text:
\section{Introduction and preliminaries}

One of the options how to study properties in certain non-commutative algebraic domains
is a commutator. For example, in  algebras the additive commutator, i.e., Lie product
$[A,B]_a=AB-BA$ is usually used and with its help some beautiful results were obtained.
Let us only mention the famous Kleinecke-Shirokov Theorem~\cite{Kleinecke,Shirokov}. In
groups   the multiplicative commutator $[A,B]_m=A^{-1}B^{-1}AB$ is used, and it is a
central tool in studying solvability of groups and hence in Galois theory of solvability
of equations by radicals~\cite{Grove}.

Additional information about non-commuting elements is obtained  by studying the
properties of a commuting graph. For example, if the commuting graphs over two finite
semisimple rings are isomorphic, then their noncommutative parts are also
isomorphic~\cite{AGHM04}. Let us remark that commuting graph can also be used  in
algebraic domains where commutator is not available, e.g., in semigroups or in semirings.

Up until now, one of the prime concerns when studying commuting graphs was calculating
its diameter~\cite{Abdollahi,AMRR06,AraKinKon10,DolObl10,GiuPope10,Mohammadian10,
Segev01}. It turned out that we obtain essentially different results if the
matrix algebra $M_n(\FF)$ is defined over an algebraically closed field $\FF$ than if it is defined over
non-closed one. While in the former case the diameter is always equal to four, provided $n\ge 3$,
in the later case the graph may be disconnected, and if it is connected
the diameter is known to be at most six. The hypothesis is that if the commuting graph is connected, its diameter
is at most five~\cite[Conjecture~18]{AMRR06}. Note that for $n=2 $ the commuting graph over any field is disconnected \cite[Remark 8]{AR06}.

In the present paper we are interested in commuting graphs of matrix algebras $M_n(\FF)$ over
algebraically closed fields $\FF$ with $n \ge 3$.
In particular we  study the maximal distances  between its
vertices. It was already proved in~\cite[Proof of Theorem~3]{AMRR06} that an
elementary Jordan matrix is  always at the maximal distance (i.e., four) from its
transpose. In the present paper we  show that the maximal distance cannot be achieved when
one of the matrices is derogatory. However, if both $A$ and $B$ are non-derogatory we
construct an invertible matrix $S$ so that $A$ and $S^{-1}BS$ are at the distance four.
We also show that there exist an infinite collection
of matrices, pairwise at the maximal distance. Next, we describe rank-one
matrices as the ones which are not at the maximal distance from any derogatory matrix. A
similar result classifies semisimple (i.e., diagonalizable) matrices. Our paper concludes
with a specific example of matrix algebra over algebraically non-closed field, such that the diameter of its commuting
graph is greater than four.

Let us briefly recall some standard definitions and notations. Unless explicitly stated
otherwise, $\FF$ is an algebraically closed field of an arbitrary characteristics.
Further, $M_{m, n}(\FF)$ is the space of $m\times n$ matrices over $\FF$ with a standard
basis $E_{ij}$, and $M_n(\FF)=M_{n, n}(\FF)$ is the matrix algebra with identity~$\Id$.
Let $e_1,\dots,e_n$ be the standard basis of column vectors in $\FF^n$ (i.e., of $n\times
1$ matrices). Given an integer $k\ge 2$  denote by $J_k(\mu)=\mu\Id_k+\sum_{i=1}^{k-1}
E_{i(i+1)}\in M_k(\FF)$ the upper-triangular elementary Jordan cell with $\mu$ on its
main diagonal, and  let $J_1(\mu)= \mu \in \FF$. We  write shortly $J_k = J_k(0)$. Matrix
$B$ is a conjugated matrix of $A$ if $B=S^{-1}AS$ for some invertible matrix $S$. As
usual, $A^{\tr}$ is a transpose of $A\in M_n(\FF)$ and $\rk A$ its rank.

For a matrix algebra $M_n(\FF)$ over a field $\FF$ its commuting graph $\Gamma(M_n(\FF))$
is a simple graph (i.e., undirected and loopless), with the vertex set consisting of all
non-scalar matrices. Two vertices $X, Y$ form an edge  $X \path  Y$ if the
corresponding matrices are different and commute, i.e., if $X \ne Y$ and  $XY=YX$. The
sequence of successive connected vertices $X_0\path  X_1$, $ X_1 \path  X_2$, ...,
$X_{k-1} \path X_{k}$ is  a path of length $k$ and is denoted by $X_0 \path  X_1 \path
\ldots \path X_k$. The distance $d(A,B)$ between  vertices $A$ and $B$ is the length of
the shortest path between them. The diameter of the graph is the maximal distance between
any two vertices of the graph.

Given a subset $\Omega \subseteq M_n(\FF)$, let
$$\C(\Omega)=\{X\in M_n(\FF);\;\;AX=XA \mbox{ for every } A \in \Omega \}$$
be its centralizer. If $\Omega = \{ A \}$ then we write shortly $\C(A) = \C(\{ A \})$. In
graph terminology, the set of all non-scalar matrices from the centralizer of $A$ is equal to the set of all vertices $X$ such that
$d(A,X) \le 1$.  Note that $\FF\Id\in \C(A)$ for any matrix~$A$ and that, by  a double centralizer theorem,
$\C(\C(A))=\FF[A]$ (see~\cite[Theorem~2, pp.~106]{Wed} or~\cite{lagerstrom}).
We remark that in
different articles  a centralizer is also called  a commutant and  is denoted by
$A'=\C(A)$.

A centralizer induces two natural relations on $M_n(\FF)$. One is
the equivalence relation, defined by $A\sim B$ if $\C(A)=\C(B)$.
We  call any such two matrices equivalent. The other relation is a
preorder given by $A\prec B$ if $\C(A)\subseteq \C(B)$. It was
already observed that  minimal  and  maximal matrices in this
poset are of special importance, see for
example~\cite{Dolinar_Semrl,Semrl,Dol-Kuz}. Recall that a matrix
$A$ is minimal if $\C(X) \subseteq \C(A)$ implies $\C(X)=\C(A)$.
It was shown in~\cite[Lemma 3.2]{Semrl} that the matrix $A$ is
minimal if and only if it is nonderogatory, which means that each
of its eigenvalue has geometric multiplicity one, which is further
equivalent to the fact that its Jordan canonical form is  equal to
$J=J_{n_1}(\lambda_1)\oplus\dots\oplus J_{n_k}(\lambda_k)$, with
$\lambda_i\ne\lambda_j$ for $i\ne j$. In this case,
$$\C(J)=\FF[J_{n_1}(\lambda_1)]\oplus\dots\oplus\FF[J_{n_k}(\lambda_k)]=\FF[J],$$
where $\FF[X]$ is an $\FF$-algebra generated by $X$, see~\cite[Theorem~1, pp. 105]{Wed}
or~\cite[Theorem~3.2.4.2]{Horn-Johnson}.

Recall also that a non-scalar matrix $A$ is maximal if $\C(A) \subseteq
\C(X)$ implies $\C(A)=\C(X)$ or $X$ is a scalar matrix. It is known (see~\cite[Lemma~4]{Dolinar_Semrl} and also~\cite[Lemma~3.1]{Semrl}) that
a matrix is maximal if and only if it is equal to $\alpha\Id+\beta P$ or $\alpha\Id+\beta N$,
where $P^2=P$ is a non-scalar idempotent, $N\ne0$ is square-zero (i.e., $N^2=0$), and a scalar $\beta$ is
nonzero. It should be noted that the proof of this fact was done only for the field of
complex numbers, but can be repeated almost unchanged in an arbitrary algebraically closed field.

\section{Results}

Throughout this section, with an exception of the last example, $\FF$ is an algebraically closed field and $n \ge 3$.
We start with three technical lemmas which will be needed in the sequel. First we observe
that every matrix commutes with a rank-one matrix.

\begin{lemma}\label{lem:rk-1'=all
For every matrix $A \in M_n(\FF)$ there exists a rank-one matrix $R\in M_n(\FF)$ with $d(A,R) \le 1$.
\end{lemma}
\begin{proof}
Given any $A \in M_n(\FF)$, it suffices to show that~$A$ commutes with
at least one matrix of rank one. Since $\FF=\overline{\FF}$, the
matrix~$A$ has at least one eigenvalue $\lambda$.
So, we may assume without
loss of generality that $A$ is singular, otherwise we would
consider $A - \lambda \Id$. Now, let $ x $ and $ y
$ be nonzero vectors in the kernels of $A$ and $A^{\tr}$,
respectively. Then, $R= x  y ^{\tr}$ is a rank-one matrix with
$AR=(A x ) y ^{\tr}=0= x (A^{\tr} y )^{\tr}=RA$.
\end{proof}

Using Lemma~\ref{lem:rk-1'=all} we can give an alternative proof of the already known
fact about the diameter of a commuting graph~\cite[Corollary~7]{AMRR06}.
\begin{corollary}\label{cor:maximal-distance-in-graph=4}
The distance between any two matrices in the commuting graph is at most four.
\end{corollary}
\begin{proof}
Let $A$ and $B$ be arbitrary matrices. By Lemma~\ref{lem:rk-1'=all} there exist rank-one
matrices $R_1= xf^{\tr}\in \C(A)$ and $R_3= y  g^{\tr}\in \C(B)$.
Since $n\ge 3$ we can find a nonzero $z\in\FF^n$ with $f^{\tr}z=0=g^{\tr}z$ and  a nonzero  $h\in\FF^n$
with $ h^{\tr}x =0=h^{\tr}y $. Then for a rank-one matrix $R_2= x h^{\tr}$ we obtain
 $A=R_0\path  R_1\path  R_2\path R_3\path R_4=B$.
\end{proof}

\begin{lemma}\label{lem:splosna-o-preseku-komuatntov
Let~$A=J_{k_1}\oplus J_{k_2}\in M_{k_1+k_2}(\FF)$ be a nilpotent matrix with two Jordan cells
of sizes~$k_1,k_2\ge 1$. Then $d(A,R) \le 2$ for an arbitrary rank-one $R\in M_{k_1+k_2}(\FF)$.
\end{lemma}
\begin{proof}
If $k_1=k_2=1$ then $A$ is a zero matrix and the conclusion is then imminent. Otherwise,
$k_1\ge2$ or $k_2\ge 2$. Let $k=k_1 + k_2$. It is elementary that the matrix $Z=x_1
E_{1k_1} +x_2 E_{1k}+x_3E_{(k_1+1)k_1}+x_4E_{(k_1+1)k}$ commutes with $A$ for any choice
of $x_1,x_2,x_3,x_4\in\FF$. Actually, $ZA = AZ = 0$. Moreover, the matrix $Z$ is non-scalar,
except when~$x_1=x_2=x_3=x_4=0$. Therefore, it suffices to show that for an arbitrary
rank-one matrix $R$ there exist $x_1,\dots,x_4 \in \FF$, such that at least one of them
is nonzero and $ZR=RZ = 0$. To this end, write $R= a  b ^{\tr}$ for some column vectors
$ a =(a_1,\dots,a_k)^{\tr}$ and $ b =(b_1,\dots,b_k)^{\tr}$. Then $ZR=RZ=0$ is
equivalent to $Z a =Z^{\tr} b =0$, hence we must solve a homogeneous system of four linear
equations
\begin{equation}\label{eq:2jordan-blocks-commute-wth-rk1}
\begin{aligned}
x_1 a_{k_1}+ x_2 a_k &=0,\\
x_3 a_{k_1} +x_4 a_k &=0,\\
x_1 b_1+ x_3 b_{k_1+1}&=0,\\
x_2 b_1 +x_4 b_{k_1+1}&=0,
\end{aligned}
\end{equation}
$x_1,x_2,x_3,x_4$ unknown. The corresponding matrix of coefficients is equal to
$$\begin{bmatrix}
 a_{k_1} & a_k & 0 & 0 \\
 0 & 0 & a_{k_1} & a_k \\
 b_1 & 0 & b_{k_1+1} & 0 \\
 0 & b_1 & 0 & b_{k_1+1}
\end{bmatrix}$$
and it is easy to check that it is always singular. Therefore the system~(\ref{eq:2jordan-blocks-commute-wth-rk1})
has a nontrivial solution. This solution defines a non-scalar matrix $Z$, which commutes
with $A$ and $R$, so $d(A,R) \le 2$ in $\Gamma(M_{k_1+k_2}(\FF))$.
\end{proof}
\begin{lemma}\label{lem:A'capR'
Suppose $A$ is not minimal. Then $d(A,R) \le 2$ for an arbitrary rank-one matrix $R\in M_n(\FF)$.
\end{lemma}
\begin{proof}
Using conjugation we might assume~$A$ is already in its Jordan form. Since it is not
minimal, hence it is derogatory, at least two Jordan cells contain the same eigenvalue.
Let $k_1, k_2\ge 1$ be their sizes. Define also $k=k_1+k_2$. Moreover,
$\C(A)=\C(A-\lambda\Id)$ so we may also assume that these two Jordan cells are nilpotent
and that $A=J_{k_1}\oplus J_{k_2}\oplus \tilde{A}$.  It is elementary that
$$\C\bigl(J_{k_1}\oplus J_{k_2}  \bigr)\oplus (\FF\Id_{n-k})\subseteq \C(A).$$

Now, let $R= x  y ^{\tr}$ be an arbitrary rank-one matrix.
Decompose $ x = x _1\oplus  x _2\in\FF^{k}\oplus
\FF^{n-k}$ and $ y = y _1\oplus  y _2\in\FF^{k}\oplus
\FF^{n-k}$. We claim that there exists a non-scalar matrix
$\widehat{Z} \in \C(J_{k_1}\oplus J_{k_2} )$ satisfying simultaneously
$\widehat{Z} x _1=\lambda  x _1$ as well as $\widehat{Z}^{\tr} y
_1=\lambda y _1$ for some $\lambda\in\FF$. In fact, this is
trivial when $ x_1= y_1=0$. Otherwise we let
$$\widehat{R}=\begin{cases}
 x _1 y _1^{\tr};&  x _1, y _1\ne0\\
 e _1 y _1^{\tr}; &  x _1=0\\
 x _1 e _1^{\tr}; &  y _1=0
\end{cases}\in M_{k}(\FF),$$
where $ e _1\in \FF^{k}$  is the first vector of the standard basis. By
Lemma~\ref{lem:splosna-o-preseku-komuatntov} there exists at least one non-scalar matrix
$\widehat{Z}\in M_{k}(\FF)$ which commutes  with $\widehat{R}$ as well as with
$J_{k_1}\oplus J_{k_2}\in M_{k}(\FF)$. Therefore, if $ x _1, y _1\ne0$, then $\widehat{Z} x
_1 y _1^{\tr}= x _1(\widehat{Z}^{\tr} y _1)^{\tr}$ and we obtain $\widehat{Z} x _1=\lambda x _1$,
and $\widehat{Z}^{\tr} y _1=\lambda  y _1$ for some $\lambda \in \FF$. If $x_1 = 0$, then
similarly as above $\widehat{Z} e _1=\lambda e _1$, and $\widehat{Z}^{\tr} y _1=\lambda  y _1$.
Obviously $\widehat{Z} x_1=\lambda x_1$. Likewise we argue  if $ y _1=0$.

With the help of~$\widehat{Z}$ we define $Z=\widehat{Z}\oplus
\lambda\Id_{k}\in\C(J_{k_1}\oplus J_{k_2}  )\oplus (\FF\Id_{n-k})\subseteq
\C(A)$. Clearly, $Z x =\widehat{Z} x _1\oplus \lambda x _2=\lambda x $, and similarly,
$Z^{\tr} y =\lambda y $, so~$Z$ commutes with $R= x  y ^{\tr}$ and with $A$.
\end{proof}

 Akbari, Mohammadian, Radjavi, and Raja
proved in~\cite[Lemma 2]{AMRR06} that, for matrices of size $n\ge 3$, the diameter of the
commuting graph is at most four (see also Corollary~\ref{cor:maximal-distance-in-graph=4}
above) and that $d(J,J^{\tr})=4$, thus showing that the diameter of the commuting graph
of matrix algebra over algebraically closed fields is equal to four. It is
well-known~\cite[p.~134]{Horn-Johnson} that the transpose of a matrix is conjugate to the
original, so \cite[Lemma 2]{AMRR06} implies that the maximal distance from $J$ to some of
its conjugates is equal to four. Our next lemma will strengthen their result by
considering maximal distances between an arbitrary minimal matrix $A\in M_n(\FF)$ and
matrices from conjugation orbit $\{S^{-1}BS;\;\;S\hbox{ invertible}\}$ of another minimal
matrix $B\in M_n(\FF)$. Recall that a minimal matrix is conjugate to
$\bigoplus\limits_{i=1}^{k} J_{n_i}(\lambda_i), $ where $\lambda_i \ne \lambda_j$ for $i
\ne j$, and where $(n_1,n_2,\ldots,n_k)$ is a partition of $n$. We will  show below that
for any two given partitions of $n$, we can find two minimal matrices with their Jordan
forms corresponding to these two partitions, at  distance four. One of the matrices is
already in its Jordan canonical form, while the other  is a matrix, conjugated to its
Jordan canonical form by an invertible matrix with all of its minors  nonzero. Such
invertible matrix is for example a Cauchy matrix $\big[\frac{1}{x_i-y_j}\big]_{ij}$ (see
\cite{Schechter}).

\begin{theorem}\label{thm:d=4
 Let $S$ be any matrix with all of its minors  nonzero. For any two minimal matrices
 $A=\bigoplus\limits_{i=1}^{k} J_{n_i}(\lambda_i)\in M_n(\FF)$
 and $B=\bigoplus\limits_{i=1}^{l} J_{m_i}(\mu_i)\in M_n(\FF)$,
 we have  $d(A ,S^{-1} B S)=4$.
\end{theorem}

\begin{proof}
  Assume erroneously that  $A$ and $B$, as defined in Lemma, are not at the maximal distance, i.e.,
   $d(A,S^{-1} B S) \leq 3$.
  Since $\C(A)=\C(\alpha A)$ for all nonzero $\alpha \in \FF$, we can lengthen every path by adding vertices
 which correspond to scalar multiples of matrices. So, there  exists a path $A\path X\path Y\path S^{-1} B S$ of length $3$
  in
  $\Gamma(M_n(\FF))$. We can assume
  without loss of generality that $X$ and $Y$ are maximal matrices. Namely, if  $X$ is not
  maximal, then
  there exists a maximal $X' \succ X$, and since $A,Y \in \C(X) \subseteq \C(X')$, we could consider a path
  $A\path X'\path Y\path S^{-1} B S$ of length 3. Likewise for $Y$.

  We will show that no two maximal matrices $X\in \C(A)$ and $Y\in \C(S^{-1} B S)$  commute
  and thus obtain  a contradiction to the assumption    $d(A,S^{-1} B S) \leq 3$.
  Since all maximal matrices are equivalent either to a square-zero matrix or to an idempotent matrix we will
  consider three  cases.

  First, let us assume that both $X$ and $Y$ are square-zero but nonzero.
  Since $A=\bigoplus_{i=1}^{k} J_{n_i}(\lambda_i)$
 and $B=\bigoplus_{i=1}^{l} J_{m_i}(\mu_i)$ are minimal,
  so $\lambda_i\ne \lambda_j$ and $\mu_i\ne\mu_j$ for $i\ne j$, we have
  that $$X=T_1\oplus T_2 \oplus\ldots\oplus T_k\quad\hbox{ and }\quad Y=S^{-1}(T'_1\oplus T'_2 \oplus\ldots\oplus T'_l)S,$$
  where all $T_i$ and $T'_j$ are upper triangular Toeplitz matrices.
  Clearly then $\Im X=\Lin \{e_{\sigma(1)},e_{\sigma(2)},\ldots,e_{\sigma(r)}\}$ for
  some permutation  $\sigma$ of length $n$ and integer $r$,
 $1\leq r \leq \frac{n}{2}$.
  Moreover,  by the block-Toeplitz  structure of $X\ne0$ there exist indices $t$ and $s$ such that  $Xe_t=\alpha e_s\ne0$.
  For the sake of simplicity let us denote $T=T'_1\oplus T'_2 \oplus\ldots\oplus T'_l$. Now, if $YX=XY$, we would have that
  $S^{-1}TSX e_t \in \Im X$. This would imply,
  $$\alpha T S e_s  \in
    S(\Im X)=\Lin\{Se_{\sigma(1)},Se_{\sigma(2)},\ldots,Se_{\sigma(r)}\}$$
  which  is clearly possible
   if and only if the rank of the $n\times r$ matrix
  $M=\left[ \frac{1}{\alpha}Se_{\sigma(1)},\frac{1}{\alpha}Se_{\sigma(2)},\ldots,\frac{1}{\alpha}Se_{\sigma(r)}
    \right]$
  is the same as the rank of the augmented matrix $\left[ M \, \vert \, TSe_s\right]$. However, we will show that
  this is not the case. Since all  minors of $S$ are nonzero, its $s$-es column $Se_s$ has no zero entries and as such cannot
  be annihilated by a nonzero block-Toeplitz matrix $T$.
  Note that $T$ is  also square-zero and so it has at least $ \frac{n}{2}$ zero rows.
  Recall that $r\le \frac{n}{2}$, consequently there exists an $(r+1)\times (r+1)$ submatrix of the augumented
  matrix, having in the
  last column exactly $r$ zeros and one nonzero element. By expanding this $(r+1)\times (r+1)$  minor
  by the last column, we observe that it is equal to a multiple of an $r \times r$ minor of matrix $M$
  which is
  equal to $(\frac{1}{\alpha})^r$ times an $r \times r$ minor of $S$. By the assumption, every minor of $S$ is nonzero and so
    $r+1=\rk \left[ M \, \vert \, TSe_s\right] > \rk M=r$. This  implies
   $TS e_s \notin S(\Im X)$, a contradiction.

 Second, suppose   a non-scalar idempotent $X\in\C(A)$ commutes with a non-scalar square-zero $Y\in\C(S^{-1}BS)$. Without loss of generality,
 $r=\rk X \leq \frac{n}{2}$, otherwise take $\Id-X$ instead of $X$. So,
 $X=\sum_{i=1}^r E_{\sigma(i)\sigma(i)} $ and
  $Y=S^{-1}TS$,   where $\sigma$ and $T=T'_1\oplus T'_2 \oplus\ldots\oplus T'_l$ are as above.
  Define  $t=\sigma(1)$.  Similarly as before, if $YX=XY$ we would have that
  $S^{-1}TSXe_t\in \Im X=\Lin \{e_{\sigma(1)},e_{\sigma(2)},\ldots,e_{\sigma(r)}\}$,
  or, equivalently, $TSe_t\in \Lin \{Se_{\sigma(1)},Se_{\sigma(2)},\ldots,Se_{\sigma(r)}\}$. We proceed
  as in the first case to obtain a contradiction.

  By the symmetry the only case remaining is the case when
  $X$ and $Y$ are both non-scalar idempotents. Write
  $X=\sum_{i=1}^r E_{\sigma(i)\sigma(i)} $ and
  $Y=S^{-1} P S$ for $P=\sum_{i=1}^s E_{\tau(i)\tau(i)}$.
   Without loss of generality, $r,s \leq \frac{n}{2}$, since otherwise we would substitute
   $X$  by $\Id-X$ or  $Y$ by $\Id-Y$.
  Again,   take  $t=\sigma(1)$. If $YX=XY$ then
  $S^{-1}PSXe_t\in \Im X=\Lin \{e_{\sigma(1)},e_{\sigma(2)},\ldots,e_{\sigma(r)}\}$,
  or, equivalently, $PSe_t\in \Lin \{Se_{\sigma(1)},Se_{\sigma(2)},\ldots,Se_{\sigma(r)}\}$.
  Since $\rk P \leq \frac{n}{2}$, it follows that the vector $PSe_t$ has at least $\frac{n}{2}$ zero
  entries. Note that $\frac{n}{2} \geq r$ and $Se_t$ is the $t$-th column of $S$, so it has no zero entries. This
   gives
  $PSe_t\ne0$, a contradiction as in the first case.

This shows $d(A,S^{-1} B S) \geq 4$. But the  diameter of commuting graph is equal to four (see~\cite[Lemma~2]{AMRR06}),
hence $d(A,S^{-1} B S) =4$.
\end{proof}

\begin{remark}The matrix $S^{-1}BS$ from Theorem~\ref{thm:d=4} can be rather complicated. In a special case, when
$A$ is nilpotent we can take $A=J_n$ to achieve that $d(J_n,B)=4$ for any  companion
matrix $B$ in the lower-triangular form. This can be seen by
 a slight adaptation of the proof of~\cite[Lemma 2]{AMRR06}. For convenience we sketch the main points of the
proof. First, it suffices to prove that each maximal $D\in\C(J_n)=\FF[J_n]$ satisfies
$\C(B)\cap\C(D)=\FF\Id$. We may further assume $D=\sum_{i=r}^{n-1}d_i J_n^i$,
$\frac{n}{2}\le r\le n-1$, is square-zero and hence write it as a $3\times 3$ block
matrix with  block at position $(1,3)$ being invertible upper-triangular Toeplitz of size
$(n-r)\times (n-r)$, while all the rest blocks are zero. If $Z\in\C(B)\cap\C(D)$ then in
particular it commutes with $D$. By direct computation using block-matrix structure we
see that the $(n,1)$ entry of $Z$ is zero. However, $Z\in\C(B)$ and since companion
matrices are nonderogatory we have $Z=\sum_{i=0}^{n-1}\lambda_i B^i$. By considering the
images of basis vectors we see that $B^i=\left[\begin{matrix}
 0_{i, (n-i)} & \bigstar_{i, i}\\
\Id_{n-i}& \bigstar_{(n-i), i}
\end{matrix}\right]$. Since $(n,1)$ entry of $Z$ is $0$ we see that
$\lambda_{n-1}=0$. Proceeding inductively we see that $\lambda_i=0$ for every
$i=(n-1),\dots,1$, whence $Z$ is scalar.
\end{remark}

By Theorem~\ref{thm:d=4} there exist different types of matrices which are at the maximal
distance. Next we show that we can find infinitely many  matrices which are in  the
commuting graph pairwise at the maximal distance. Actually, we find an induced graph
which is a tree with an internal vertex and all of its leaves at distance two from the
internal vertex.

\begin{theorem}
There exist an infinite
family of matrices
$(X_\alpha)_\alpha\in M_n(\FF)$ and a rank-one  matrix $Z$ such that
$d(X_\alpha,X_\beta)=4$ for $\alpha\ne \beta$ and $d(X_\alpha,Z)=2$
for all $\alpha$.
\end{theorem}
\begin{proof}
We consider three cases separately.

{\bf Case  \bm{$n=3$}}. Choose $Z=E_{11}$ and let the infinite family consist
of rank one
nilpotent matrices
$$R_\alpha=(0,1,\alpha)^{\tr}(0,\alpha,-1), \quad \alpha \in \FF.$$
It is easy to see that each member commutes with $E_{11}$ and that the elements of the family are pairwise at distance two.
For each index
$\alpha \in \FF$ choose a nilpotent $X_\alpha$ such that $X_\alpha ^2=R_\alpha
$. Since $n=3$
all non-scalar matrices, which commute  with  $X_\alpha$ are equivalent to $X_\alpha$ or to   $X_
\alpha^2=R_\alpha$. Therefore, as
$d(X_\alpha^2,X_\beta^2)=2$ for $\alpha\ne\beta$, we see that $d(X_
\alpha,X_\beta)=4$ for
$\alpha\ne\beta$.

  {\bf Case \bm{$n=4$}}.  Choose $\lambda\in\FF\setminus\{0,1\}$. For nonzero
$\alpha\in\FF$ consider rank-one nilpotent matrix
$N_\alpha=(0,\lambda,\lambda\alpha,\lambda)^{\tr}(0,-\alpha,1,0)$ and rank-one idempotent
$P_\alpha=(0,1,\alpha,0)^{\tr}(0,1,0,-1).$
It is a straightforward calculation that all these matrices are pairwise non-commutative but they all commute with $E_{11}$, hence
\begin{equation}\label{eq:n=4}
d(N_\alpha,N_\beta)=d(P_\alpha,P_\beta)=d(N_\alpha,P_\beta)=2
\end{equation}
for every $\alpha\ne \beta\in\FF\setminus\{0\}$. Moreover, there exists a conjugation
such that $S_\alpha^{-1} N_\alpha S_\alpha=E_{13}$ and $S_\alpha^{-1} P_\alpha
S_\alpha=E_{44}$, for example, take
$$S_\alpha=\begin{bmatrix}
 0 & 1 & 0 & 0 \\
 \lambda  & 0 & 0 & -1 \\
 \alpha  \lambda  & 0 & 1 &
   -\alpha  \\
 \lambda  & 0 & 0 & 0
\end{bmatrix}.$$
Then, for each
$\alpha$ we can find a minimal matrix $X_\alpha=S_\alpha (J_3\oplus 1)S_\alpha^{-1}$
with $X_\alpha\prec P_\alpha$ and
$X_\alpha\prec N_\alpha$.

We claim that $d(X_\alpha,X_\beta)=4$. In fact, if a maximal matrix $M$ satisfies $M
\succ X_{\alpha}$, then,
 up to equivalence, either $M=N_\alpha$, or $M=P_\alpha$.
Hence, if $X_\alpha\path  Y_{\alpha,\beta}\path Z_{\alpha,\beta}\path X_{\beta}$ would be
a path of length three, connecting $X_\alpha$ and $X_\beta$ for $\alpha \ne \beta$, then
we may assume without loss of generality that $Y_{\alpha,\beta}$ and $Z_{\alpha,\beta}$
are maximal matrices (see the proof of Theorem~\ref{thm:d=4}). Hence $Y_{\alpha,\beta}$
is equivalent either to $N_{\alpha}$ or $P_{\alpha}$ and $Z_{\alpha,\beta}$ is equivalent
either to $N_{\beta}$ or $P_{\beta}$. This contradicts equation~\eqref{eq:n=4}, so
$d(X_\alpha,X_\beta)\ge 4$ for $\alpha\ne\beta$. Observe that one of the paths from
$X_\alpha$ to $X_\beta$ is $X_\alpha\path  N_\alpha \path  E_{11} \path N_\beta\path
X_\beta$.

{\bf Case \bm{$n\ge 5$}}. Let $A=\diag(\lambda_1,\dots,\lambda_n)$ where $
\lambda_i$ are
pairwise distinct. Consider an infinite family of rank one nilpotent
matrices $R_\alpha$ indexed by scalars  $\alpha\in\FF$:
$$R_\alpha=R+\alpha\tilde{R};\quad\; R=xf^{\tr},\tilde{R}= xg^{\tr},\; x=
\left[\begin{smallmatrix}
1 \\
\vdots\\[2mm]
1\\
\end{smallmatrix}\right],\;f=\left[\begin{smallmatrix}
2-n\\
1 \\
\vdots\\[2mm]
1\\
0
\end{smallmatrix}\right]
,\;g=\left[\begin{smallmatrix}
1-n\\
1 \\
\vdots\\[2mm]
1\\
1
\end{smallmatrix}\right],$$
 Note that $S_\alpha=\Id+R_\alpha
$ is invertible with
  $S_\alpha^{-1}=\Id-R_\alpha$, and that $S_\alpha^{-1}S_\beta
=(\Id-R_\alpha)(\Id+R_\beta) =\Id+(\beta-\alpha)\tilde{R}$. Let
us define for every $\alpha\in \FF$ the matrix
$X_\alpha=S_\alpha AS_\alpha^{-1}.$
We will prove that $d(X_\alpha,X_\beta)=4$ for $\alpha\ne\beta$.

Note first that the distance is invariant for simultaneous conjugation. So, we may
replace $(X_\alpha,X_\beta)$ with $(S_\alpha^{-1}X_\alpha S_\alpha,S_\alpha^{-1}X_\beta
S_\alpha)=(A,SAS^{-1})$, where $S=S_\alpha^{-1}S_\beta=\Id+(\beta-\alpha)\tilde{R}$.
Now, to prove $d(A,SAS^{-1})=4$ it suffices to show that, given any non-scalar matrices
$D_1\in \C(A)$ and $SD_2S^{-1}\in S\C(A)S^{-1}$, they do not commute.

By the choice of minimal $A$, $\C(A)$ consists of diagonal matrices
only, hence $D_1$ and
$D_2$ are diagonal. Assume erroneously that $D_1$ and $SD_2S^{-1}$ do
commute, i.e., that
$D_1(SD_2S^{-1})=(SD_2S^{-1})D_1$, or equivalently,
$$D_1(\Id+\tilde{x} g^{\tr})D_2(\Id-\tilde{x} g^{\tr})
=(\Id+\tilde{x} g^{\tr})D_2(\Id-\tilde{x} g^{\tr})D_1,$$
where  $\tilde{x}=(\beta-\alpha)x$.
Since diagonal matrices commute,   we get after expansion and
simplification
\begin{multline}\label{eq:star}
(D_1\tilde{x})(D_2g)^{\tr}-(D_1D_2\tilde{x})g^{\tr}-
(g^{\tr}D_2\tilde{x})\cdot(D_1\tilde{x}) g^{\tr}\\
=\tilde{x}(D_1D_2g-(g^{\tr}D_2\tilde{x})\cdot D_1g)^{\tr}-
(D_2\tilde{x})(D_1g)^{\tr}.
\end{multline}
  Notice  that an eigenvector  of a  non-scalar diagonal matrix has at least one nonzero entry.
Hence, $\tilde{x}=(\beta-\alpha)(1,\dots,1)^{\tr}$ and $g=(1-n,1,\dots,
1,1)^{\tr}$ can not be
eigenvectors of a non-scalar diagonal matrix. In particular, $g$ and
$D_2g$ are linearly
independent and so there exists a vector $y$ such that $g^{\tr}y=0$ and
$(D_2g)^{\tr}y=1$. Post-multiplying both sides  of equation~
\eqref{eq:star} with $y$, we
now have
$D_1\tilde{x}=\mu \tilde{x}+\nu D_2\tilde{x}$, $\mu=(D_1D_2g-
(g^{\tr}D_2\tilde{x}) D_1g)^{\tr}y$
and $\nu=-(D_1g)^{\tr}y$.
  We infer that $(D_1-\nu D_2)\tilde{x}=\mu \tilde{x}$, hence $(D_1-
\nu D_2)$ is a scalar matrix because
$\tilde{x}$ has all its entries nonzero.
Thus,
$D_1=\lambda\Id+\nu D_2$ for some $\lambda$. This
simplifies the starting
equation $D_1(SD_2S^{-1})=(SD_2S^{-1})D_1$ into
$$D(SDS^{-1})=(SDS^{-1})D;\qquad D=D_2,$$
wherefrom also the derived equation~(\ref{eq:star}) simplifies into
\begin{multline}\label{eq:multline2}
(D \tilde{x})(D g)^{\tr}-(D^2 \tilde{x})g^{\tr}-(g^{\tr}D \tilde{x})
\cdot(D \tilde{x})
g^{\tr}\\
=\tilde{x}(D^2 g-(g^{\tr}D \tilde{x})\cdot D g)^{\tr}-(D \tilde{x})(D
g)^{\tr}.
\end{multline}
  By the similar arguments as above we find a vector $z$ such that
$g^{\tr}z=0$ and
$(Dg)^{\tr}z=1$, and continuing along the lines we see that
$$D\tilde{x}=((D^2g)^{\tr}z-(g^{\tr}D \tilde{x}))\cdot \tilde{x}-D
\tilde{x}.$$
If $\mathrm{char}\,\FF\ne2$ then the above equation implies that  $
\tilde{x}$ is an
eigenvector of a diagonal matrix $D$ which is possible only when $D_2=D
$ is scalar, a
contradiction.

However, if $\mathrm{char}\,\FF=2$ then we choose a vector, still
named $z$, such that
$g^{\tr}z=1$ and $(Dg)^{\tr}z=0$. Similarly as above, this simplifies
equation~\eqref{eq:multline2} into
$$D^2\tilde{x}+(g^{\tr}D \tilde{x})\cdot
D\tilde{x}=\mu\tilde{x};\qquad\mu=(D^2{g})^{\tr}z.$$ Arguing as above, $D^2+(g^{\tr}D
\tilde{x})D-\mu\Id=0$. Thus, $D_2=D$, being non-scalar diagonal,  has exactly two
distinct eigenvalues: $d_1$ and $d_2$ (with multiplicities $k$ and $n-k $, respectively),
because it is annihilated by a quadratic polynomial $p(\lambda)= \lambda^2+(g^{\tr}D
\tilde{x})\lambda-\mu=(\lambda-d_1)(\lambda-d_2)$. With no loss of generality we assume
that $d_1=D_{1,1}$.  Then, comparing the coefficients in characteristics 2, gives
$d_1+d_2=(g^{\tr}D \tilde{x})=(\beta-\alpha) \bigl(
(1-n)d_1+(k-1)d_1+(n-k)d_2\bigr)=(\beta-\alpha)(n-k)(d_1+d_2)$. Since
$\mathrm{char}\,\FF=2$ and  $D$ is not a scalar matrix, we can divide by $d_1+d_2$ to
obtain $(\beta-\alpha)(n-k)=1$. Observe that in characteristic two, $(\beta-\alpha)(n-k)$
is either equal to 0 or $\beta-\alpha$ and since $(\beta-\alpha)(n-k)=1$, we have that
$\beta-\alpha=(\beta-\alpha)(n-k)=1$. and thus $\beta=\alpha+1$. Clearly, we can choose
an infinite subset of indices
$\mathfrak{A}=\{0,\alpha_1,\alpha_1+\alpha_2,\alpha_1+\alpha_2+
\alpha_3,\dots\}\subset\FF$ such that $\alpha-\beta\ne 1$ for $\alpha,\beta\in
\mathfrak{A}$. For this subset, $d(X_\alpha,X_\beta)=4$.\medskip

To prove the rest, observe that the rank-one matrix $$(\Id+ x (f+
\alpha
g)^{\tr})e_1e_1^{\tr}(\Id- x (f+\alpha g)^{\tr})=S_\alpha E_{11} S_
\alpha^{-1}$$
commutes with $X_\alpha=S_\alpha AS_\alpha^{-1}$. Now, since $n\ge 5$
there exists a
nonzero vector $w$ with $w^{\tr}e_1=0=w^{\tr} x =w^{\tr}f=w^{\tr}g$.
Then, a rank-one
matrix $Z=ww^{\tr}$ commutes with $S_\alpha E_{11} S_\alpha^{-1}$
which gives the path
$$X_\alpha\path  S_\alpha E_{11} S_\alpha^{-1}\path  Z.$$
Hence, $d(X_\alpha,Z)\le 2$ for every $\alpha$. Actually, no shorter
path exists, because
otherwise, we could join the shorter path for some $\alpha$ with the
above path for some
other index $\beta$ to obtain that $d(X_\alpha,X_\beta)\le 3$, a
contradiction.
\end{proof}

We next proceed with the classification of matrices which are equivalent to rank-one matrices.
In this classification we will need the following lemma.

\begin{lemma}\label{lem:nonminimal}
Let $n\ge 4$.
Suppose $A \in M_{n}(\FF)$ is
  \begin{enumerate}
   \item[(i)] either a maximal matrix with $2 \leq \rk A \leq n-2$, or
   \item[(ii)] a nilpotent matrix with $A^3=0$ and $\rk(A^2)=1$.
  \end{enumerate}
Then there exists a nonminimal  matrix  $X$, such that $d(A,X) \geq 3$.
\end{lemma}

\begin{proof}
(i)  As already observed in Preliminaries, a  maximal matrix $A$  is either a square-zero
matrix or an idempotent, up to equivalence. Let $k=\rk A$.

If $A$ is square-zero, then $2 \leq k \leq \frac{n}{2}$. We define
$s_{\ell}=(1,1,\dots,1)^{\tr}\in \FF^\ell$ and
 $z_{2\ell}= (0,1,0,1,\dots,0,1)^{\tr}\in \FF^{2\ell}$.
   Also, let $N_{2\ell}=\bigoplus_{i=1}^\ell J_2^{\tr} \in M_{2\ell}(\FF)$. Note that $N_{2\ell}^2=0$ and
  $\rk N_{2\ell} =\ell$.
It is easy to see that a matrix
\begin{equation} \label{eq:matrixA}
\begin{bmatrix}
       N_{2k-2} & 0 & z_{2k-2} \\
       0 & 0_{n-2k+1, n-2k+1} & s_{n-2k+1}\\
   0 &   0  & 0_{1, 1}
  \end{bmatrix}
\end{equation}
is a square-zero of  rank $k$, hence conjugate to $A$. So, we can assume without loss of generality
that $A$ is already in the form \eqref{eq:matrixA}.

  Next, let us define a matrix $X=J_2 \oplus 0_1 \oplus D$, where $D$ is a diagonal matrix
  with $n-3$ distinct nonzero diagonal entries. Clearly, $X$ is nonminimal.
   We will prove that $d(A,X)\geq 3$, i.e.,  any matrix that commutes with $A$ and $X$ is a scalar
   matrix.

  First, let us assume  $k=2$. It is easy to see that every matrix  $B\in\C(X)$ can be decomposed in the following way
  \begin{equation}\label{eq:commX}
   B= \left[ \begin{matrix}
    T & S_1 & 0_{2, 1} \\
    S_2 & D' & 0_{n-3, 1} \\
    0_{1, 2} & 0_{1, n-3} & \lambda
   \end{matrix}
   \right]\,
  \end{equation}
  where  $T= \left[ \begin{matrix}
   a & b  \\
   0 & a
  \end{matrix}
  \right] \in M_2(\FF)$, $D'=\mathrm{diag}(d_3,d_4,\ldots,d_{n-1})\in M_{n-3}(\FF)$, $S_1\in M_{2, n-3}(\FF)$ has the only
  nonzero entry in its upper left corner, $S_2 \in M_{n-3, 2}(\FF)$ has the only
  nonzero entry in its upper right corner, and $\lambda\in\FF$. Note that the blocks in the decomposition of $B$ correspond to the blocks in the decomposition of $A$.
  Suppose $B \in \C(X)$ also commutes with  $A= \left[ \begin{matrix}
       N_2 & 0 & z_2 \\
       0 & 0_{n-3, n-3} & s_{n-3}\\
       0 & 0& 0_{1, 1}
  \end{matrix}
  \right]$. Then, $N_2S_1=0$ and $S_2N_2=0$ imply that $S_1=0$ and $S_2=0$.
  Moreover, from $D's_{n-3}=\lambda s_{n-3}$ and $Tz_2=\lambda z_2$  we easily see that $D'=
  \lambda\Id_{n-3}$ and
  $T=\lambda\Id_2$. Thus, $B=\lambda\Id$, so $d(A,X)\geq 3$.

  Now, let us consider the case $k\geq 3$. Again we decompose every $B\in \C(X)$
  to the blocks that correspond to the block decomposition of $A$:
  \begin{equation*}
   B= \left[ \begin{matrix}
    B_1 & 0_{2k-2, n-2k+1} & 0_{2k-2, 1} \\
    0_{n-2k+1, 2k-2} & D' & 0_{n-2k+1, 1} \\
    0_{1, 2k-2} & 0_{1, n-2k+1} & \lambda
   \end{matrix}
   \right]\,
  \end{equation*}
  where  $B_1= \left[ \begin{matrix}
   a & b & c_1   \\
   0 & a & 0\\
   0 & c_2 & d_3
  \end{matrix}
  \right] \oplus \mathrm{diag}(d_4,d_5,\ldots,d_{2k-2}) \in M_{2k-2}(\FF)$ and
   $D' \in M_{n-2k+1}(\FF)$ is a diagonal
  matrix. Note that $B$ is the same as in~\eqref{eq:commX} but decomposed in a different way.
  Suppose $B$ also commutes with  $A$ as defined in \eqref{eq:matrixA}.
Similarly as before, we have $D's_{n-2k+1}=\lambda s_{n-2k+1}$ and thus
$D'=\lambda\Id_{n-2k+1}$.
  Moreover, it is straightforward that  from $B_1z_{2k-2}=\lambda z_{2k-2}$ and $N_{2k-2}B_1=B_1N_{2k-2}$
  we obtain $B_1=\lambda\Id_{2k-2}$. This completes the proof that  $d(A,X)\geq 3$.

 If $A$ is an idempotent, then  $\rk(\Id-A)=n-\rk A$. Since $A$ and
    $\Id-A$ are equivalent, we can thus assume
 without loss of generality that $A$ is of rank $k$ with  $ \frac{n}{2} \leq k \leq n-2$. Let
 $W$ be a $k \times (n-k)$ matrix with the only nonzero elements being
  $$W_{1,n-k}=W_{2,n-k-1}=W_{3,n-k-2}=\ldots=W_{n-k,1}=W_{k,1}=W_{k,n-k}=1\, .$$
  Note that, if $k=\frac{n}{2}$,  the rows $k$ and $n-k$ coincide. Using an appropriate conjugation we can additionally assume that
$$  A= \begin{bmatrix}
   \Id_k & W\\
   0_{n-k, k} & 0_{n-k, n-k}
  \end{bmatrix}.$$
Let us define a nonminimal matrix $X=J_k \oplus 0_1 \oplus \Id_{n-k-1}$. We
 will prove that $d(A,X)\geq 3$, i.e.,   any matrix $B \in \C(A) \cap \C(X)$ is a
 scalar matrix. It is a straightforward calculation that
 $$\C(A)=\left\{
  \left[ \begin{matrix}
   M & MW-WN\\
   0_{n-k,k} & N
  \end{matrix}
  \right]; \;  M \in M_{k}(\FF), N \in M_{n-k}(\FF)
 \right\}$$
 and that $\C(X)$ consists of all matrices of the form
  $$ \left[ \begin{matrix}
   U & s & 0_{k, n-k-1}\\
   v^{\tr} & \lambda & 0_{1, n-k-1}\\
   0_{n-k-1, k}&   0_{n-k-1, 1}  & Y
  \end{matrix}
  \right],$$
  where  $U$ is an upper triangular Toeplitz $k \times k$ matrix,
  $s=  (s_1 , 0 , \ldots ,0  )^{\tr} \in\FF^k$,
  $v=  ( 0 , \ldots ,0 , v_k )^{\tr}\in\FF^k$, $Y = \big[y_{ij}\big]_{2\le i,j\le n-k}\in M_{n-k-1}(\FF)$, and $\lambda\in\FF$.

Suppose $B=\left[ \begin{matrix}
   M & MW-WN\\
   0 & N
  \end{matrix}
  \right] \in \C(A) \cap \C(X)$. It follows that $M= \sum_{i=1}^k m_i J_k^{i-1}$,
  $N=\lambda \oplus Y$ and $(MW-WN)_{ij}=0$ except possibly for $i=j=1$.

 By equations
 \begin{align*}
 0&=(MW-WN)_{k,1}=m_1-\lambda, \\
 0&=(MW-WN)_{k,n-k}=m_1-y_{n-k,n-k},\\
 0&=(MW-WN)_{i,n-k}=m_{k-i+1}\qquad \hbox{ for all $i$ with }(k-1)\ge i\ge (n-k)
 \end{align*}
 it follows that
 $m_1=y_{n-k,n-k}=\lambda $ and $m_2=m_3=\ldots=m_{2k-n+1}=0$. Moreover, by
 $0=(MW-WN)_{i,1}=m_{k-i+1}$ for $i=2,3,\ldots,n-k-1$, it follows that
 $m_{2k-n+2}=\ldots=m_{k-1}=0$. Now, equation $0=(MW-WN)_{1,n-k}=m_{k}$ completes the
 proof that $M=\lambda  \Id_k$.

 We proceed by
  $0=(MW-WN)_{i,n-k-i+1}=m_1-y_{n-k-i+1,n-k-i+1}$ for  $i=2,\ldots,n-k-1$ and
  $0=(MW-WN)_{i,j}=-y_{n-k-i+1,j}$ for  $i=1,2,\ldots,n-k-1$ and $j=2,3,\ldots,n-k$,
 such that $i+j \ne n-k+1$. It follows that $N=\lambda\Id_{n-k}$ and $(MW-WN)=0$.
 Thus, $B=\lambda \Id$ and $d(A,X) \geq 3$.

(ii) Let $A$ be a nilpotent matrix such that $A^3=0$ and $\rk(A^2)=1$. We may assume $A$
is already in its Jordan canonical form, i.e.,
$$A=J_3\oplus \bigoplus_{i=1}^kJ_2\oplus 0_{n-3-2k}.$$
The centralizer of $A$ is contained in the set of matrices of the form
 $B=\left[ \begin{matrix}
   T & S_1 \\
   S_2 & V
  \end{matrix}
  \right] $, where $T=t_0 \Id_3+t_1J_3+t_2J_3^2 \in M_3(\FF)$,
  $V  \in M_{n-3}(\FF)$, and where the first column of
  $S_2  \in M_{n-3, 3}(\FF)$  as well as the last row of
  $S_1  \in M_{3, n-3}(\FF)$ contain only zero entries.

 Now, let us define the nonminimal matrix $X=1 \oplus 0 \oplus J_{n-2}$ and take any
 $B \in \C(A) \cap \C(X)$. Since $B\in \C(X)$, its  off-diagonal entries on the first row
 and the first column
 are all zero. Comparing with the above form for $B$ we deduce that $T=t_0 \Id$. Moreover,
 $B\in \C(X)$ also implies that the bottom-right $(n-2) \times (n-2)$ block of $B$ is
 upper triangular Toeplitz matrix, which is moreover equal to $t_0' \Id_{n-2}$ for some $t_0'\in\FF$ by the fact that the third row of
 $S_1$ vanishes. Actually, $t_0=t_0'$ because a $3\times 3$ block $T$ overlaps with
 $(n-2)\times (n-2)$ bottom right block. Further, $B\in\C(X)$ implies that the only
 possible off-diagonal nonzero entries in the second row and column lie at positions $(2,n)$, and
 $(3,2)$. Actually, $B_{32}=T_{32}=0$, while from $B\in \C(A)$ we deduce that if
 $B_{2n}\ne0$ then also $B_{1(n-1)}\ne0$, which would contradict the fact that the first row
 of $B$ has zero off-diagonal entries. Hence, $B_{2n}=B_{32}=0$ and so $B= t_0\Id$ is a
 scalar and therefore $d(A,X) \geq 3$.
\end{proof}

\begin{theorem}\label{thm:classification-of-rk1}
The following statements are equivalent for a non-scalar matrix $R$.
\begin{itemize}
\item [(i)] $R$ is equivalent to a matrix of rank one.
\item[(ii)] $d(R,X)\le 2$ for every nonminimal matrix $X$.
\end{itemize}
\end{theorem}

\begin{proof}
 If $n=3$, then every nonminimal matrix is equivalent to a rank-one matrix,
 so we may assume that $n\ge4$.

 To prove that $(i)\Longrightarrow (ii)$, we can assume without loss of generality that $\rk R=1$.
 Let  $X$ be an arbitrary nonminimal matrix. Then $d(R,X)\le 2$ by
 Lemma~\ref{lem:A'capR'}.

 $\neg(i)\Longrightarrow \neg(ii)$. Suppose $R$ is not equivalent to a rank-one matrix.
 Note that there exists at least one maximal matrix $M\succ R$. In fact,  $M=p(R)$
 for some polynomial $p$. Moreover, we can assume that every maximal $M\succ R$ is either a nonzero square-zero matrix
 or a non-scalar idempotent. Hence $1 \le \rk M \le n-1$. Note that $\rk M = n-1$ implies $M$ is an idempotent and therefore it
 is equivalent to a maximal matrix of rank one. So we can assume that $1 \le \rk M \le n-2$.

 If for a maximal $M\succ R$ we have $2\le \rk M \le n-2$, then by Lemma
 \ref{lem:nonminimal}   there exists a nonminimal matrix $X$ with $d(M,X)\ge 3$. Hence also
  $d(R,X)\ge 3$ because $\C(R)\subseteq\C(M)$.

 Otherwise, every maximal matrix $M\succ R$ is equivalent to a rank-one matrix. This
 implies that (i) $R$ is  either equivalent to a nilpotent matrix with exactly one Jordan block of
 dimension $3$ and all other
 cells of dimension at most $2$, or (ii) $R$ is equivalent to a matrix whose Jordan structure is equal to
  $1 \oplus J_2 \oplus 0_{n-3}$, or (iii) $R$ is equivalent to a matrix whose Jordan structure is equal to $1 \oplus J_3\oplus\bigoplus_{i=1}^k J_2 \oplus 0_{n-3-2k}$.
 In the first case, Lemma~\ref{lem:nonminimal} assures that there exists
  a nonminimal $X$ with $d(R,X)\ge 3$. In the case (ii) we have, modulo conjugation, $R=1 \oplus J_2 \oplus 0_{n-3}$. It is easy to see that
  $X=J_2\oplus J_{n-2}$ is nonminimal and $d(R,X)\ge 3$. In case (iii) we have, modulo conjugation,
  $R\prec R'=0 \oplus J_3\oplus\bigoplus_{i=1}^k J_2 \oplus 0_{n-3-2k}$. Again, Lemma~\ref{lem:nonminimal}
  gives a nonminimal matrix~$X$ with $d(R',X) \geq 3$, so also $d(R,X)\ge 3$.
\end{proof}

In the previous theorem rank-one matrices are classified with the help of matrices which
are not minimal. We next classify minimal matrices as the ones which maximize the
distance in a commuting graph.
\begin{theorem}
The following are equivalent for a matrix $A\in M_n(\FF)$.
\begin{itemize}
\item [(i)] $A$ is minimal.
\item[(ii)] There exists a matrix $X$ such that $d(A,X)=4$.
\end{itemize}
\end{theorem}
\begin{proof}
$(i)\Longrightarrow(ii)$. This follows from Theorem~\ref{thm:d=4}.

$\neg(i)\Longrightarrow\neg(ii)$.
Let~$A$ be a non-minimal matrix, and let $X$ be any matrix. By Lemma~\ref{lem:rk-1'=all},
there exists a rank-one matrix
$R$ with $d(X,R)\le 1$.
By Theorem~\ref{thm:classification-of-rk1} we have $d(A,R)\le2$, so triangle inequality gives \mbox{$d(A,X)\le 3$}.
\end{proof}

\begin{remark}
Combining the previous two theorems yields that $R$ is equivalent to rank-one matrix
if and only if $d(R,X)\le 2$ for every  matrix $X$ such that $d(X,Z)\le 3$,
for all $Z\in M_n(\FF)$.
\end{remark}

Semisimple matrices can also be classified using the distance in the commuting graph. Before doing that we need two lemmas.
\begin{lemma}\label{lem:diagonalizable1}
Suppose a minimal matrix $B\in M_n(\FF)$ is semisimple. Then
for
any $Y\path X \path B$ there exists a minimal matrix $M$ with $Y\path  M\path  X$.
\end{lemma}
\begin{proof}
Assume with no loss of generality that $B$ is diagonal. Then,
every $X\in \C(B)$ is also diagonal. Using simultaneous
conjugation on $(B,X)$ we may further assume that
$X=\lambda_1\Id_{n_1}\oplus\dots\oplus\lambda_k \Id_{n_k}$, with
$\lambda_1,\dots,\lambda_k$ pairwise distinct and
$n_1,\dots,n_k\ge1$. Now, since $Y$ commutes with $X$ we have that
$Y\in \C(X)=M_{n_1}(\FF)\oplus\dots\oplus M_{n_k}(\FF)$.
Consequently, $Y=Y_1\oplus \dots\oplus Y_k$ is block-diagonal and
we may find an invertible block-diagonal matrix
$S=S_1\oplus\dots\oplus S_k$ such that $S^{-1}XS=X$ and
$S^{-1}YS=\bigoplus_{i=1}^k S_i^{-1}Y_iS_i$ is in Jordan
upper-triangular form; say $S^{-1}YS=\bigoplus_{i=1}^s
J_{m_i}(\mu_i)$, with $m_i\ge 1$, $s \ge k$. Then we can choose
distinct $\nu_1,\dots,\nu_s\in\FF$, such that the matrix
$M=S\bigoplus_{i=1}^s J_{m_i}(\nu_i)S^{-1}$ is neither equal to $X$ nor $Y$.
Also, since $\nu_1,\dots,\nu_s$ are distinct, $M$ is nonderogatory,
hence minimal, and it commutes with $X$ and with $Y$.
\end{proof}

\begin{lemma}\label{lem:diagonalizable2}
 Suppose a minimal $B\in M_n(\FF)$ is not semisimple. Then there exist
matrices $X,Y$ with $Y\path X\path  B$, but such that no minimal matrix commutes with
both $X$ and $Y$.
\end{lemma}
\begin{proof}
With no loss of generality assume $B$ is already in its upper-triangular Jordan form,
$B=J_{n_1}(\lambda_1)\oplus\dots\oplus J_{n_k}(\lambda_k)$ with $\lambda_1, \ldots, \lambda_{n_k}$ distinct
and $n_1\ge2$. Define
$X=E_{1n_1}=J_{n_1}^{n_1-1}\oplus 0_{n-n_1}\in \C(B)$ and for an arbitrary
$k\in\{1,\dots,n\}\setminus\{1,n_1\}$ define $Y=E_{1k}$. Clearly, $X$ commutes with $Y$.
Let us show that no minimal $A$ commutes with both $X$ and $Y$. Assume $A=\bigoplus_{j=1}^s J_{n_j}(\mu_j)$ is already in its Jordan canonical form.
Since $X\in \C(A)$ is of rank one, it follows that $X\in\FF J_{n_{j_1}}^{n_{j_1}-1}$
for some $j_1$, and likewise  $Y\in\FF J_{n_{j_2}}^{n_{j_2}-1}$ for some $j_2$. However, $\rk(X+Y)=1$ and so $j_1=j_2$,
which gives $X$ and $Y$ must be linearly dependent, a contradiction.
\end{proof}

\begin{theorem}
Let $A\in M_n(\FF)$ be a non-scalar matrix. Then the following are
equivalent.
\begin{itemize}
\item [(i)] $A$ is semisimple.
\item[(ii)]  There exists a minimal $B\in\C(A)$ such that for any $Y\path     X\path
    B$ there exists a minimal matrix $M$ with $Y\path M\path X$.
\end{itemize}
\end{theorem}
\begin{proof}
$(i) \Longrightarrow (ii)$. Assume without loss of generality that $A$ is already
diagonal. Choose distinct scalars $\mu_1,\dots,\mu_n$ to form a minimal matrix
$B=\diag(\mu_1,\dots,\mu_n)$ which clearly commutes with $A$.  Then, (ii) follows from
Lemma~\ref{lem:diagonalizable1}.

 $\neg(i) \Longrightarrow \neg(ii)$. Choose any minimal $B$
which commutes with non-semisimple $A$ (at least one does exist, for example, if
$A=S\bigoplus_{i=1}^k J_{n_i}(\lambda_i)S^{-1}$, then for distinct scalars
$\lambda_1,\dots,\lambda_k$ matrices $A$ and $B=S\bigoplus_{i=1}^k J_{n_i}(\lambda_i)S^{-1}$ commute).
Since $\C(B)=\FF[B]$ it follows that $A\in \FF[B]$ which implies that $B$ itself is not semisimple. It now follows
from Lemma~\ref{lem:diagonalizable2} that there exist $X,Y$ with $Y\path X\path B$, but no
minimal matrix commutes with both of them. So (ii) does not hold.
\end{proof}

Let us conclude  with an example of a connected commuting graph  over  algebraically non-closed field
with the diameter strictly larger than~$4$.
\begin{example} The commuting graph for $M_9(\ZZ_2)$ is connected with
diameter at least $5$.

Note that $\ZZ_2$ permits only one field extension  of degree $n=9$, and this is the
Galois field $GF(2^9)$ which  contains $GF(2^3)$ as the only proper intermediate
subfield. So, by~\cite[Theorem 6]{Akbari_Bidkhori_Mohammadian} the commuting graph of
$M_9(\ZZ_2)$ is connected. To see that its diameter is at least $5$, consider an
irreducible polynomial
$m(\lambda)=\lambda^9+\lambda^8+\lambda^4+\lambda^2+1\in\ZZ_2[\lambda]$ and let
$\widehat{A}=C(m)\in M_9(\ZZ_2)$ be its companion matrix. Since $\widehat{A}$ has a
cyclic vector, $\C(\widehat{A})=\ZZ_2[\widehat{A}]$ by a well known Frobenius result on
dimension of centralizer~(see for example
\cite[Corollary~1]{Akbari_Bidkhori_Mohammadian}), and this is a field extension of
$\ZZ_2$~\cite[Theorem 4.14, pp. 472]{Joshi}  of index $n=9$. Actually, $\C(\widehat{A})$
is isomorphic to $GF(2^9)$ by the uniqueness of field extensions for finite fields. In
the sequel we will identify the two.

Since the field extension $\ZZ_2\subset GF(2^9)$ contains only $GF(2^3)$ as a proper
intermediate subfield, we see that each  $X\in\C(\widehat{A})\setminus GF(2^3)$ satisfies
$\ZZ_2[X]=\ZZ_2[\widehat{A}]=\C(\widehat{A})$ and in particular $X$ and $\widehat{A}$ are
polynomials in each other so they are equivalent. Moreover, each non-scalar
$\widehat{Y}\in GF(2^3)$ satisfies $\ZZ_2[\widehat{Y}]=GF(2^3)$, because no proper
intermediate subfields exist between $\ZZ_2\subset GF(2^3)$, and in particular,
$\C(\widehat{Y}_1)=\C(\widehat{Y}_2)$ for any two non-scalar
$\widehat{Y}_1,\widehat{Y}_2\in GF(2^3)\subset GF(2^9)=\C(\widehat{A})$.

There exists a polynomial $p$ so that $\widehat{Y}=p(\widehat{A})\in GF(2^3)\setminus\{0,1\}$. As
the field $GF(2^3)$ contains no idempotents other than $0$ and $1$ we see that the
rational canonical form of $\widehat{Y}$ consists only of cells which correspond to powers of the same
irreducible polynomials. Likewise, the field contains no nonzero nilpotents, so each cell
of $\widehat{Y}$ corresponds to the same irreducible polynomial, raised to power $1$.
Moreover, $GF(2^3)$ has no subfields other that $\ZZ_2$, so $\ZZ_2[\widehat{Y}]=GF(2^3)$ and
hence the minimal polynomial of $\widehat{Y}\in GF(2^3)$ has degree $[GF(2^3):\ZZ_2]=3$. This
polynomial is relatively prime to its derivative, so in a splitting field, $\widehat{Y}$ has
three distinct eigenvalues. It easily follows that $\widehat{Y}$ is conjugate to a matrix
$C\oplus C\oplus C$, with $C$ being a $3\times 3$ companion matrix of some irreducible
polynomial of degree $3$. Let $S_1$ be an invertible matrix such that
$\widehat{Y}=S_1^{-1}(C \oplus C \oplus C)S_1$ and define
$$A=S_1\widehat{A}S_1^{-1}.$$ Clearly then $p(A)=S_1\widehat{Y}S_1^{-1}=C\oplus C\oplus C$ and
it  follows that
\begin{equation}\label{eq:exa}
\C(p(A))=\begin{bmatrix}
\ZZ_2[C] &\ZZ_2[C] &\ZZ_2[C]\\
\ZZ_2[C] &\ZZ_2[C] &\ZZ_2[C]\\
\ZZ_2[C] &\ZZ_2[C] &\ZZ_2[C]
\end{bmatrix}.
\end{equation} Since $\ZZ_2[\widehat{Y}]=GF(2^3)$ we obtain $\ZZ_2[C]=GF(2^3)$.

 Consider a $3\times 3$ block
matrix
\begin{equation*}\label{eq:N}
N=\begin{bmatrix}
E_{13} &0 &0\\
0 &0 &E_{13}\\
E_{32} & 0 &0
\end{bmatrix}, \quad E_{13},E_{13},E_{32}\in M_3(\ZZ_2).
\end{equation*}
It is immediate that
$N^3=0$, so $\Id+N$ is invertible. Define
$$B=(\Id+N)A(\Id+N)^{-1}.$$
We will show that $d(A,B) \ge 5$.

 Suppose there exists a path $A \path V \path Z \path W \path B$ of length 4.
 Note that $V\in
GF(2^3)\subset\C(A)$. Otherwise, if  $V\in \C(A)\setminus GF(2^3)$ then $\C(V)=\C(A)$ and
such $V$ has exactly the same neighbours as $A$. Since $B=(\Id+N)A(\Id+N)^{-1}$, it follows
$W=(\Id+N)U(\Id+N)^{-1}$ for some $U\in GF(2^3)\subset\C(A)=(\Id+N)^{-1}\C(B)(\Id+N)$.
 Recall that any two non-scalar elements in $GF(2^3)$ have
the same centralizer. So in particular we might take $U=V=p(A)=C\oplus C\oplus C$ where
polynomial $p$ was defined before. For any $Z\in \C(V)\cap\C((\Id+N)V(\Id+N)^{-1})$ we
have
$$Z=(\Id+N)\widehat{Z}(\Id+N)^{-1},\qquad Z, \widehat{Z}\in \C(V)$$
and hence, by postmultiplying with $(\Id+N)$ and rearranging,
\begin{equation}\label{eq:exa-2ndeq}
Z-\widehat{Z}=N\widehat{Z}-ZN.
\end{equation}
Let us write $Z=\bigl[Z_{ij}  \bigr]_{1\le i,j\le3}$ and $\widehat{Z}=\bigl[\widehat{Z}_{ij}
\bigr]_{1\le i,j\le3}$ as $3\times 3$ block matrices and by \eqref{eq:exa} we have that
 $Z_{ij},\widehat{Z}_{ij}\in\ZZ_2[C]=GF(2^3)\subseteq
M_3(\ZZ_2)$, hence each of them is either zero or invertible. Then
(\ref{eq:exa-2ndeq}) implies
$$\bigl[Z_{ij}-\widehat{Z}_{ij}   \bigr]_{ij}=
\begin{bmatrix}
-Z_{11}E_{13}-Z_{13}E_{32}+E_{13}\widehat{Z}_{11}&E_{13}\widehat{Z}_{12}&E_{13}\widehat{Z}_{13}-Z_{12}E_{13}\\
-Z_{21}E_{13}-Z_{23}E_{32}+E_{13}\widehat{Z}_{31}&E_{13}\widehat{Z}_{32}&E_{13}\widehat{Z}_{33}-Z_{22}E_{13}\\
-Z_{31}E_{13}-Z_{33}E_{32}+E_{32}\widehat{Z}_{11}&E_{32}\widehat{Z}_{12}&E_{32}\widehat{Z}_{13}-Z_{32}E_{13}
\end{bmatrix}.
$$
Observe that each block on the left side belongs to $\ZZ_2[C]=GF(2^3)\subseteq
M_3(\ZZ_2)$, and so is either zero or invertible. On the other
hand, on the right side, each block in the last two columns  has rank at most two. We
deduce that  the last two columns on both sides are zero. In particular, comparing  the
second columns we see that $\widehat{Z}_{12}=Z_{12}=0$ and $\widehat{Z}_{32}=0$, so
$Z_{22}=\widehat{Z}_{22}$, and $Z_{32}=\widehat{Z}_{32}=0$. Putting this in the
above equation and simplifying,  the last column then gives $\widehat{Z}_{13}=0$, so
$Z_{13}=\widehat{Z}_{13}=0$, $Z_{23}=\widehat{Z}_{23}$, and
$Z_{33}=\widehat{Z}_{33}$. Also, comparing the   $(2,3)$ positions gives
$$0=Z_{23}-\widehat{Z}_{23}=E_{13}\widehat{Z}_{33}-\widehat{Z}_{22}E_{13}=e_1({\widehat{Z}_{33}}^{\tr}e_3)^{\tr}-
\widehat{Z}_{22}e_1e_3^{\tr}.$$  Moreover,
$\widehat{Z}_{33}^{\tr}e_3=\lambda e_3$ and $\widehat{Z}_{22}e_1=\lambda e_1$, $\lambda \in \ZZ_2$. Since
$\widehat{Z}_{33},\widehat{Z}_{22}\in\ZZ_2[C]$ and every vector is cyclic for $C$ we see
that $\widehat{Z}_{33}=\widehat{Z}_{22}=\lambda\Id_3$.
The matrix equation therefore simplifies to
$$\begin{bmatrix}
 Z_{11}-\widehat{Z}_{11} & 0 & 0 \\
 Z_{21}-\widehat{Z}_{21} & 0 & 0 \\
 Z_{31}-\widehat{Z}_{31} & 0 & 0
\end{bmatrix}=\begin{bmatrix}
   -Z_{11} E_{13} + E_{13}\widehat{Z}_{11} & 0 & 0 \\
    -Z_{21} E_{13}-\widehat{Z}_{23} E_{32} + E_{13} \widehat{Z}_{31}& 0 & 0   \\
  -Z_{31} E_{13}-\lambda  E_{32} +   E_{32}\widehat{Z}_{11} & 0 & 0
\end{bmatrix}.$$
Comparing the position $(1,1)$ gives by similar arguments as above that
$\widehat{Z}_{11}=Z_{11}=\mu\Id_3$. Inserting this into the equation we see after
rearrangement that the rank of the block at position $(3,1)$ is equal to $\rk((\mu -\lambda )
E_{32}-Z_{31}E_{13})\le 2$, which forces the two blocks at position $(3,1)$ to be
zero, i.e., $Z_{31}-\widehat{Z}_{31}=0=(\mu -\lambda ) E_{32}-Z_{31}E_{13}= (\mu
-\lambda ) e_3e_2^{\tr}-Z_{31}e_1e_3^{\tr}$. We immediately get
$Z_{31}=\widehat{Z}_{31}=0=(\mu -\lambda )$. Therefore,
$Z_{11}=Z_{22}=Z_{33}=\lambda\Id_3$. Finally, comparing the $(2,1)$
positions gives
$$Z_{21}-\widehat{Z}_{21}=-Z_{21}E_{13}-\widehat{Z}_{23}E_{32},$$
and arguing as above, $Z_{21}=\widehat{Z}_{21}=0$. Hence, $Z$ is scalar.
So, $\C(V)\cap \C(W)$ contains only scalars,
which gives that $d(A,B)\ge 5$.

\end{example}


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*a painful reminder of the Jim Crow South.
Call it what you will, but don't call it "removed." A headline and story in the July 17 Daily Progress made things sound pretty dire for fans of preserving the noteworthy entrance.
The city "lost a historical landmark during renovation work in May on the theater," read the Progress story. But in fact, the historic entrance is still there, as it always was. It is merely the rusty old fire escape that was removed.
"I don't know how the removal of the fire escape became the removal of the minority entrance," says George Ray, vice president of restoration on the Paramount board. "When I hear we tore down the old minority entrance, it's simply not accurate."
The black entrance will be totally restored to the way it was "the day it opened," says Ray, and the board plans to memorialize that segment of the Paramount's history.
So did the removed fire escape have any historical significance? No, says Ray.
In fact, its removal was approved by two entities that issue tax credits for historical preservation: the National Park Service and the State Historic Resources Department, as well as by the Board of Architectural Review, notes Ray.
The old entrance still has the power to pull emotions. Growing up in Charlottesville in the 1960s, Audrey Oliver had to use the "colored" entrance. "It bothers me when I go by there," Oliver said in a story in a local newsletter. "I don't like to walk there."


text:
Business TPP to deal big blow to Vietnam livestock industry
TPP to deal big blow to Vietnam livestock industry
By Bao Van, Thanh Nien News
Thursday, October 22, 2015 09:31 Email Print
How Vietnam should ride the new foreign investment wave that comes with TPP
Why Vietnam could be the biggest winner of the TPP
The biggest winner from TPP trade deal may be Vietnam
TPP will bring more benefits than problems to Vietnam: trade minister
Vietnam cries fowl as U.S. chicken suit shows TPP flip side
Australian cows are brought into Vietnam via a seaport. Beef, chicken, and pork from Australia, Japan and the US have flooded supermarkets around the country recently. Photo: Quang Thuan
Minister of Trade Vu Huy Hoang fears the livestock industry will face the biggest challenge from the Trans-Pacific Partnership (TPP), which will eliminate import tax on meat that stands currently at 15-40 percent.
Many of its members, including the US, Canada, Australia, and New Zealand, are big exporters of animal products and are sure to export large volumes to Vietnam.
Nguyen Duc Thanh, director of the Vietnam Center for Economic and Policy Research (VEPR) at the Vietnam National University in Hanoi, said the industry is less competitive mainly because of its heavy dependence on imported animal feed and remedial production technology, which have raised production costs for livestock firms.
Vietnam annually imports some 11 million tons of animal feed at a cost of more than US$3 billion, according to the Vietnam Animal Feed Association.
Animal feed accounts for up to 70 percent of breeding firms' production costs. The poor quality of breed stock, small scale of production, and the lack of advanced technology also increase costs.
It costs $2.08 to produce a kilogram of pork in Vietnam, much higher than the $1.41 in the US, and $2.53 for beef compared to $1.77 in Australia.
Van Duc Muoi, general director of Vissan, a leading meat processor, said: "Eighty five percent of livestock is now bred on small-sized farms and only 15 percent is bred on large-scale farms."
He said though Vietnam has the world's third largest number of sows – 4.5 million – it only ranks 15th-16th in pork output.
"Low productivity makes Vietnamese pork prices higher than that of other countries, and so it will be difficult for local pork to compete with imports after the TPP."
Tong Xuan Chinh, deputy head of the Animal Husbandry Department, said the increasing competition from imports would force the livestock industry to fix its weaknesses, failing which it would not survive.
Imports dominate
Livestock breeding companies are facing increased competition from imported cattle and poultry products. Beef, chicken, and pork from Australia, Japan and the US have flooded supermarkets around the country.
Their prices are even lower than those of local products, and their safety and hygiene standards are thought to be superior, and hence an increasing number of customers are opting for them.
Pork imported from Europe, the US and Canada costs 15-20 percent lower than local products. Thus, food producers and restaurants too use imported meat.
An executive at food producer Duc Viet said his firm uses pork imported from Canada and the US to make hotdogs and sausages. "Using imported products ensures our competitiveness."
Muoi of Vissan said the livestock industry has been badly hit by imported meat.
Imported cattle and poultry meat will become even more dominant after the pact comes into force, and some food producers plan to build cold storages for imported meats.
Livestock firms are facing huge losses due to an oversupply caused by excessive domestic production and growing imports.
Their situation is expected to worsen when the TPP comes into effect. After the conclusion of the TPP talks early this month, the participating nations would review legal procedures to prepare for the official signing, Hoang said, adding that the processes often takes at least 18 months.
Unable to compete with chicken imported from US, the Southeastern Poultry Association's members reported a combined loss of VND500 billion ($22.61 million) in the last six months.
Bankruptcy is unavoidable for many of them if they do not receive more support from the government.
According to the General Department of Customs, Vietnam imported 122,000 tons of beef, pork and poultry meat worth $194 million last year, two thirds of which was poultry. The country also imported nearly 181,000 live cows and buffaloes for slaughter.
Talking about the risk of failure the local animal husbandry industry could face after the TPP comes into effect, economist Pham Chi Lan said Vietnam should abandon some weak industries so that it could prioritize the development of other, more competitive industries.
Minister Hoang said Vietnam would have to restructure its agricultural sector by adopting technologies and increasing the scale of cultivation and livestock breeding to reduce costs and control quality.
TPP trade husbandry
Japanese fashion retailer Miniso comes to Vietnam
The chain plans 12 stores this year and 200 by 2021
Uber must pay tax for drivers in Vietnam: finance ministry
German companies eye young solar energy market in Vietnam
Emerging markets in longest slump since june as volatility jumps
Vinamilk seeks to buy second U.S. dairy company in growth push

redpajama_set_name: RedPajamaCommonCrawl


text:
Illustration by Matt Blease
Recycling Is Broken. Now What?
By Michele Bianchi
Patagonia is no stranger to the difficulty of throwing stuff away. We take back 100 percent of the gear you return for recycling through our Worn Wear program. In 2018, we recycled 6,797 pounds of products. But we can't recycle or repair everything you send us. Some of it was just too well-loved during use (you're totally using it right — keep at it). Other products smell too bad to be repurposed (thank you, dirtbags, for keeping the spirit — and bacteria — alive). If we can't find a market or if there isn't yet a technology to repurpose this gear, we have to choose between sending it to a landfill, sending it to the incinerator or holding on to it until we find a better solution. Since we chose the latter, our stockpile is taking up a room in our Reno warehouse. A lot of room, actually.
Our own pile of unusable clothes could very well be a metaphor for the ocean of trash we're drowning ourselves in as a global community. In 2015, in the United States alone, we generated 262 million tons of municipal solid waste (MSW). That's the equivalent of 17.5 million conventional school buses packed at full capacity with passengers. Only 91 million tons, or 34.7 percent, of that was recycled and composted. The rest ended up in landfills and combustion energy recovery, a dirty process that turns waste into heat, electricity or fuel. And in the past year, we've seen how China's ban on importing most recyclables has revealed the blue bin as magical thinking.
"We have to choose between sending it to a landfill, sending it to the incinerator or holding on to it until we find a better solution. Since we chose the latter, our stockpile is taking up a room in our Reno warehouse."
Typically, everything that you put into your blue bin gets collected and sent to a Materials Recovery Facility (MRF) where it gets sorted, compacted, shredded or melted, then turned into bales heading for domestic or foreign mills. Unless there's a nearby mill that's willing to accept your MRF's recyclable materials, it's heading somewhere else. Until recently, one of those "somewhere else" places was 7,000 miles away in China. But in January 2018, China stopped importing 24 kinds of solid wastes, including postconsumer plastics, unsorted wastepaper and waste textile materials. India followed suit, and in March 2019, completely banned solid plastic waste and scrap imports.
Give us your dirty and your torn. Not all gear you send us for repairs or recycling can get a second life. The ones that are too dirty or destroyed are being piled up in our distribution center in Reno, Nevada. And that pile is growing. Photo: Ken Etzel
The bans had an immediate seismic effect: Western countries started stockpiling recycled waste, diverting it to landfills and incinerators, or shipping it to countries in southeast Asia, like Malaysia, Thailand and Vietnam, as well as Turkey and South Korea. The increase in incineration of "recycled" stuff added to last year's record-high, human-caused carbon emissions. Animals and birds ranging from ravens and gulls to baboons and bears have all noticed, too, drawing closer and closer to cities to dig through our growing pile of trash.
Fixing the problem
There's a hidden footprint to every product that is being made, bought and then thrown away. According to the World Resources Institute, for every can of garbage at the curb, there are 87 cans' worth of materials that come from the extraction industries — such as timber, agriculture, mining and petroleum — that turn natural resources into finished products.
Annie Leonard, Greenpeace USA executive director and creator of The Story of Stuff, an award-winning documentary about our consumption-crazed culture, believes we need to think beyond recycling.
"That's not to say we shouldn't recycle," says Leonard. "But we can no longer look at it as the fix-all.
The best thing we can do for the planet is get more use out of stuff we already own, cutting down on consumption. If you have gear in need of repair, our Worn Wear program will help you fix it. Photo: Kern Ducote
"Recycling is what we do when we're out of options to avoid, repair or reuse the product first," says Leonard. "First, reduce: Don't buy what you don't need. Then, repair: Fix stuff that still has life in it. Also, reuse and share. Then, only when you've exhausted those options, recycle."
Some people and cities are living that ethos by going zero waste. The city of San Francisco is striving to become zero waste by 2030 through programs like the mandatory composting collection and a zero-waste textile initiative that turns used clothes into toys, insulation and carpeting. But living a zero-waste life can be more difficult than it should be. And not for the reasons you might think.
Take an everyday product we're all familiar with: toothpaste. In 2018, one company alone sold 80.7 million tubes of toothpaste in the United States, according to Statista. But these toothpaste tubes (like similar squeeze tubes) are made up of at least three different materials — such as plastic, paperboard, aluminum or another metal — that are hard to separate. The result? They can't be recycled through your basic municipal recycling systems.
If you don't want your toothpaste tube to end up in a landfill, you could check Earth911.com to find a recycling solution in your area. Or you could start making your own toothpaste with help from zero-waste communities online. Or you could question where all the goods and services you use come from and make better choices.
Through our Worn Wear program, Patagonia repaired 100,000 pieces of clothing in 2018. Keeping clothing in use just nine extra months can reduce the related carbon, water and waste footprints by up to a third. Photo: Donnie Hedden
Monica Wilson, associate director at GAIA (Global Alliance for Incinerator Alternatives), thinks we should be demanding more from our favorite brands. "If a product is designed to be used over and over, that's what will happen. If it's designed to be used, thrown away, trashed and then burned, then that's the fate it will have," she says.
Wilson is not alone in her thinking. Zero-waste designers believe product design should be inspired by nature, where materials are recycled in a circular loop. But so much of what is being made today is built to be used and then thrown away. In 2016, the average person bought 60 percent more items of clothing and kept them for about half as long as 15 years prior, according to a study by McKinsey.
"Designers need to think about what happens to a product at the end of its life," Wilson says. "Companies are saying they are investing millions of dollars in recycling. Wouldn't it be better if they designed their products better?"
So far this year, none of the gear you sent back to Patagonia for fixing or recycling ended up in the recycling bin. And that's a good thing. We repaired, upcycled, used toward DIY-repair education workshops, donated or resold on Worn Wear all your previously loved gear. And still, that's not enough. Just take a tour of our Reno warehouse.
To considerably shrink our stockpile of unusable gear in Reno, we are also investing in solutions that close the loop in the garment industry supply chain. The toughest nut to crack is the process of mechanically or chemically recycling products made of mixed fabrics (like a cotton and spandex T-shirt) into materials that can be given a second life. Our goal is to use only recycled and renewable ones (any natural material that can be grown or harvested on an annual basis, like wool or organic cotton) in our products by 2025.
Should we continue to recycle? Wilson says we definitely have to keep at it, but there are other things we need to do, too: "More people need to get involved in demanding better systems and asking more from brands."
The recycling rate for all textiles was 15.3 percent in 2015, with 2.5 million tons recycled, according to the EPA. What doesn't get recycled ends up in landfills or gets incinerated, which releases emissions and pollutants into our air. Photo: Tim Davis
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Fehling Instruments is a traditional family owned and family run company with more than thirty years of experience in the medical business.
FEHLING INSTRUMENTS is constantly striving for excellence in function and economy of products. This objective is achieved by continuous innovation in materials, mechanics and design. Customer satisfaction is the prevailing goal of our business. Therefore, FEHLING INSTRUMENTS provides outstanding service in addition to quality products.
FEHLING INSTRUMENTS develops, manufactures, and distributes surgical instruments, implants and single use products for use mainly in the OR. FI also provides all corresponding repair service.
The most important target markets for FEHLING INSTRUMENTS are neuro surgery (spine and brain) and thoracic, cardiac and vascular surgery.
FEHLING INSTRUMENTS' quality is top-ranked. This is a result of both quality construction and quality management. FEHLING INSTRUMENTS is certified according to ISO EN DIN 13485.


Here's a blog post by a talented math teacher, about a time in his life when he was bad at math, and how hard that was, and always is, and why.
This blog post has resonated in a raw, painful sort of way for me. So, for my third daily writing session, I'm going to write about it.
Please understand that this is a painful topic for me; what I've written about it could easily be described as whiny, or emo. I have lived a life of relative privilege, and I've gotten to do some things that many of my friends would give their eye teeth to do. What I've written could easily sound like I'm not aware of that privilege, and of course, sometimes that's the case.
Also, fair warning: this post is long.
I'm writing it, though, to convey what my life felt like, to me as I was living it. I mean to communicate that even someone who is good at something can still feel wracked by insecurity about it, can feel bad at it.
So, in one way or another, I've been struggling with math anxiety for most of my life.
I was a smart twelve year old kid who didn't know I had any talents, lying to my mom that I'd read the algebra textbook in my useless homeschooling time. I just read novels. I was sure I was an idiot at math.
I was a thirteen year old who had literally guessed my way through the state math test, sitting in a community college classroom waiting for the first day of "college algebra" (remedial upper high school stuff), pretty sure I was going to fail it. It took the best teacher I ever had to erase that fear, and he erased it in a matter of days.
I was fifteen, with an ego, having aced a bunch of calculus and diff-eq classes, but I was struggling with a Number Theory book given to me by a UT professor. I was worried that because my parents wouldn't let me go to Princeton, I was already the end of my math career. I was failing to understand the book on the first try, and having overdramatic "maybe this would all be solved if i was dead" type thoughts, as fifteen year olds are wont to do. I gave up on the book and became afraid of the subject.
I was sixteen, dropping my first class at UT, that same Number Theory. I dropped it because I was afraid of it, because I hadn't understood the book on the first try. I remember feeling such shame in the counselor's office, as if she had any reason in the world to judge me for it. I remember being worried that the department was going to kick me out of school for being a fraud, worried my parents were going to give up and pull me back home.
I was seventeen, and I had aced number theory and four or five other undergrad math classes, and I was taking my first grad class, Complex Analysis. Nobody told me there was an undergrad version I was supposed to take first. I tried, I tried hard, and then I gave up. I failed like the blog author failed: catatonically, silently, for months. It was a bad time for me. I stopped talking so much to my friends and family. I lost all my confidence. In some ways I never recovered from this one. All the mathy or academic things I've ever done since then give me a faint hint of anxiety, a faint hint of possible failure that I try very hard to ignore.
I was eighteen, having aced another ten math classes (they were all I took), including two more grad classes, and preparing to apply to grad schools, but struggling to do any work, pushing myself to do any work, because I still wasn't over the C in the first grad class. I worked hard, did eight hour homework sets on saturdays, but did it in a stressed out way, shouting at my family and friends on a regular basis, panicked that letting go would mean giving up. I was afraid of getting another C. Grownup me can't even imagine being afraid of that anymore.
I was nineteen, with wind back in my sails, even more grad classes aced, a research paper in pre-publication, newly arrived at Cambridge, starting to do work on an eighty year old unsolved problem (Artin's Little Conjecture), running up against it as hard as I could, still feeling so anxious about my talents, investing so much of myself in it. I told myself, regularly, that solving this problem was the litmus test of whether I could do great things in my field, whether I deserved to be there at all.
I remember asking the favorite prof about it in his office, and getting a careful "you're going to risk your doctorate" (he was right) and a dismissive "and anyway you're just a Part III student, I don't have time for you" (he was a jerk)… and that wrecked me. It's pathetic to admit, right in keeping with the blog post, but I don't think I've had a day of true mathematics ambition since that day.
This is the kind of thing that happens when you care very much about something and you are deathly afraid to fail it. Little things are greatly magnified, and you handle them poorly and derail yourself.
I was twenty, putting off preparing for my Part III exams, entirely, until the last three or four weeks, because I was sure I was going to fail. I was headed back to UT, having not gotten into MIT's grad math program, having chosen Texas from homesickness, and I was already sure I was a failure as a mathematician, just for the little reasons described above. I passed my tests and got my degree, but I would quit mathematics altogether within a year.
I was twenty one, a math grad student at UT, but completely unwilling to work. I didn't sleep well, I didn't leave the house often enough. I never ever showed up in my shared office. I stopped talking to my friends in the department (sorry, Todd, Shel, etc. It was a hard time in my life). In one of my class finals, I became so convinced I couldn't do the work, so unwilling to even try, that I angrily drew a comic ("Signs you shouldn't be a mathematician"), turned it in, and stormed out.
The department chair told me, in a kind and gentle way, that I was risking probation, and asked if I was okay. I had nothing to say; a month later I went back and asked him if I could quit, said I wanted to transfer to being an undeclared undergrad. He was kind about this. I felt so much shame. So much completely unnecessary shame.
I'd like to make an aside to point out how little any of this had to matter in my life. None of this needed to be important. None of matters as much as the first time I kissed my wife, the first time I juggled fire, the best day I've spent with my brother. None of this will matter on that future day where I first see my child. But we choose what matters, and we have to choose it again every day. Sometimes we choose wrong.
I'm still not even done.
I was twenty two, in the middle of a second undergrad in either chemistry or computer science (I tried two things, because I was worried I wouldn't even be able to do at least one of them), having dropped numerous classes in both, struggling to get out of bed and do my work, struggling to feel like I could do any useful work at all. Feeling old, feeling like a public failure, like the blog post says. Feeling old and failed, at twenty two.
I was twenty four, twisting and drifting in an internship at Amazon, feeling too old to be an intern anyway, in a team where the manager had left after three weeks and we were unmanaged. I was struggling with real professional code, convinced I couldn't make sense of it, that I was finally going to be run out of comp sci for not being an engineer.
It was a twelve week internship and I got stuck on some small stumbling blocks… and I sat doing nothing for the middle five weeks of that internship. I was just plain afraid of the anxiety that came with being stuck. It took a coworker taking me for a walk around the block and pointing out that I was in trouble of not getting anything done, and asking where I was specifically stuck and pushing to help me, took all of that, to get me to finish. I did finish with some good stuff.
I was twenty six, at Microsoft in the wrong group, feeling (or imagining) the daily contempt of the engineers around me for not being able to operate in the Hotmail environment, struggling to make myself show up at all. At night I was listening to my friends Dan and Avi talk about Stanford grad life, looking through my mental window like an urchin outside a banquet, wanting that experience back but feeling like I had closed it off to myself forever, because at this point I couldn't even remember most of what I had worked on, and because… how could I go back and fail that again? I couldn't even follow a lot of Avi's conversation without a lot of refresher and explanation and stumbling.
And then I committed to a plan to turn this around. I would go to Seattle, do a different kind of work, use the University of Washington to get back into academic life. For a while it was great! I had the promise of mathematical work (eventually) at Amazon, even turned down that same promise at Swype, and was taking a grad quantum chemistry class, something totally outside my expertise, but also totally outside my old social groups and history. I had no fear, so I did well, got the highest or maybe second highest scores in the class.
Then I failed. I couldn't handle Amazon and a grad class at the same time. I wrecked up my health by giving up exercise and eating all kinds of bad food, got heavier again, started getting exhausted at random times, as one does when one gains a lot of weight quickly, and finally I just had a really bad morning for thinking, and I completely failed my grad chemistry final. Like, bottom fourth of the class, after being the top.
The professor still really liked me, and I could have recovered for sure, taken the hit and continued the plan… but I felt that old failure anxiety, that old "I am really, secretly bad at this" anxiety. I bailed on my plan completely and utterly. Some of my friends ask whether I am considering going back to grad school… and I talk about my maturity level and whether I've aged out of math, but behind that talk I'm just frankly really damned uneasy about trying again and failing again. This math/academic/science anxiety hasn't really ever left me.
I was twenty eight, finally starting the aforementioned mathy part of my Amazon job… sitting sullenly in my corner playing Tiny Wings, or looking out the window. Unwilling, more or less completely unwilling to believe in myself again after such a bitter failure.
I was thirty, this afternoon, at Google with an awesome machine learning project, now five or six months overdue for a variety of life and personal and team and company and whatever reasons… with good, clear and reasonable steps to take to move the project forward… unwilling to do so because I think maybe I will be fired, and because I don't actually know machine learning well enough, and because I am still afraid of trying and failing.
These are some of the moments where I held myself to the most unreasonable standards, afraid that because I wasn't the prodigy I set out to be, I wasn't going to amount to anything. I still struggle not to feel that, right up to today. I'm still comparing to an imaginary ideal.
But why? The world is an amazing place, whether I am here to enjoy it or not, built on millions of beautiful incremental advances, the kind I could aspire to. There are always going to be people with talents totally beyond mine, whether I admit it or not, and ultimately it's my choice how to feel about all this, so I can't really justify feeling so bad.
I'm going to try not to feel so bad.
The man in the video has been roundly and soundly criticized for having his child in the car, a few hundred feet away from the plant. People can't imagine someone could be so stupid.
I have a different take. In fact, the video makes me feel guilty for all the other times I've held people (including myself) to impossibly high standards.
What he did is dumb, but not impossibly dumb. I could easily see myself doing that. Before the explosion, it's just a fire. A chemical plant fire, admittedly, but it's something you could forget.
Of course, we don't think we're all that special, nor do we think we've reached our potential… but we do think we have a consistent dignity, a certain nobility, some intelligence.
Every day, maybe almost every hour, we do something this profoundly stupid. Each of us. In the last hour I've forgotten that I was actually writing this piece. In the last hour, you've might have forgotten you were having a conversation. You might have put your keys on the little table behind the couch where you won't see them. Maybe bumped into a table on your way to the kitchen. Both of us probably spent a moment re-affirming our belief in something as implausible as running through walls.
I think the man in the explosion video was having such a moment. I think almost everybody I've ever judged was having such a moment, a derpy moment, a hyucksome moment.
All this, in a one minute theme song.
This is you, this is me, this is everybody, unpredictably, many times, every day, for our whole lives. With, you know, some Luke Skywalker thrown in.
Pity, not judgement. Pity, and maybe good natured laughter. That's what I keep telling myself.
Here's another game, clone of an old Popcap favorite of mine, the little known Alchemy.
Alchemy is a stone placing game where you match rune stones by color and symbol to make gold.
Here's the game, and as always, if you care, here's the source.
Oh, and here's Popcap's unpleasant Java based port of the game.
My game has four levels of increasing difficulty. I've already spent a lot of time happily play testing it; the last level is wicked hard, at least for me.
So I wrote myself a short python script to count mp3s in a folder, and to tell me which and how many mixed folders I had. Here's how I did it: countMp3s.py (and here's a version with more comments). I thought this would make a nice python tutorial for beginners with just a little bit of programming experience.
Let's take a look at the code, and then walk through it step by step.
print "Please supply a path argument; this is the folder path to your music folder"
Here, we're asking python to import a bunch of extra tools for us to use. 'os' is operating system. This gives us powers to do operating system type stuff, like listing folders, moving files, stuff like that. 'sys' is system. This gives us the power to read things from the command line. 'fnmatch' is file name match. It will let us do filename wildcards, in this case '*.mp3'.
to run our program. The first part tells the computer to run python. The next two parts are called arguments; they're extra pieces of information for python to use. The first argument is the name of our script, countMp3s.py. With this, we're telling python to run our script. The second argument is going to tell our program where to find music.
This code checks to see whether the script has been given two arguments. If it hasn't, the script asks the user to try again, and quits.
Sys.argv is the list of arguments. It should have two things: the script name, and the location of music.
Sys.argv is how we get the location of the music. The bracketed 1 means "get me thing 1 from Sys.argv". If we had said Sys.argv, we would have gotten the script name. We put the location of music in a variable named path.
Here we're creating counters for the number of mp3s, number of folders, and number of weird folders. They all start at zero.
os.walk takes a path (the location of our music) and "walks" through every sub folder. In each case, you get three things: the path to the sub folder (which we call sub_path), a list of folders inside this folder, and a list of files in this folder. Since we're in a for loop, the next few lines of code (everything which is indented) will happen repeatedly, once for every sub folder.
This one might be self explanatory. It's equivalent to saying: please give me every file in the list of files, if the file matches "*.mp3".
len(mp3files) is the length of the list of mp3files. However long that list is, that's how many mp3s are in this folder. We add this to the total mp3 count. We also add one to the total folder count.
If this folder has more than zero sub folders and more than zero mp3s, it's a weird folder (remember, that's something I'm looking to count). We print out the location of this folder, and add one to the weird folders count. Okay, we're done with the loop. The rest of this stuff happens just once, after the loop is over and everything has been counted.
We know the number of weird folders, but I also want to know the fraction. So, assuming the total number of folders isn't zero, I divide number of weird folders by total number of folders to get the fraction.
Now we just print out all the numbers. You can put a variable into the print statement by use of %s (for strings), %d (for integers), or %f (for fractional or floating point numbers).
So I have 25,191 mp3s, and just under 1% of my folders are weird. Yay!
You can do a lot of really powerful things with the 'os' tools in python. You could remove all the duplicates from your music collection. You could remove "The" from the beginning of every file name. You could copy all the folders with more than 10 files to a different location.
If you have an idea for something to do with these powers, leave me a comment. I might write another few of these tutorials based on reader input.


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Smiles, natural ones & a lot of them
Family portrait shoots can be small, say two siblings with parents, mum & the new born baby, dad and baby – or they can be big, with grandparents, cousins, dogs or cats. We love to photograph them all and capture the things that make your family special - the little looks, cheeky grins and bonds that bring you all together. We tend to capture a number of informal family pictures; we don't believe you all need to be sitting still or even facing the camera for a successful family portrait. It's more about having an enjoyable family time together with the memories being captured by our cameras and the images to be cherished year after year.
Please click the slider buttons <> to see more pics.
Our professional photography studio is one of the biggest and most well equipped photography studio in Northern Ireland and is also disabled friendly, has free car parking available for guests and baby changing facilities etc. Children are kept entertained and spirited with sweets and activities like Foosball, Darts in the short breaks during a photo-shoot. Tea, Coffee, Chilled beverages and light refreshments are always available.
For us, a family shoot involves you being together, and an experience to be enjoyed, whether that be in a formal line-up, or a picture that is full of action and movement as you play together. Our family fun portrait package is a great gift.
FAMILY FUN PACKAGE : £200
Upto Two Hour's Fun Filled Family Portrait Photography Package
in which after being welcomed to the studio with tea/coffee/soft drink,
we show you a lot of reference images and ideas for poses and styles of pictures
and shoot as per the ones of your liking.
Our Customers Love this!
A Relaxed & Enjoyable upto 2 hours session at our Studio.
Any number of poses with as many shots as possible.
Any number of clothing changes.
Gift Vouchers are available.
Two 7"x 5" Photo Frames (6"x 4" prints) worth £30.
10 edited high resolution pictures.*
*Images sent via download link over internet.
A wide range of custom printed products including Canvas, Poster & Calendars, Mugs, T-Shirts, Diaries can be ordered after the photo-shoot at very competitive prices.
© 2023 by MagicEye Design


Bristol boasts a lively and welcoming gay scene with numerous LGBT clubs, pubs and bars. Below you'll find recommended events and LGBT venues, recommended by local organisations.
Frogmore Street, just off Park Street, is buzzing with activity most nights of the week. Lively bars and nightclubs attract a mixed crowd looking for a great night out. Head to the QueenShilling - Bristol's longest established gay venue, or to OMG.
Pubs and bars in the Old Market area include The Old Market Tavern, The Phoenix, The Gin Palace, Bristol Bear Bar. The Old Market Assembly hosts quarterly LGBT+ night DTYM as well as regular LGBT+ theatre nights.


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Q: How do I enable JMX/Management Agent on running JVM? I'm trying to get JMX metrics from a currently running JVM. I'd like to be able to do this without having to enable it on the CLI when running the app. I'm trying to write a generic CLI utility, with some features missing from JConsole & JMC. Both JConsole and JMC do this. However there doesn't seem to be any documentation anywhere on how they are doing it.

A: Depends. If you just want to get it from a locally running app, you can use the attach API. Here is an example on how to use the attach API to run diagnostic commands:
http://hirt.se/blog/?p=542
You can also use the attach API to get at the stub for local JMXRMI communication.
If you want to start up the remote agent, you can do that using the jcmd and the ManagementAgent.start diagnostic command.

A: A little bit off-topic for cli app developer, but may be useful for command line users :
You can activate / deactivate  "remote listening JMX agent" of a running JVM started without anything particular regarding JMX by using the jcmd command. Constraint: command must be run on the same host and with same user than the targeted jvm
enable JMX agent:
 jcmd <jvm_pid> ManagementAgent.start \
    jmxremote.port=<listen_on_tcp_port> \
    jmxremote.authenticate=false \
    jmxremote.ssl=false

disable JMX agent:
jcmd <jvm_pid> ManagementAgent.stop

redpajama_set_name: RedPajamaStackExchange


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Oct 5 Migrants: Not much discussion on the opportunities migrants offer to NZ businesses - HT Group's research
Economy, News
HTG and social trends research company, Windshift, have just completed a major research project among migrants.
"We spent 100 hours conducting in-home interviews as well as a national survey comparing NZ born and migrant New Zealanders. We found that migrants tend to be highly educated middle class people who are savvy consumers and well disposed towards New Zealand brands. Migrants are loyal to brands that clearly demonstrate they value migrants' business. Sixty five percent are educated to degree level or above compared with just 36% of those that are New Zealand born. And 72% consider themselves financially stable or prospering, compared with 54% of New Zealand born," notes Windshift.
"There has been much discussion about immigration recently, but none has really centred on what a huge opportunity this group offers New Zealand businesses," added HT Group's MD Mike Hall-Taylor.
Most companies base their thinking around the traditional bicultural Kiwi model rather than thinking about the messages and actions that will make immigrant consumers feel embraced by and engaged with a brand, Hall-Taylor says.
"They may use diverse actors in their advertising or translate communications into different languages but it is about incorporating diverse thinking into campaigns at the outset."
The research also found that the top five Kiwi brands rated by migrants are Pak n Save, TradeMe, Whittakers, Air New Zealand and ASB.
"All were felt to be friendly, offer cost savings or make life easier for migrants. Pak n Save, for example, was recognised for having a wide range of ethnic products and for being much better value for money. ASB was acknowledged for actively makes life easier for new migrants, speaking their language and offering hassle free service," it said.
Other findings:
Migrants with poor English would avoid products and services where communication was difficult. For example food products that don't have a product picture on the packaging. Or the technicalities of insurance which would be difficult without a broker who could also act as a translator.
Often describing themselves as people who "think outside the box", compared with their friends that remained in their home country, the majority of migrants considered themselves cosmopolitan, experimental, ambitious, hard-working and optimistic about their new life.
Migrants felt the New Zealand culture allowed for a new healthier lifestyle, with 38% of long-term migrants on a low-sugar diet (compared to 22% NZ born) and almost a quarter (24%) of recent migrants eating organic (compared to 14% NZ born).
Oct 5 Debate: Candidates face-off on climate issues
Oct 5 Multilingualism: Acquisition and maintenance of heritage and community languages in multilingual Christchurch


text:
Rails.application.config.assets.version = '1.0'

Rails.application.config.opal.method_missing           = true
Rails.application.config.opal.optimized_operators      = true
Rails.application.config.opal.arity_check              = !Rails.env.production?
Rails.application.config.opal.const_missing            = true
Rails.application.config.opal.dynamic_require_severity = :ignore

# Precompile additional assets.
# application.js, application.css, and all non-JS/CSS in the app/assets
# folder are already added.
Rails.application.config.assets.precompile += %w( debug.js debug.css
                              slides.css reveal.js highlight.js)

Rails.application.config.assets.paths << "#{Gem.loaded_specs["rubyx-debugger"].gem_dir}/assets/"
Rails.application.config.assets.paths << "#{Gem.loaded_specs["rubyx-debugger"].gem_dir}/lib/"

redpajama_set_name: RedPajamaGithub