\markboth{W. H. Gustafson, D. W. Robinson, R. B. Richter, W. P. Wardlaw}{Matrix Completions over Principal Ideal Rings}

\wstoc{Matrix Completions over Principal Ideal Rings}{W. H. Gustafson, D. W. Robinson, R. B. Richter, W. P. Wardlaw}

\title{Matrix Completions over Principal Ideal Rings}
\author{William H. Gustafson}
\address{Department of Mathematics\\
Texas Tech University\\
Lubbock, Texas}
\author{Donald W. Robinson}
\address{Department of Mathematics\\
Brigham Young University\\
Provo, Utah}
\author{R. Bruce Richter}
\address{Department of Mathematics\\
University of Waterloo\\
Waterloo, Ontario}
\author{William P. Wardlaw}
\address{Department of Mathematics\\
U.S. Naval Academy\\
Annapolis, MD 21402}
%\subjclass{Primary 20E26; Secondary 20E05, 20E06}
%\keywords{}


\vskip .3truein

{\it Dedicated to 
Anthony M. Gaglione on his sixtieth birthday 
and to the memory of William H. Gustafson.}

\vskip .3truein

\begin{abstract}
We show that if  $A$  is a  $k\times n$  matrix over a principal
ideal ring  $R,$  with  $k<n$,  and if  $d$  is any element of the
ideal generated by the  $k\times k$  minors of  $A,$  then  $A$  forms
the top  $k$  rows of an  $n\times n$  matrix of determinant  $d$. 
This parallels a 1981 result of Gustafson, Moore, and Reiner, and continues
a program initiated by Hermite in 1849.  Then we use these results to
obtain an extension of a 1997 result of Richter and Wardlaw for good
matrices.
\end{abstract}

%\vskip .5truein
\newpage

\begin{center}
\textbf{1.  Introduction}
\end{center}

\medskip If  $A$  is a  $k\times n$  matrix with  $k<n,$ 
the matrix completion problem intiated by Hermite asks if  $A$  can be
completed to an  $n\times n$  matrix with prescribed determinant  $d$. 
Gustafson, Moore, and Reiner, at the beginning of \lbrack 5\rbrack, give a
brief summary of the history of the problem of completing a  $k\times n$ 
matrix with  $k<n$  over certain commutative rings to an  $n\times n$ 
matrix over the same ring with appropriate determinant. They also include
references to some of the principal players in this program initiated by
Hermite in 1849.  In our contribution below, we show in Theorem 1 that
principal ideal rings are among the rings over which this matrix completion
is always possible.  Theorem 2 states the relationships between six
properties of a  $k\times n$  matrix over a commutative ring.  It extends
a similar theorem in \lbrack 9\rbrack\  by giving a best possible exposition
of these relationships.  Finally we show in Theorem 3 that if such
completions are always possible over each ring in a given collection of
rings, then they are also always possible over the unrestricted direct
product of the collection.

 Throughout this paper,  $R$ will denote a
commutative ring with identity.  If  $A$  is a  $k\times n$  matrix
over  $R$ with  $k\leq n,$  then  $D_{k}(A)$  denotes the ideal of  $%
R$  generated by the  $k\times k$  subdeterminants of  $A$.  We say
that  $A$  has \textit{left block form} if  $A$  is equivalent over  $R$
  to a matrix  $E=\lbrack L\,\,0\rbrack$,  where  $L$  is a  $%
k\times (k+1)$  matrix over  $R$ and  $0$  is the  $k\times (n-k-1)$
 zero matrix over  $R$.  That is, there are matrices  $P\in GL(k,R)$ 
and  $Q\in GL(n,R)$  such that  $PAQ=\lbrack L\;0\rbrack$.  Note that
if  $k=n-1$  or  $k=n,$  the  $0$  block is missing and  $E=L;$ 
indeed, we can take  $A=E=L$.  

\bigskip

\begin{center}
\textbf{2.  Results}
\end{center}

\medskip

 The following lemma was proved but not explicitly stated in \lbrack
5\rbrack, and was used to prove their main result.  For the sake of
completeness, we include a proof here.



\bigskip\noindent \textbf{Lemma.} {\em  Let  $A$  be a  $k\times n$  matrix over the
commutative ring $R$ with identity,  let  $k<n,$  and let  $d\in
D_{k}(A)$.  If  $A$  has left block form over  $R,$  then  $A$ 
enlarges to an  $n\times n$  matrix  $A^{\ast }$  over  $R$ whose
determinant is  $d$  and whose top  $k$  rows form the matrix  $A$.}

\bigskip

{\em Proof.}   Let  $P\in GL(k,R)$  and  $Q\in GL(n,R)$  be such that  $%
PAQ=E=\lbrack L\;0\rbrack$,  where  $L$  is  $k\times (k+1)$. 
Clearly,  $D_{k}(A)=D_{k}(E)=D_{k}(L)$.  Let  $c_{j}=(-1)^{k+1+j}\det
(L_{j}),$  where  $L_{j}$  is the  $k\times k$  submatrix of  $L$ 
obtained by deleting the  $j$th  column of  $L$.  Thus we can write  $%
d\in D_{k}(A)$  as a linear combination  $d=\sum a_{j}c_{j}=\det (L^{\ast
}),$  where  $L^{\ast }$  is the  $(k+1)\times (k+1)$  matrix obtained
from  $L$  by adding  $\lbrack a_{1}\,a_{2}\,\cdots \,a_{k+1}\rbrack$
as its last row.  Let  $p=\det (P)$  and  $q=\det (Q),$  and multiply
the last row of  $L^{\ast }$  by the unit  $pq$  to obtain the matrix  $%
M^{\ast }$  with  $\det (M^{\ast })=pdq$.  Now let  $E^{\ast }$  be the
direct sum of  $M^{\ast }$  and the  $(n-k-1)\times (n-k-1)$  identity
matrix  $I_{n-k-1};$  thus,

\begin{center}
$E^{\ast }=\left[ 
\begin{array}{cc}
M^{\ast } & 0 \\
0 & I_{n-k-1}
\end{array}
\right] =\left[ 
\begin{array}{c}
E \\ 
F
\end{array}
\right] $
\end{center}

\noindent is an  $n\times n$  matrix over  $R$ with  $\det (E^{\ast })=pdq$ 
whose first  $k$  rows form the matrix  $E=PAQ$.  It follows that the
matrix  

\begin{center}
$A^{\ast }=\left[ 
\begin{array}{cc}
P^{-1} & 0 \\ 
0 & I_{n-k-1}
\end{array}
\right] E^{\ast }Q^{-1}=\left[ 
\begin{array}{c}
P^{-1}EQ^{-1} \\ 
FQ^{-1}
\end{array}
\right] =\left[ 
\begin{array}{c}
A \\
A^{\prime }
\end{array}
\right] $
\end{center}

\noindent has  $\det (A^{\ast })=d$  and its first  $k$  rows form the matrix  $A$.  \hfill$\Box$

\bigskip


 Our first main result is

\bigskip

\noindent \textbf{Theorem 1.} {\em  Suppose that  $R$ is either a Dedekind
domain or a principal ideal ring, and that  $A$  is a  $k\times n$ 
matrix over  $R$ with  $k<n$.  If  $d$  is any element of  $%
D_{k}(A), $  then there is an  $n\times n$  matrix  $A^{\ast }$  over  
$R$  with determinant  $\det (A^{\ast })=d$  whose first  $k$  rows
form the matrix  $A$.}

\bigskip
{\em Proof.}  In view of our lemma, we need only establish that  $A$  has a
left block form over  $R$ for each of the two cases.  

When  $R$ is a Dedekind domain, Theorem 1 is the main result of
[5], where they proved a lemma that every  $k\times n$ 
matrix over a Dedekind domain  $R$ has a left block form over  $R$. 
They comment that this lemma was established in a more general form by Levy
\lbrack 6\rbrack\  in 1972.

 When  $R$ is a principal ideal ring, W. C. Brown shows in \lbrack
2, Thm.\ 15.24, p.\ 194\rbrack, that every matrix over  $R$ has a Smith
normal form.  When  $A$  is  $k\times n$  over  $R$ with  $k<n,$ 
its Smith normal form is a left block form for  $A$  over  $R$. \hfill$\Box$

\bigskip

 Since every principal ideal domain is also a Dedekind domain, Theorem
1 only extends the result of \lbrack 5\rbrack\  when  $R$ is a principal
ideal ring with nonzero divisors of zero.

 We were especially interested in the connection between Theorem 1 and
the 1997 result \lbrack 9\rbrack\  regarding good matrices.  In \lbrack
9\rbrack,  $R$ was a commutative ring with identity and an  $r\times n$
 matrix  $A$  over  $R$ was defined to be \textit{left good} if, for
every vector  $\mathbf{x}$  in  $R^{1\times r},$  the ideal  $(\mathbf{x%
}A)$  generated by the entries in the vector  $\mathbf{x}A$  is the same
as the ideal   $(\mathbf{x})$  generated by the entries of the vector  $%
\mathbf{x}$.  Our lemma allows us to extend the Main Theorem of \lbrack
9\rbrack\  to our second main result.

\bigskip

\noindent\textbf{Theorem 2.} {\em\ Consider the following statements about an  $%
r\times n$  matrix  $A$  over the commutative ring $R$ with identity.
\begin{enumerate}
\item  The rows of  $A$  extend to a basis of  $R^{1\times n}$.

\item  $A$  can be enlarged to a matrix  $A^{\ast }\in GL(n,R)$.

\item  $A$  has a Smith normal form  $\lbrack I_{r}\,\,0\rbrack$.

\item  $A$  has a right inverse over  $R$.

\item  $D_{r}(A)=R$.

\item  $A$  is left good.
\end{enumerate}
Then  
\begin{itemize}
\item[(a)]  The statements  (1), (2), and  (3)   are
equivalent over any commutative ring $R$ with identity.

\item[(b)]  The statements  (4), (5), and  (6)   are equivalent over
any commutative ring $R$ with identity.

\item[(c)]  The statement  (3)  implies the statement  (4)  but in
general they are not equivalent.

\item[(d)]  If  $A$  has left block form then all six statements are
equivalent.
\end{itemize}}


\bigskip
{\em Proof.}  Theorem 2 (a), (b), and (c) was proved in \lbrack 9\rbrack , except
for the implications (2) $\Rightarrow $ (3) and (5) $\Rightarrow $
(4), and the fact that (4) $\nRightarrow $ (3).

 The statement (2) means that there is an  $(n-r)\times n$  matrix  
$A^{\prime }$  over  $R$ and an  $n\times n$  matrix  $B^{\ast }$ 
over  $R$ such that

\begin{center}
$A^{\ast }=\left[ 
\begin{array}{c}
A \\ 
A^{\prime }
\end{array}
\right] $
\end{center}

\noindent and  $A^{\ast }B^{\ast }=I$  is the  $n\times n$  identity matrix.  But
then it is clear that  $AB^{\ast }=\lbrack I_{r}\,\,0\rbrack$  is a Smith
normal form for  $A$.  That is, $(2)\Rightarrow(3)$. 

The implication  $(5)\Rightarrow (4)$  is immediate from
 \lbrack 8, Cor.\ I.28, p.\ 84\rbrack .  However, for the sake of
completeness we give the following elementary proof.  If  $M$  is any  $%
m\times n$  matrix over  $R$ and  $\mathbf{v}=(c_{1},\,\cdots
\,,\,c_{r})$  a vector of column indices of  $M,$  so that  $1\leq
c_{j}\leq n,$  we let  $M(\mathbf{v})$  denote the  $m\times r$ 
submatrix of  $M$  whose  $j$th column is the $c_{j}$th column of  $M$.
 It is easy to see that if  $I_{n}$  is the  $n\times n$  identity
matrix, then  $M(\mathbf{v})=M\,I_{n}(\mathbf{v})$.  Now each  $r$-subset
 $\{c_{1},\,\cdots \,,\,c_{r}\}$  of  $\{1,\,\,2,\,\,\cdots \,,\,n\}$  
with  $1\leq c_{1}<c_{2}<\cdots <c_{r}\leq n$  corresponds uniquely to a
vector  $\mathbf{v}=(c_{1},\,\cdots \,,\,c_{r}),$  and we can number these
vectors (perhaps lexicographically)  $\mathbf{v}_{1},\,\mathbf{v}%
_{2},\,\cdots \,,\,\mathbf{v}_{N}$  with  $N={n\choose r}
 $.  Let  $d_{j}=\det (A_{j})$  with  $A_{j}=A(\mathbf{v}_{j}).$
 Then  $D_{r}(A)=R$  implies  $1=\sum b_{j}d_{j}$  for scalars  $%
b_{1},\,b_{2},\,\cdots \,,b_{N}$  in  $R$.  Now, for each  $%
j=1,\,2,\,\cdots \,,N,$  let  $B_{j}$  be the  $n\times r$  matrix  $%
B_{j}=I_{n}(\mathbf{v}_{j})Adj(A_{j}),$  and let  $B$  be the  $n\times
r $  matrix  $B=\sum b_{j}B_{j}$.  Then

\begin{eqnarray*}
AB &= &\sum b_{j}AB_{j}=\sum b_{j}AI_{n}(\mathbf{v}_{j})Adj(A_{j})\\ &=&\sum
b_{j}A_{j}Adj(A_{j})=\sum b_{j}d_{j}I_{r}=I_{r}
\end{eqnarray*}

\medskip
\noindent and  (4)  $A$  has a right inverse  $B$.  That is,  (5) $%
\Rightarrow $ (4).

 The following example from \lbrack 4\rbrack, attributed to Kaplansky
in \lbrack 1, p.\ 7\rbrack, shows  that  (4)  $\nRightarrow $  (3)  in
Theorem 2 (c), and hence the result of Theorem 2 (c)  is the best possible.
 Let  $R$ be the ring of polynomials in  $x,$ $y,$ $z$  over the real
numbers modulo the ideal generated by  $x^{2}+y^{2}+z^{2}-1.$  This is the
ring of polynomial functions on the standard  $2$-sphere in  $3$-space. 
The  $1\times 3$  matrix  $A=\lbrack x\,\,\,\,y\,\,\,\,z\rbrack$  has a
right inverse  $A^{T}.$  If it had a Smith normal form  $\lbrack
1\,\,\,\,0\,\,\,\,0\rbrack$,  then there would be a matrix  $Q\in GL(3,R)$
 such that  $AQ=\lbrack 1\,\,\,\,0\,\,\,\,0\rbrack$.  Assume such a  $Q$
 exists with last column  $\mathbf{q}=\lbrack f\,\,\,\,g\,\,\,\,h\rbrack
^{T}$.  Then  $A$ $\mathbf{q=\,}xf+yg+zh=0$  for all points on the  $2$-%
Sphere.  Thus   $\mathbf{q}$  provides a tangent vector field to the  $%
2$-sphere which, because of independence of the columns of  $Q,$  is never
zero on the  $2$-sphere.  But no such vector field exists, as is shown in
\lbrack 3, p.\ 70\rbrack.  This contradiction shows  (4)  $\nRightarrow $
 (3).  (In fact, the same argument shows directly that  $A$  does not
have a left block form, since that would require an invertible  $Q$  with
 $AQ=\lbrack u\,\,\,\,v\,\,\,\,0\rbrack$.)  This completes the proof of
Theorem 2 (a), (b), and (c).

 To establish  (d), we first observe that when  $r=n,$  the
implication  (4)  $\Rightarrow $  (2)  is a tautology.  Then we use
our lemma to show that  (5)  $\Rightarrow $  (2)  when  $r<n$ 
and  $A$  has a left block form over  $R.$  It is clear from  (5) 
that  $1\in R=D_{r}(A).$  By our lemma,  $A$  can be enlarged to a
matrix  $A^{\ast }$  with determinant  $1$  when  $r<n$.  It is well
known that a matrix over a commutative ring  $R$ with identity is
invertible over  $R$ if and only if its determinant is a unit in  $R$. 
(See \lbrack 7, Thm.\ 50, p.\ 158\rbrack.)  Hence  (5)  $\Rightarrow $
 (2).  \hfill$\Box$


\bigskip We remark that if  $A$  is an  $(n-1)\times n$  matrix over  $R$%
,  then it is already in left block form, so statements  (1) - (6)  of
Theorem 2 are equivalent.  In particular, if  $A$  has a right inverse,
then it extends to an  $n\times n$  matrix which is invertible over  $R$.
 (The latter was shown using an outer product argument in \lbrack 9\rbrack
.)  

Recall that in the proof of Theorem 1 we observed that if  $R$ was
either a principal ideal ring or a Dedekind domain, then every  $r\times n$
 matrix over  $R$ with  $r\leq n$  had a left block form.  Thus we
have the following corollary to Theorem 2.



\bigskip\noindent\textbf{Corollary.} {\em \  If  $R$ is a principal ideal ring or a
Dedekind domain, then statements  (1) - (6)  of Theorem 2 are equivalent.}

\bigskip

 This corollary extends the Main Theorem of \lbrack 9\rbrack\  from
principal ideal rings to rings which are either principal ideal rings or are
Dedekind domains.  Our next theorem allows further extension of the class
of rings for which certain properties mentioned above hold.

 Let  $R$ be a commutative ring with identity.  Then  $R$ has
property  \textbf{L}  if every  $r\times n$  matrix  $A$  over  $R$ 
with  $r\leq n$  has a left block form over  $R$.  $R$ has property  
\textbf{C}  if every  $r\times n$  matrix  $A$  over  $R$ with  $r<n
$  has, for each  $d\in D_{r}(A)$  an  $n\times n$  completion  $%
A^{\ast }$  with  $\det (A^{\ast })=d$.  $R$ has property  \textbf{G}
 if statements  (1) - (6)  of Theorem 2 are equivalent for every  $%
r\times n$  matrix  $A$  over  $R$ with  $r\leq n$.  Note that 
\textbf{L} $\Rightarrow $ \textbf{C} $\Rightarrow $ \textbf{G}, by
our Lemma and Theorem 2.

\bigskip

\noindent \textbf{Theorem 3.}  {\em Let  \textbf{P}  be any one of the properties 
\textbf{ L, C, }or  \textbf{G}, and let  $R=\oplus _{j}R_{j}$  be the
unrestricted direct sum of the commutative rings  $R_{j}\,\,\,(j\in J),$ 
where each  $R_{j}$  has identity  $1_{j}$.  Then  $R$ has property  
\textbf{P}  if and only if each  $R_{j}$  has property  \textbf{P}.}

\bigskip

{\em Proof.}  We consider  $R$ to be an internal direct sum, so each  $R_{j}$ 
is a subring and an ideal of  $R$.  For each  $a\in R,$  $a_{j}=a1_{j}$
 denotes the projection of  $a$  into  $R_{j};$  we call  $a_{j}$ 
the  $j$-component of  $a$.  Thus  ($a_{j})_{k}=0$  if  $j\neq k$ 
and  $(a_{j})_{j}=a_{j}$  for all  $j,\,\,k\in J$.  If  $A$  is a
matrix over  $R,$  then we let  $A_{j}=1_{j}A$  be the matrix of the
same size over  $R_{j}$  obtained by replacing each entry in  $A$  by
its  $j$-component.  We write  $A_{j}^{\prime }$  to denote a matrix
chosen with entries in  $R_{j},$  to distinguish it from the  $j$-%
component   $A_{j}=1_{j}A$  obtained  from a matrix  $A$  already
chosen with entries in  $R$.  In the proofs below, we will often define a
matrix  $A$  over  $R$ by first specifying a matrix  $A_{j}^{\prime }$
 over  $R_{j}$  for each  $j\in J,$  and letting  $A$  be the matrix
of the same size over  $R$ with  $j$-component  $A_{j}=1_{j}A=A_{j}^{%
\prime }$.



 Now suppose that  $R$ has property  \textbf{L}  and that  $%
A_{j}^{\prime }\in (R_{j})^{r\times n}$  with  $r\leq n$.  Since  $%
A_{j}^{\prime }\in R^{r\times n},$  there are matrices  $P\in GL(r,R)$ 
and  $Q\in GL(n,R)$  such that  $PA_{j}^{\prime }Q=E=\lbrack
L\,\,\,0\rbrack$,  with  $L\in R^{r\times (r+1)}$.  But  $A_{j}^{\prime
}=1_{j}A^{\prime }$  implies that  $PA_{j}^{\prime }Q$ $=P(1_{j}A^{\prime
})Q$ $=(1_{j}P)(A_{j}^{\prime })(1_{j}Q)$ $=P_{j}A_{j}^{\prime }Q_{j}$ $=E$ $%
=E_{j}$ $=\lbrack L_{j}\,\,\,0\rbrack$  with  $P_{j}\in GL(r,R_{j})$ 
and  $Q_{j}\in GL(n,R_{j})$.  Thus,  $R_{j}$  has property  \textbf{L}.

 On the other hand, suppose that for each  $j\in J,$  $R_{j}$  has
property  \textbf{L}, and that  $A\in R^{r\times n}$  with  $r\leq n$. 
Then  $A_{j}\in (R_{j})^{r\times n}$  for each  $j\in J,$  and so there
are matrices  $P_{j}^{\prime }\in GL(r,R_{j})$  and  $Q_{j}^{\prime }\in
GL(n,R_{j})$  such that  $P_{j}^{\prime }A_{j}Q_{j}^{\prime }=\lbrack
L_{j}^{\prime }\,\,\,0\rbrack$  with  $L_{j}^{\prime }\in
(R_{j})^{r\times (r+1)}$.  Let  $P\in R^{r\times r}$  be the matrix with
 $j$-component  $1_{j}P=P_{j}^{\prime }$  and let  $Q\in R^{n\times n}$
 be the matrix  with  $j$-component  $1_{j}Q=Q_{j}^{\prime }$  for
every  $j\in J$.  It is easy to see that  $P\in GL(r,R),$  $Q\in
GL(n,R), $  and  $PAQ=\lbrack L\,\,\,0\rbrack$  with  $L\in R^{r\times
(r+1)}$  such that  $1_{j}L=L_{j}^{\prime }$  for each  $j\in J$. 
Thus,  $R$ has property   \textbf{L}.



 Now suppose that  $R$ has property   \textbf{C}  and that  $%
A_{j}^{\prime }\in (R_{j})^{r\times n}$  with  $r<n$  and  $d\in
D_{r}(A_{r}^{\prime })$.  Since  $A_{j}^{\prime }\in R^{r\times n},$ 
there is an  $A^{\ast }\in R^{n\times n}$  whose first  $R$ rows form
the matrix  $A_{j}^{\prime }$  and with determinant  $\det (A^{\ast })=d$%
.  But the first  $R$ rows of  $1_{j}A^{\ast }=(A^{\ast })_{j}\in
(R_{j})^{n\times n}$  also form the matrix  $A_{j}^{\prime }$  and  $%
\det ((A^{\ast })_{j})=d=d_{j}$.  Thus,  $R_{j}$  has property  \textbf{C%
}.

 On the other hand, suppose that for each  $j\in J,$  $R_{j}$  has
property  \textbf{C},  $A\in R^{r\times n}$  with  $r<n,$  and  $d\in
D_{r}(A)$.  For each   $j\in J,$  $1_{j}A=A_{j}\in (R_{j})^{r\times n}$
 has  $1_{j}d=d_{j}\in D_{r}(A_{j})$  and has an  $n\times n$ 
completion  $(A_{j})^{\ast }$  over  $R_{j}$  with  $\det
((A_{j})^{\ast })=d=d_{j}$.  Let  $A^{\ast }$  be the $ n\times n$ 
matrix over  $R$ with  $1_{j}A^{\ast }=(A^{\ast })_{j}=(A_{j})^{\ast }$
 for each   $j\in J$.  Since  $\det ((A^{\ast })_{j})=d_{j}$  for each
  $j\in J,$  it follows that  $\det (A^{\ast })=d$.  Since the first  $%
r$  rows of  $(A^{\ast })_{j}$  form the matrix  $A_{j}$  for each   $%
j\in J,$  it follows that the first  $R$ rows of  $A^{\ast }$  form
the matrix  $A$.  That is,  $A^{\ast }$  is the  $n\times n$ 
completion of  $A$  with determinant  $d$.  Hence,  $R$ has property
 \textbf{C.}

 Suppose  $R$ has property  \textbf{G}  and that  $A_{j}^{\prime
},$ $(B_{j}^{\prime })^{T}\in (R_{j})^{r\times n}$  satisfy  $%
A_{j}^{\prime }B_{j}^{\prime }=(I_{r})_{j},$  which is statement  (4)  of
Theorem 3 for the ring  $R_{j}$\vspace{1pt}$.$  Let  $E=\lbrack
I_{r}\,\,\,0\rbrack{}  -\lbrack I_{r}\,\,\,0\rbrack  _{j},$  $A=E+A_{j}^{\prime
},$  and  $B=E^{T}+B_{j}^{\prime }$.  Note that  $A_{i}=\lbrack
I_{r}\,\,\,0\rbrack_{i}$  if  $i\neq j$,  $A_{j}=A_{j}^{\prime }$,  and
similarly for  $B$.  Then  $AB=I_{r}$  shows that  $A$  satisfies 
(4)  for the ring  $R$.  Since  $R$ has property  \textbf{G, } $A$ 
must also satisfy  (2), so  $A$  has an invertible completion  $A^{\ast
} $  over  $R$.  It follows that  $(A^{\ast })_{j}\in GL(n,R_{j})$  is
the  $n\times n$  completion of  $A_{j}=A_{j}^{\prime }$  over  $R_{j}$%
.  Thus,  (4)  $\Rightarrow $  (2)  in  $R_{j},$  so  $R_{j}$ 
has property  \textbf{G.}

 Finally, suppose for each  $j\in J$  that  $R_{j}$  has property
 \textbf{G}  and that  $A,$  $B^{T}\in R^{r\times n}$  satisfy  $%
AB=I_{r}$.  Then  $A_{j}B_{j}=(I_{r})_{j}$  for each   $j\in J,$  and
so property  \textbf{G}  ensures that each  $A_{j}$  can be completed to
an  $(A_{j})^{\ast }\in GL(n,R_{j})$.  Now let  $A^{\ast }$  be the  $%
n\times n$  matrix over  $R$  with  $j$-component  $1_{j}A^{\ast
}=(A^{\ast })_{j}=(A_{j})^{\ast }$.  Then   $A^{\ast }\in GL(n,R)$  and
its first  $R$ rows form the matrix  $A$.  Thus  (4)  $%
\Rightarrow $  (2)  in  $R,$  so  $R$ has property   \textbf{G.}
 \hfill$\Box$

\bigskip
\begin{thebibliography}{9}

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