Equivalence of Matrices in a Principal Ideal Domain

Math 520B Handout: November 11, 2005

Let R denote a given principal ideal domain.

Definition. Two m \times n matrices A, B in R will be called equivalent if there exist matrices U \in GL_{m}(R) and V \in GL_{n}(R) such B = U A V. To indicate that A and B are equivalent one may write A ~ B.

Observe that the ideal in R generated by the entries of A and the ideal generated by the entries of B are the same when A and B are equivalent. Since R is a principal ideal domain, it follows that the entries of A and the entries of B share the same greatest common divisors inasmuch as these greatest common divisors serve as single generators for these ideals.

By rank of a matrix A in R one understands the rank of A when it is regarded as a matrix in the fraction field of R.

Lemma 1.   If a and b are non-zero entries sharing either a row or a column in an m \times n matrix over R, then there is an equivalent matrix having a greatest common divisor of a and b as entries.

Proof. The case where they share a row is the transpose of the case where they share a column. If they share a column one may narrow the scope to that column and the two rows that are involved, i.e., it is essentially a question about the case m = 2, n = 1. If Ra + Rb = Rd, then one may choose e, f \in R such that ea + fb = d. If a' = a/d and b' = b/d, then

 
(
f
-b'
a'
)
 
(
a
b
)
  =  
(
d
0
)
    . 

Theorem 1.   Every m \times n matrix in R of rank r is equivalent to a matrix C for which C_{ii} = c_{i} for 1 <= i <= r and C_{ij} = 0 for all other pairs (i, j) where the non-zero entries c_{i} are successively divisible, i.e., c_{i} | c_{i+1} for 1 <= i <= r-1.

Proof. Let k = max(m, n). Use induction on k. The result is trivially true if k = 1 or if the given matrix A = 0. Assume k > 1. Among the non-zero entries in all of the matrices equivalent to A there is an entry in one of those matrices having the minimum number of prime factors occuring among those entries. Let m be an entry having the said minimum number of prime factors, and replace A, if necessary, by an equivalent matrix in which m is an entry. Since any entry may be moved to position (1, 1) using row and column operations, replacing A again, if necessary, by an equivalent matrix, one may assume that m is the (1, 1) entry of A. By the lemma, in view of the choice of m, m must divide all entries in the first row and the first column of A. For each entry in the first column of A other than the m in position (1, 1), performing an elementary row operation on A, hence replacing A by an equivalent matrix, will zero that entry. Likewise elementary column operations will zero entries in the first row of A beyond the (1, 1) position. Thus, one may assume that the m in position (1, 1) is the only non-zero entry in either the first row or the first column of A. By the inductive hypothesis the (m-1)\times(n-1) matrix A_{1} formed by deleting the first row and the first column of A satisfies U_{1} A_{1} V_{1} = C_{1} where the only non-zero entries in C_{1} are successively divisible elements c_{2}, …, c_{r} in positions (1, 1), … (r-1, r-1) of C_{1}. Taking

 U  =  
(
0
0
U_{1}
)
    and      V  =  
(
0
0
V_{1}
)
 

one obtains

 U A V  =  C 

with the only non-zero entries being C_{11} = m, C_{22} = c_{2}, …, C_{rr} = c_{r}. There is still, however, the question of whether m divides c_{2}. Let d be a greatest common divisor of m and c_{2}, and let em + fc_{2} = d. Replacing the first row of C with the sum of itself and the second row multiplied by f and then replacing the second column of that by the sum of itself and the first column multiplied by e yields a matrix equivalent to C, hence equivalent to A, having the entry d = em + fc_{2}. Since d divides m but, in view of the choice of m, has no fewer prime factors than m, one sees that m is the product of a unit in R with d. Therefore, m divides c_{2} since d divides c_{2}.