}+e_{n-s-r-1}\\},$

i.e., the image of $N$ is the subspace generated by

$S=\\{e_{1}\\}\cup\\{e_{n-s+i}+e_{n-2s-r-1+i}\quad|\quad i=1,\dots,s\\}.$

Let us see that $N(S)=0$. Clearly, $N(e_{1})=0$; moreover, if $i=1,\dots,s-1$,
we have

$\displaystyle N(e_{n-s+i}+e_{n-2s-r-1+i})=$
$\displaystyle=-e_{n-s+i+1}-e_{n-2s-r+i}+e_{n-2s-r+i}+e_{n-s+i+1}=0$

and $N(e_{n}+e_{n-s-r-1})=-e_{1}+e_{1}=0$. Therefore, $N^{2}=0$.

Since the matrix $Q_{n-2s-r,n-s-r-1}$ satisfies
$Q_{n-2s-r,n-s-r-1}^{2}=\operatorname{Id}$, we deduce that $N^{\prime}_{s,r}$
is also a zero-square matrix, as required. ∎

In Theorem 3.10 we will calculate the characteristic polynomials of the
matrices

$T_{s,r}:=A-N_{s,r}\quad\hbox{ and }\quad
T^{\prime}_{s,r}:=A-N^{\prime}_{s,r},$

defined in 3.3. Since characteristic polynomials are invariant under
similarity, our argument will consist on transforming by similarity the
matrices $T_{s,r}$ and $T^{\prime}_{s,r}$ in a finite number of steps into
matrices of the form $T_{s^{\prime},r^{\prime}}$ or
$T^{\prime}_{s^{\prime},r^{\prime}}$, $s^{\prime}=0,1$ or 2. We will first
prove two descending lemmas. In each of these descending lemmas, we will
decrease the first subscript of $T_{s,r}$ or of $T^{\prime}_{s,r}$ by three
units and increase the second subscript by three units. Afterwards, it will
only remain to explicitly calculate the characteristic polynomial of the
matrices $T_{s^{\prime},r^{\prime}}$ and $T^{\prime}_{s^{\prime},r^{\prime}}$,
$s^{\prime}=0,1,2$.

######  Lemma 3.5 (From $T_{s,r}$ to $T^{\prime}_{s-3,r+3}$).

Let $\mathbb{F}$ be a field and let $A\in\mathbb{M}_{n}(\mathbb{F})$ be a
nilpotent matrix consisting on a single Jordan block of rank bigger than or
equal to $\frac{n}{2}$ and $s$ Jordan blocks of size $1$. Suppose that $s\geq
3$ and let $r\in\mathbb{N}\cup\\{0\\}$ such that $1+2s+r<n$. Let $T_{s,r}$ and
$T^{\prime}_{s,r}$ be the matrices defined in 3.3. Then, the matrix $T_{s,r}$
is similar to the matrix $T^{\prime}_{s-3,r+3}$.

###### Proof.

For $v=s+r+1$ and

$\displaystyle T_{1}:$ $\displaystyle=P_{n-s+3,n-s+3-v}(-1)\cdot T_{s,r}\cdot
P_{n-s+3,n-s+3-v}(1)$ $\displaystyle T_{2}:$
$\displaystyle=P_{n-s+2,n-s+2-v}\cdot T_{1}\cdot P_{n-s+2,n-s+2-v}$
$\displaystyle T_{3}:$ $\displaystyle=P_{n-s+1-v,n-s+1}(-1)\cdot T_{2}\cdot
P_{n-s+1-v,n-s+1}(1),$ $\displaystyle T_{4}:$
$\displaystyle=Q_{n-s+2-v,n-s+2}\cdot T_{3}\cdot Q_{n-s+2-v,n-s+2},$

we will show that $T_{4}=T^{\prime}_{s-3,r+3}$.

Along this proof, let us identify matrices with the endomorphisms acting on
the canonical basis $\\{e_{1},\dots,e_{n}\\}$ of $\mathbb{F}^{n}$. Let us
denote by $P$ the following matrix

$P:=P_{n-s+3,n-s+3-v}(1)\cdot P_{n-s+2,n-s+2-v}\cdot P_{n-s+1-v,n-s+1}(1)\cdot
Q_{n-s+2-v,n+r+3-v}.$

With this notation in hand, we have $T_{4}=P^{-1}T_{s,r}P$. Let us calculate
$P(e_{i})$, $i=1,\dots,n$:

  * $P(e_{i})=e_{i}$ for $i=1,2,\dots,n-s-v$,

  * $P(e_{n-s-v+1})=e_{n-s-v+1}$,

  * $P(e_{n-s-v+2})=-e_{n-s+2}$,

  * $P(e_{n-s-v+3})=-e_{n-s-v+3}-e_{n-s+3}$,

  * $P(e_{i})=-e_{i}$ for $i=n-s-v+4,\dots,n-v$ (if $n-s-v+4\leq n-v$),

  * $P(e_{i})=-e_{i}$ for $i=n-v+1,\dots,n-s-1$ (if $n-v+1\leq n-s-1$),

  * $P(e_{n-s})=-e_{n-s}$ (if $s\neq 3$),

  * $P(e_{n-s+1})=-e_{n-s+1}-e_{n-s+1-v}$,

  * $P(e_{n-s+2})=-e_{n-s+2-v}$ and

  * $P(e_{i})=e_{i}$ for $i=n-s+3,\dots,n$.

Let us see how $T_{4}=P^{-1}T_{s,r}P$ acts on the canonical basis: to do so
let us divide the set $\\{1,2,\dots,n\\}$ into the following subsets relative
to the definition of $T^{\prime}_{s-3,r+3}$:

$\begin{array}[]{ll}\Delta^{\prime}_{1}:=\\{1,\dots,n-s-v+3\\}&\Delta^{\prime}_{2}:=\\{n-s-v+4,\dots,n-v-1\\}\\\
\Delta^{\prime}_{3}:=\\{n-v,\text{ if }s\neq
3\\}&\Delta^{\prime}_{4}:=\\{n-v+1,\dots,n-s+2\\}\\\
\Delta^{\prime}_{5}:=\\{n-s+3,\dots,n-1\\}&\Delta^{\prime}_{6}:=\\{n\\}\end{array}$

(notice that if $s=3$, then $\Delta^{\prime}_{2}=\emptyset$,
$\Delta^{\prime}_{3}=\emptyset$ and $\Delta^{\prime}_{5}=\emptyset$; and if
$s=4$, then $\Delta^{\prime}_{2}=\emptyset$).

$\bullet$ $T_{4}$ acting on elements of $\Delta^{\prime}_{1}$:

  * (1)

$T_{4}(e_{i})=P^{-1}T_{s,r}P(e_{i})=P^{-1}T_{s,r}(e_{i})=P^{-1}(e_{i+1})=e_{i+1}$
for $i=1,2,\dots,n-s-v$,

  * (2)

$T_{4}(e_{n-s-v+1})=P^{-1}T_{s,r}P(e_{n-s-v+1})=P^{-1}T_{s,r}(e_{n-s-v+1})=P^{-1}(-e_{n-s+2})=e_{n-s-v+2}$,

  * (3)

$T_{4}(e_{n-s-v+2})=P^{-1}T_{s,r}P(e_{n-s-v+2})=P^{-1}T_{s,r}(-e_{n-s+2})=P^{-1}(-e_{n-s+3}-e_{n-s-v+3})=e_{n-s-v+3}$,

  * (4)

$T_{4}(e_{n-s-v+3})=P^{-1}T_{s,r}P(e_{n-s-v+3})=P^{-1}T_{s,r}(-e_{n-s+3}-e_{n-s-v+3})=P^{-1}(-e_{n-s+4}-e_{n-s-v+4}+e_{n-s+4})=P^{-1}(-e_{n-s-v+4})=e_{n-s-v+4}$.

$\bullet$ $T_{4}$ acting on elements of $\Delta^{\prime}_{2}$ (when
$\Delta_{2}^{\prime}\neq\emptyset$):

  * (5)

$T_{4}(e_{i})=P^{-1}T_{s,r}P(e_{i})=P^{-1}T_{s,r}(-e_{i})=P^{-1}(e_{i+1+v})=e_{i+1+v}$
for $i=n-s-v+4,\dots,n-v-1$.

$\bullet$ $T_{4}$ acting on elements of $\Delta^{\prime}_{3}$ (when
$\Delta_{3}^{\prime}\neq\emptyset$):

  * (6)

$T_{4}(e_{n-v})=P^{-1}T_{s,r}P(e_{n-v})=P^{-1}T_{s,r}(-e_{n-v})=P^{-1}(e_{1}-e_{n-v+1})=e_{1}+e_{n-v+1}$.

$\bullet$ $T_{4}$ acting on elements of $\Delta^{\prime}_{4}$:

  * (7)

$T_{4}(e_{i})=P^{-1}T_{s,r}P(e_{i})=P^{-1}T_{s,r}(-e_{i})=P^{-1}(-e_{i+1})=e_{i+1}$
for $i=n-v+1,\dots,n-s-1$,

  * (8)

$T_{4}(e_{n-s})=P^{-1}T_{s,r}P(e_{n-s})=P^{-1}T_{s,r}(-e_{n-s})=P^{-1}(-e_{n-s+1}-e_{n-s+1-v})\\\
=e_{n-s+1}$,

  * (9)

$T_{4}(e_{n-s+1})=P^{-1}T_{s,r}P(e_{n-s+1})=P^{-1}T_{s,r}(-e_{n-s+1}-e_{n-s+1-v})=\\\
P^{-1}(-e_{n-s+2}-e_{n-s+2-v}+e_{n-s+2})=P^{-1}(-e_{n-s+2-v})=e_{n-s+2}$,

  * (10)

$T_{4}(e_{n-s+2})=P^{-1}T_{s,r}P(e_{n-s+2})=P^{-1}T_{s,r}(-e_{n-s+2-v})=P^{-1}(e_{n-s+3})=e_{n-s+3}$.

$\bullet$ $T_{4}$ acting on elements of $\Delta^{\prime}_{5}$ (when
$\Delta_{5}^{\prime}\neq\emptyset$):

  * (11)

$T_{4}(e_{n-s+3})=P^{-1}T_{s,r}P(e_{n-s+3})=P^{-1}T_{s,r}(e_{n-s+3})=P^{-1}(e_{n-s+4}+e_{n-s+4-v})=e_{n-s+4}-e_{n-s+4-v}$,

  * (12)

$T_{4}(e_{i})=P^{-1}T_{s,r}P(e_{i})=P^{-1}T_{s,r}(e_{i})=P^{-1}(e_{i+1}+e_{i+1-v})=e^{\prime}_{i+1}-e^{\prime}_{i+1-v}$
for $i=n-s+4,\dots,n-1$.

$\bullet$ $T_{4}$ acting on elements of $\Delta^{\prime}_{6}$:

  * (13)

$T_{4}(e_{n})=P^{-1}T_{s,r}P(e_{n})=P^{-1}T_{s,r}(e_{n})=P^{-1}(e_{1})=e_{1}$,

which proves that $T_{4}=P^{-1}T_{s,r}P$ behaves on the canonical basis in the
same way as $T^{\prime}_{s-3,r+3}$ so $T_{4}=T^{\prime}_{s-3,r+3}$. ∎

Following the same arguments as in the above proof we also obtain the
following descending lemma.

######  Lemma 3.6 (From $T^{\prime}_{s,r}$ to $T_{s-3,r+3}$).

Let $\mathbb{F}$ be a field and let $A\in\mathbb{M}_{n}(\mathbb{F})$ be a
nilpotent matrix consisting on a single Jordan block of rank bigger than or
equal to $\frac{n}{2}$ and $s$ Jordan blocks of size $1$. Suppose that $s\geq
3$ and let $r\in\mathbb{N}\cup\\{0\\}$ such that $1+2s+r<n$. Let $T_{s,r}$ and
$T^{\prime}_{s,r}$ be the matrices defined in 3.3. Then the matrix
$T^{\prime}_{s,r}$ is similar to the matrix $T_{s-3,r+3}$. Indeed, if
$v=s+r+1$ and

$\displaystyle T^{\prime}_{1}:$ $\displaystyle=P_{n-s+3,n-s+3-v}(1)\cdot
T^{\prime}_{s,r}\cdot P_{n-s+3,n-s+3-v}(-1)$ $\displaystyle T^{\prime}_{2}:$
$\displaystyle=P_{n-s+2,n-s+2-v}\cdot T^{\prime}_{1}\cdot P_{n-s+2,n-s+2-v}$
$\displaystyle T^{\prime}_{3}:$ $\displaystyle=P_{n-s+1-v,n-s+1}(1)\cdot
T^{\prime}_{2}\cdot P_{n-s+1-v,n-s+1}(-1),$ $\displaystyle T^{\prime}_{4}:$
$\displaystyle=Q_{n-v-s+3,n-s+1}\cdot T^{\prime}_{3}\cdot Q_{n-v-s+3,n-s+1},$

we obtain that $T^{\prime}_{4}=T_{s-3,r+3}$.

After these two descending lemmas, let us compute the characteristic
polynomials of $T_{s,r}$ and $T^{\prime}_{s,r}$ when $s=0,1$ or $2$.

######  Remark 3.7 (Case $s=0$).

Let $\mathbb{F}$ be a field and let $A\in\mathbb{M}_{n}(\mathbb{F})$ be a
nilpotent matrix consisting on a single Jordan block of rank bigger than or
equal to $\frac{n}{2}$ and no Jordan blocks of size $1$ ($s=0$). For any
$r<n-1$ we have that
$T_{0,r}=T^{\prime}_{0,r}=\sum_{t=1}^{n-1}e_{t+1,t}+e_{1,n}$, and its
characteristic polynomial is exactly $x^{n}-1$.

######  Lemma 3.8 (Case $s=1$).

Let $\mathbb{F}$ be a field and let $A\in\mathbb{M}_{n}(\mathbb{F})$ be a
nilpotent matrix consisting on a single Jordan block of rank bigger than or
equal to $\frac{n}{2}$ and one Jordan block of size $1$. Let $r\geq 0$ be such
that $r+3<n$. Let us consider the matrices $T_{1,r}=A-N_{1,r}$ and
$T^{\prime}_{1,r}=A-N^{\prime}_{1,r}$. Then, the characteristic polynomials
are

  * •

$p_{T_{1,r}}(x)=x^{n}+x^{r+2}-x^{n-r-2}-1=(x^{r+2}-1)(x^{n-r-2}+1)$

  * •

$p_{T^{\prime}_{1,r}}(x)=x^{n}-x^{r+2}+x^{n-r-2}-1=(x^{r+2}+1)(x^{n-r-2}-1)$.

###### Proof.

We will use the following general remark: suppose we are computing the
determinant of a matrix of the form

$\displaystyle B=\left(\begin{array}[]{ccccc}x&0&\cdots&0&1\\\
-1&x&\cdots&&0\\\ \vdots&\ddots&\ddots&\vdots&\vdots\\\ 0&0&\ddots&x&0\\\
0&0&\cdots&-1&x\\\ \end{array}\right)+a\,e_{i,j}$

where $a$ is a non-zero element at position $(i,j)$, $j>i$. If we multiply $B$
on the left by $P_{i,j+1}(a)$ (we add to row $i$ the row $j+1$ multiplied by
$a$), we get a new matrix that has a zero at position $(i,j)$ and has $ax$ at
position $(i,j+1)$. Since $|P_{i,j+1}(a)|=1$ we get

$\displaystyle\det\left(\left(\begin{array}[]{ccccc}x&0&\cdots&0&1\\\
-1&x&\cdots&&0\\\ \vdots&\ddots&\ddots&\vdots&\vdots\\\ 0&0&\ddots&x&0\\\
0&0&\cdots&-1&x\\\ \end{array}\right)+a\,e_{i,j}\right)=$
$\displaystyle\det\left(\left(\begin{array}[]{ccccc}x&0&\cdots&0&1\\\
-1&x&\cdots&&0\\\ \vdots&\ddots&\ddots&\vdots&\vdots\\\ 0&0&\ddots&x&0\\\
0&0&\cdots&-1&x\\\ \end{array}\right)+ax\,e_{i,j+1}\right)$

i.e., when computing the determinant of $B$, nonzero elements at positions
$(i,j)$, $j>i$, can be moved to the right by multiplying by $x$ at each step.

By a similar argument, multiplying $B$ on the right by the elementary matrix
$P_{i-1,j}(a)$ (adding to column $j$ the column $i-1$ multiplied by $a$), we
also have that

$\displaystyle\det\left(\left(\begin{array}[]{ccccc}x&0&\cdots&0&1\\\
-1&x&\cdots&&0\\\ \vdots&\ddots&\ddots&\vdots&\vdots\\\ 0&0&\ddots&x&0\\\
0&0&\cdots&-1&x\\\ \end{array}\right)+a\,e_{i,j}\right)=$
$\displaystyle\det\left(\left(\begin{array}[]{ccccc}x&0&\cdots&0&1\\\
-1&x&\cdots&&0\\\ \vdots&\ddots&\ddots&\vdots&\vdots\\\ 0&0&\ddots&x&0\\\
0&0&\cdots&-1&x\\\ \end{array}\right)+ax\,e_{i-1,j}\right)$

i.e., when computing the determinant of $B$, nonzero elements at positions
$(i,j)$, $j>i$, can be moved upwards to position $(i-1,j)$ by multiplying at
each step by $x$.

Notice that

$\displaystyle T_{1,r}$
$\displaystyle=\sum_{t=1}^{n-1}e_{t+1,t}+e_{1,n}+e_{n-r-2,n-1}-e_{1,n-r-2}.$

Applying the argument above to the determinant of the matrix
$x\operatorname{Id}-T_{1,r}$, we can move the nonzero element at position
$(n-r-2,n-1)$ towards position $(1,n)$ obtaining $-x^{n-r-2}$ and we can move
the non-zero element at position $({1,n-r-2})$ towards position $(1,n)$
obtaining $x^{r+2}$, getting at the end the determinant of a matrix of the
form

$\displaystyle\left(\begin{array}[]{cccccc}x&0&0&\cdots&0&-x^{n-r-2}+x^{r+2}-1\\\
-1&x&0&\cdots&&0\\\ 0&-1&x&\cdots&0&0\\\
\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\\ 0&0&0&\ddots&x&0\\\
0&0&0&\cdots&-1&x\\\ \end{array}\right).$

Finally, by the Laplace expansion of the determinant along the first row we
get that the characteristic polynomial of $T_{1,r}$ is given by
$x^{n}-x^{n-r-2}+x^{r+2}-1$.

Similarly, since

$T^{\prime}_{1,r}=\sum_{t=1}^{n-1}e_{t+1,t}+e_{1,n}-e_{n-r-2,n-1}+e_{1,n-r-2},$

its characteristic polynomial is $x^{n}+x^{n-r-2}-x^{r+2}-1$. ∎

We continue our work with the next technical assertion.

######  Lemma 3.9 (Case $s=2$).

Let $\mathbb{F}$ be a field and let $A\in\mathbb{M}_{n}(\mathbb{F})$ be a
nilpotent matrix consisting on a single Jordan block of rank bigger than or
equal to $\frac{n}{2}$ and two Jordan blocks of size $1$. Let $r\geq 0$ be
such that $r+5<n$. Let us consider the matrices $T_{2,r}=A-N_{2,r}$ and
$T^{\prime}_{2,r}=A-N^{\prime}_{2,r}$. Then, the characteristic polynomials
are

  * •

$p_{T_{2,r}}(x)=x^{n}+x^{r+3}-x^{n-r-3}-1=(x^{n-r-3}+1)(x^{r+3}-1)$

  * •

$p_{T^{\prime}_{2,r}}(x)=x^{n}-x^{r+3}+x^{n-r-3}-1=(x^{n-r-3}-1)(x^{r+3}+1)$.

###### Proof.

Let $\\{e_{1},\dots,e_{n}\\}$ be the canonical basis of $\mathbb{F}^{n}$. Let

$P:=P_{n,n-r-3}\cdot Q_{n-r-3,n-r-3}$

and consider the new basis

$\\{e^{\prime}_{i},|\ i=1,\dots,n\\},\hbox{ where each
$e^{\prime}_{i}:=P(e_{i})$.}$

This new basis is just a reordering of the canonical basis and a change of
sign of one of its vectors:

$\\{e^{\prime}_{i},|\
i=1,\dots,n\\}=\\{e_{1},e_{2},\dots,e_{n-r-4},-e_{n}\\}\cup\\{e_{n-r-3},\dots,e_{n-1}\\}.$

Recall that

$\displaystyle T_{2,r}=\sum_{t=1}^{n-r-5}e_{t+1,t}-e_{n,n-r-4}+e_{1,n}$
$\displaystyle+\sum_{t=n-r-3}^{n-1}e_{t+1,t}-e_{1,n-r-3}+e_{n-r-4,n-2}+e_{n-r-1,n-1}.$

Then, the matrix $P^{-1}\cdot T_{2,r}\cdot P$ is of the form

$\left(\begin{array}[]{cccc|cccc}0&0&0&-1&*&*&*&*\\\ 1&0&0&0&*&*&*&*\\\
0&\ddots&0&0&*&*&*&*\\\ 0&0&1&0&*&*&*&*\\\ \hline\cr 0&0&0&0&0&0&0&1\\\
0&0&0&0&1&0&0&0\\\ 0&0&0&0&0&\ddots&0&0\\\ 0&0&0&0&0&0&1&0\\\
\end{array}\right)$

and its characteristic polynomial, which coincides with the characteristic
polynomial of $T_{2,r}$, is the product of the characteristic polynomials of
the two diagonal blocks, i.e., it is precisely $(x^{n-r-3}+1)(x^{r+3}-1).$

Similarly, if we consider $P^{\prime}:=P_{n,n-r-3}\cdot Q_{n,n}$ and the new
basis

$\\{e^{\prime\prime}_{i},|\ i=1,\dots,n\\},\ \hbox{where each
$e^{\prime\prime}_{i}:=P^{\prime}(e_{i})$,}$

then,

$\\{e^{\prime\prime}_{i},|\
i=1,\dots,n\\}=\\{e_{1},e_{2},\dots,e_{n-r-4},e_{n}\\}\cup\\{e_{n-r-3},\dots,-e_{n-1}\\}.$

Thus, the matrix $(P^{\prime})^{-1}\cdot T^{\prime}_{2,r}\cdot P^{\prime}$ is
of the form

$\left(\begin{array}[]{cccc|cccc}0&0&0&1&*&*&*&*\\\ 1&0&0&0&*&*&*&*\\\
0&\ddots&0&0&*&*&*&*\\\ 0&0&1&0&*&*&*&*\\\ \hline\cr 0&0&0&0&0&0&0&-1\\\
0&0&0&0&1&0&0&0\\\ 0&0&0&0&0&\ddots&0&0\\\ 0&0&0&0&0&0&1&0\\\
\end{array}\right)$

and its characteristic polynomial, which coincides with the characteristic
polynomial of $T^{\prime}_{2,r}$, is the product of the characteristic
polynomials of the two diagonal blocks, i.e., it is precisely
$(x^{n-r-3}-1)(x^{r+3}+1).$ ∎

We thus arrive at the following result in which we compute the characteristic
polynomials of the matrices $T_{s,r}=A-N_{s,r}$ and
$T^{\prime}_{s,r}=A-N^{\prime}_{s,r}$. They depend on the equivalence class of
$s$ modulo 3.

######  Theorem 3.10.

Let $\mathbb{F}$ be a field, $n\in\mathbb{N}$, and let
$s\in\mathbb{N}\cup\\{0\\}$ such that $2(s+1)\leq n$. Let
$A\in\mathbb{M}_{n}(\mathbb{F})$ be a nilpotent matrix consisting on a single
Jordan block and $s$ Jordan blocks of size $1$. Then,

  1. (1)

If $s=3\alpha$ for some $\alpha\geq 0$ and we take any $r$ such that
$1+2s+r<n$, then the characteristic polynomial of $T_{s,r}=A-N_{s,r}$ is

$p_{T_{3\alpha,r}}(x)=x^{n}-1.$

  2. (2.1)

If $s=3\alpha+1$ for some even $\alpha\geq 0$ and we take any $r$ such that
$1+2s+r<n$ then the characteristic polynomial of $T_{s,r}=A-N_{s,r}$ is given
by

$p_{T_{3\alpha+1,r}}(x)=p_{T_{1,r+3\alpha}}(x)=(x^{r+3\alpha+2}-1)(x^{n-r-3\alpha-2}+1),$

and the characteristic polynomial of $T^{\prime}_{s,r}=A-N^{\prime}_{s,r}$ is
given by

$p_{T^{\prime}_{3\alpha+1,r}}(x)=p_{T^{\prime}_{1,r+3\alpha}}(x)=(x^{r+3\alpha+2}+1)(x^{n-r-3\alpha-2}-1).$

  3. (2.2)

If $s=3\alpha+1$ for some odd $\alpha\geq 1$, and we take any $r$ such that
$1+2s+r<n$ then the characteristic polynomial of $T_{s,r}=A-N_{s,r}$ is given
by

$p_{T_{3\alpha+1,r}}(x)=p_{T^{\prime}_{1,r+3\alpha}}(x)=(x^{r+3\alpha+2}+1)(x^{n-r-3\alpha-2}-1),$

and the characteristic polynomial of $T^{\prime}_{s,r}=A-N^{\prime}_{s,r}$ is
given by

$p_{T^{\prime}_{3\alpha+1,r}}(x)=p_{T_{1,r+3\alpha}}(x)=(x^{r+3\alpha+2}-1)(x^{n-r-3\alpha-2}+1).$

  4. (3.1)

If $s=3\alpha+2$ for some even $\alpha\geq 0$ and we take any $r$ such that
$1+2s+r<n$, then the characteristic polynomial of $T_{s,r}=A-N_{s,r}$ is given
by

$p_{T_{3\alpha+2,r}}(x)=p_{T_{2,r+3\alpha}}(x)=(x^{n-r-3\alpha-3}+1)(x^{r+3\alpha+3}-1),$

and the characteristic polynomial of $T^{\prime}_{s,r}=A-N^{\prime}_{s,r}$ is
given by

$P_{T^{\prime}_{3\alpha+2,r}}(x)=p_{T^{\prime}_{2,r+3\alpha}}(x)=(x^{n-r-3\alpha-3}-1)(x^{r+3\alpha+3}+1).$

  5. (3.2)

If $s=3\alpha+2$ for some odd $\alpha\geq 0$ and we take any $r$ such that
$1+2s+r<n$, then the characteristic polynomial of $T_{s,r}=A-N_{s,r}$ is given
by

$p_{T_{3\alpha+2,r}}(x)=p_{T^{\prime}_{2,r+3\alpha}}(x)=(x^{n-r-3\alpha-3}-1)(x^{r+3\alpha+3}+1),$

and the characteristic polynomial of $T^{\prime}_{s,r}=A-N^{\prime}_{s,r}$ is
given by

$p_{T^{\prime}_{3\alpha+2,r}}(x)=p_{T_{2,r+3\alpha}}(x)=(x^{n-r-3\alpha-3}+1)(x^{r+3\alpha+3}-1).$

###### Proof.

Notice that the condition $2(s+1)\leq n$ is just equivalent to $A$ having rank
bigger or equal to $\frac{n}{2}$.

(1) Suppose that $s=3\alpha$ for some $\alpha\geq 0$. Let us consider
$T_{3\alpha,r}=A-N_{3\alpha,r}$. If $\alpha\geq 1$, applying Lemma 3.5 to
$T_{3\alpha,r}$ we obtain the matrix $T^{\prime}_{3(\alpha-1),r+3}$, which is
similar to $T_{3\alpha,r}$. If $\alpha-1\geq 1$ then use Lemma 3.6 to get
another similar matrix $T_{3(\alpha-2),r+6}$, and repeating this process, in
$\alpha$ steps we will end with either $T_{0,r+3\alpha}$ or with
$T^{\prime}_{0,r+3\alpha}$, which are both equal to the matrix
$\sum_{t=1}^{n-1}e_{t+1,t}+e_{1,n}$ (see Remark 3.7), whose characteristic
polynomial is $x^{n}-1$. Thus the characteristic polynomial of $T_{s,r}$ is
$x^{n}-1$.

(2.1) Suppose that $s=3\alpha+1$ for some even $\alpha\geq 0$. Let us consider
$T_{s,r}$. If $\alpha>1$, as in (1) we can use Lemma 3.5 and Lemma 3.6 several
times to obtain, in $\alpha$ steps, a similar matrix $T_{1,r+3\alpha}$;
therefore, the characteristic polynomial of $T_{s,r}$ coincides with the
characteristic polynomial of $T_{1,r+3\alpha}$, which is
$(x^{r+3\alpha+2}-1)(x^{n-r-3\alpha-2}+1)$ by Lemma 3.8. Similarly, if we
begin with $T^{\prime}_{s,r}$ and we use Lemmas 3.6 and 3.5, we will end up in
$\alpha$ steps with the similar matrix $T^{\prime}_{1,r+3\alpha}$, whose
characteristic polynomial is $(x^{r+3\alpha+2}+1)(x^{n-r-3\alpha-2}-1)$ by
Lemma 3.8.

(2.2) Suppose that $s=3\alpha+1$ for some odd $\alpha\geq 1$. If we start with
$T_{s,r}$ and apply Lemmas 3.5 and 3.6, after $\alpha$ steps we will end up
with the similar matrix $T^{\prime}_{1,r+3\alpha}$, and if we begin with
$T^{\prime}_{s,r}$ we will end up with $T_{1,r+3\alpha}$.

(3.1) and (3.2) follow as (1) and (2.1) and (2.2) above, taking into account
the characteristic polynomials of $T_{2,r+3\alpha}$ and
$T^{\prime}_{2,r+3\alpha}$ calculated in Lemma 3.9. ∎

As three important consequences, we derive the following:

######  Corollary 3.11.

Let $\mathbb{F}$ be a field, suppose that $n\in\mathbb{N}$ is even, and let
$s\in\mathbb{N}\cup\\{0\\}$ such that $2(s+1)\leq n$. Let
$A\in\mathbb{M}_{n}(\mathbb{F})$ be a nilpotent matrix consisting on a single
Jordan block and $s$ Jordan blocks of size $1$. If we take
$r=\frac{n}{2}-s-1$, the characteristic polynomial of $T_{s,r}=A-N_{s,r}$ and
the characteristic polynomial of $T^{\prime}_{s,r}=A-N_{s,r}$ coincide and
they are $x^{n}-1$. In particular, $T_{s,r}^{n}=\operatorname{Id}$ so that $A$
decomposes as the sum of a torsion matrix $T_{s,r}$ and a zero-square matrix
$N_{s,r}$.

###### Proof.

If $s\equiv_{3}0$, since the characteristic polynomial of $T_{s,r}$ is always
$x^{n}-1$ for any $r$ such that $1+2s+r<n$, this also is true if we take
$r=\frac{n}{2}-s-1$.

If $s\equiv_{3}1$ or $s\equiv_{3}2$ and we take $r=\frac{n}{2}-s-1$, we obtain
that the characteristic polynomial of $T_{s,r}$ is
$(x^{\frac{n}{2}}-1)(x^{\frac{n}{2}}+1)=x^{n}-1$. ∎

######  Corollary 3.12.

Let $\mathbb{F}$ be a field, suppose that $n\in\mathbb{N}$ is odd, and let
$s\in\mathbb{N}\cup\\{0\\}$ such that $2(s+1)\leq n$. Let
$A\in\mathbb{M}_{n}(\mathbb{F})$ be a nilpotent matrix consisting on a single
Jordan block and $s$ Jordan blocks of size $1$.

  1. (1)

If $s\equiv_{3}0$ and we take any $r$ such that $1+2s+r<n$, then $N_{s,r}$
such that the characteristic polynomial of $T_{s,r}=A-N_{s,r}$ is $x^{n}-1$.

  2. (2)

If $s\equiv_{3}1$ or $s\equiv_{3}2$, and:

    1. (a)

if $n=4m+1$, $\alpha$ is even, $s=3\alpha+1$ or $s=3\alpha+2$, and we take
$r=2m-s$, then $N_{s,r}$ is such that the characteristic polynomial of
$T_{s,r}=A-N_{s,r}$ is $(x^{2m}+1)(x^{2m+1}-1)$;

    2. (b)

if $n=4m+1$, $\alpha$ is odd, $s=3\alpha+1$ or $s=3\alpha+2$, and we take
$r=2m-s$, then $N^{\prime}_{s,r}$ is such that the characteristic polynomial
of $T^{\prime}_{s,r}=A-N^{\prime}_{s,r}$ is $(x^{2m}+1)(x^{2m+1}-1)$;

    3. (c)

if $n=4m+3$, $\alpha$ is even, $s=3\alpha+1$ or $s=3\alpha+2$, and we take
$r=2m-s$, then $N_{s,r}$ is such that the characteristic polynomial of
$T_{s,r}=A-N_{s,r}$ is $(x^{2m+2}+1)(x^{2m+1}-1)$;

    4. (d)

if $n=4m+3$, $\alpha$ is odd, $s=3\alpha+1$ or $s=3\alpha+2$, and we take
$r=2m-s$, then $N^{\prime}_{s,r}$ is such that the characteristic polynomial
of $T^{\prime}_{s,r}=A-N^{\prime}_{s,r}$ is $(x^{2m+2}+1)(x^{2m+1}-1)$.

###### Proof.

It is an automatic consequence of Theorem 3.10. ∎

######  Corollary 3.13.

Let $\mathbb{F}$ be a field of zero characteristic, suppose that
$n\in\mathbb{N}$ is odd, and let $s\in\mathbb{N}\cup\\{0\\}$ such that
$2(s+1)\leq n$. Let $A\in\mathbb{M}_{n}(\mathbb{F})$ be a nilpotent matrix
consisting on a single Jordan block and $s$ Jordan blocks of size $1$. Then,
there exists a zero-square matrix $N$ and $d\geq 1$ such that
$(A-N)^{d}=\operatorname{Id}$, i.e., $A$ decomposes as the sum of a torsion
matrix and a zero-square matrix.

###### Proof.

Following the cases of the above corollary, we have:

  * -

if we are in case (1), take $N=N_{s,r}$ and $d=n$; clearly
$(A-N)^{n}=\operatorname{Id}$;

  * -

if we are in case (2)(a), take $r=2m-s$ and $N=N_{s,r}$; then if
$d=lcm(4m,2m+1)$, we obtain $(A-N)^{d}=\operatorname{Id}$, because the
polynomials $x^{2m}+1$ and $x^{2m+1}-1$ have no common roots, so the matrix
$A-N$ has $n$ different eigenvalues (in the algebraic closure of
$\mathbb{F}$), all of them roots of unity $\xi$ with $\xi^{4m}=1$ or
$\xi^{2m+1}=1$;

  * -

if we are in case (2)(b), take $r=2m-s$ and $N=N^{\prime}_{s,r}$; then if
$d=lcm(4m,2m+1)$, we obtain $(A-N)^{d}=\operatorname{Id}$ arguing as in case
(2)(a);

  * -

if we are in case (2)(c), take $r=2m-s$ and $N=N_{s,r}$; then if
$d=lcm(4m+2,2m+1)$, we obtain $(A-N)^{d}=\operatorname{Id}$ arguing as in case
(2)(a);

  * -

if we are in case (2)(d), take $r=2m-s$ and $N=N^{\prime}_{s,r}$; then if
$d=lcm(4m+2,2m+1)$, we obtain $(A-N)^{d}=\operatorname{Id}$ arguing as in case
(2)(a).

∎

We now have all the ingredients necessary to establish our chief result in
this section.

######  Theorem 3.14.

Any nilpotent matrix in $\mathbb{M}_{n}(\mathbb{F})$ can be written as the sum
of a torsion matrix plus a square-zero matrix if, and only if, its rank is at
least $\frac{n}{2}$.

###### Proof.

According to the above Theorem 2.1, a necessary condition to express
$A\in\mathbb{M}_{n}(\mathbb{F})$ as the sum $T+N$, where $T$ is a torsion
matrix and $N^{2}=0$, is that the rank of $A$ is greater than or equal to
$\frac{n}{2}$, because torsion matrices must be invertible. Conversely, to
prove the sufficiency, suppose that the rank of
$A\in\mathbb{M}_{n}(\mathbb{F})$ is no less than $\frac{n}{2}$.

If $\mathbb{F}$ is a field of prime characteristic, then the result follows by
Corollary 2.8 given above. Suppose from now on that $\mathbb{F}$ is a field of
characteristic zero. By the primary rational canonical decomposition of any
nilpotent matrix, we can assume without loss of generality that $A$ is the
direct sum of its Jordan blocks.

If $A$ only contains Jordan blocks of size $r>1$, each of them can be
decomposed separately in the following way

$\left(\begin{array}[]{cccc}0&0&\dots&0\\\ 1&0&&\vdots\\\ &\ddots&\ddots&\\\
0&&1&0\\\ \end{array}\right)=\left(\begin{array}[]{cccc}0&0&\dots&1\\\
1&0&&\vdots\\\ &\ddots&\ddots&\\\ 0&&1&0\\\
\end{array}\right)+\left(\begin{array}[]{cccc}0&\dots&\dots&-1\\\
\vdots&&&\vdots\\\ \vdots&&&\\\ 0&\dots&\dots&0\\\ \end{array}\right)$

where the first matrix is a torsion matrix and the second one is zero-square.

Otherwise, since the rank of $A$ is greater than or equal to $\frac{n}{2}$, we
can reorder the blocks of matrix $A$ – which just corresponds to a reordering
of the elementary divisors of $A$ – as follows: we follow each Jordan block of
size $r>1$ by $s$ ($s\leq r-2$) blocks of size $1$. If $r+s$ is even, the
combined block of size $r+s$ can be decomposed as the sum of a torsion matrix
and a zero-square matrix by virtue of Corollary 3.11. If, however, $r+s$ is
odd, we elementarily see that the combined block of size $r+s$ can be
decomposed as the sum of a torsion matrix and a zero-square matrix in view of
Corollary 3.13. ∎

The next commentaries are worthwhile to show that the above theorem is
somewhat true in more general settings.

######  Remark 3.15.

It is well known (see, for example, [10], [16] or [17]) that every nilpotent
matrix over a division ring is still similar to its Jordan form, i.e., it is
similar to a direct sum of Jordan blocks, all of them associated to 0. Thus,
Theorem 3.14 is still valid for nilpotent matrices over division rings.

The above result suggests the following question:

Open Question 2: Given a fixed index of nilpotence $k\geq 2$, is it true that
any nilpotent matrix in $\mathbb{M}_{n}(\mathbb{F})$ can be written as the sum
of a torsion matrix and a nilpotent matrix of index less than or equal to $k$
if, and only if, its rank is at least $\frac{n}{k}$?

Funding: The first-named author (Peter V. Danchev) was supported in part by
the Bulgarian National Science Fund under Grant KP-06 No. 32/1 of December 07,
2019, the second-name author (Esther García) was partially supported by Ayuda
Puente 2022, URJC. The three authors were partially supported by the Junta de
Andalucía FQM264.

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