text:
\begin{document}

\begin{center}
{\Large
Spectrum of Rota---Baxter operators}

V. Gubarev
\end{center}

\begin{abstract}
We prove that the spectrum of every Rota---Baxter operator of weight~$\lambda$
on a~unital algebraic (not necessarily associative) algebra over a field of characteristic zero
is a subset of $\{0,-\lambda\}$.
For a finite-dimensional unital algebra the same statement is shown to hold without a restriction
on the characteristic of the ground field.
Based on these results, we define the Rota---Baxter $\lambda$-index
$\rb_\lambda(A)$ of an algebra $A$ as the infimum of the degrees
of minimal polynomials of all Rota---Baxter operators of weight~$\lambda$ on~$A$.
We calculate the Rota---Baxter $\lambda$-index for the matrix algebra $M_n(F)$, $\charr F = 0$:
it is shown that $\rb_\lambda(M_n(F)) = 2n-1$.

\medskip
{\it Keywords}:
Rota---Baxter operator, matrix algebra, simple Jordan algebra of a~bilinear form,
matrix Cayley---Dickson algebra, Rota---Baxter module.
\end{abstract}

\section{Introduction}

Given an algebra $A$ over a ground field~$F$ and a scalar $\lambda\in F$,
a~linear operator $R\colon A\rightarrow A$ is called a Rota---Baxter operator
(RB-operator, for short) of weight~$\lambda$ on $A$ if the following identity holds for all $x,y\in A$
\begin{equation}\label{RB}
R(x)R(y) = R( R(x)y + xR(y) + \lambda xy ).
\end{equation}
An algebra $A$ endowed with an RB-operator is called Rota---Baxter algebra (RB-algebra).

Rota---Baxter operators appeared in the work of G.~Baxter~\cite{Baxter} in 1960
in the study of fluctuation theory as a formal generalization of integration by parts formula.
Further, they were studied by G.-C.~Rota~\cite{Rota} and others \cite{Atkinson,Miller66,Cartier}.
The deep connections between solutions of different versions
of the classical Yang---Baxter equation in mathematical physics
and RB-operators were discovered in~\cite{BelaDrin82,Aguiar00,FardThesis}.
The  subject of double algebras is also tightly connected to Rota---Baxter operators,
see \cite{DoublePoissonFree,DoubleLie}.
Since 2000, the connection between RB-algebras and so called Loday algebras
(prealgebras and postalgebras) has been studied~\cite{Aguiar00,GolubchikSokolov,BBGN2011,Embedding}.

Rota---Baxter operators have a broad area of application
in symmetric polynomials, shuffle algebra, etc.~\cite{Atkinson,GuoMonograph,GuoKeigher}.

There is a plenty of papers devoted to a partial or complete
classification of Rota---Baxter operators on a given small-dimensional algebra:
$\mathrm{sl}_2(F)$~\cite{Kolesnikov,sl2,sl2-0},
$\mathrm{sl}_3(F)$~\cite{KonovDissert},
$M_2(F)$~\cite{Aguiar00-2,BGP,Mat2},
$M_3(F)$~\cite{GonGub,Sokolov},
the Grassmann algebra $\mathrm{Gr}_2$~\cite{BGP},
and others, see the survey~\cite{Unital} and the references therein.

The classifications obtained in these works are often bulky and hard to observe,
so there is a need to investigate the general properties of Rota---Baxter operators
at least on finite-dimensional (and close to them) algebras.
In an attempt of such a study, the author proved in~\cite{Unital} the following result.
Given a unital algebraic associative (alternative, Jordan) algebra $A$
over a~field of characteristic zero,
every RB-operator of weight zero on $A$ is nilpotent.
Moreover, if $A$ is algebraic of a bounded degree, then
there exists $N$ such that $R^N = 0$ for every RB-operator~$R$ of weight zero on $A$.
(By an algebraic algebra in the nonassociative case we mean such an algebra
that every finite subset generates a~finite-dimensional subalgebra.)

The statement above leads to a new invariant of an algebra~$A$:
the Rota---Baxter index (RB-index) $\rb(A)$ defined
as the lower bound of nilpotency indices of all RB-operators of weight zero on~$A$.
This invariant was calculated for different finite-dimensional
algebras~\cite{Unital,Gub2017}, including the matrix algebra $M_n(F)$.

In the present work, we generalize and extend this result in different directions.
First, we consider nonzero weight case of Rota---Baxter operators (when $\lambda\neq0$).
Further, we remove the restriction on a variety of algebras,
we do note assume that $A$ is associative, alternative, or Jordan.
Finally, we state the result in the case of a~field of any characteristic.

Before stating the main result of the work, we need a new definition.
Given an algebra $A$, denote the set of all RB-operators of weight~$\lambda$ on~$A$ by $\RB_\lambda(A)$.
Define the Rota---Baxter $\lambda$-index ($\RB(\lambda)$-index) of $A$ as
$$
\rb_\lambda(A) = \min\{n\in\mathbb{N}\mid \mbox{for all }R\in \RB_\lambda(A)\
 \mbox{exists } k:\ R^k(R+\lambda\id)^{n-k} = 0\}.
$$
If such a number does not exist, then put $\rb_\lambda(A) = \infty$.

If $\lambda = 0$, then $\rb_0(A)$ coincides with
the earlier defined Rota---Baxter index $\rb(A)$~\cite{Unital}.

In~\S3 (Theorem~1, over a~field of characteristic zero)
and in~\S4 (Theorem~2, over a~field of positive characteristic),
we prove

{\bf Main Theorem}.
Let $A$ be a unital finite-dimensional algebra over a field~$F$. Let us fix $\lambda\in F$.

a) Given an RB-operator $R$ of weight~$\lambda$ on $A$,
there exist $k,l\geq0$ such that $R^k(R+\lambda\id)^l = 0$. In particular,
$\Spec(R)\subset\{0,-\lambda\}$.

b) There exists a natural~$n$ such that for every RB-operator~$R$
of weight~$\lambda$ on $A$ one can find $0\leq k\leq n$ such that
$R^k(R+\lambda\id)^{n-k}(1) = 0$ holds.
Moreover, $\rb_\lambda(A)\leq 2n$.

When $\charr F = 0$, the part a) of the Theorem holds even if $A$ is algebraic (see~\S3).

In~\S5 (Theorem~3), we state another important property of Rota---Baxter operators.
Given a unital algebraic algebra $A$ and an RB-operator~$R$ of nonzero weight on~$A$,
the minimal polynomial of the element $R(1)$ has no multiple roots.
This result allows us to extend and reprove some results from~\cite{Unital}
including the diagonalizibilty of $R(1)$ for an RB-operator~$R$ of nonzero weight
on the matrix algebra $M_n(F)$ over a field of characteristic zero.

In~\S6, we compute the Rota---Baxter $\lambda$-index for different algebras:
a sum of fields, the matrix algebra $M_n(F)$, its subalgebra $t_n(F)$ of not strictly upper-triangular matrices,
the simple Jordan algebra $J_n(f)$ of a nondegenerate symmetric bilinear form~$f$,
and the matrix (split) Cayley---Dickson algebra $C(F)$.

Let us collect all results obtained earlier (see~\cite{BGP,Gub2017,Unital}) and
in the current work about $\RB(\lambda)$-index of particular algebras in the following table (see the next page).
By $\mathrm{Gr}_n$ we denote the Grassmann algebra
of a vector space $V = \Span\{e_1,\ldots,e_n\}$.
The sign ``(?)'' denotes the hypothesis,
see Conjectures~1 and~2 in~\S6.2.

In~\S7, we apply the Main Theorem for Rota---Baxter modules (\S7.1)
and parameterized Rota---Baxter operators (\S7.2).

Finally, in~\S7.3, we generalize the construction which is already known
for different varieties of algebras~\cite{GuoMonograph,preLie,Burde}.
Given an algebra~$A$ of a~variety $\Var$ and an RB-operator~$R$ on~$A$,
one get a countable sequence of algebras~$A_i$ belonging to the same variety~$\Var$.
Moreover, all of $A_i$ considered with~$R$ are Rota---Baxter algebras.

We also apply the Main Theorem to state the structure of the limiting algebra $A_\infty$
(see~\S7.3).

\begin{table}[t]
\begin{center}
Table: the Rota---Baxter $\lambda$-index of different algebras
\end{center}
\begin{tabular}{c|c|c|c}
Algebra & Field & $\rb_0$ & $\rb_\lambda$, $\lambda\neq0$ \\
\hline
$F^n$ & any & 1 & $n$ \\
$M_n(F)$ & $\charr(F) = 0$ & $2n-1$ & $2n-1$ \\
$t_n(F)$ & $\charr(F) = 0$ & $n$ (?) & $2n-1$ \\
$\mathrm{Gr}_n$ & $\charr(F) = 0$ & ? & 2 \\
$J_n(f)$ & $\bar{F} {=} F$, $\charr(F) {\neq}2,3$ & $\begin{cases}
2, & n=3  \\
3, & n>3 \end{cases}$ & $\begin{cases}
2, & 2\!\not|\,n  \\
3, & 2\,|\,n
\end{cases}$\\
$C(F)$ & $\bar{F} {=} F$, $\charr(F) {\neq}2,3$ & 3 & 3 \\
$A = \bigoplus\limits_{i=1}^k M_{n_i}(F)$ & $\charr(F) = 0$ &
$2\max\limits_{i=1,\ldots,k}\{n_i\} {-} 1$\,(?) &
$\sum\limits_{i=1}^k n_i {+} \max\limits_{i=1,\ldots,k}\{n_i\} - 1$\,(?)
\end{tabular}
\end{table}

\section{Preliminaries}

Trivial RB-operators of weight $\lambda$ are zero operator and $-\lambda\id$.

{\bf Lemma 1}~\cite{GuoMonograph}.
Given an RB-operator $P$ of weight $\lambda$,

(a) the operator $-P-\lambda\id$ is an RB-operator of weight $\lambda$,

(b) the operator $\lambda^{-1}P$ is an RB-operator of weight 1, provided $\lambda\neq0$.

Given an algebra $A$, let us define a map $\phi$ on the set of all RB-operators on $A$
as $\phi(P)=-P-\lambda(P)\id$, where $\lambda(P)$ denotes the weight of an RB-operator~$P$.
It is clear that $\phi^2$ coincides with the identity map.

{\bf Lemma 2}~\cite{BGP}.
Given an algebra $A$, an RB-operator $P$ on $A$ of weight $\lambda$,
and $\psi\in\Aut(A)$, the operator $P^{(\psi)} = \psi^{-1}P\psi$
is an RB-operator of weight~$\lambda$ on~$A$.

The same result is true when $\psi$ is an antiautomorphism of $A$, i.e.,
a~bijection from $A$ to $A$ satisfying $\psi(xy) = \psi(y)\psi(x)$ for all $x,y\in A$;
e.g., transpose on the matrix algebra.

{\bf Lemma 3}~\cite{GuoMonograph}.
Let an algebra $A$ split as a vector space
into a direct sum of two subalgebras $A_1$ and $A_2$.
An operator $P$ defined as
\begin{equation}\label{Split}
P(a_1 + a_2) = -\lambda a_2,\quad a_1\in A_1,\ a_2\in A_2,
\end{equation}
is RB-operator of weight~$\lambda$ on~$A$.

Let us call an RB-operator from Lemma~3 as
{\it splitting} RB-operator with subalgebras $A_1,A_2$.
Note that the set of all splitting RB-operators on
an algebra $A$ is in bijection with all decompositions of~$A$
into a~direct sum of two subalgebras $A_1,A_2$.

{\bf Lemma 4}~\cite{BGP}.
Let $A$ be a unital algebra and let $R$ be an RB-operator
of nonzero weight~$\lambda$ on $A$.
If $R(1)\in F$, then $R$ is splitting with one of subalgebras containing~1.

The following construction of RB-operators of nonzero weight generalizes Lemma~3.

{\bf Lemma 5}~\cite{Guil}.
Let an algebra $A$ be a direct sum of subspaces $A_-,A_0,A_+$,
moreover, $A_\pm,A_0$ are subalgebras of $A$,
and $A_\pm$ are $A_0$-modules.
If $R_0$ is an RB-operator of weight~$\lambda$ on $A_0$,
then an operator $P$ defined as
\begin{equation}\label{RB:SubAlg2}
P(a_-+a_0+a_+) = R_0(a_0) - \lambda a_+,\quad
a_{\pm}\in A_{\pm},\ a_0\in A_0,
\end{equation}
is an RB-operator of weight~$\lambda$ on~$A$.

Let us call an RB-operator of nonzero weight
defined by~\eqref{RB:SubAlg2} as triangular-splitting one
provided that at least one of $A_-,A_+$ is nonzero.

{\bf Lemma 6}.
Let $A$ be a unital algebra and $R$ be an RB-operator of weight~$\lambda$ on~$A$.
a) Then the subalgebra $S$ generated by the set $\{R^k(1) \mid k\in\mathbb{N}\}$
is commutative and associative.

b) \cite[\S3.2.2]{GuoMonograph} for all $k,l\in\mathbb{N}$
the following formula holds,
\begin{equation}\label{R^k(1)R^l(1)Main}
R^k(1)R^l(1)
 =  \sum\limits_{j=0}^{\min\{k,l\}}\lambda^j\binom{k}{j}\binom{k+l-j}{k}R^{k+l-j}(1).
\end{equation}

{\sc Proof}.
a) We prove that the associator
$$
h(k,l,m)
 = (R^k(1),R^l(1),R^m(1))
 = (R^k(1)R^l(1))R^m(1) - R^k(1)(R^l(1)R^m(1))
$$
is zero for all natural $k,l,m$ by induction on $r = k+l+m$.
For $r = 0$, it is true. Suppose that the statement holds true for all $p<r$.
Note that $h(k,l,m) = 0$ if at least one of the numbers $k,l,m$ is zero.
So, we may assume that $k,l,m\geq1$.

On the one hand, we have
\begin{multline*}
(R^k(1)R^l(1))R^m(1) \\
 = R( R^{k-1}(1)R^l(1) + R^k(1)R^{l-1}(1) + \lambda R^{k-1}(1)R^{l-1}(1))R^m(1) \\
 = R[ (R^{k-1}(1)R^l(1))R^m(1) + (R^k(1)R^{l-1}(1))R^m(1) + \lambda (R^{k-1}(1)R^{l-1}(1))R^m(1) \\
  + R( R^{k-1}(1)R^l(1) + R^k(1)R^{l-1}(1) + \lambda R^{k-1}(1)R^{l-1}(1))R^{m-1}(1) \\
  + \lambda (R^{k-1}(1)R^l(1))R^{m-1}(1)
  + \lambda(R^k(1)R^{l-1}(1))R^{m-1}(1)
  + \lambda^2 (R^{k-1}(1)R^{l-1}(1))R^{m-1}(1) ].
\end{multline*}
On the other hand,
\begin{multline*}
R^k(1)(R^l(1)R^m(1)) \\
 = R^k(1) R( R^{l-1}(1)R^m(1) + R^l(1)R^{m-1}(1) + \lambda R^{l-1}(1)R^{m-1}(1)) \\
 = R[ R^k(1)(R^{l-1}(1)R^m(1)) + R^k(1)(R^l(1)R^{m-1}(1)) + \lambda R^k(1)(R^{l-1}(1)R^{m-1}(1)) \\
  + R^{k-1}(1) R(  R^{l-1}(1)R^m(1) + R^l(1)R^{m-1}(1) + \lambda R^{l-1}(1)R^{m-1}(1) ) \\
  + \lambda R^{k-1}(1)(R^{l-1}(1)R^m(1))
  + \lambda R^{k-1}(1)(R^l(1)R^{m-1}(1))
  + \lambda^2 R^{k-1}(1)(R^{l-1}(1)R^{m-1}(1)) ].
\end{multline*}
Applying the induction hypothesis and~\eqref{RB},
we compute $h(k,l,m) = R(\Delta)$, where
\begin{multline*}
\Delta
 = R^{k-1}(1)R^l(1)R^m(1) + R^k(1)R^{l-1}(1)R^m(1) + \lambda R^{k-1}(1)R^{l-1}(1)R^m(1) \\
 + R^{k}(1)R^l(1)R^{m-1}(1) + \lambda R^{k-1}(1)R^l(1)R^{m-1}(1) \\
 + \lambda R^k(1)R^{l-1}(1)R^{m-1}(1) + \lambda^2 R^{k-1}(1)R^{l-1}(1)R^{m-1}(1) \\ \allowdisplaybreaks
 - R^k(1)R^{l-1}(1)R^m(1) - R^k(1)R^l(1)R^{m-1}(1) - \lambda R^k(1)R^{l-1}(1)R^{m-1}(1) \\
 - R^{k-1}(1)R^l(1)R^{m}(1) - \lambda R^{k-1}(1)R^{l-1}(1)R^m(1) \\
 - \lambda R^{k-1}(1)R^l(1)R^{m-1}(1)
  - \lambda^2 R^{k-1}(1)R^{l-1}(1)R^{m-1}(1)
 = 0.
\end{multline*}

Finally, we prove that $R^k(1)R^l(1) = R^l(1)R^k(1)$ by induction on $r = k+l$.
For $r=0$, it is trivial. Suppose commutativity holds for all $p<r$.
If $k = 0$ or $l = 0$, then $R^k(1)R^l(1) = R^l(1)R^k(1)$.
Assume that $k,l\geq1$. We have
\begin{multline*}
R^k(1)R^l(1) - R^l(1)R^k(1) \\
 = R( R^{k-1}(1)R^l(1) + R^k(1)R^{l-1}(1) + \lambda R^{k-1}(1)R^{l-1}(1) \\
 - R^{l-1}(1)R^k(1) - R^l(1)R^{k-1}(1) - \lambda R^{l-1}(1)R^{k-1}(1) )
 = 0
\end{multline*}
by the induction hypothesis. \hfill $\square$

\section{Main theorem in characteristic zero}

Let $A$ be a unital associative algebra and $R$ be an RB-operator of weight~$\lambda$ on~$A$.
It is known~\cite{GuoMonograph} that the following formula
\begin{equation}\label{Stirling1}
n! R^n(1)  = \sum\limits_{k=1}^n (-1)^{n-k}\lambda^{n-k}s(n,k)(R(1))^k
\end{equation}
holds. Here $s(n,k)$ is a~Stirling number of the first kind.
By Lemma~6, we may remove the restriction on~$A$ to be associative.

{\bf Lemma 7}.
Let $A$ be a unital algebraic algebra over a field of characteristic zero
and let $R$ be an RB-operator of a nonzero weight~$\lambda$ on $A$.
Then there exist $r,s$ such that $(R+\lambda\id)^r R^s(1) = 0$.

{\sc Proof}.
Denote $a = R(1)$. We put $\lambda = -1$ for simplicity.
Since $A$ is algebraic, there exists
a minimal polynomial $m$ for $a$ of the degree $n$.
The formula~\eqref{Stirling1} allows us
to present~$m$ in the form
\begin{equation}\label{MinPolWeightNonzero}
m(a) = R^n(1) + \alpha_{n-1}R^{n-1}(1) + \ldots + \alpha_1 R^1(1) + \alpha_0 = 0,\quad \alpha_i\in F.
\end{equation}
Applying the formula
\begin{equation}\label{R^k(1)R(1)}
R^k(1)R(1) = (k+1)R^{k+1}(1) + \lambda k R^k(1)
\end{equation}
holding by Lemma~6b, we calculate
\begin{multline}\label{MinPolWeightNonzeroXR(1)}
0 = m(a)R(1)
 = (n+1)R^{n+1}(1) + n(\alpha_{n-1}-1)R^n(1) + (n-1)(\alpha_{n-2}-\alpha_{n-1})R^{n-1}(1) \\
 +\ldots + \alpha_0 R^1(1).
\end{multline}
Subtracting~\eqref{MinPolWeightNonzeroXR(1)} from $(n+1)R(m(a))$, we obtain
\begin{equation}\label{MinPolWeightNonzeroXR(1)Reduced}
0 = (\alpha_{n-1}+n)R^n(1) + (2\alpha_{n-2}+(n-1)\alpha_{n-1})R^{n-1}(1) + \ldots + (n\alpha_0 + \alpha_1)R^1(1).
\end{equation}
We have two cases.

{\sc Case 1}: all coefficients of~\eqref{MinPolWeightNonzeroXR(1)Reduced} are zero.
Then it is easy to derive the formulas
$$
\alpha_{n-1} = - n,\quad
\alpha_{n-2} = \frac{n(n-1)}{2},\quad \ldots,\quad
\alpha_{n-i} = (-1)^i \binom{n}{i},\quad \ldots,\quad
\alpha_0 = (-1)^n.
$$
So, we get the equality $(R-\id)^n(1) = 0$, and we are done.

\newpage
{\sc Case 2}: not all coefficients of~\eqref{MinPolWeightNonzeroXR(1)Reduced} are zero.
By minimality of $m$, it means that $m$ and~\eqref{MinPolWeightNonzeroXR(1)Reduced} are proportional.
Find a minimal $k$ such that $\alpha_0 = \ldots = \alpha_{k-1} = 0$ and $\alpha_k\neq0$.
Thus, coefficients by $R^i(1)$ in $km(a)$ and~\eqref{MinPolWeightNonzeroXR(1)Reduced} are equal.
Again, we write down the equalities
\begin{gather*}
\alpha_{k+1} = -(n-k)\alpha_k,\quad
\alpha_{k+2} = \frac{(n-k)(n-k-1)}{2}\alpha_k,\quad \ldots,\\
\alpha_{n-i} = (-1)^{n-k-i}\binom{n-k}{i}\alpha_k,\quad \ldots,\quad
\alpha_{n-1} = (-1)^{n-k-1}(n-k)\alpha_k.
\end{gather*}
Thus, we get
$\frac{(-1)^{n-k}}{\alpha_k}m(a) = (R-\id)^{n-k}R^k(1) = 0$. \hfill $\square$

{\bf Lemma 8}.
Let $A$ be an algebra and $R$ be an RB-operator of weight~$\lambda$ on $A$.

a) If $\lambda = 0$, then $\ker(R^k)$ is an $\Imm(R^k)$-module for each $k\geq1$.

b)~\cite{Burde} If $\lambda \neq 0$, then $\ker(R^k)$ is an ideal in $\Imm(R+\lambda\id)^k$.

{\sc Proof}.
We prove Lemma~8 by induction on $k$. The following equality
\begin{equation}\label{kerRnIdeal}
R(x(R(y)+\lambda y))  = R(x)R(y) - R(R(x)y)
\end{equation}
holds by~\eqref{RB} for all $x,y\in A$.
If $x\in \ker(R)$, then $x\Imm(R+\lambda\id)\in \ker(R)$.
Analogously, $\Imm(R+\lambda\id)x\in \ker(R)$.

Suppose we have proved Lemma~8 for all numbers less or equal to~$k$.
Let $x\in\ker(R^{k+1})$, $y\in A$ and $z = R(y)$.
Then by~\eqref{kerRnIdeal} we get
$$
R^{k+1}(x(R+\lambda\id)^{k+1}(y))
 = R^k[ R(x)(R+\lambda\id)^k(z) - R(R(x)(R+\lambda\id)^k(y)) ].
$$
By the induction hypothesis,
$x\Imm(R+\lambda\id)^{k+1}\in \ker(R^{k+1})$.
Analogously, we obtain $\Imm(R+\lambda\id)^{k+1}x\in \ker(R^{k+1})$. \hfill $\square$

Given an algebra $A$, denote the set of all RB-operators of weight~$\lambda$ on~$A$ by $\RB_\lambda(A)$.
Define Rota---Baxter $\lambda$-index (RB($\lambda$)-index) of $A$ as
$$
\rb_\lambda(A) = \min\{n\in\mathbb{N}\mid \mbox{for all }R\in \RB_\lambda(A)\
 \mbox{exists } k:\ R^k(R+\lambda\id)^{n-k} = 0\}.
$$
If such a number does not exist, then put $\rb_\lambda(A) = \infty$.

When $\lambda = 0$,
$$
\rb_0(A) = \min\{n\in\mathbb{N}\mid  R^n = 0\mbox{ for all }R\in \RB_0(A)\}
$$
which coincides with the earlier defined Rota---Baxter index of $A$~\cite{Unital}
(it was denoted there as $\rb(A)$).

Now, we prove the Main Theorem over a field of characteristic zero, generalizing~\cite{Unital}.

{\bf Theorem 1}.
Let $A$ be a unital algebraic algebra over
a field $F$ of characteristic zero. Let us fix $\lambda\in F$.

a) Given an RB-operator $R$ of weight~$\lambda$ on $A$,
there exist $k,l\geq0$ such that $R^k(R+\lambda\id)^l = 0$. In particular,
$\Spec(R)\subset\{0,-\lambda\}$.

b) Suppose that $A$ is algebraic of the restricted degree $N$, then
there exists a natural $n\leq N$ such that for every RB-operator
$R$ of weight~$\lambda$ on $A$ one can find $0\leq k\leq n$
such that
$R^k(R+\lambda\id)^{n-k}(1) = 0$ holds.
Moreover, $\rb_\lambda(A)\leq 2n$.

{\sc Proof}.
Let $\lambda = 0$. In~\cite{Unital}, it was proved that
$R^n(1) = 0$  for some $n\in\mathbb{N}$ under the condition that $A$ is power-associative.
By Lemma~6, we may avoid this condition.
When $A$ is algebraic of the restricted degree~$N$, then $n\leq N$.
Let $x\in \Imm(R^n)$. By Lemma~8a, $1\cdot x = x \in \ker(R^n)$.
Thus, $R^{2n}(A) = (0)$, we are done.

Let $\lambda\neq0$. We may assume that $\lambda = -1$.
Since $A$ is algebraic, we may consider the minimal (unital)
polynomial $f(x)$ of the element $R(1) = a$ of the degree $n$.
By Lemma~7, there exists $k$ such that
$R^k(R-\id)^{n-k}(1) = 0$.
When $A$ is algebraic of the restricted degree~$N$, then $n\leq N$.

We want to prove that $1\in \ker(R^k)\oplus \ker(R-\id)^{n-k}$.
First, it is easy to check that $\ker(R^k)\cap \ker(R-\id)^{n-k} = (0)$.
Second, let $e_\gamma$, $\gamma\in\Gamma$, and $f_\delta$, $\delta\in\Delta$,
be a~linear basis of $\ker(R^k)$ and of $\ker(R-\id)^{n-k}$ respectively.
We may complete the set
$\{e_\gamma\mid \gamma\in\Gamma\}\cup \{f_\delta\mid \delta\in\Delta\}$
to a basis of $A$ by adding vectors $\{g_\pi\mid \pi\in\Pi\}$.
So, we represent 1 as
$$
1 = 1_e + 1_f + 1_g
$$
meaning that $1_e\in \Span\{e_\gamma\}$, $1_f\in \Span\{f_\delta\}$,
and $1_g\in \Span\{g_\pi\}$. Let us show that $1_g = 0$ and so,
$1\in \ker(R^k)\oplus \ker(R-\id)^{n-k}$. Indeed,
$R^k(1) = R^k(1_f) + R^k(1_g)$, and $R^k(1_f)\in \ker(R-\id)^{n-k}$,
since $\ker(R-\id)^{n-k}$ is $R$-invariant.
As $(R-\id)^{n-k}(R^k(1)) = 0$, we conclude that
$R^k(1_g) \in \ker(R-\id)^{n-k}$.
The map $R^k$ is invertible on $\ker(R-\id)^{n-k}$, so, $1_g\in \ker(R-\id)^{n-k}$
and $1_g = 0$.

Finally, let $x\in \Imm R^{n-k}(R-\id)^k$.
By Lemma~8b,
$$
(\ker(R^k)\oplus \ker(R-\id)^{n-k})\cdot\Imm R^{n-k}(R-\id)^k
 \subset \ker(R^k)\oplus \ker(R-\id)^{n-k}.
$$
So, $1\cdot x = x \in \ker(R^k)\oplus \ker(R-\id)^{n-k}$.
As a consequence, $R^n(R-\id)^n(A) = (0)$.
\hfill $\square$

{\bf Remark 1}.
Let us show that all conditions of Theorem~1 are necessary.

\begin{itemize}
\item
If $A$ is not unital, then we may consider an algebra $A$ with zero product.
Then every linear map on $A$ is an RB-operator of a weight~$\lambda$.
So, we may find a linear map satisfying $\Spec(R)\not\subset\{0,-\lambda\}$.

\item
If $A$ is not algebraic, then $A = F[x]$ with the RB-operator $R(x^n) = \frac{x^{n+1}}{n+1}$
is a~counterexample to the conclusion of Theorem~1 when $\lambda = 0$.
Regardless of weight~$\lambda$, we may consider $A$ equal
to the free commutative unital RB-algebra generated by
a non-empty set $X = \{1\}$ (see~\cite{GuoKeigher}),
it is also a~counterexample to the conclusion of Theorem~1.

\item
If $\charr (F) = p > 0$, then let $A$ be
the free commutative unital RB-algebra of weight~0 generated by unit.
Due to~\cite{GuoKeigher}, $A = \bigoplus\limits_{i\geq0} Fe_i$ with the product
$e_i e_j = \left(\binom{i+j}{i}\!\! \mod p\right)e_{i+j}$
and the RB-operator~$R$ acting as follows, $R(e_i) = e_{i+1}$.
Note that $A$ is algebraic.
Thus, the conclusion of Theorem~1 does not hold.
\end{itemize}

\section{Main theorem in positive characteristic}

{\bf Lemma 9}.
Let $A$ be a finite-dimensional unital algebra over a field~$F$ of positive characteristic~$p$.
Then given an RB-operator $R$ of weight~$\lambda$ on $A$,
there exist $k,l\geq0$ such that $R^k(R+\lambda\id)^l = 0$.

{\sc Proof}.
\underline{Let $\lambda = 0$}. Since $A$ is finite-dimensional, there exists a minimal $l$ such that
\begin{equation}\label{ThmMain:p>0,lambda=0-Assum}
R^l(1) + \alpha_{l-1}R^{l-1}(1) + \ldots + \alpha_1 R(1) = 0,\quad \alpha_i\in F.
\end{equation}
It is well-known that unit does not lie in the image of $R$,
thus we take a zero free coefficient in~\eqref{ThmMain:p>0,lambda=0-Assum}.
Multiplying~\eqref{ThmMain:p>0,lambda=0-Assum} by $R(1)$
and applying the formula~\eqref{R^k(1)R(1)} for $\lambda=0$, we get
\begin{equation}\label{ThmMain:p>0,lambda=0-1}
(l+1)R^{l+1}(1) + l\alpha_{l-1}R^l(1) + \ldots + 2\alpha_1 R^2(1) = 0.
\end{equation}
Acting on~\eqref{ThmMain:p>0,lambda=0-Assum} by $R$, we have
\begin{equation}\label{ThmMain:p>0,lambda=0-2}
R^{l+1}(1) + \alpha_{l-1}R^l(1) + \ldots + \alpha_1 R^2(1) = 0.
\end{equation}
From~\eqref{ThmMain:p>0,lambda=0-1} and~\eqref{ThmMain:p>0,lambda=0-2}, we conclude
$$
\alpha_{l-1}R^l(1) + 2\alpha_{l-2}R^{l-1}(1) + \ldots + (l-1)\alpha_1 R^2(1) = 0.
$$
Since $l$ was chosen as a minimal number satisfying~\eqref{ThmMain:p>0,lambda=0-Assum},
we have two cases.

{\sc Case 1}.
The vectors $(1,\alpha_{l-1},\ldots,\alpha_2,\alpha_1)$
and $(\alpha_{l-1},2\alpha_{l-2}\ldots,(l-1)\alpha_1,0)$ are proportional
(with some nonzero coefficient).
It means that $\alpha_1 = \alpha_2 = \ldots = \alpha_{l-1} = 0$ and $R^l(1) = 0$, as required.

{\sc Case 2}.
The vector $(\alpha_{l-1},2\alpha_{l-2}\ldots,(l-1)\alpha_1,0)$ is zero.
If $l\leq p$, then we derive $\alpha_1 = \alpha_2 = \ldots = \alpha_{l-1} = 0$ and $R^l(1) = 0$.
If $l>p$, then only coefficients $\alpha_{l-p},\alpha_{l-2p},\ldots$ may be nonzero.
Anyway, we can find an equality
\begin{equation}\label{ThmMain:p>0,lambda=0-Fine}
R^{k_1}(1) + \beta_2 R^{k_2}(1) + \ldots + \beta_s R^{k_s}(1) = 0,\quad k_1>k_2>\ldots>k_s,\
p|k_1,\ldots,p|k_s
\end{equation}
with a minimal number of nonzero coefficients (here $s$).
If $s=1$, we are done.

Suppose that $s>1$. Consider $\Delta = k_1-k_s = p^tq$, where
$(p,q) = 1$ and $t\geq1$.
Acting by the operator $R^{\delta}$ on~\eqref{ThmMain:p>0,lambda=0-Fine}
for a number $\delta\in p\mathbb{N}$, we may get
$$
k_1 = p^N - p^t,\quad
k_s = p^N - p^t - \Delta
$$
for some sufficiently big~$N$.

Now, we mutiply~\eqref{ThmMain:p>0,lambda=0-Fine} by $R^{p^t}(1)$.
Applying Lemma~6b for $\lambda = 0$
$$
R^a(1)R^b(1) = \binom{a+b}{a}R^{a+b}(1),
$$
we have
$$
R^{k_1}(1)R^{p^t}(1) = 0,\quad
R^{k_s}(1)R^{p^t}(1) = \binom{p^t(p^{N-t}-q)}{p^t}R^{p^N - \Delta}(1),
$$
where the coefficient $\binom{p^t(p^{N-t}-q)}{p^t}$ is nonzero by the Lucas' Theorem.
So, we have found an expression like~\eqref{ThmMain:p>0,lambda=0-Fine}
with less number of nonzero coefficients. It is a contradiction, so $s = 1$, and we are done.

\underline{Let $\lambda$ be nonzero}, so we fix $\lambda = -1$.
Since $A$ is finite-dimensional, there exists an expression~$m$ of the form
$$
R^n(1) + \alpha_{n-1}R^{n-1}(1) + \ldots + \alpha_1 R^1(1) + \alpha_0 = 0, \quad \alpha_i\in F,
$$
of the minimal degree $n$.
\newpage

Analogously to the proof of Lemma~7, we derive the formula
$$
q = (\alpha_{n-1}+n)R^n(1) + (2\alpha_{n-2}+(n-1)\alpha_{n-1})R^{n-1}(1) + \ldots + (n\alpha_0 + \alpha_1)R^1(1) = 0.
$$
If all $\alpha_i$ are zero, $R^n(1) = 0$ and we are done by Lemma~8.

Otherwise, find a minimal~$s$ such that $\alpha_s\neq0$.
So, $q = sm$, and we have the relations
\begin{equation}\label{ThmMain:p>0,system}
\begin{gathered} \allowdisplaybreaks
s\alpha_{s+1} = (n-s)\alpha_s + (s+1)\alpha_{s+1},\\
s\alpha_{s+2} = (n-s-1)\alpha_{s+1} + (s+2)\alpha_{s+2},\\
\ldots, \\
s\alpha_{n-1} = 2\alpha_{n-2} + (n-1)\alpha_{n-1},\\
s = \alpha_{n-1} + n.
\end{gathered}
\end{equation}
Define $n-s = tp + k$ for $0\leq k<p$ and $t\geq0$.
Due to the system~\eqref{ThmMain:p>0,system}, we have
\begin{multline*}
m = \alpha_s\left( R^s(1) - \binom{k}{1}R^{s+1}(1) + \binom{k}{2}R^{s+2}(1)
 \ldots + (-)^k \binom{k}{k}R^{s+k}(1) \right) \\
 + \alpha_{s+p}\left( R^{s+p}(1) -\ldots + (-)^k \binom{k}{k}R^{s+p+k}(1) \right) \\
 + \ldots
 + (-1)^k\left( R^{n-k}(1) - \ldots + (-1)^k \binom{k}{k}R^n(1) \right) \\
 = R^s(R-\id)^k ( \alpha_s + \alpha_{s+p} R^p + \ldots + (-1)^k R^{tp} )(1) = 0.
\end{multline*}
Acting on the last expression by the operator
$R^\delta(R-\id)^{\varepsilon}$
with suitable $\delta,\varepsilon$, we get the equality of the form
\begin{equation}\label{ThmMain:p>0,equ}
R^{pk_0} (R-\id)^p ( R^{pk_1} + \beta_2 R^{pk_2} + \ldots + \beta_e R^{pk_e})(1) = 0,
\end{equation}
where $k_1>k_2>\ldots>k_e$, $k_0>0$, and $\beta_i\neq0$, $i=2,\ldots,e$.

Since $(R-\id)^p = R^p -\id$ modulo $p$,
the equality~\eqref{ThmMain:p>0,equ} can be rewritten in the form
\begin{equation}\label{ThmMain:p>0,equ2}
R^{pm_1}(1) + \gamma_2 R^{p(m_1-m_2)}(1) + \ldots + \gamma_f R^{p(m_1-m_f)}(1) = 0,
\end{equation}
where $0<m_2<m_3<\ldots <m_f<m_1$.

We may assume, that $m_1 = p^{c-1}$ for some $c\geq1$.
Let $d$ be any natural not less than~$c$.
Acting by the operator $R^{p^d - p^c}$ on~\eqref{ThmMain:p>0,equ2},
we get the equation
\begin{equation}\label{ThmMain:p>0,equ3}
R^{p^d}(1) + \gamma_2 R^{p^d - pm_2}(1) + \ldots + \gamma_f R^{p^d -pm_f}(1) = 0.
\end{equation}

By Lemma~6b, we have for $\lambda = -1$
\begin{equation}\label{R^k(1)R^l(1)Nonzero}
R^{ap}(1)R^{bp}(1)
 =  \sum\limits_{c=0}^{\min\{a,b\}}(-1)^{cp}\binom{ap}{cp}\binom{p(a+b-c)}{ap}R^{p(a+b-c)}(1)
\end{equation}
over a field of characteristic~$p$.

With the help of the formula~\eqref{R^k(1)R^l(1)Nonzero} and the Lucas' Theorem,
we calculate
\begin{multline}
0 = (R^{p^d}(1) + \gamma_2 R^{p^d - pm_2}(1) + \ldots + \gamma_f R^{p^d -pm_f}(1))R^{p^d}(1) \\
 = E(d) := 2R^{2p^d}(1) + (-1)^{p^d}R^{p^d}(1) + \sum\limits_{j=2}^f\gamma_j \binom{2p^d-pm_j}{p^d} R^{2p^d - pm_j}(1).
\end{multline}
By the pigeonhole principle, we may find such $d_1>d_2\geq c$ that
$\binom{2p^{d_1}-pm_j}{p^{d_1}} = \binom{2p^{d_2}-pm_j}{p^{d_2}}$ for all $j=2,\ldots,f$.
Thus,
$$
R^{2p^{d_1}-2p^{d_2} }(E(d_2)) - E(d_1)
 = (-1)^p R^{p^{d_1} }( R^{ p^{d_1} - p^{d_2} } - \id )(1) = 0.
$$
Fix $\Delta = p^{d_1-1} - p^{d_2-1}$.
Then we have
\begin{equation}\label{ThmMain:p>0,equ4}
R^{p^k}(1) - R^{p^k-\Delta p}(1) = 0
\end{equation}
for all sufficiently big $k$.
Multiplying~\eqref{ThmMain:p>0,equ4} by $R^{p^k}(1)$,
we obtain
\begin{equation}\label{ThmMain:p>0,equ5}
2R^{2p^k}(1) + (-1)^p R^{p^k}(1) - \binom{2p^k-\Delta p}{p^k}R^{2p^k-\Delta p}(1) = 0.
\end{equation}
Acting by $\binom{2p^k-\Delta p}{p^k}R^{p^k}$ on~\eqref{ThmMain:p>0,equ4}
and subtracting the result from~\eqref{ThmMain:p>0,equ5}, we get
$$
\left(2 - \binom{2p^k-\Delta p}{p^k}\right)R^{2p^k}(1) + (-1)^p R^{p^k}(1) = 0.
$$
If the number in the brackets is zero, we are done.
Otherwise, we find an expression of the form
\begin{equation}\label{ThmMain:p>0,equ6}
R^{2p^k}(1) - lR^{p^k}(1) = 0,\quad l\in\mathbb{Z}^*_p.
\end{equation}
If $\charr(F) = p = 2$, then
$$
R^{2p^k}(1) - R^{p^k}(1)
 = R^{p^k}(R-\id)^{p^k}(1) = 0,
$$
and have proved the statement.

Suppose that $p>2$.
It is easy to show that consecutive applications of~\eqref{ThmMain:p>0,equ6}
imply the equality
\begin{equation}\label{ThmMain:p>0,equ7}
0
 = R^{p^{k+1} }(1) - l^{p-1}R^{p^k}(1)
 = R^{p^{k+1} }(1) - R^{p^k}(1) .
\end{equation}
Finally, with the help of~\eqref{ThmMain:p>0,equ7} we compute
\begin{multline*}
0 = (R^{p^{k+1} }(1) - R^{p^k}(1))^2 \\
  = 2R^{2p^{k+1} }(1) - R^{p^{k+1} }(1)
 - 2R^{p^{k+1} + p^k }(1)
 + 2R^{2p^k}(1) - R^{p^k}(1) \\
 = 2( R^{2p^k}(1) - R^{p^k}(1))
 = 2R^{p^k}(R-\id)^{p^k}(1),
\end{multline*}
so, $R^{p^k}(R-\id)^{p^k}(1) = 0$, and we are done. \hfill $\square$

{\bf Theorem 2}.
Let $A$ be a unital finite-dimensional algebra over
a field $F$ of positive characteristic. Let us fix $\lambda\in F$.

a) Given an RB-operator $R$ of weight~$\lambda$ on $A$,
there exist $k,l\geq0$ such that $R^k(R+\lambda\id)^l = 0$. In particular,
$\Spec(R)\subset\{0,-\lambda\}$.

b) There exists a natural~$n$ such that for every RB-operator~$R$
of weight~$\lambda$ on $A$ one can find $0\leq k\leq n$ such that
$R^k(R+\lambda\id)^{n-k}(1) = 0$ holds.
Moreover, $\rb_\lambda(A)\leq 2n$.

{\sc Proof}.
a) It is Lemma~9.

b) We may find such~$n$ by Lemma~9 and by the condition that $A$ is finite-dimensional.
For the bound $\rb_\lambda(A)\leq 2n$, we use Lemma~8 and repeat the end of the proof of Theorem~1.
\hfill $\square$

\section{Miminal polynomial of the image of unit}

Define polynomials
\begin{equation}\label{H-Poly}
H_{r,s}(x) = (x-r)(x-r+1)\ldots x(x+1)\ldots (x+s-1)
\end{equation}
for natural $r,s\geq0$.

{\bf Lemma 10}.
Let $A$ be a unital algebra and let $R$ be an RB-operator of weight~$-1$ on~$A$.
Denote $a = R(1)$. Then for all $r,s\geq0$,
\begin{equation}\label{ROnHrs}
(r+s+1)R(H_{r,s}(a)) = H_{r,s+1}(a).
\end{equation}

{\sc Proof}.
Let us prove~\eqref{ROnHrs} by induction on $r$.
For $r= 0$, we want to state the formula
\begin{equation}\label{R^n(a)}
(n+1)!R^n(a) = H_{0,n+1}(a)
\end{equation}
by induction on $n$.
For $n = 0$, we have $a = a$, it is true.
Suppose that we have proved that
$n!R^{n-1}(a) = H_{0,n}(a)$ or, equivalently,
$a^n = n!R^{n-1}(a) + f(a)$,
where
$f(x) = x^n - H_{0,n}(x)$.

By~\eqref{R^k(1)R(1)}, we calculate
\begin{multline*}
a^{n+1}
 = a^n\cdot a
 = (n!R^n(1)+f(a))R(1) \\
 = (n+1)!R^{n+1}(1) - n\cdot n!R^n(1) + af(a) \\
 = (n+1)!R^{n+1}(1) - nH_{0,n}(a) + a(a^n - H_{0,n}(a)) \\
 = (n+1)!R^{n+1}(1) + a^{n+1} - (a+n)H_{0,n}(a).
\end{multline*}
So, $(n+1)!R^{n+1}(1) = H_{0,n+1}(a)$ as required.
Thus, we get~\eqref{ROnHrs} for $r = 0$.

Suppose we have proved~\eqref{ROnHrs} for all numbers less than~$r$.
Represent
$$
H_{r,s}(x)
 = H_{r-1,s+1}(x) - (r+s)H_{r-1,s}(x).
$$
Applying the induction hypothesis for the last equality, we get
\begin{multline*}
(r+s+1)R(H_{r,s}(a))
 = (r+s+1)R(H_{r-1,s+1}(a)) - (r+s+1)(r+s)R(H_{r-1,s}(a)) \\
 = H_{r-1,s+2}(a) - (r+s+1)H_{r-1,s+1}(a)
 = H_{r,s+1}(a),
\end{multline*}
as required. \hfill $\square$

\newpage
{\bf Lemma 11}.
Let $A$ be a unital algebra and let $R$ be an RB-operator of weight~$-1$ on~$A$.
Denote $a = R(1)$. Then for all $r,s\geq0$,
\begin{equation}\label{R'^rR^s}
(r+s)!(R-\id)^r R^s(1) = H_{r,s}(a).
\end{equation}

{\sc Proof}.
We prove the statement by induction on $r$.
For $r=0$, it follows from~\eqref{R^n(a)}.
Suppose we have proved~\eqref{R'^rR^s} for all numbers less than~$r$.
Applying the induction hypothesis and~\eqref{ROnHrs},
we get the following equalities
\begin{multline*}
(r+s)!(R-\id)^r R^s(1)
 = (R-\id)(r+s)H_{r-1,s}(a) \\
 = H_{r-1,s+1}(a) - (r+s)H_{r-1,s}(a)
 = H_{r-1,s}(a)(a+s-(r+s))
 = H_{r,s}(a).
\end{multline*}
So, Lemma is proved. \hfill $\square$

{\bf Theorem 3}.
Let $A$ be a unital algebra over a field $F$
and let $R$ be an RB-operator of weight~$-1$ on $A$.
Denote $a = R(1)$.

a) If $\charr(F) = 0$ and $A$ is algebraic, then there exist $r,s$ such that $H_{r,s}(a) = 0$.

b) If $\charr(F) = p>0$, then $H_{0,p}(a) = 0$.

\noindent
In particular, we get that the minimal polynomial of $a$ under the conditions
has no multiple roots.

{\sc Proof}.
a) It follows from Lemma~7 and Lemma~11.

b) We are done by~\eqref{R'^rR^s}. \hfill $\square$

Now, we get easily the following result which requires in~\cite{Unital} several pages.

{\bf Corollary 1}~\cite[Thm.~4.17]{Unital}.
Given a field $F$ of characteristic~0, and an RB-operator of weight~$-1$ on $M_n(F)$,
the matrix $R(1)$ is diagonalizable and the set of diagonal elements in an appropriate basis
has the form $\{-s,-s+1,\ldots,0,1,\ldots,r-1,r\}$
for some natural numbers $r,s$ such that $r+s+1\leq n$.

We call an algebra $A$ as a nil-algebra if for every $x\in A$
there exists $k$ such that $x^k = 0$ for each bracketing of the nonassociative monomial $x^k$.
We generalize and simplify the proof of another statement from~\cite{Unital}.

{\bf Corollary 2}~\cite[Thm.~4.8]{Unital}.
Let $A$ be a unital algebra over a field~$F$ and
$A = F\cdot1\oplus N$ (direct vector-space sum), where $N$ is a nil-algebra.
Then each Rota---Baxter operator $R$ of nonzero weight on $A$ is splitting
and (up to $\phi$) we have $R(1) = 0$.

{\sc Proof}.
Denote $a = R(1)$. If $a\in N$, then $a^k = 0$ for some $k$.
By Theorem~3, we get $a = 0$. Then by Lemma~4, $R$ is splitting.

If $a\not\in N$, then $a = \alpha\cdot1 + n$, where $\alpha\in F$ and $n\in N$.
So, there exists $k$ such that $(a-\alpha\cdot1)^k = 0$.
By Theorem~3, $a-\alpha\cdot1 = 0$ or $a\in F$.
By Lemma~4, $R$ is splitting and $\phi(R)(1) = 0$.
\hfill $\square$

By~Corollary~2, we have $\rb_\lambda(\mathrm{Gr}_n) = 2$ for the Grassmann algebra~$\mathrm{Gr}_n$~\cite{Unital}.

{\bf Corollary 3}.
Let $A$ be a unital algebra over a field $F$ of characteristic zero
and let $R$ be an RB-operator of weight~$\lambda$ on~$A$.
Denote by $S$ a subalgebra generated by the set $\{R^k(1) \mid k\in\mathbb{N}\}$.

a) If $\dim(S) = \infty$, then $S$ is a free commutative RB-algebra of weight~$\lambda$ generated by~1.

b) If $\dim(S) = n <\infty$ and $\lambda = 0$, then
$S$~is isomorphic to the quotient of $F[x]$ by the ideal generated by $x^n$.
Moreover, $R$ acts on $S$ as follows, $R(x^i) = \frac{x^{i+1}}{i+1}$ for $i=0,\ldots,n-1$.

c) If $\dim(S)<\infty$, $\lambda = -1$, $F$ is algebraically closed,
and $H_{r,s}(x)$ is a minimal polynomial of $R(1)$, then $S\cong F^{r+s}$.

{\sc Proof}.
a) It follows directly.

b) It follows from Theorem~1 and formula~\eqref{Stirling1}.

c) Note that $S$ is isomoprhic to the quotient of $F[x]$ by the ideal generated
by the polynomial $H_{r,s}$ of degree $n = r + s$ which has no multiple roots.
So, $S$ is semisimple. Since $F$ is algebraically closed, $S\cong F^{r+s}$.
\hfill $\square$

The notion of capacity plays a crucial role in the structural theory of Jordan algebras~\cite{Taste}.
Let us generalize it for all unital algebras (of every variety).
Given a unital algebra~$A$, define a capacity of $A$ as the maximal number of pairwise
orthogonal idempotents in $A$. If capacity of $A$ equals~$n$, then there exist
$e_1,\ldots,e_n\in A$ such that $e_i^2 = e_i$ and $e_i e_j = 0$ if $i\neq j$.

{\bf Corollary 4}.
Let $A$ be a unital algebraic algebra over an algebraically closed field $F$ of characteristic zero.
If capacity of $A$ equals $n<\infty$, then $\rb_\lambda(A)\leq 2n$ for $\lambda\neq0$.

{\sc Proof}.
Let $R$ be an RB-operator of nonzero weight~$\lambda$ on $A$.
By Theorem~1a, there exist $k,l$ such that $R^k(R+\lambda\id)^l = 0$ on~$A$.
In particular, it means that the subalgebra~$S$ in~$A$ generated by the set
$\{R^j(1) \mid j\in\mathbb{N}\}$ is finite-dimensional.
By Corollary~3c, $S\cong F^m$ for some $m$.
Clearly, $m\leq n$ by the definition of capacity.
So, minimal polynomial of $P(1)$ for every RB-operator~$P$ of weight $\lambda$ on~$A$
has degree not greater than~$n$.
By the proof of Theorem~1b, $\rb_\lambda(A)\leq 2n$.
\hfill $\square$

\section{Rota-Baxter-index of different algebras}

\subsection{Sum of fields}
Let $F^n = Fe_1\oplus Fe_2\oplus \ldots \oplus Fe_n$
denote a direct sum of $n$ copies of a field $F$.
Here $e_i^2 = e_i$.

{\bf Proposition 1}.
$\rb_\lambda(F^n) = \begin{cases}
1, & \lambda = 0, \\
n, & \lambda \neq 0.
\end{cases}$

{\sc Proof}.
For $\lambda = 0$, it follows from Corollary~5.7~\cite{Unital}.

Let $\lambda\neq0$ and let $R$ be an RB-operator of weight~$\lambda$ on $F^n$.
By the dimensional reasons and the Main Theorem,
$R^k(R+\lambda\id)^{n-k} = 0$ for some $0\leq k\leq n$.
On the other hand, define $R$ on $F^n$ as follows:
$R(e_i) = e_{i+1}+\ldots+e_n$.
It is the RB-operator of weight~1 on~$F^n$.
Moreover, $R^n = 0$ and $R^{n-1}\neq0$. \hfill $\square$

\subsection{Matrix algebra}

Let us generalize the RB-operator (M1) on $M_2(F)$ from~\cite[Thm. 4.13]{Unital}
and the RB-operator 1-I on $M_3(F)$ from~\cite[Thm. 3]{GonGub}.

{\bf Example 1}.
A linear map $R$ defined on $M_n(F)$ as follows,
$$
R(e_{ij})
 = \begin{cases}
  \sum\limits_{k\geq1}e_{i+k,j+k}, & i\geq j, \\
- \sum\limits_{k\geq0}e_{i-k,j-k}, & i<j,
   \end{cases}
$$
is an RB-operator of weight~1 on $M_n(F)$.
Moreover, $R^n(R+\id)^{n-1} = 0$ but we have
$(R(R+\id))^{n-1}\neq0$ and $R^n(R+\id)^{n-2}\neq0$.

The following theorem was stated in~\cite{Unital} in the case of $\lambda = 0$.

{\bf Theorem 4}.
Let $F$ be a field of characteristic zero. Then
$\rb_\lambda(M_n(F)) = 2n-1$ for all $\lambda\in F$.

{\sc Proof}.
For $\lambda = 0$, it was proved in~\cite{Unital}.

Let $\lambda\neq0$, we may assume that $\lambda = 1$.
By Example~1, $\rb_\lambda(M_n(F)) \geq 2n-1$.

Let $R$ be an RB-operator of weight~$-1$ on $M_n(F)$.
By the Main Theorem, $\Spec(R)\subset\{0,1\}$.
So, we extend the action of $R$ to the algebra $M_n(\bar{F})$.
We want to show that either $R^n(R-\id)^{n-1} = 0$ or $R^{n-1}(R-\id)^n = 0$.
Thus, it is enough to state this for $R$ over $\bar{F}$.

Define $a = R(1)$.
If the degree of the minimal polynomial $m_t(a)$ of~$a$ is less than~$n$, then
by Theorem~1, $(R(R-\id))^{n-1} = 0$, and we are done.
Suppose that $\deg(m_t(a)) = n$ and $m_t(a)$ coincides with
the characteristic polynomial of $a$.
By Corollary~1, $a$ is conjugate to a diagonal matrix
$D = \diag\{-r,-r+1,\ldots,s-2,s-1\}$ with $r + s = n$.
By Lemma~1 from \cite{GonGub}, all subspaces
$V_q = \Span\{e_{ij}\mid i-j = q\}$ are $R$-invariant.
Since all $V_q$ for $q\neq0$ have dimension less or equal to $n-1$,
we have $(R(R-\id))^{n-1} = 0$ on each of such $V_q$.
Further, $V_0$ which is the subalgebra of diagonal matrices from $M_n(\bar{F})$
has the dimension equal to $n$. So, either $R^n = 0$ or $(R-\id)^n$ or
$(R(R-\id))^{n-1} = 0$ on~$V_0$. In any case, $\rb_\lambda(M_n(\bar{F})) = 2n-1$
as well as of $M_n(F)$. \hfill $\square$

Recall that $t_n(F)$ denotes the subalgebra of not strictly upper-triangular matrices from $M_n(F)$.

{\bf Corollary 5}.
Let $F$ be a field of characteristic zero and let $\lambda\in F$ be nonzero.
Then $\rb_\lambda(t_n(F)) = 2n-1$.

{\sc Proof}.
For simplicity, let $\lambda = 1$.
The restriction of the RB-operator $R$ from Example~1 on $t_n(F)$ implies
$\rb_\lambda(t_n(F)) \geq 2n-1$.
We may assume that $F$ is algebraically closed, since
$\rb_\lambda(t_n(F))\leq \rb_\lambda(t_n(\bar{F}))$.
If the degree of the minimal polynomial $m_x(a)$ of $a = R(1)$
is less than $n$, we are done by Theorem~1.
Suppose that $\deg(m_x(a)) = n$.

Find $\psi\in\Aut(M_n(F))$ such that
$R^{(\psi)}(1)$ equals to the $D = \diag\{-r,-r+1,\ldots,s-2,s-1\}$ with $r + s = n$.
Now, the RB-operator $P = R^{(\psi)}$ acts on the subalgebra $\psi(t_n(F))$.
Define $V_q = \Span\{e_{ij}\mid i-j = q\}$.
Applying the proof of Lemma~1 from \cite{GonGub},
we conclude that a subspace $V_q\cap \psi(t_n(F))$ is $P$-invariant for every~$q$.
It remains to repeat the proof of Theorem~4. \hfill $\square$

{\bf Conjecture 1}.
Let $F$ be a field of characteristic zero. Then $\rb_0(t_n(F)) = n$.

Let us extend Conjecture 5.24 from~\cite{Unital} on the case of nonzero weight.

{\bf Conjecture 2}.
Let $A = \bigoplus\limits_{i=1}^k M_{n_i}(F)$
be a semisimple finite-dimensional associative algebra
over a field $F$ of characteristic zero. Then

a) $\rb_0(A) = 2\max\limits_{i=1,\ldots,k}\{n_i\} - 1$.

b) $\rb_\lambda(A) = \sum\limits_{i=1}^k n_i + \max\limits_{i=1,\ldots,k}\{n_i\} - 1$ for nonzero~$\lambda$.

The following example shows that there exists an RB-operator reaching the bound from Conjecture~2b.

Let $D_n$ denote the subalgebra of all diagonal matrices in $M_n(F)$
and $L_n$ ($U_n$) the set of all strictly lower (upper) triangular
matrices in $M_n(F)$.

{\bf Example 2}.
We define a triangular-splitting RB-operator $R$ of nonzero weight~$\lambda$ on
the algebra $A = \bigoplus\limits_{i=1}^k M_{n_i}(F)$ with the subalgebras
$A_0 = D$, $A_- = U$, $A_+ = L$, where
$$
D = \bigoplus\limits_{i=1}^k D_{n_i},\quad
L = \bigoplus\limits_{i=1}^k L_{n_i},\quad
U = \bigoplus\limits_{i=1}^k U_{n_i},
$$
and $R$ is defined on $D \cong F^N$ for $N = \sum\limits_{i=1}^k n_i$
in the same way as the RB-operator from Example~1.
So, $x^N(x+\lambda)^{n_0-1} = 0$ is the minimal polynomial for~$R$.
Here $n_0 = \max\limits_{i=1,\ldots,k}\{n_i\}$.

\subsection{Simple Jordan algebra of a bilinear form}
Let $J = J_{n+1}(f) = F1\oplus V$ be a direct vector-space sum of $F$
and finite-dimensional vector space~$V$, $\dim V = n > 1$,
and $f$ be a nondegenerate symmetric bilinear form on~$V$.
Under the product
\begin{equation}\label{FormProduct}
(\alpha\cdot1 + a)(\beta\cdot1 + b)
 = (\alpha\beta+f(a,b))\cdot1 + (\alpha b +\beta a),\quad
\alpha,\beta\in F,\ a,b\in V,
\end{equation}
the space $J$ is a simple Jordan algebra \cite{Nearly}.
We assume that $\charr F\neq2,3$.

Let us choose a basis $e_1$, $e_2$, \ldots, $e_n$ of $V$ such that
the matrix of the form $f$ in this basis is diagonal with
elements $d_1,d_2,\ldots,d_n$ on the main diagonal.
As $f$ is nondegenerate, $d_i\neq0$ for each $i$.

{\bf Example 3}~\cite{BGP}.
Let $J_{2n}(f)$, $n\geq2$, be the simple Jordan algebra of bilinear form~$f$
over an algebraically closed field~$F$ with $\charr(F)\neq2,3$.
Let $R$ be a linear operator on $J_{2n}(f)$
defined by a matrix $(r_{ij})_{i,j=0}^{2n-1}$
in the basis $1,e_1,e_2,\ldots,e_n$
with the following nonzero entries
\begin{gather*}
r_{00} = -3,\quad r_{01} = \sqrt{d_1},\quad r_{10}
 = -\frac{1}{\sqrt{d_1}},\quad r_{jj} = -1,\ j=1,\ldots,2n-1, \\
r_{i\,i+1} = \frac{d_{i+1}}{d_i}\sqrt{ -\frac{d_i}{d_{i+1}} }, \quad
r_{i+1\,i} = - \sqrt{ -\frac{d_i}{d_{i+1}} },\ i=2,\ldots,2n-2.
\end{gather*}
Then $R$ is an RB-operator of weight~$2$ on $J_{2n}(f)$.
Moreover, $R(R+2\id)^2 = 0$ but $R(R+2\id)\neq0$ and $(R+2\id)^2 \neq 0$.

{\bf Proposition 2}.
Let $\lambda\in F$ be nonzero and let $F$ be an algebraically closed field
with $\charr F\neq2,3$. Then
$\rb_\lambda(J_n(F))
 = \begin{cases}
2, & n \mbox{ is odd}, \\
3, & n \mbox{ is even}.
\end{cases}$

{\sc Proof}.
When $n$ is odd, the statement follows from \cite[Thm.\,4.1]{BGP}.

Suppose that $n$ is even. By Example~3, we have $\rb_\lambda(J_n(F))\geq3$.
Since $F$ is algebraically closed, we may assume that all $d_i$ equal to 1.
Let $R$ be an RB-operator of weight~$\lambda$ on $J_n(F)$.
Assume that $R$ is defined by a matrix $(r_{ij})_{i,j=0}^n$
in the basis $1,e_1,e_2,\ldots,e_n$.
Due to the proof of Theorem~4.1 from~\cite{BGP}, we have
the following relations
\begin{gather*}
r_{ii} = -\frac{\lambda}{2},\quad
r_{kl} = - r_{lk},\quad
r_{0i} = zr_{i0},\\
\sum\limits_{p=1}^n \bar{r}_{p0}^2 = \frac{\lambda^2}{4},\quad
\sum\limits_{p=1}^n \bar{r}_{pi}^2 = \bar{r}_{0i}^2,\quad
\sum\limits_{p=1}^n \bar{r}_{pk}\bar{r}_{pl}
 = \bar{r}_{0k}\bar{r}_{0l},\quad
\sum\limits_{p=1}^n \bar{r}_{pi}\bar{r}_{p0}
 = -\frac{\lambda z}{2}\bar{r}_{0i},
\end{gather*}
where $i,k,l>0$ and $k\neq l$,
and we have three subcases:

(I) $z = 1$, $\bar{r}_{00} = -\lambda/2$,

(II) $z = -1$, $\bar{r}_{00} = \lambda/2$,

(III) $z = -1$, $\bar{r}_{00} = -3\lambda/2$.

It is easy to check that an RB-operator $R$ from (I) is splitting.
Further, $\phi$ maps RB-operators from (II) to (III) and vice versa.
Finally, an RB-operator from (III) satisfies the equation
$R(R+\lambda\id)^2 = 0$. \hfill $\square$

\subsection{Matrix Cayley---Dickson algebra}

Cayley---Dickson algebras are known to be only simple nonassociative alternative
finite-dimensional algebras, they are either division algebras or so called split ones.
The first ones have only trivial RB-operators~\cite{BGP}.
Let us study the $\rb(\lambda)$-index of the matrix (split)
Cayley---Dickson algebra $C(F)$, which is simple alternative 8-dimensional algebra.

Recall the definition of the product on $C(F) = M_2(F)\oplus v M_2(F)$.
We extend the product from
$M_2(F)$ on $C(F)$ as follows:
\begin{equation}\label{CD-product}
a(vb) = v(\bar{a}b),\quad (vb)a = v(ab),\quad
(va)(vb) = b\bar{a},\quad a,b\in M_2(F).
\end{equation}
Here\, $\bar{}$\, denotes the involution on $M_2(F)$
which maps a matrix $a = (a_{ij})\in M_2(F)$ to the matrix
$\begin{pmatrix}
a_{22} & - a_{12} \\
-a_{21} & a_{11} \end{pmatrix}$.

Given an algebra $A$ with the product $\cdot$, define the operations $\circ$, $[,]$ on
the space $A$:
$$
a\circ b=a\cdot b+b\cdot a,\quad [a,b]=a\cdot b-b\cdot a.
$$
We denote the space $A$ endowed with $\circ$ as $A^{(+)}$
and the space $A$ endowed with $[,]$ as~$A^{(-)}$.

{\bf Lemma 12}.
Let $A$ be an algebra. Then

(a)~\cite{BGP}
If $R$ is an RB-operator of weight~$\lambda$ on~$A$,
then $R$ is an RB-operator of weight~$\lambda$ on $A^{(\pm)}$.

(b) $\rb_\lambda(A)\leq \rb_\lambda(A^{\pm})$.

{\sc Proof}.
(b) It follows from the definition of $\rb(\lambda)$-index of an algebra and (a).
\hfill $\square$

{\bf Proposition 3}.
Let $F$ be an algebraically closed field with $\charr(F)\neq2,3$.
Then $\rb_\lambda(C(F)) = 3$ for all $\lambda\in F$.

{\sc Proof}.
When $\lambda = 0$, it was proved in \cite[Thm.\,5.26]{Unital}.

Let $\lambda\neq0$. It is known~\cite[p.\,57]{Nearly} that
$C(F)^{(+)}$ is isomorphic to the simple Jordan algebra of
a~bilinear form. By Proposition~2 and Lemma~12b, $\rb_\lambda(C(F))\leq3$.

Consider a triangular-splitting RB-operator $R$ of weight~1 on $C(F)$
with subalgebras $A_0 = \Span\{e_{11},e_{22},ve_{11},ve_{22}\}$,
$A_+ = \Span\{e_{12},ve_{12}\}$, and $A_- = \Span\{e_{21},ve_{21}\}$.
Further, $R$ is defined on $A_0$ again as a triangular-splitting
RB-operator with subalgebras
$B_0 = \Span\{e_{11},e_{22}\}$,
$B_+ = \Span\{ve_{11}\}$, and
$B_- = \Span\{ve_{22}\}$. Finally, $R$ is defined on $B_0\cong F\oplus F$
as follows: $R(e_{11}) = e_{22}$, $R(e_{22}) = 0$.
Thus, $R^2(R+\id) = 0$ on the entire $C(F)$
but $R^2\neq0$ and $R(R+\id)\neq0$.
So, $\rb_\lambda(C(F)) = 3$. \hfill $\square$

\section{Applications}

\subsection{Rota---Baxter modules}

Let $A$ be an algebra and let $R$ be an RB-operator $R$ of weight~$\lambda$ on $A$.
An $A$-bimo\-dule~$M$ with a linear map $p\colon M\to M$ is called
a~Rota---Baxter bimodule over $(A,R)$ and simply an $(A,R)$-bimodule~\cite{RBModules},
if the following equalities hold for all $x\in A$ and for all $m\in M$,
\begin{equation}\label{RBModule}
\begin{gathered}
R(x)p(m) =  p( R(x)m + xp(m) + \lambda xm ),\\
p(m)R(x) =  p( mR(x) + p(m)x + \lambda mx ).
\end{gathered}
\end{equation}
We call an $(A,R)$-bimodule $(M,p)$ a unital one, if
$M$ is a unital $A$-bimodule.

The relations~\eqref{RBModule} are equivalent to the condition that
$R+p$ is an RB-operator of weight~$\lambda$ on $A\ltimes M$.

{\bf Proposition 4}.
Let $A$ be a unital algebraic algebra over a field $F$ of characteristic zero
and let $R$ be an RB-operator of weight~$\lambda$ on $A$.
Also, $(M,p)$ is a unital $(A,R)$-bimo\-dule.
Then

a) If $\lambda = 0$, then $p$ is nilpotent on $M$.

b) If $\lambda\neq0$, then there exist $s,t$ such that
$p^s(p+\lambda\id)^t = 0$ on $M$. So, $M = M_0\oplus M_1$,
where $M_0 = \ker(p^s)$ and $M_1 = \ker(p+\lambda\id)^t$.
For $k,l$ such that $R^k(R+\lambda\id)^l = 0$ define
$A_0 = \ker(R^k)$ and $A_1 = \ker(R+\lambda\id)^l$.
We get that $(M_i,p)$ is a (not necessary unital) $(A_i,R)$-bimodule for $i=0,1$.

c) $\rb_\lambda(A\ltimes M)\leq 2\rb_\lambda(A)$ for all $\lambda\in F$.

{\sc Proof}.
a) By Theorem~1a, $R+p$ is nilpotent on $A\ltimes M$, so $p$ is nilpotent on $M$.

b) By Theorem~1a, there exist $n,m$ such that $(R+p)^n (R+p+\lambda\id)^m = 0$ on $A$.
So, $p^n(p+\lambda\id)^m = 0$ on $M$, and
$M = M_0\oplus M_1$ for $M_0 = \ker(p^s)$ and $M_1 = \ker((p+\lambda\id)^t )$.
Also, $\ker(R+p)^n$ and $\ker(R+p+\lambda\id)^m$ are subalgebras of $A\ltimes M$.
It implies that $(M_0,p)$ is an $(A_0,R)$-bimodule and $(M_1,p)$ is an $(A_1,R)$-bimodule.

c) Suppose that $\rb_\lambda(A)<\infty$.
Then there exists a minimal $n$ such that for every RB-operator
$R$ of weight~$\lambda$ on $A$ one can find
$0\leq k\leq n$ satisfying the equality
$R^k(R+\lambda\id)^{n-k}(1) = 0$.
On the one hand, $n\leq\rb_\lambda(A)$ by the definition of the $\RB(\lambda)$-index.
On the other hand, since
$(R+p)^k(R+\lambda\id)^{n-k}(1) = R^k(R+\lambda\id)^{n-k}(1) = 0$,
we get $\rb_\lambda(A)\leq 2n$ by Theorem~1b.
\hfill $\square$

{\bf Remark 2}.
The same result holds if $A$ is a unital finite-dimensional
algebra over any field (not necessary of characterstic zero).

\subsection{Parameterized Rota---Baxter operators}

The study of the parameterized associative Yang---Baxter equation
was initiated by A.~Polishchuk~\cite{Polishchuk} in 2002.
Further, the definition of parameterized associative Yang---Baxter equation
of weight $\lambda$ appeared~\cite{FardThesis,Anghel}.
Given an associative algebra~$A$, a linear map
$r\colon F\to A\otimes A$ is called a solution of
the parameterized associative Yang---Baxter equation
of weight~$\lambda$ if the equation
\begin{equation}\label{AYBE}
r_{13}(z_{13})r_{12}(z_{12})
 - r_{12}(z_{12})r_{23}(z_{23})
 + r_{23}(z_{23})r_{13}(z_{13})
 + \lambda r_{13}(z_{13}) = 0
\end{equation}
holds for all $z_1,z_2,z_3\in F$.
Here $z_{ij} = z_i-z_j$ and given a decomposition
$r(z) = \sum\limits_i a_i\otimes b_i$,
$$
r_{12} = \sum a_i\otimes b_i\otimes 1,\quad
r_{13} = \sum a_i\otimes 1\otimes b_i,\quad
r_{23} = \sum 1\otimes a_i\otimes b_i.
$$

It is known~\cite{FardThesis} that a solution $r$ of~\eqref{AYBE}
gives rise to a parameterized Rota---Baxter operator $R$ of weight~$\lambda$ on $A$.
More precisely, a linear map
$R\colon F\otimes A\to A$ is called parameterized Rota---Baxter operator $R$
of weight~$\lambda$ on $A$ if $R$ satisfies the relation
\begin{equation}\label{RBparam}
R_b(x)R_c(y) = R_c( R_{b-c}(x)y ) + R_b( xR_{c-b}(y) )  + \lambda R_b(xy),
\end{equation}
where $b,c\in F$ and $x,y\in A$.
By $R_b(x)$ we denote the element $R(b,x)\in A$.

So, a parameterized RB-operator $R$ on $A$ is a family of linear operators
$R_b\colon A\to A$, $b\in F$.
Note that $R_0$ is a usual RB-operator of weight~$\lambda$ on $A$.

Let us apply the results of Theorem~1 to get some information about operators $R_b$.

{\bf Proposition 5}.
Let $A$ be a unital algebraic associative algebra over a field of characteristic zero.
Let $R$ be a parameterized RB-operator of weight~$\lambda$ on $A$. Then

a) If $\lambda = 0$, then there exists $n$ such that
$(R_b(1))^n = 0$ for all $b\in F$,

b) If $\lambda  = -1$, then there exist $n,m$
such that $H_{n,m}(R_b(1)) = 0$ for all $b\in F$.

{\sc Proof}.
a) By Theorem~1a, there exists $n$ such that $R^n_0(1) = 0$.
By~\eqref{RBparam}, we get $(R_b(1))^{n+1} = R_b(R_0^n(1)) = 0$.

b) Denote $a = R_0(1)$ and $c = R_b(1)$.
Note that by~\eqref{RBparam},
$$
R_b(x)R_b(y) = R_b(R_0(x)y + xR_0(y) - xy).
$$
So, the right-hand side has the form
$R_b(Z)$, where $Z$ is a linear combination of the products
involving only the action of $R_0$.
The same holds for the product of every number of elements $R_b(x_i)$.
This fact allows us analogously to the formula
$$
(R_0(1))^n
 = a^n
 = R_0\left( \sum\limits_{k=1}^n k! S(n,k)(-1)^{n-k} R_0^{k-1}(1) \right),
$$
from~\cite[\S3.3.2]{GuoMonograph} to state the fomula
$$
(R_b(1))^n
 = c^n
 = R_b\left( \sum\limits_{k=1}^n k! S(n,k)(-1)^{n-k} R_0^{k-1}(1) \right).
$$
Here $S(n,k)$ denotes a Stirling number of the second kind.

So, if we have an equility
$g(a) = R_0(q(a))$ for some polynomials $g,q\in F[x]$,
then we may derive the equality $g(c) = R_b(q(a))$.
In particular, applying Lemma~10, we get
$$
\frac{H_{r,s+1}(a) }{r+s+1}
 = R_0( H_{r,s}(a) ),\quad
\frac{H_{r,s+1}(c) }{r+s+1}
 = R_b( H_{r,s}(a) ).
$$
By Theorem~1a and Lemma~11, there exist $r,s$ such that $H_{r,s}(a) = 0$.
Thus, $H_{r,s+1}(c) = 0$, we are done.
\hfill $\square$

Suppose that $R$ is a parameterized RB-operator of weight~$\lambda$ on $M_n(\mathbb{C})$.
So, Proposition~5 says that if $\lambda = 0$, then
matrices $R_b(1)$ are nilpotent for all $b\in F$.
If $\lambda \neq0$, then matrices $R_b(1)$ are diagonalizable for all $b\in F$.

\subsection{Derived Rota---Baxter algebras}

Let us generalize the following construction already known
for different varieties of algebras, see~\cite{GuoMonograph,preLie,Burde}.

Let $A$ be an algebra of a variety $\Var$ and let $R$ be an RB-operator of weight~$\lambda$ on $A$.
Denote by $V$ the vector space $A$.
We define by induction the algebra $A_i$, $i\geq0$, on $V$ as follows
\begin{equation}\label{DerivedProduct}
\begin{gathered}
x\circ_0 y = xy,\\
x\circ_{i+1} y = R(x)\circ_i y + x\circ_i R(y) + \lambda x\circ_i y.
\end{gathered}
\end{equation}
It is known~\cite{BBGN2011,Embedding} that $R$ is an RB-operator
of weight~$\lambda$ on all $A_i$ and all algebras $A_i$ belong to the same variety $\Var$.
By the definition, $R$ and $-(R+\lambda\id)$ are homomorphisms from $A_n$ to $A_{n-1}$,
and we get compositions of such homomorphisms,
$$
\ldots \xrightarrow[-R-\lambda \id]{R} A_{i+1} \xrightarrow[-R-\lambda \id]{R} A_i \xrightarrow[-R-\lambda \id]{R} A_{i-1}
 \xrightarrow[-R-\lambda \id]{R} \cdots  \xrightarrow[-R-\lambda \id]{R} A_0.
$$

Let us call $A_i$ as the $i$th derived Rota---Baxter algebra of the RB-algebra $A$.
If there exists a~minimal natural number $m$ such that $A_n\cong A_m$ for all $n>m$,
then we define the limiting derived algebra $A_\infty$ as $A_m$.
Otherwise, $A_\infty$ is not defined.

The following result generalizes Corollary~4.10 from~\cite{preLie} in the case of weight zero.

{\bf Proposition 6}.
Let $A$ be a unital algebraic algebra over a field $F$ of characteristic zero.
Let $R$ be an RB-operator of weight~$\lambda$ on $A$. Then $A_\infty$ is well-defined.

a) If $\lambda = 0$, then $A_\infty$ has trivial product.

b) If $\lambda\neq0$, then
$A_\infty \cong \ker(R^n)\oplus \ker(R+\lambda\id)^n$ for sufficiently big $n$.

{\bf Proof}.
a) By Theorem~1, we know that $R^n = 0$ for some natural $n$.
By~\eqref{DerivedProduct}, we get
\begin{equation}\label{DerivedProductZero}
x\circ_k y = \sum\limits_{i=0}^k \binom{k}{i}R^i(x)R^{k-i}(y).
\end{equation}
Thus, $x\circ_i y = 0$ for all $i\geq2n-1$.

b) By Theorem~1, there exists $n$ such that $R^n(R+\lambda\id)^n = 0$.
So, $\ker(R^n)$ and $\ker(R+\lambda\id)^n$ are ideals in $A_t$ for all $t\geq n$
as kernels of the homomorphisms.
So, $A_t = \ker(R^n)\oplus \ker(R+\lambda\id)^n$ for all $t\geq n$.
It remains to note that $R$ is invertible on $\ker(R+\lambda\id)^n$
as well as $R+\lambda\id$ on $\ker(R^n)$.
Thus,
$$
\langle \ker(R^n),\circ_s\rangle \cong \langle \ker(R^n),\circ_n\rangle,\quad
\langle \ker(R+\lambda\id)^n,\circ_s\rangle \cong \langle \ker(R+\lambda\id)^n,\circ_n\rangle
$$
for all $s\geq0$. It implies that $A_t\cong A_n$ for all $t\geq n$.
\hfill $\square$

{\bf Remark 3}.
The same result holds if $A$ is a unital finite-dimensional
algebra over any field (not necessary of characterstic zero).

{\bf Question}.
Let $A = M_n(F)$. What is a minimal $m$ such that $A_m$
has trivial product for every RB-operator $R$ of weight zero on $A$?

\section*{Acknowledgements}

Author is grateful to the participants of the scientific seminar
in Altai State Pedagogical University (Barnaul)
for the helpful discussion.
Author is thankful to P. Kolesnikov for the helpful remarks.

Author is supported by the Program of fundamental scientific researches
of the Siberian Branch of Russian Academy of Sciences, I.1.1, project 0314-2019-0001.


text:
\begin{document}
\title{GCD matrices, posets, and nonintersecting paths
}

\author{Ercan Altinisik}
\address{Department of Mathematics, Faculty of Arts and Sciences, Gazi
  University, 06500 Teknikokullar - Ankara, TURKEY}
\email{ealtinisik@gazi.edu.tr}

\author{Bruce E. Sagan}
\address{Department of Mathematics, Michigan State University,
East Lansing, MI 48824-1027, USA}
\email{sagan@math.msu.edu}

\author{Naim Tuglu}
\address{Department of Mathematics, Faculty of Arts and Sciences, Gazi
  University, 06500 Teknikokullar - Ankara, TURKEY}
\email{naimtuglu@gazi.edu.tr}

\date{\today}
\keywords{determinant, GCD matrix, nonintersecting paths, poset,
}
\subjclass{ Primary 11C20; Secondary 05E99, 11A25, 15A36
}

	\begin{abstract}
We show that with any finite partially ordered set, $P$, one can
associate a matrix whose determinant factors nicely.  As corollaries,
we obtain a number of results in the literature about GCD matrices
and their relatives.  Our main theorem is proved combinatorially using
nonintersecting paths in a directed graph.
	\end{abstract}

\maketitle

\section{Introduction}
\label{i}

Let $\bbP$ denote the positive integers and suppose we are given a
subset $S=\{a_1,\ldots,a_n\}$ of $\bbP$.  The corresponding {\it GCD
matrix} is $(S)=(s_{ij})$ where $s_{ij}=(a_i,a_j)$, the greatest
common divisor of $a_i$
and $a_j$.  We say that $S$ is {\it factor closed\/} if given any
$a_i\in S$ and any divisor $d|a_i$ then $d\in S$.
H. J. S. Smith~\cite{smi:vca}
proved the following beautiful result about the determinant of $(S)$.
\bth[Smith]
\label{smith}
If $S=\{a_1,\ldots,a_n\}$ is factor closed then
\beq
\label{(S)}
\det (S)=\phi(a_1)\cdots\phi(a_n)
\eeq
where $\phi$ is Euler's totient function.\qed
\eth

Since Smith's pioneering paper, a host of related results have
appeared in the literature.  For a survey with references, see the
paper of Haukkanen, Wang, and Sillanp\"a\"a~\cite{hwp:sd}.  We will
show that many of these are special cases of a general determinantal
identity associated with any finite partially ordered set $P$ (see
Theorem~\ref{main} below).  This result can be proved by factoring the matrix
constructed from $P$, but we choose to give a combinatorial proof
based on counting families of nonintersecting paths in acyclic digraphs.  This
technique is due to Lindstr\"om~\cite{lin:vri} and independently
Karlin~\cite{kar:cpa}. It was later rediscovered and popularized by
Gessel and Viennot~\cite{gv:bdp,gv:dpp}.  We will use the approach in
Stembridge's paper~\cite{ste:npp}.

The rest of this article is structured as follows.  The next section
will be devoted to  proving our main theorem and giving some
preliminary applications.  In Section~\ref{ms}, we specialize to the
case where $P$ is a meet semilattice.  This permits us to invert the
sums appearing in the general case.  Section~\ref{lcmcs} investigates
what happens to the determinant if the set $S$ is lower closed (the
partially ordered set generalization of being factor closed) or meet closed.
We end with a section containing comments and an open question.

\section{The main theorem}
\label{mt}

We will first review the digraph machinery which we will need to prove
our main theorem.  Explanations of any undefined terms from graph
theory can be
found in the texts of Harary~\cite{har:gt} or Chartrand and
Lesniak~\cite{cl:gd}.  For definitions of concepts about
posets (partially ordered sets), the reader can consult Stanley's
book~\cite{sta:ec1}.

Let $D$ be a finite acyclic digraph with vertices $V$ and arcs $A$.
Let $R$ be a commutative ring with identity and suppose we are given a  function (weighting)
$\om:A\ra R$.
Then any directed path $p:\ v_0v_1\ldots v_k$ has an associated weight
$$
\om(p)=\prod_{i=1}^k \om(v_{i-1}v_i).
$$
We also let
$$
\om(v_0,v_k)=\sum_p \om(p)
$$
where the sum is over all directed paths $p$ from $v_0$ to $v_k$.  Note
that this sum is finite because $D$ is finite and acyclic.

Now suppose we are given two disjoint sets of vertices
$V'=\{v_1',\ldots,v_n'\}$ and $V''=\{v_1'',\ldots,v_n''\}$ in $V$.  Consider
an $n$-tuple of directed paths $\pi=(p_1,\ldots,p_n)$ where $p_i$ goes
from $v_i'$ to $v_i''$ for all $i$, $1\le i\le n$.  Then we assign
weights to $\pi$ and to the pair $(V',V'')$ in a way analogous to the
one used in the preceding paragraph
$$
\om(\pi)=\prod_{i=1}^n \om(p_i)\qmq{and} \om(V',V'')=\sum_{\pi}\om(\pi)
$$
where the sum is over all $\pi$ where no two of the paths in the
$n$-tuple intersect.

Finally, given a permutation $g$ in the symmetric group $S_n$, we
define $\pi_g=(p_1,\ldots,p_n)$ to be an $n$-tuple of directed paths
such that $p_i$ goes from $v_i'$ to $v_{g(i)}''$ for all $i$.  So the
$n$-tuples considered in the previous paragraph would have $g=e$
where $e$  is the identity permutation.  If any pair of
paths in $\pi_g$ intersect then we call the $n$-tuple {\it
intersecting\/}  and {\it nonintersecting\/} otherwise.
We can now state Stembridge's theorem~\cite{ste:npp} about enumerating
nonintersecting paths.

\bth [Stembridge]
\label{paths}
Let $D$ be a finite acyclic digraph with disjoint vertex sets
$V'=\{v_1',\ldots,v_n'\}$ and $V''=\{v_1'',\ldots,v_n''\}$ such that
any $\pi_g$ is intersecting if $g\neq e$.   Let
$(D)=(d_{ij})$ be the matrix with
$$
d_{ij}=\om(v_i',v_j'').
$$
Then
$$
\det (D)=\om(V',V'').\qed
$$
\eth

Now let $P$ be a finite poset and consider the {\it incidence
algebra\/} $I(P,R)$ of $P$ over $R$ which consists of all functions
$F:P\times P\ra R$ such that $F(a,b)=0$ unless $a\le b$.  The identity
element of $I(P,R)$ is the Kronecker delta function
$$
\de(a,b)=\case{1}{if $a=b$,}{0}{otherwise.}
$$
We will also need the {\it zeta function\/} of $I(P,R)$ defined by
$$
\zeta(a,b)=\case{1}{if $a\le b$,}{0}{otherwise.}
$$
The zeta function is invertible and its inverse is called the M\"obius
function $\mu$ of $I(P,R)$.  In other words, $\mu$ is the unique
function in $I(P,R)$ satisfying
$$
\sum_{a\le c\le b} \mu(a,c)=\sum_{a\le c\le b} \mu(c,b)=\de(a,b).
$$

Now fix a linear ordering of $P$ which can be used to index the rows and
columns of a matrix $M$.  Since we will be taking determinants, it will
not matter which linear order is used and we will merely say that the
matrix is indexed by $P$.
We can now state and prove our main theorem.
\bth
\label{main}
Let $P$ be a finite poset and let $F,G\in I(P,R)$.
Let $(P)_{FG}$ be the  matrix indexed by $P$ with entries
\beq
\label{pab}
p_{ab}=\sum_{c\in P} F(c,a)G(c,b).
\eeq
Then
$$
\det (P)_{FG}=\prod_{a\in P} F(a,a)G(a,a).
$$
\eth
\proof
Construct a digraph $D$ as follows.  For the vertices of $D$, take
three copies of the elements of $P$ which we denote by $P'$, $P''$,
and $P'''$.
Now put an arc from $a'\in P'$  to $c'''\in P'''$ if and only if $a\ge c$ in
$P$.  Dually, put an arc from $c'''\in P'''$ to $b''\in P''$ if and only if
$c\le b$.  Finally, give weights to the arcs by
$$
\om(a',c''')=F(c,a) \qmq{and} \om(c''',b'')=G(c,b).
$$

Consider paths $p$ from $a'\in P'$ to $b''\in P''$.  Clearly all such
paths have the form $p: a',c''',b''$ where $c\le a$ and $c\le b$ in
$P$.  So if we take $V'=P'$ and $V''=P''$ then we have $(D)=(P)_{FG}$ since
$$
d_{a'b''}=\om(a',b'')
=\sum_{c\le a,\ c\le b } F(c,a)G(c,b) =p_{ab}.

$$
The last step follows from the fact that a term in $p_{ab}$
corresponding to some $c$ not satisfying the given inequalities is zero.

We need to show that $D$ satisfies the hypotheses of
Theorem~\ref{paths}.  By construction, $D$ is acyclic.  Now
take any $n$-tuple of paths $\pi_g$ which is
nonintersecting.  We will show that this forces $g=e$.  First we claim
that the path starting at any $a'\in P'$ must then follow the arc to
$a'''\in P'''$.  Suppose this is not true for some $a'$ and let
$c'''$ be the next vertex on the path.  Then
we must have $a>c$ in $P$.  Now consider the path of $\pi_g$ starting
at $c'$.  Since it cannot intersect the previous path, it must go to
some $d'''$ with $c>d$.  Continuing in this fashion, we can construct
an infinite decreasing chain in $P$, contradicting the assumption that
$P$ is finite.  So our claim is true.  By a dual argument, one can show
that the path starting at $a'$ must continue from $a'''$ to $a''$ and
so $g=e$.  Furthermore, we have shown that this family $\pi_e$ is the only
nonintersecting path family.   So, by the way we have defined the weights and
Theorem~\ref{paths},
$$
\det (D) = \om(\pi_e)=\prod_{a\in P} F(a,a)G(a,a).\qed
$$

We should note that this theorem can also be proved
algebraically, either as a corollary to a result of
D. A. Smith~\cite[Corollary 2]{smi:bfa} about the incidence algebra of
$P$, or by directly factoring the matrix for $P$ as $(P)_{FG}=M_F^t M_G$ where
$M_F$ and $M_G$ are the matrices corresponding to the incidence
algebra elements $F$ and $G$, respectively, and $t$ denotes transpose.

\thicklines
\setlength{\unitlength}{1pt}
\bfi
\bpi(50,80)(0,-10)
\Gda \Gag \Ggg
\Gdaag  \Gdagg
\put(30,-10){\makebox(0,0){$a$}}
\put(0,70){\makebox(0,0){$b$}}
\put(60,70){\makebox(0,0){$c$}}
\epi
\capt{A poset $P$}\label{poset}
\efi

As an example of Theorem~\ref{main}, consider the poset $P$ whose
Hasse diagram is shown in Figure~\ref{poset}.  Then using the linear
order $a,b,c$ one obtains
$$
\barr{l}
\hs{-2pt}\det
  \left(
  \barr{ccc}
  \Faa G(a,a) &\Faa G(a,b)             &\Faa G(a,c)\\[5pt]
  \Fab G(a,a) &\Fab G(a,b)+\Fbb G(b,b) &\Fab G(a,c)\\[5pt]
  \Fac G(a,a) &\Fac G(a,b)             &\Fac G(a,c)+\Fcc G(c,c)
  \earr
  \right)\\[30pt]
\qquad=F(a,a)G(a,a)F(b,b)G(b,b)F(c,c)G(c,c)
\earr
$$

As a first application, we note a corollary of Theorem~\ref{main}
which simultaneously generalizes a theorem of Apostol~\cite{apo:apg}
and one of Daniloff~\cite{dan:ctf}.  Let $f,g$ be arbitrary functions
from the poset $P$ to the ring $R$.  Then on substituting $F(a,b)f(a)$
and $G(a,b)g(a)$ for $F(a,b)$ and $G(a,b)$, respectively, we
immediately have the following result.
\bco
\label{FfGg}
Let $P$ be a finite poset.  Let $(\Pb)$ be the matrix
indexed by $P$ with entries
$$
\pb_{ab}=\sum_{c\in P} F(c,a)f(c)G(c,b)g(c).
$$
Then
$$
\det (\Pb)=\prod_{a\in P} F(a,a)f(a)G(a,a)g(a).\qed
$$
\eco

Now consider the poset defined by using the divisor ordering on
$P_n\stackrel{\rm def}{=}\{1,2,\ldots,n\}$.
Apostol's theorem is obtained by specializing the previous
result to the case where $P=P_n$,
$F(a,b)=\zeta(a,b)$, $G(a,b)=G(b/a)$ for any function $G:P_n\ra R$, and
$g(a)=1$ for all $a\in P$.  In this case the determinant becomes
$$
\det (\Pb_n)=f(1)f(2)\cdots f(n) G(1)^n.
$$
With this evaluation in hand, Apostol showed that letting $f(a)=a$ for
$a\in P_n$  and $G(a,b)=\mu(b/a)$ where $\mu$ is the usual
number-theoretic M\"obius function gives
$$
\det (c(a,b)) = n!
$$
where $c(a,b)$ is Ramanujan's sum.

To obtain Daniloff's theorem, let $P=P_n$, $g(a)=1$ for all $a\in P_n$,
and $F(a,b)=G(a,b)=\Om_k(b/a)$  where
$$
\Om_k(a)=\case{a^{1/k}}{if $a^{1/k}\in \bbP$,}{0}{else.}
$$
Now Corollary~\ref{FfGg} becomes Daniloff's result,
$$
\det (\Pb_n)=f(1)f(2)\cdots f(n).
$$
Note that one obtains  the same evaluation not just for $\Om_k$, but also
for any functions $F,G:P\ra R$ such that $F(a)=G(a)=1$ for all $a\in P$.

\section{Meet semilattices}
\label{ms}

Now suppose that our poset is a meet semilattice $L$ so that every pair of
elements $a,b\in L$ have a greatest lower bound or {\it meet\/}
$a\mt b$.  Note that in this case the sum~\ree{pab} can be restricted
to $c\le a\mt b$.  This special case of Theorem~\ref{main} was
discovered by Haukkanen, Wang, and Sillanp\"a\"a~\cite{hwp:sd}.  They
also showed that by further specialization one could obtain a theorem
of Jager~\cite{jag:uai} concerning a unitary analogue of~\ree{(S)},
Smith's evaluation~\cite{smi:vca} of the LCM determinant, as well as
a number of other results.

It would be nice to be able to compute the value of
determinants where each entry is a single term, as in Smith's original
case, rather than a sum.
Since~\ree{pab} now consists of a sum over all $c$ below a certain
element in $L$, one can use M\"obius inversion to accomplish this.
Thus we can prove
the following theorem of Lindstr\"om~\cite{lin:ds} as a corollary to
Theorem~\ref{main}.
\bth[Lindstr\"om]
\label{main2}
Let $L$ be a finite meet semilattice and let $f\in I(L,R)$.
Let $(L)_f$ be the matrix indexed by $L$ with entries
$$
l_{ab}=f(a\mt b,a)
$$
Then
\beq
\label{Lf}
\det (L)_f=\prod_{a\in L} \left(\sum_{c\in L}\mu(c,a)f(c,a)\right).
\eeq
\eth
\proof
Define a function $F\in I(L,R)$ by
$$
F(a,b)=\sum_{c\le a} \mu(c,a)f(c,b)
$$
if $a\le b$ and $F(a,b)=0$ otherwise.  By M\"obius inversion, this is
equivalent to
$$
f(a,b)=\sum_{c\le a} F(c,b).
$$
It follows that
$$
l_{ab}=f(a\mt b,a)=\sum_{c\le a\mt b} F(c,a)
=\sum_{c\in L} F(c,a)\zeta(c,b).
$$
So applying Theorem~\ref{main} we obtain
$$
\det (L)_f=\prod_{a\in L} F(a,a)\zeta(a,a)=
\prod_{a\in L} \left(\sum_{c\in L}\mu(c,a)f(c,a)\right). \qed
$$

This result can also be found as an exercise in Stanley's
book~\cite[Chapter 3, Exercise 37]{sta:ec1}.  The special
case where $f(a,b)$ depends only on $a$ was proved independently by
Wilf~\cite{wil:hdm}.  Smith himself~\cite{smi:vca} noted that this
theorem holds under the further assumption that $L$ is a
factor closed subset of $\bbP$.

To see how Smith's determinant~\ree{(S)} follows from
Theorem~\ref{main2}, just let $f(a,b)=a$ for all $a\in S$.  Then
$s_{ij}=f(a_i\mt a_j, a_i)$ and, since $S$ is factor closed,
$$
\sum_{c\in S}\mu(c,a)f(c,a)=
\sum_{c|a}\mu(a/c)\ c =\phi(a).
$$

\section{Lower-closed and meet-closed sets}
\label{lcmcs}

In work on analogues of Theorem~\ref{smith}, there are two conditions
that are commonly imposed on the set $S$.  If $S\sbe P$ for some poset
$P$, we say that $S$ is {\it lower closed\/} or a {\it lower order
ideal\/} if $a\in S$ and $b\le a$ in $P$ implies $b\in S$.  This
corresponds exactly to $S$ being factor closed if it is a subset of
$\bbP$ ordered by division.  If, in particular, our poset is a
meet semilattice $L$ and $S$ is lower closed, then $S$ is also a meet
semilattice with the same M\"obius function as $L$.  So our previous
results cover this case without change.

If $S\sbe L$ for a meet semilattice one can also talk about $S$ being
{\it meet closed\/} which means that if $a,b\in S$ then $a\mt b\in S$
where the meet is taken in $L$.  Again, $S$ is also a meet semilattice
and so Theorem~\ref{main2} still applies and in fact generalizes a
result of Bhat~\cite{bha:gcd} who considered the case when $f$ is a
function of only one argument.

However, if $S\sbe L$ is meet closed we will show that there is also an
analogue of~\ree{Lf} which involves the M\"obius function of $L$,
rather than that of $S$.  This result will generalize a theorem of
Haukkanen~\cite{hau:mmp} who, as in Bhat's paper, only looked at
functions of a
single variable.  Specializing yet further to $L=P_n$ and the function
$f(a)=a$ for all $a\in P_n$, one obtains a
theorem of Beslin and Ligh~\cite{bl:ags}.

To state our result, it will be convenient to have some notation.
Let $\ell$ be a linear extension of the partial order on $S$
so that if  $\ell=a_1,a_2,\ldots,a_n$ then $a_i<a_j$ implies $i<j$.
If $d\in L$ then we will write
$d\lte a_i$ if $d\le a_i$ and $d\not\le a_j$ for any $j<i$.
\bth
Let $L$ be a finite meet semilattice and let $f\in I(L,R)$.  Suppose
$S\sbe L$ is meet closed and fix a linear extension
$\ell=a_1,a_2,\ldots,a_n$ of $S$.  Then
$$
\det(S)_f=\prod_{i=1}^n
\left(\sum_{d\lte a_i}\ \sum_{c\in L}\mu(c,d)f(c,a_i)\right)
$$
\eth
\proof
We will use unsubscripted variables for elements of $L$ which are not
necessarily in $S$.  Define $F\in I(L,R)$ as in the proof of
Theorem~\ref{main2} so that we have
\beq
\label{f}
f(a_i,a_j)=\sum_{d\le a_i} F(d,a_j).
\eeq
Also define $\Fh\in I(S,R)$ by
$$
\Fh(a_i,a_j)=\sum_{d\lte a_i}\ \sum_{c\in L}\mu(c,d) f(c,a_j)
=\sum_{d\lte a_i} F(d,a_j)
$$
if $a_i\le a_j$ and $\Fh(a_i,a_j)=0$  otherwise.  If we can show that
$$
f(a_i,a_j)=\sum_{a_k\le a_i} \Fh(a_k,a_j)
$$
then the rest of the proof will follow as in the demonstration of
Theorem~\ref{main2}.   So, by the definition of $\Fh$, it suffices to
show
\beq
\label{f2}
f(a_i,a_j)=\sum_{a_k\le a_i}\ \sum_{d\lte a_k} F(d,a_j).
\eeq
We will do this by showing that there is a one-to-one correspondence
between the terms in~\ree{f2} and those in~\ree{f}.

First note that each $d\in L$ occurs at most once in~\ree{f} and at
most once in~\ree{f2} (since $d\lte a_k$ for at most one $a_k\in S$).
If $d$ occurs in~\ree{f2} then $d\le a_k\le a_i$ and so $d$ occurs
in~\ree{f}.  Conversely, if $d$ occurs in~\ree{f} then $d\le a_i$ and
so we must have $d\lte a_k$ for some $a_k$ with $k\le i$.  But now
$d\le a_i\mt a_k = a_l$ for some $l$ since $S$ is meet closed.
Furthermore, $a_l\le a_k$ implies $l\le k$, and so $l=k$ since
$d\lte a_k$.  It follows that $d\lte a_k$ where $a_k=a_l\le a_i$.
Thus $d$ occurs in~\ree{f2} and we have finished the proof.\qed

We should note that Breslin and Ligh~\cite{bl:gcd} have derived a
formula for $(S)_f$ for any subset $S\sbs\bbP$ in terms of an arbitrary
lower-closed set containing $S$.  Since this identity expresses
$(S)_f$ as a product of two (not necessarily square) matrices, one can
appy the Cauchy-Binet formula to it, as done in Li's
paper~\cite{li:dgc}, to obtain $\det (S)_f$ as a sum of determinants.
This approach was also generalized to meet semi-lattices in an article
of Haukkanen~\cite{hau:mmp}.

\section{Comments and an open question}

Our results can also yield other
information about the matrices $(P)_{FG}$ and $(L)_f$.  The following
theorem is an example of this.  In it, we specialize the ring to be
the complex numbers $\bbC$ for simplicity, although more general rings
can be used.

\bth
Let $P$ be a finite poset and let $F,G\in I(P,\bbC)$.
\ben
\item  The matrix $(P)_{FG}$ is invertible if and only if
  $F(a,a),G(a,a)\neq0$ for all $a\in P$.
\item  The matrix $(P)_{FG}$ is positive definite if and only if
  $F(a,a)G(a,a)>0$ for all $a\in P$.
\een
\eth
\proof
Part (1) follows immediately from Theorem~\ref{main} and the fact that
a matrix over $\bbC$ is invertible if and only if its determinant is
nonzero.

For part (2), let the total order used to index the rows and columns
of $P$ be a linear extension of $P$.  Then each principal
submatrix of $P$ is indexed by a lower-closed subset $S$ of $P$.  It
follows that the submatrix indexed by $S$ is exactly $(S)_{FG}$.  But
now we are done by Theorem~\ref{main} again, since a matrix is
positive definite if and only if the principal subdeterminants are all
positive.  \qed

If $L$ is a finite lattice, then one can compute the
determinant of a matrix involving joins (least upper bounds) $a\jn b$
of elements $a,b\in L$ by working in the dual of $L$ so that joins
become meets.
For general results about join and meet matrices, see the paper of
Korkee and Haukkanen~\cite{kh:mjm}.

However, there are factorizations of such determinants which
we have been unable to obtain by our methods.  As an example, we
consider the matrix $T_n(q)$ of chromatic joins introduced by Birkhoff
and Lewis~\cite{bl:cp} to study the chromatic polynomial of planar maps.
To define this matrix, let $q$ be a formal parameter.
Let $\Pi_n$ denote the lattice of partitions of $\{1,2,\ldots,n\}$
ordered by refinement and let $NC_n$ denote the lattice of noncrossing
partitions of the same set.  (An excellent survey about noncrossing
partitions can be found in the article of Simion~\cite{sim:np}.)
Then Tutte~\cite{tut:ble} showed that the matrix of chromatic joins
could be defined as
$$
T_n(q)=\left(q^{\blk(a\jn_{\Pi_n} b)}\right)_{a,b\in NC_n}
$$
where $\blk(a)$ is the number of blocks (subsets) in the partition $a$.

Tutte~\cite{tut:mcj} also derived a product formula for the determinant
of $T_n(q)$ in terms of Beraha polynomials.  (See also
Dahab~\cite{dah:ble}.)  Letting $\fl{\cdot}$ denote the floor or round
down function, the {\it $n$th Beraha polynomial\/} is defined to be
$p_0(q)=0$ for $n=0$, and for $n\ge1$
$$
p_n(q) =\sum_{i=0}^{\fl{n/2}} (-1)^i{n-i-1\choose i}q^{\fl{n/2}-i}.
$$
Using the version of Tutte's formula in the paper of Copeland,
Schmidt, and Simion~\cite{css:ntd} gives
$$
\det T_n(q) = q^{{2n-1\choose n}}   \prod_{m=1}^{n-1}
\left[\frac{p_{m+2}(q)}{qp_m(q)}\right]^{\frac{m+1}{n}{2n\choose n-m-1}}.
$$
The same paper also contains a related determinant-product identity
which the authors
note that they were unable to prove using Lindstr\"om's Theorem
(Theorem~\ref{main2} above), although other proofs exist.  It would be
interesting to find a way to apply the machinery of this paper to
these identities.

\medskip

{\it Acknowledgement.}  We would like to thank Vic Reiner for helpful
discussions and, in particular, referring us to the paper of Copeland,
Schmidt, and Simion~\cite{css:ntd}.

\begin{\bib}{99}

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\end{document}

config: arxiv


TITLE: Universal property of the Grassmanian
QUESTION [1 upvotes]: Does anybody have a good reference on the Grassmanian and its universal property?
I am reading this paper on Quot schemes: https://arxiv.org/abs/math/0504590
Where the Grassmanian is constructed, but its representability and its universal quotient are "exercises". In particular exercise (2) in chapter 1.

REPLY [2 votes]: I suggest you:

Eisenbud, Harris - The Geometry of Schemes, (2000) Springer Verlag, paragraph III.2.7;
Eisenbud, Harris - 3264 & All That, Intersection Theory in Algebraic Geometry, chapters 3 and 4 (click);
Görtz, Wedhorn - Algebraic Geometry I, (2010) Vieweg+Teubner Verlag, paragraphs from 8.4 to 8.10;
Kleiman S. L. - Geometry on Grassmannians and Applications to Splitting Bundles and Smoothing Cycles, Pubblications Mathématiques de l'I.H.É.S., 36 (1969) 281-297 (clack);
Vakil - FOAG (December 29 2015 version), paragraph 16.7 (clock).


TITLE: Intuition for graphing Sine/Cosine
QUESTION [8 upvotes]: So I'm working my way through some basic trig (Khan Academy) - I'm trying to get a better intuition for what graphs of sine and cosine represent.
I've seen the nice unit circle animations that do explain it fairly well.
But would it be fair to say that:
The period of the curve of the sine or cosine function is how fast one full rotation of the circle happens. So as the period decreases I imagine the circle or "wheel" speeding up.
The amplitude of the curve of the sine or cosine function is the radius of this imaginary circle. So as the amplitude of the cure increases so does the size of the circle or "wheel".
So I could imagine speeding up and increases the size of a wheel by decreasing the period and increasing the amplitude of this circle.
Thanks!

REPLY [0 votes]: Technically, the $\sin()$ and $\cos()$ functions always have an argument consisting of an angle (i.e. speed of the angle changing does not matter), they always have a period of $2\pi$ and have an amplitude of $1$.  We can change the period of the function by scaling the argument.  For example $\sin(2\pi x)$ has a period of $1$.  We change the amplitude by multiplying by a constant.  For example $2\sin(2\pi x)$ produces an amplitude of 2.