Page 119 - Matrices theory and applications

102

Furthermore, if M = P D Q is another decomposition with these two prop-
erties, the scalars d j and d are associated. Up to invertible elements, they

j
are thus unique.
Definition 6.2.1 For this reason, the scalars d 1 ,... ,d r (r =min(n, m))
are called the invariant factors of M.
Proof 6. Matrices with Entries in a Principal Ideal Domain; Jordan Reduction
Uniqueness: for k ≤ r, let us denote by D k (N) the gcd of minors of order
k of the matrix N. From Corollary 2.1.1, we have D k (M)= D k (D)=
D k (D ). It is immediate that D k (D)= d 1 ··· d k (because the minors

of order k are either null, or products of k terms d j with distinct
subscripts), so that

d 1 ··· d k = u k d ··· d , 1 ≤ k ≤ r,

k
1

∗
for some u k ∈ A . Hence, d 1 and d are associated. Since A is an
1
integral domain, we also have d = u −1 u k−1 d k .Inother words, d k

k k

and d are associated.
k
Existence: We see from the above that the d j ’s are determined by the
equalities d 1 ··· d j = D j (M). In particular, d 1 is the gcd of the entries
of M. Hence the first step consists in finding a matrix M ,equivalent

to M,suchthat m 11 is equal to this gcd.
To do so, we construct a sequence of equivalent matrices M (p) ,with M (0) =
(p) (p−1) (p−1)
M, such that m divides m .Given thematrix N := M ,we
11 11
distinguish four cases:
1. n 11 divides n 11 ,... ,n 1,j−1 , but does not divide n 1j .Then d :=
gcd(n 11 ,n 1j )reads d = un 11 + vn 1j . Let us define w := −n 1j /d
and z := n 11 /d and let us define a matrix Q ∈ GL m (A)by:
• q 11 = u, q j1 = v, q 1j = w, q jj = z,
l
• q kl = δ ,otherwise.
k
Then M (p) := M (p−1) Q is suitable, because m (p) = d|n 11 = m (p−1) .
11 11
2. n 11 divides each n 1j ,aswell as n 11 ,... ,n i−1,1 , but does not divide
n i1 . This case is symmetric to the previous one. Multiplication on
the right by a suitable P ∈ GL n (A) furnishes M (p) ,with m (p) =
11
(p−1)
gcd(n 11 ,n i1 )|m .
11
3. n 11 divides each n 1j and each n i1 , but does not divide some n ij with
i, j ≥ 2. Then n i1 = an 11 . Let us define a matrix P ∈ GL n (A)by
• p 11 = a +1,p i1 =1,p 1i = −1,p ii =0;
l
• p kl = δ ,otherwise;
k
If we then set N = PN,wehave n = n 11 and n =(a+1)n 1j −n ij .

11 1j
We have thus returned to the first case, and there exists an equiv-