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\begin{document}
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\noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=1.75]{BJMA.jpg}}
\noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent
Banach J. Math. Anal. 2 (2008), no. 2, 107--120\\
$\frac{\rule{4.55in}{0.05in}}{{}}$\\
{\footnotesize
\textcolor[rgb]{0.65,0.00,0.95}{\textsc{\textbf{\large{B}}anach
\textbf{\large{J}}ournal of \textbf{\large{M}}athematical
\textbf{\large{A}}nalysis}}\\
ISSN: 1735-8787 (electronic)\\
\textcolor[rgb]{0.00,0.00,0.84}{\textbf{http://www.math-analysis.org }}\\
$\frac{{}}{\rule{4.55in}{0.05in}}$}\\[.5in]}

\title[Spectral Radius Preservers]{Spectral radius preservers of \\
products of nonnegative matrices$^{*}$}

\author[S. Clark, C.K. Li, L. Rodman]
{Sean Clark$^1$, Chi-Kwong Li$^2$$^*$ and Leiba Rodman$^3$}
\address{$^{1, 2, 3}$ Department of Mathematics,
College of William and Mary, Williamsburg, VA 23187-8795, USA.}
\email{\textcolor[rgb]{0.00,0.00,0.84}{siclar@wm.edu},
\textcolor[rgb]{0.00,0.00,0.84}{ckli@math.wm.edu},
\textcolor[rgb]{0.00,0.00,0.84}{lxrodm@math.wm.edu}}

\dedicatory{This paper is dedicated to Professor Josip E. Pe\v{c}ari\'{c}\\
\vspace{.5cm} {\rm Submitted by M. Bre\v sar}}

\subjclass[2000]{Primary 15A18; Secondary 15A04, 15A60.}

\keywords{Preserver, spectral radius, Jordan triple product,
nonnegative matrices.}

\date{Received: 13 April; Accepted: 17 June 2008.
\newline \indent $^{*}$ Corresponding author}

\begin{abstract} A characterization of nonlinear spectral radius
preserving maps is obtained for the usual and triple products of
nonnegative matrices.
\end{abstract}

\maketitle

\section{Introduction}
Preserver problems concern the characterization of maps on matrices
or operators leaving invariant a certain function, a certain subset
or a certain relation. Earlier studies focused on {\em linear} maps
with these properties. The literature on this subject is extensive;
see, for example,  \cite{LP, S} and monographs \cite{Mo, ODA, Su}.
Recently, researchers have studied  preservers under mild
assumptions. For instance, for a given function $\nu$ on a matrix
set ${\bf M}$ with a binary operator $A\circ B$, researchers study
maps $f: {\bf M} \rightarrow {\bf M}$ satisfying $\nu(f(A)\circ
f(B)) = \nu(A\circ B)$ for all $A, B \in {\bf M}$; \cite{CLS, DS,
TW} is a small selection of recent work on the topic. There has been
interest in studying such problem when $\nu(A)$ is the spectrum, the
peripheral spectrum, the numerical radius, the spectral norm, etc.
Moreover, these problems have also been considered for in more
general contexts such as functions on operator algebras or uniform
algebras \cite{Mo, LT, RTT}; the latter two papers served as a
motivation for our study of peripheral spectrum preservers.

It is worth noting that
even without the linearity assumption, the preservers often end up
to be linear and have certain ``standard'' or ``expected'' form.
Even though the statements of results look similar to those of
linear preservers,
researchers often have to develop new techniques to
solve the preserver problems with mild assumption.
In some cases, one may get unexpected preservers, which lead
to better understanding and insight to the subjects under consideration.

The purpose of this paper is to characterize preservers of
the spectral radius, spectrum and peripheral spectrum
of the product or the Jordan triple product on entrywise nonnegative matrices.
We note that the literature on preservers in the context of
entrywise nonnegative matrices is meager; see \cite{Mi}.

\medskip
The following notation will be used throughout the paper:

$M_n^+$ stands for the set of $n \times n$ real matrices with
nonnegative entries.

We assume $n\geq 2$  throughout.

$E_{ij}\in M_n^+$ the matrix unit: $1$ in the $(i,j)$th position and zeros
everywhere else.

$r(A)$ the spectral radius of a matrix $A$.

$\sigma (A)$ the spectrum (the set of eigenvalues) of a matrix $A$.

$\sigma_p (A)=\sigma (A) \cap \{\lambda \in \mathbb{C}\, :\, |\lambda|=r(A)\} $
the peripheral spectrum of $A$.

$\tr(A)$ the trace of $A$.

$p_A(t)={\rm det}\, (tI-A)$ the characteristic polynomial of $A$.

$A^t$ the transpose of $A$.

$\diag(x_1,\ldots ,x_n)$ diagonal matrix with $x_1,\ldots, x_n$ on
the diagonal (in that order)

$\calP\subseteq M_n^+$ the group of permutation matrices.

$\calP\calD\subseteq M_n^+ $ the group of matrices of the form
$PD$ where $P$ is a permutation matrix and $D$ is a diagonal matrix with
positive entries on the diagonal. Note that $A\in M_n^+$ has the property that
$A$ is invertible and $A^{-1}\in M_n^+$ if and only if $A\in \calP\calD$
(see, e.g., \cite{LR} for a proof).

\medskip
Here is our main theorem.
\begin{theorem}\label{main0} For $A, B \in M_n^+$,
let $A*B$ denote the usual product $A*B = AB$ or the Jordan triple
product $A*B = ABA$. Then the following statements
${\rm (1)}$ - ${\rm (4)}$ are equivalent for a surjective function
$f:M_n^+ \ \longrightarrow \ M_n^+$:

\begin{itemize}
\item[${\rm (1)}$]
\begin{equation}\label{maincondition} r(A*B)=r(f(A)*f(B)),
\qquad (A,B\in M_n^+).\end{equation}
\item[${\rm (2)}$]
\begin{equation*} \sigma_p (A*B)=\sigma_p(f(A)*f(B)),
\qquad (A,B\in M_n^+).\end{equation*}
\item[${\rm (3)}$]
\begin{equation*} \sigma (A*B)=\sigma (f(A)*f(B)),
\qquad (A,B\in M_n^+).\end{equation*}
\item[${\rm (4)}$]
There exists a matrix $Q\in \calP\calD$
such that either
$$ f(A)=Q^{-1}AQ, \qquad  (A\in M_n^+), $$
or
$$f(A)=Q^{-1}A^tQ, \qquad  (A\in M_n^+).
$$
\end{itemize}
\end{theorem}

Note that in Theorem \ref{main0} the function $f$ is not assumed
to be linear or multiplicative \it a priori. \rm

The result of Theorem \ref{main0} for $A*B = A+B$
was obtained in \cite{LR} without the surjective assumption.
It would be interesting to remove the surjective assumption in
Theorem \ref{main0}. We are not able to do that at present.

Since for $A\in M_n^+$ we always have $r(A)\in \sigma_p (A)$, the
implications (3) $\Longrightarrow$ (2) $\Longrightarrow $ (1) are trivial.
Also, (4) $\Longrightarrow$ (3) is easy to verify.
It remains to show that
(1) implies (4). We first present some preliminary and auxiliary results
in Section 2. In particular, we prove a function
$f:M_n^+\rightarrow M_n^+$ having some special properties on
matrix units will have the nice form described in
Theorem  \ref{main0} (4).
Then we show that a function $f: M_n^+ \rightarrow M_n^+$ satisfying
Theorem \ref{main0} (1) will possess the special properties
on matrix units, and hence $f$ has the form in
Theorem \ref{main0} (4). This is done in Sections 3 and 4
for the usual product and Jordan triple product, respectively.

\section{Preliminaries}
In this section we present some known results and easy  observations
that will be often used, sometimes without explicit  reference,
throughout the paper. We list several well-known properties of
nonnegative matrices and their spectral radii (see, for example,
\cite[Theorem 8.4.5]{HJ} or \cite{BP}).

The following two observations are useful when considering the triple product.
Let
$\sqrt{E_{ij}}=\begin{cases}
    E_{ik}+E_{kj}: k\ne i,j & \text{if } i\ne j \\
    E_{ii} & \text{if } i=j
\end{cases}$ for which a trivial calculation shows $\sqrt{E_{ij}}^2=E_{ij}$.
Clearly our choice of the specific $k$ in the above definition does not matter
so long as it respects our constraint in each case. Note that this construction
requires $n \ge 3$, so $n = 2$ will be covered separately.

\medskip

Since $r((BA)B)=r(B(BA))=r(B^2A)=r(AB^2)$,
we will use the following three equivalent conditions
for the triple product interchangeably:
\begin{eqnarray*}
r(BAB)=r(f(B)f(A)f(B)) \\
r(B^2A)=r(f(B)^2f(A)) \\
r(AB^2)=r(f(A)f(B)^2).
\end{eqnarray*}

\begin{lemma} \label{bijective}
Let $f:M_n^+ \ \longrightarrow \ M_n^+$ be a map that satisfies
{\rm (\ref{maincondition})} and is surjective. Assume further $n\ge 3$
if $A*B$ is the Jordan triple product.
Then $f$ is bijective. \end{lemma}

\begin{proof}
Since we assume surjectivity, we will prove injectivity.
Suppose  $A,B\in M_n^+$ satisfy $f(A)=f(B)$.
For any $(i,j)$ pair with
$1 \le i,j \le n$,
since $AE_{ij}$ has all columns zero except for the $j$th column, and
the $j$th column of  $AE_{ij}$ is just the $i$th column of $A$, we have
$r(A\sqrt{E_{ij}}^2)=r(AE_{ij})=a_{ji}$.
Similarly $r(B\sqrt{E_{ij}}^2)=r(BE_{ij})=b_{ji}$. Then by our
spectral radius conditions,
$$a_{ji}=r(AE_{ij})=r(f(A)f(E_{ij}))=r(f(B)f(E_{ij}))=r(BE_{ij})=b_{ji},
$$
or
$$a_{ji}=r(f(A)f(\sqrt{E_{ij}}^2))=r(f(B)f(\sqrt{E_{ij}})^2)
=r(B\sqrt{E_{ij}}^2)=b_{ji}.$$
Thus, $A = B$.
\end{proof}

\medskip

\begin{remark}\label{bijecremark}
Since $f$ is a bijection, it is simple to observe that its inverse
$f^{-1}$  fulfills (\ref{maincondition}) if $f$ does,
i.e., $$r(f(A)*f(B))=r(A*B)=r(f^{-1}(f(A))*f^{-1}(f(B))). $$
\end{remark}

\medskip
The following observations will be used throughout our discussion.

\begin{lemma}\label{PreLem}
Assume that the function $f: M_n^+ \rightarrow M_n^+$
satisfies condition {\rm (1)} in Theorem ${\rm \ref{main0}}$. Then:


{\rm (a)} For any $A \in M_n^+$ we have $r(A) = r(f(A))$.

{\rm (b)}  $A\in M_n^+$ is nilpotent if and only if $f(A)$ is nilpotent.

{\rm (c)}  If  $A\in M_n^+$ is nilpotent, i.e., $r(A)=0$,
then all diagonal elements of $A$ and $f(A)$ are zeros.

{\rm (d)} If  in addition the range of $f$ contains a matrix with
positive entries, then
$A$ is nonzero if and only if $f(A)$ is nonzero.
\end{lemma}

\begin{proof}
Condition (a) follows from
setting $A=B$ in (\ref{maincondition}).
Condition (b) follows trivially from (a).

Condition (c) follows from nilpotency and nonnegativity. A nilpotent matrix
has all zero eigenvalues, so the sum of all eigenvalues, and equivalently the
trace, is zero. Since the trace is the sum of the diagonal entries, all of
which are nonnegative, we finally obtain that the diagonal entries must all be
zero.   By (b), we get the conclusion on $f(A)$.

For condition (d), let $A \in M_n^+$, and let $X\in M_n^+$
be the  matrix with all entries equal to $1/n$. Then $X^2 = X$,
and $\tr(A*X) =  \sum_{i=1}^n\sum_{j=1}^n a_{ij}/n.$
Hence, if $A \ne 0$, then $\tr(A*X) \ne 0$ and hence
$0 \ne r(A*X) = r(f(A)*f(X))$. It follows that $f(A) \ne 0$.
Similarly, if $f(A) \ne 0$ and if $f(Y) = Z$, where $Z$ is a matrix
with all entries positive in the  range of $f$,
then $\tr (f(A)*Z)\neq 0$. So $0 \ne r(f(A)*Z) = r(A*Y)$ and hence $A \ne 0$.
\end{proof}

\begin{theorem}\label{UnitTheorem}
Let $f:M_n^{\rm +}\rightarrow M_n^{\rm +}$, $n\geq 3$, be such that
\begin{equation} \label{apr51} r(A)=r(f(A))\quad (A\in M_n^+),
\end{equation} and let
${\cal N}=\{(i,j): 1\le i, j \le n\}$.
Assume that there exist a permutation $\tau$ on the set $\cal N$
and a collection of positive numbers $\{\gamma_{ij}: (i,j) \in {\cal N}\}$
that satisfies
\begin{enumerate}
\item $\gamma_{ij}=1/\gamma_{ji}$,
\item $\tau(i,j)= (p,q) \Rightarrow \tau(j,i)=(q,p)$,
\end{enumerate}
and $f$ has the property that
$$f(\displaystyle\sum_{(i,j)\in S} E_{ij})
=\sum_{(i,j) \in S}\gamma_{ij}E_{\tau(i,j)}$$ for all $S \subseteq
{\cal N},\ S \ne \emptyset$. Then there exists a matrix $Q\in
\calP\calD$ such that either
\begin{equation}\label{apr55} f(E_{ij})
=Q^{-1}E_{ij}Q,\qquad ((i,j) \in {\cal N}),\end{equation} or
\begin{equation}\label{apr56} f(E_{ij})=Q^{-1}E_{ij}^tQ,
\qquad ((i,j) \in {\cal N}). \end{equation}
\end{theorem}
\begin{proof}
Let
$F_{ij} = f(E_{ij})$ for $1 \le i, j \le n$.
We adjust our map $f$ via $X \mapsto P^tf(X)P$ for a suitable permutation
matrix $P$ so that $F_{jj} = E_{jj}$ for $j = 1, \dots, n$.
We proceed in 4 steps:

\medskip\noindent
{\bf Step 1.} We show that $F_{ij} = \gamma_{ij}E_{ij}$
or $F_{ij} = E_{ji}/\gamma_{ij}$. We may assume $i\neq j$.
\medskip
Let $F_{ij}=\gamma_{ij}E_{pq}$, and assume $p \ne i, j$.
Consider $A=E_{ij}+E_{ji}+E_{pp}$.
Then clearly
$r(A)=1$. But $f(A)=\gamma_{ij}E_{pq}+E_{qp}/\gamma_{ij}+E_{pp}$, and
$p_{f(A)}(t)=t^n-t^{n-1}-t^{n-2}=t^{n-2}(t^2-t-1)$,
so $r(f(A))=\frac{1+\sqrt{5}}{2}>1=r(A)$, a contradiction.
Similarly, assume $q \ne i,j$, and we reach the same
contradiction. In view of (\ref{apr51}), we cannot have $p=q$.
Therefore $F_{ij}=\gamma_{ij}E_{ij}$ or $F_{ij}=E_{ji}/\gamma_{ij}$.

\medskip\noindent
{\bf Step 2.} We show that $F_{1j} = \gamma_{1j}E_{1j}$
and $F_{j1} = E_{j1}/\gamma_{1j}$, after possible replacement
of $f$ by a map of the form
$X \mapsto f(X)^t$.
\medskip

We may assume that $F_{12} = \gamma_{12} E_{12}$
and $F_{21} = E_{21}/\gamma_{21}$; otherwise, use the map
$X \mapsto f(X)^t$.
Now consider $A=E_{12}+E_{2j}+E_{j1}$ for $3 \le j \le n$.
Then $r(A)=1=r(f(A))$.
But $$f(A)=F_{12}+F_{2j}+F_{j1}=\gamma_{12}E_{12}+F_{2j}+F_{j1}, $$
and if $F_{2j}=\gamma_{j2}E_{j2}$ or $F_{j1}=\gamma_{1j}E_{1j}$, then
$f(A)$ is nilpotent, and $r(f(A))=0$, a contradiction.
So $F_{j1}=E_{j1}/\gamma_{1j}$ which gives us $F_{1j}=\gamma_{1j}E_{1j}$.


\medskip\noindent
{\bf Step 3.} We show that $F_{ij} = \gamma_{ij}E_{ij}$
and $F_{ji} = E_{ji}/\gamma_{ij}$.
\medskip

We only need to consider the case $1<i<j \le n$.
So let $A=E_{1i}+E_{ij}+E_{j1}$, and we have
$$f(A)=F_{1i}+F_{ij}+F_{j1}=\gamma_{1i}E_{1i}+F_{ij} + E_{j1}/\gamma_{1j}.$$
But again, if $F_{ij}=E_{ji}/\gamma_{ij}$
then $f(A)$ is nilpotent. But then $r(A)=1>0=r(f(A))$, a contradiction.
Therefore, $f(E_{ij})=\gamma_{ij}E_{ij}$ and $f(E_{ji})=E_{ji}/\gamma_{ij}$.

\medskip\noindent
{\bf Step 4.} We show $\gamma_{ij}=\gamma_{1j}/\gamma_{1i}$.
\medskip

Consider the same matrix $A=E_{1i}+E_{ij}+E_{j1}$, with $r(A)=1=r(f(A))$.
But $f(A)=\gamma_{1i}E_{1i}+\gamma_{ij}E_{ij} + E_{j1}/\gamma_{1j}$, so
$$p_{f(A)}(t)=t^n-\gamma_{1i}\gamma_{ij}/\gamma_{1j}t^{n-3}
=t^{n-3}(t^3-\gamma_{1i}\gamma_{ij}/\gamma_{1j}). $$
Then $$\sigma(f(A))=\{0, \sqrt[3]{\gamma_{1i}\gamma_{ij}/\gamma_{1j}},
\omega\sqrt[3]{\gamma_{1i}\gamma_{ij}/\gamma_{1j}},
\omega^2\sqrt[3]{\gamma_{1i}\gamma_{ij}/\gamma_{1j}}\} $$
(zero is present only if $n>3$),
where $\omega$ is the primitive cubic root of 1.
Since $r(f(A))=1$, we have
$\sqrt[3]{\gamma_{1i}\gamma_{ij}/\gamma_{1j}}=1$,
so $\gamma_{1i}\gamma_{ij}/\gamma_{1j}=1$.
Our conclusion follows.

\medskip
Now replace $f$ by the map
$X \mapsto Df(X)D^{-1}$ with $D = \diag(1, \gamma_{12}, \dots, \gamma_{1n})$.
Then we have $f(E_{ij})=F_{ij}= E_{ij}$ for all
$1 \le i,j \le n$. Therefore, reversing our
modifications, for $Q=PD \in \calP\calD$, then $f$ must be of the form
(\ref{apr55}) or of the form (\ref{apr56}).
\end{proof}

\section{The usual product}
This section concerns the proof of  ``(1) $\Rightarrow$ (4)'' of
Theorem \ref{main0} for the usual product $A*B = AB$.

For the rest of this section, we always assume that $f$
is a bijective (cf. Lemma \ref{bijective}) map on $M_n^+$ that satisfies
\begin{equation}\label{oct61} r(AB)=r(f(A)f(B)),
\qquad (A,B\in M_n^+).\end{equation}

Let us now define a set of matrices useful for our proof.

\begin{definition} \label{friendsdefinition}
For every $A \in M_n^+$ define
$\cF(A) = \{X \in M_n^+: r(AX) > 0\}$.

\end{definition}

The next few results examine and exploit the relationships between these sets
and the bijectivity of our function to extract relationships between a matrix
and its image.

\noindent
\begin{lemma}\label{Friend} Suppose $A = [a_{ij}]\in M_n^+$,
$B = [b_{ij}] \in M_n^+$.
\begin{itemize}
\item[{\rm (a)}] If there is $(i,j)$ pair such that
$b_{ij} > 0 = a_{ij}$ then $\cF(B) \setminus \cF(A)$ is non-empty.
\item[{\rm (b)}]
The inclusion $\cF(A) \subseteq \cF(B)$ holds if and only if
$b_{ij} > 0$ whenever $a_{ij} > 0$, i.e.,
there is $\gamma > 0$ such that $\gamma B - A \in M_n^+$.
\item[{\rm (c)}] The following two conditions are equivalent:

{\rm (c.1)}
$\cF(A) = \cF(B)$.

{\rm (c.2)}
$a_{ij} = 0$ if and only
if $b_{ij} = 0$.
\end{itemize}
\end{lemma}
\begin{proof}
(a) Let $a_{ij}=0$, and $b_{ij}>0$. Let $X = E_{ji}$.
Then
$$r(BX)=b_{ij} > 0 = a_{ij} = r(AX).$$
So $X\in \cF(B)$ and $X \notin \cF(A)$.
The result follows.

\medskip
(b) The necessity follows from (a):
if $a_{ij}>0=b_{ij}$, then $\exists X \in \cF(A)\setminus \cF(B)$.

To prove the sufficiency,
assume that $b_{ij} > 0$ whenever $a_{ij} > 0$. Then
there is $\gamma > 0$ such that
$\gamma B - A \in M_n^+$. So $\gamma B \ge A$ (entrywise inequality),
and then $\gamma BX \ge AX$ and $r(\gamma BX)\ge r(AX)$ for all $X\in M_n^+$.
(We use here the well known monotonicity property of the spectral radius,
see, e.g., \cite[Theorem 8.1.18]{HJ} or \cite{BP}.)
Thus, for any $X \in \cF(A)$,
$$r(\gamma BX) =\gamma r(BX) \ge r(AX) > 0.$$
So $r(BX) > 0$, thus $X \in \cF(B)$.
It follows that $\cF(A) \subseteq \cF(B)$.

(c) Note that $\cF(A)=\cF(B)$ if and only if
$\cF(A) \subseteq \cF(B) \subseteq \cF(A)$.
By (b), this is equivalent to any one of the following conditions:

\medskip
(i) $b_{ij}>0$ if and only if $a_{ij}>0$.
\qquad
(ii) $b_{ij}=0$ if and only if  $a_{ij}=0$.
\end{proof}

\begin{corollary} \label{FriendCor} \it A matrix $X \in M_n^+$ has exactly $k$
nonzero entries if and only if there is a sequence of matrices $X_1, \dots,
X_{k}, \ldots X_{n^2}$ in $M_n^+$ with $X_k = X$
such that $\cF(X_i)$ is proper non-empty
subset of $\cF(X_{i+1})$ for $i = 1, \dots, n^2-1$. \end{corollary}
\begin{proof}
If $X_k = X$ has exactly $k$ nonzero entries, we can replace zero entries with
nonzero entries one at a time to get $X_{k+1}, \dots, X_{n^2}$. Similarly, we
can replace nonzero entries with zeros one at a time to get the required
$X_{k-1}, X_{k-2}, \dots, X_1$. Observe that since we only replaced $k-1$
nonzero entries with zeros, $X_{1}\ne 0$. So this construction yields the
desired sequence.

Conversely, if $X = X_1, X_2, \dots, X_{n^2}$ have the described property,
then $X_1 \ne 0$ because $\cF(X_1)$ is non-empty. Moreover, by Lemma
\ref{Friend}, $X_{i+1}$ has at least one more nonzero entry than $X_i$. It
follows that $X_i$ must have exactly $i$ nonzero entries for each $i$, so the
result follows. \end{proof}

Note that for $A \in M_n^+$, we have
\begin{eqnarray*}
\cF(f(A))&=&\{X\in M_n^+: r(f(A)X)>0\}=\{X\in M_n^+: r(Af^{-1}(X))>0\}\\
&=&\{f(Y)\in M_n^+: r(AY)>0\}=f(\cF(A)).
\end{eqnarray*}
Thus, we have the following.

\begin{lemma}\label{Friend2} \it If $A\in M_n^+$,
then $\cF(f(A))=f(\cF(A))$.
\end{lemma}

\begin{corollary}\label{FriendCor2}
A matrix $X \in M_n^+$ has exactly $k$ nonzero entries
if and only if $f(X)$ has exactly $k$ nonzero entries.
\end{corollary}
\begin{proof}
Let $X\in M_n^+$ such that $X$ has exactly $k$ nonzero entries.
By Corollary \ref{FriendCor} there exist
$X_1, \dots, X_{n^2}$ in $M_n^+$ with $X_k = X$ such that
$\cF(X_i)$ is proper non-empty subset of
$\cF(X_{i+1})$ for $i = 1, \dots, n^2-1$.
By Lemma \ref{Friend2}, we have
$$\cF(f(X_i))=f(\cF(X_i))\subseteq f(\cF(X_{i+1}))=\cF(f(X_{i+1})),$$
and the inclusion is strict in view of bijectivity of $f$.
Thus, $f(X_1), \dots, f(X_{n^2})$ is a sequence
satisfying Corollary \ref{FriendCor}. So,
$f(X_k)=f(X)$ has $k$ nonzero entries.
Applying the above proof to $f^{-1}$ in place of $f$
(see Remark \ref{bijecremark})
we see  that $f^{-1}(X)$ has $k$ nonzero entries.
\end{proof}

\medskip

This concludes our direct involvement with our (\ref{friendsdefinition}) sets.
Now we will use our obtained results to characterize the image of another set
of useful matrices: the matrix units.

\begin{lemma} \label{lem2}
Let $f(E_{ij}) = F_{ij}$ for $1 \le i, j \le n$. Then:

{\rm (a)} For $i \ne j$, we have
$F_{ij}=\gamma_{ij}E_{pq}$ for some $1 \le p,q \le n$, $p \ne q$, where
$\gamma_{ij}>0$.

{\rm (b)} $\{F_{11}, \dots, F_{nn}\} = \{E_{11}, \dots, E_{nn}\}$.

{\rm (c)} If $i\neq j$, then
$$ F_{ij}=\gamma_{ij}E_{pq} \qquad \Longrightarrow \qquad
F_{ji}=\gamma_{ij}^{-1}E_{qp}; $$
thus $\gamma_{ji}=\gamma_{ij}^{-1}$.

{\rm (d)} There is a permutation $\tau$ of $\{(i,j): 1 \le i, j \le n\}$
with the properties that
\begin{equation}\label{apr61}
\tau(i,j)=(p,q)\qquad \Longrightarrow \qquad \tau(j,i)=(q,p) \end{equation}
and $F_{ij}=\gamma_{ij} E_{\tau(i,j)}$ for all pairs $(i,j)$, $1\leq i,j\leq n$
(we take $\gamma_{ii}=1$ for $i=1,2,\ldots, n$).
%E_{\tau(i,j)}^t/\gamma_{ij}\}, \qquad 1 \le i < j \le n.$$

{\rm (e)} For $f([a_{kl}])=[x_{kl}]$,
we have $\gamma_{ij} a_{ij} =x_{\tau(i, j)}$.
\end{lemma}
\begin{proof}
For (a) let $i\ne j$. From Corollary \ref{FriendCor2},
$F_{ij}$ has exactly one
nonzero entry. But $r(F_{ij})=r(E_{ij})=0$, so this nonzero entry is not on
the diagonal, thus $F_{ij}=\gamma_{ij}E_{pq}$ for some positive
$\gamma_{ij}$, $p \ne q$.

For (b), by Corollary \ref{FriendCor2} $F_{ij}$ has exactly one
nonzero entry for all $i,j$. For $i=j$, since
$r(F_{ii})=r(E_{ii})=1$, this nonzero entry must be on the diagonal,
and it must be 1. So $F_{ii}=E_{pp}$ for some $p$. Furthermore,
since $r(F_{ii}F_{kk})=r(E_{ii}E_{kk})=0$ for $i \ne k$, no two
$F_{ii}$'s can have the same nonzero position, so we get the desired
result.

\medskip
For (c), let $i\ne j$. So $F_{ij}=\gamma_{ij}E_{pq}$ for some
$(p,q)$, $p\neq q$. Since $r(F_{ij}F_{ji})=r(E_{ij}E_{ji})=1$, then
the nonzero entry of $F_{ji}$ must be in the transposed position to
the nonzero entry of $F_{ij}$ to get a nonzero entry on the
diagonal. Furthermore, these entries therefore must be inverse to
each other. Thus $F_{ji}=E_{qp}/\gamma_{ij}$.

Since for $(p,q)\not\in \{(i,j),(j,i)\}$ we have
$$r(F_{ij}F_{pq})=r(E_{ij}E_{pq})=0=r(E_{ji}E_{pq})=r(F_{ji}F_{pq}),$$ it is
clear that no other $F_{pq}$ shares its nonzero position with $F_{ij}$ or
$F_{ji}$. Then our $\tau$ can be defined by $\tau(i,j)=(p,q)$ if
$F_{ij}=\gamma_{ij}E_{pq}$, and is bijective by our above discussion, so it is
the permutation required by (d). Property (\ref{apr61}) holds in view of (c).

Finally for (e), note that
\begin{eqnarray*}
a_{ij} &=& r(AE_{ji}) = r([x_{ij}]f(E_{ji}))
=r\left([x_{ij}]\gamma_{ji}E_{\tau(j,i)}\right)  \\
&=& r\left([x_{ij}]\gamma_{ij}^{-1}E^t_{\tau(i,j)}\right)
=\gamma_{ij}^{-1} x_{\tau(i,j)}
\end{eqnarray*}
by (d).
It follows that $x_{\tau(i,j)}= \gamma_{ij}a_{ij}$.
\end{proof}

\begin{corollary}\label{lem3}
Let ${\cal N}=\{(i,j): 1\le i,j \le n\}$, and let $S\subseteq {\cal
N}$, $S\ne \emptyset$. Then
$$f(\displaystyle\sum_{(i,j)\in S}E_{ij})=\sum_{(i,j)\in S}F_{ij}.$$
\end{corollary}
\begin{proof}
Let $A=\displaystyle\sum_{(i,j)\in S}E_{ij}=[a_{ij}]$, and $f(A)=[x_{ij}]$.

First observe that
$a_{ij}=1$ if $(i,j)\in S$ and $a_{ij}=0$ otherwise.
{}From (e) we have that $\gamma_{ij}a_{ij}=x_{\tau(i,j)}$.
Then $x_{\tau(i,j)}=\gamma_{ij}$ if $(i,j)\in S$ and
$x_{\tau(i,j)}=0 $ otherwise. Representing $f(A)$ as a combination of matrix units, we get
$$f(A)=\displaystyle\sum_{(i,j)\in {\cal N}}x_{ij}E_{ij}=
\sum_{\tau(i,j)\in S} \gamma_{ij} E_{\tau(i,j)}=
\sum_{(i,j) \in S} F_{ij}. $$
\end{proof}

We are now ready to present the  proof
of Theorem \ref{main0} for the usual product.

\begin{proof}
Recall that the bijective map $f:M_n^+ \ \longrightarrow \ M_n^+$ has the property (\ref{oct61}). If $n=2$, then
Theorem \ref{main0} follows easily from Lemma \ref{lem2} and Corollary \ref{lem3}.
Thus suppose $n>2$.
By Lemma \ref{lem2} and Corollary \ref{lem3}, the hypotheses of Theorem \ref{UnitTheorem} are satisfied. Thus, either (\ref{apr55}) or (\ref{apr56}) holds.
For $f(E_{ij})=Q^{-1}E_{ij}Q$, define $\hat{f}_1(A)=Qf(A)Q^{-1}$,
and for $f(E_{ij})=Q^{-1}E_{ij}^tQ$,
define $\hat{f}_2(A)=\hat{f}_1(A)^t$.
Then $\hat{f}_1, \hat{f}_2:M_n^+ \ \longrightarrow \ M_n^+$
since $Q,f(A),Q^{-1}\in M_n^+$.
Also, for all $A,B\in M_n^+$, $$r(AB)=r(f(A)f(B))=r(Qf(A)(Q^{-1}Q)f(B)Q^{-1})
=r(\hat{f}_1(A)\hat{f}_1(B)),$$ and
$$r(AB)=r(\hat{f}_1(A)\hat{f}_1(B))
=r(\hat{f}_1(B)^t\hat{f}_1(A)^t)
=r(\hat{f}_2(B)\hat{f}_2(A))=r(\hat{f}_2(A)\hat{f}_2(B)),$$
thus we may use our previous machinery on these functions.

Trivially, $\hat{f}_1(E_{ij})=\hat{f}_2(E_{ij})=E_{ij}$.
Applying Lemma \ref{lem2} to $\hat{f}_k$,
we have $\gamma_{ij}=1$ and $\tau$ the identity permutation.
Then for $A=[a_{ij}]\in M_n^+$ and
$\hat{f}_k(A)=[x_{ij}]$, we apply part (e) of that lemma to get
$x_{ij}=x_{\tau(i,j)}=\gamma_{ij}a_{ij}=a_{ij}$, and so $\hat{f}_k(A)=A$.
Thus, $f(A)=Q^{-1}\hat{f}_1(A)Q=Q^{-1}AQ$
or $f(A)=Q^{-1}\hat{f}_2(A)^{t}Q=Q^{-1}A^{t}Q$.
\end{proof}

\section{The Triple Product}
This section concerns the proof of  ``(1) $\Rightarrow$ (4)'' of
Theorem \ref{main0} for the Jordan triple product $A*B = ABA$. For
the rest of this section, we always assume that the surjective map
$f$ on $M_n^+$ satisfies
\begin{equation}\label{radcon'} r(ABA)=r(f(A)f(B)f(A)),
\qquad (A,B\in M_n^+).\end{equation} Note that by Lemma
\ref{bijective} $f$ is automatically bijective if $n>2$.

We first treat the special case when $n = 2$.

\begin{lemma} Suppose $n = 2$ and $A*B = ABA$. Then the implication
``{\rm (1)} $\Rightarrow$ {\rm (4)}'' of Theorem {\rm  \ref{main0}}
holds.
\end{lemma}

\begin{proof}
We divide the proof into several assertions.
We will use the observation that $A \in M_2^+$ is a non-zero
nilpotent if and only if it has exactly one non-zero
entry at the off-diagonal position.

\medskip\noindent
{\bf Assertion 1}
$\{f(E_{11}), f(E_{22})\} =\{E_{11}, E_{22}\}$.
\medskip

To see this, let $f(E_{ii})=[x_{pq}]$.
Observe that since all diagonal entries of a nilpotent matrix are 0,
$r(E_{ii}^2N)=r(E_{ii}N)=0$ for any nilpotent matrix $N$.
So for $N$ such that $f(N)=E_{12}$, we have $r(f(E_{ii})^2E_{12})=r(E_{ii}^2N)=0$. But
$$f(E_{ii})^2E_{12}=\begin{pmatrix} 0 & x_{11}^2+x_{12}x_{21} \\
0 & x_{21}(x_{11}+x_{22}) \end{pmatrix}, $$
so $x_{21}(x_{11}+x_{22})=0$.
By similar argument, we see that $x_{12}(x_{11}+x_{22})=0$.

Assume that $x_{11}+x_{22}=0$. Then
$$f(E_{ii})^2={\rm diag}\, (x_{12}x_{21}, x_{12}x_{21})=x_{12}x_{21}I. $$
But for $i \ne j$,
$$0=r(E_{ii}E_{jj})=r(f(E_{ii})^2f(E_{jj}))
=r(x_{12}x_{21}f(E_{jj}))=x_{12}x_{21}r(f(E_{jj}))=x_{12}x_{21}.$$
This is impossible, for then $r(f(E_{ii}))=0\ne1=r(E_{ii})$
which is a contradiction.
So we must have $x_{21}=x_{12}=0$.

Thus,
$f(E_{11})={\rm diag}\, (x_{11}, x_{22})$ and $
f(E_{22})={\rm diag}\, (y_{11}, y_{22})$ for some nonnegative numbers $x_{11},x_{22},
y_{11}, y_{22}$.
But then
$$f(E_{11})^2f(E_{22})=\begin{pmatrix} x_{11}^2y_{11} & 0 \\
0 & x_{22}^2y_{22} \end{pmatrix}.$$ Since
$0=r(E_{ii}E_{jj})=r(f(E_{ii})^2f(E_{jj}))$, we have
$x_{11}^2y_{11}+x_{22}^2y_{22}=0$. Now, since $x_{11}+x_{22}>0$ and
$y_{11}+y_{22}>0$, $f(E_{11})$ and $f(E_{22})$ must have exactly one
nonzero entry on the diagonal in different positions, and that
nonzero entry must be 1. This completes the proof of Assertion 1.
\medskip

Replacing $f$ by the map $X \mapsto P^tf(X)P$ for a suitable permutation
matrix $P$, we may assume that $f(E_{ii})=E_{ii}$.
Additionally, up to transposition, $f(E_{12})=\begin{pmatrix}0 & \gamma \\
0 & 0\end{pmatrix}$ for $\gamma \geq 0$. Observe that since $E_{ij}$
is nonzero, we have $\gamma > 0$. We will assume this is the case since if it
is not we can instead consider the map $X\rightarrow f(X)^{\rm t}$.

After these modifications, we can proceed to prove the following.

\medskip\noindent
{\bf Assertion 2} \textit{Let $A=[a_{ij}]$ and $f(A)=[f_{ij}]$. Then
$f_{ii}=a_{ii}$.}
\medskip

To see this,
simply consider $f_{ii}=r(E_{ii}^2f(A))=r(E_{ii}^2A)=a_{ii}$.

\medskip\noindent
{\bf Assertion 3}

\textit{Let $X=\begin{pmatrix}1&1 \\ 0&0\end{pmatrix}$,
$Y=\begin{pmatrix}0&0 \\ 1&1\end{pmatrix}$. Then
\begin{equation}\label{apr57} f(X)=E_{11}+\gamma E_{12}
\qquad \mbox{and}\qquad  f(Y)={\gamma}^{-1} E_{21}+E_{22}.
\end{equation}}

Let $f(X)=[x_{ij}]$, $f(Y)=[y_{ij}]$. Then $x_{11}=1=y_{22}$,
$x_{22}=0=y_{11}$ by the previous assertion.
But then $$f(X)^2=\begin{pmatrix}1+x_{12}x_{21} & x_{12} \\
x_{21}&x_{12}x_{21}\end{pmatrix}, \qquad
f(Y)^2=\begin{pmatrix} y_{12}y_{21} & y_{12} \\
y_{21} & 1+y_{12}y_{21} \end{pmatrix}, $$
which gives us
$$0=r(X^2E_{22})=r(f(X)^2E_{22})=x_{12}x_{21}$$
and $$0=r(Y^2E_{11})=r(f(Y)^2E_{11})=y_{12}y_{21}. $$
Now $$0=r(X^2E_{12})=r(f(X)^2\gamma E_{12})=\gamma x_{21}, $$ so $x_{21}=0$.
Similarly, $$1=r(Y^2E_{12})=r(f(Y)^2\gamma E_{12})
=\gamma y_{21}, $$ so $y_{21}=\gamma^{-1}$ and $y_{12}=0$.

Finally, observe $X^2Y=X$, so $$ 1=r(X^2Y)=r(f(X)^2f(Y))=r(f(X)f(Y)). $$  But
$f(X)f(Y)=\begin{pmatrix}{\gamma}^{-1}x_{12}&x_{12} \\ 0&0\end{pmatrix}$,
therefore, $x_{12}=\gamma$, giving us the desired result (\ref{apr57}).

We can now modify our function by $X \rightarrow Df(X)D^{-1}$ where
$D={\rm diag}\, ({\gamma}^{-1}, 1)$ so $f(X)=X$ and $f(Y)=Y$. With
this additional modification, we can complete the proof of our lemma
by proving the following.

\medskip\noindent
{\bf Assertion 4} \textit{$f(A)=A$ for all $A\in M_2^+$}.

Let $A=[a_{ij}]$ and $f(A)=[x_{ij}]$.
Then by Assertion 2 $x_{ii}=a_{ii}$.
Furthermore, $$X^2A=XA=\begin{pmatrix} a_{11}+a_{21}& a_{12}+a_{22}  \\
0&0\end{pmatrix}$$
and $$f(X)^2f(A)=f(X)f(A)=\begin{pmatrix} a_{11}+x_{21}& x_{12}+a_{22}
\\ 0&0\end{pmatrix}, $$
so $$a_{11}+x_{21}=r(X^2f(A))=r(X^2A)=a_{11}+a_{21}, $$ thus $a_{21}=x_{21}$.

Repeating the same calculation with $Y$ yields $a_{12}=x_{12}$.
Thus $a_{ij}=x_{ij}$, so $f(A)=A$.
\end{proof}

Now, we turn to the case when $n > 2$.
\begin{remark}
It is clear from our consideration of $n=2$ that the arguments in
Section 3 are not directly extendable to the triple product.
However, we shall adapt the same approach and modify it as needed to
obtain our new result for $n \ge 3$. For those results having
similar proofs exhibited in Section 3, we shall restate the results
but often suppress the proof.
\end{remark}

For every $A \in M_n^+$ define
$$\widetilde\cF(A) = \{X \in M_n^+: r(AX^2) > 0\}.$$

\medskip
\noindent
\begin{lemma}\label{FriendTrip} Suppose $A = [a_{ij}], B = [b_{ij}] \in M_n^+$.
\begin{itemize}
\item[{\rm (a)}] If there is $(i,j)$ pair such that
$b_{ij} > 0 = a_{ij}$ then
$\widetilde\cF(B) \setminus \widetilde\cF(A)\neq \emptyset$.
\item[{\rm (b)}]
The inclusion $\widetilde\cF(A) \subseteq \widetilde\cF(B)$
holds if and only if
$b_{ij} > 0$ whenever $a_{ij} > 0$, i.e.,
there is $\gamma > 0$ such that $\gamma B - A \in M_n^+$.
\item[{\rm (c)}] The following two conditions are equivalent:

{\rm (c.1)}
$\widetilde\cF(A) = \widetilde\cF(B)$.

{\rm (c.2)}
$a_{ij} = 0$ if and only
if $b_{ij} = 0$.
\end{itemize}
\end{lemma}
\begin{proof}
(a) Let $a_{ij}=0$, and $b_{ij}>0$. Let $X = \sqrt{E_{ji}}$. Then
$r(BX^2)=b_{ij} > 0 = a_{ij} = r(AX^2)$, so $X\in \widetilde\cF(B)$
and $X \notin \widetilde\cF(A)$. The result follows.

(b) The necessity follows from (a): if $a_{ij}>0=b_{ij}$, then there
exists $X \in \widetilde\cF(A)\setminus \widetilde\cF(B)$.

To prove the sufficiency,
assume that $b_{ij} > 0$ whenever $a_{ij} > 0$. Then
there is $\gamma > 0$ such that $\gamma B - A \in M_n^+$. So $\gamma B \ge A$,
so $r(\gamma BX)\ge r(AX)$ for all $X\in M_n^+$.
Thus, for any $X \in \widetilde\cF(A)$,
$$r(\gamma BX^2) =\gamma r(BX^2) \ge r(AX^2) > 0, $$
thus $X \in \widetilde\cF(B)$.
It follows that $\widetilde\cF(A) \subseteq \widetilde\cF(B)$.

Finally, (c) follows easily from (b).
\end{proof}

\begin{corollary} \label{FriendTripCor} \it A matrix $X \in M_n^+$
has exactly $k$ nonzero entries
if and only if there is a sequence of matrices
$X_1, X_2, \dots X = X_k, \dots, X_{n^2}$ in $M_n^+$
such that $\widetilde\cF(X_i)$ is proper non-empty subset
of $\widetilde\cF(X_{i+1})$ for
$i = 1, \dots, n^2-1$.
\end{corollary}
\begin{proof}
Similar to that of Corollary \ref{FriendCor}.
 \end{proof}

Note that for $A \in M_n^+$, we have
\begin{eqnarray*}
\widetilde\cF(f(A))
&=& \{X\in M_n^+: r(f(A)X^2)>0\} =\{X\in M_n^+: r(A(f^{-1}(X))^2)>0\}\\
&=& \{f(Y)\in M_n^+: r(AY^2)>0\}=f(\widetilde\cF(A)).
\end{eqnarray*}
So, we have the following.

\begin{lemma}\label{FriendTrip2}
If $A\in M_n^+$, then
$\widetilde\cF(f(A))=f(\widetilde\cF(A))$.
\end{lemma}

\begin{corollary}\label{FriendTripCor2}
A matrix $X \in M_n^+$ has exactly $k$ nonzero entries
if and only if $f(X)$ has exactly $k$ nonzero entries.
\end{corollary}
\begin{proof} Similar to that of
Corollary \ref{FriendCor2}.
\end{proof}

\begin{lemma} \label{Triplem2}
Let $f(E_{ij}) = F_{ij}$ for $1 \le i, j \le n$.

{\rm (a)} If $i\neq j$, then $F_{ij}=\gamma_{ij}E_{pq}$ for some $1 \le p,q \le n$,
with $p \neq q$ and $\gamma_{ij}>0$.

{\rm (b)} We have $\{F_{11}, \dots, F_{nn}\} = \{E_{11}, \dots, E_{nn}\}$.

{\rm (c)} There is a permutation $\tau$ of $\{(i,j): 1 \le i, j \le n\}$
such that
$F_{ij}=\gamma_{ij}E_{\tau(i,j)}$ and $F_{ji}
=\gamma_{ji}E_{\tau(i,j)}^t$, for all pairs $(i,j)$; moreover, $\tau$
satisfies the property:
$$ \tau (i,j)=(p,q) \qquad \Longrightarrow \qquad \tau (j,i)=(q,p). $$

{\rm (d)} For $f([a_{ij}])=[x_{ij}]$, we have
$\gamma_{ij} a_{ij} =x_{\tau(i, j)}$.
\end{lemma}

\begin{proof}
For (a), let $i\ne j$. By Corollary \ref{FriendTripCor2},
$F_{ij}$ has
exactly one nonzero entry. But $r(F_{ij})=r(E_{ij})=0$, so this nonzero entry
is not on the diagonal, thus $F_{ij}=\gamma_{ij}E_{pq}$ for some
$\gamma_{ij}>0$, $p \ne q$.

For (b), again by Corollary \ref{FriendTripCor2} $F_{ii}$ has
exactly one nonzero entry, and since $r(F_{ii})=r(E_{ii})=1$, this
nonzero entry must be on the diagonal, and it must be equal  $1$. So
$F_{ii}=E_{pp}$ for some $p$. Furthermore, since
$r(F_{ii}F_{kk})=r(E_{ii}E_{kk})=0$ for $i \ne k$, no two $F_{ii}$'s
can have the same nonzero position, so we get the desired result.

For (c), let $i\ne j$. By Lemma \ref{FriendTrip} (a) we have that
$\widetilde\cF(E_{ij})\not\subseteq\widetilde\cF(E_{ji})$ and vice
versa, so by Lemma \ref{FriendTrip2} we have
$\widetilde\cF(F_{ij})\not\subseteq\widetilde\cF(F_{ji})$. It is
clear then that $F_{ij}$ and $F_{ji}$ do not have the same nonzero
position. However, consider $A=E_{ij}+E_{ji}$. Then by our lemmas,
we know that $\widetilde\cF(E_{ij})\subseteq\widetilde\cF(A)$ and
$\widetilde\cF(E_{ji})\subseteq\widetilde\cF(A)$, so it follows that
$\widetilde\cF(F_{ij})\subseteq\widetilde\cF(f(A))$ and
$\widetilde\cF(F_{ji})\subseteq\widetilde\cF(f(A))$. By Corollary
\ref{FriendTripCor2}, $f(A)$ must have exactly two nonzero entries.
But by Lemma \ref{FriendTrip} (b), the nonzero positions of $F_{ij}$
and $F_{ji}$ must lie in the nonzero positions of $f(A)$, so each
must occupy a distinct nonzero position of $f(A)$. (It is not
possible for $F_{ij}$ and $F_{ji}$ to have the same nonzero
position; indeed, if they did, then we would have
$\widetilde\cF(F_{ij})= \widetilde\cF(F_{ji})$ and
$$ f(\widetilde\cF(E_{ij}))=\widetilde\cF (f(E_{ij}))=
\widetilde\cF(F_{ij})=\widetilde\cF(F_{ji})=\widetilde\cF (f(E_{ji}))
=f(\widetilde\cF(E_{ji})), $$
which gives a contradiction because $f$ is bijective and
$\widetilde\cF(E_{ij})\neq \widetilde\cF(E_{ji})$.)
Furthermore, since $r(f(A))=r(A)=1$, the nonzero positions of $A$ must form a cycle, and so
must be transposed of each other.

Thus if $F_{ij}=\gamma_{ij}E_{pq}$ then it must be the case that
$F_{ji}=\gamma_{ji}E_{qp}=\gamma_{ji}E_{pq}^t$. By our previous
considerations it is clear that each $F_{ij}$ has a unique nonzero position
$(p,q)$ with respect to one another so we may define a bijection
$\tau(i,j)=(p,q)$ accordingly, giving us the required permutation.

Finally for (d), we temporarily return to our square roots. By
Corollary \ref{FriendTripCor2},
for $i \ne j$ we have that $f(\sqrt{E_{ij}})$ has
exactly two nonzero entries, so either $f(\sqrt{E_{ij}})^2=0$ or
$f(\sqrt{E_{ij}})^2$ has exactly 1 nonzero entry (note that $f(\sqrt{E_{ij}})$
is nilpotent by Lemma \ref{PreLem}). Since
$$r(F_{ji}f(\sqrt{E_{ij}})^2)=r(E_{ji}\sqrt{E_{ij}}^2)=1, $$ it must be the
latter case. Moreover, if $F_{ji}=\gamma_{ji}E_{\tau(j,i)}$, then it is clear
that $f(\sqrt{E_{ij}})^2=E_{\tau(i,j)}/\gamma_{ji}$.
Now let $A=[a_{ij}]$ and $f(A)=[x_{ij}]$. Fix $i\neq j$, and
consider $r(A\sqrt{E_{ji}}^2)=r(AE_{ji})=a_{ij}$ and
$$r(A\sqrt{E_{ji}}^2)=r(f(A)f(\sqrt{E_{ji}})^2)
=r([x_{ij}]E_{\tau(j,i)}/\gamma_{ij})=x_{\tau(i,j)}/\gamma_{ij}. $$
Therefore, $x_{\tau(i,j)}=\gamma_{ij}a_{ij}$. In the case $i=j$,
$\sqrt{E_{ii}}=E_{ii}$, so
$$a_{ii}=r(AE_{ii})=r(AE_{ii}^2)=r([x_{ij}]F_{ii})
=r([x_{ij}]E_{\tau(i,i)})=x_{\tau(i,i)}.$$
\vskip -.3in \end{proof}

\begin{corollary}\label{Triplem3}
Let ${\cal N}=\{(i,j): 1\le i,j \le n\}$, and let $S\subseteq {\cal N}$,
$S\ne \emptyset$. Then
$$f(\displaystyle\sum_{(i,j)\in S}E_{ij})=\sum_{(i,j)\in S}F_{ij}.$$
\end{corollary}

The proof is completely analogous to that of Corollary \ref{lem3}.

We note the following equalities:
\begin{equation*} \gamma_{ji}=1/\gamma_{ij}, \qquad (i,j, \quad 1\leq i,j \leq n). \end{equation*} Indeed,
$f(E_{ij}+E_{ji})=F_{ij}+F_{ji}$, so
$$\gamma_{ij}\gamma_{ji}=r(F_{ij}+F_{ji})=r(E_{ij}+E_{ji})=1.$$

We are now ready to prove the implication ``(1) $\Rightarrow$ (4)''
in  Theorem  \ref{main0} for the Jordan triple product.

\begin{proof}
Recall that $f:M_n^+ \ \longrightarrow \ M_n^+$ has the property
(\ref{radcon'}). Using Corollary \ref{Triplem3} and Theorem
\ref{UnitTheorem}, we see that $f$ must satisfy either (\ref{apr55})
or (\ref{apr56}). For $f(E_{ij})=Q^{-1}E_{ij}Q$, define
$\hat{f}_1(A)=Qf(A)Q^{-1}$, and for $f(E_{ij})=Q^{-1}E_{ij}^tQ$,
define $\hat{f}_2(A)=\hat{f}_1(A)^t$. Then $\hat{f}_1,
\hat{f}_2:M_n^+ \ \longrightarrow \ M_n^+$ since $Q,f(A),Q^{-1}\in
M_n^+$. Also, for all $A,B\in M_n^+$,
\begin{eqnarray*}
r(AB^2)
&=&r(f(A)f(B)^2)=r(Qf(A)f(B)^2Q^{-1})=r(Qf(A)(Q^{-1}Q)f(B)^2Q^{-1})\\
&=&r((Qf(A)Q^{-1})(Qf(B)Q^{-1})^2)
=r(\hat{f}_1(A)\hat{f}_1(B)^2),
\end{eqnarray*}
and
\begin{eqnarray*}
r(AB^2)&=&r(\hat{f}_1(A)\hat{f}_1(B)^2)
=r((\hat{f}_1(B)^t)^2\hat{f}_1(A)^t) \\
&=&r(\hat{f}_2(B)^2\hat{f}_2(A))=r(\hat{f}_2(A)\hat{f}_2(B)^2).
\end{eqnarray*}
Trivially, $\hat{f}_1(E_{ij})=\hat{f}_2(E_{ij})=E_{ij}$. Applying
Lemma \ref{Triplem2} to $\hat{f}_k$ we have $\gamma_{ij}=1$ and
$\tau(i,j)=(i,j)$. Now the proof is completed exactly as in the case
of the usual product.
\end{proof}

\textbf{Acknowledgement.} The first author is supported in part by
the REU supplement of the NSF grants DMS 0456625 and DMS 0600859;
the second author is supported in part by NSF grant DMS 0600859 and
a HK RGC grant and the third author is supported in part by NSF
grant DMS 0456625.
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