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6.2. Invariant Factors of a Matrix
103
(p)
(p)
,with m
=gcd(n ,n )= gcd(n 11 ,n ij )|n 11 =
alent matrix M
1j
11
11
(p−1)
m
.
11
(p)
:=
4. n 11 divides all the entries of the matrix N.In that case, M
(p−1)
M
.
(p)
It is essential to observe that in the first three cases, m
11 is not associated
(p−1)
to m , though it divides it.
11
(p)
From Proposition 6.1.2, the elements of the sequence m are pair-
11
p≥0
wise associated, once p is large enough. We are then in the last of the
(q) (q) (q) (q)
four cases above: m 11 divides all the m ij ’s. We have m i1 = a i m 11 and
(q) (q)
m = b j m .Thenlet P ∈ GL n (A)and Q ∈ GL m (A) be the matrices
1j 11
defined by:
• p ii =1, p i1 = −a i if i ≥ 2, p ij =0 otherwise,
• q jj =1, q 1j = −b j if j ≥ 2, q ij =0 otherwise.
The matrix M := PM (q) Q is equivalent to M (q) ,hence to M.It has the
form
m 0 ··· 0
0
M = . ,
. . M
0
where m divides all the entries of M . Obviously, m = D 1 (M )= D 1 (M).
Having shown that every matrix M is equivalent to a matrix of the form
described above, one may argue by induction on the size of M (that is,
on the integer r =min(n, m)). If r =1, wehavejust provedthe claim.
If r ≥ 2 and if the claim is true up to the order r − 1, we apply the
induction hypothesis to the factor M ∈ M (n−1)×(m−1) (A)inthe above
reduction: there exist P ∈ GL n−1 (A)and Q ∈ GL m−1 (A)such that
P M Q is quasi-diagonal, with diagonal entries d 2 ,... ,d r ordered by
d l |d l+1 for l ≥ 2. From the uniqueness step, d 2 = D 1 (M ). Since m divides
the entries of M ,wehave m|d 2 . Let us then define P = diag(1,P )
and Q = diag(1,Q ), which are invertible: P M Q is quasi-diagonal, with
diagonal entries d 1 = m, d 2 ,... , a nondecreasing sequence (according to
the division in A). Since M is equivalent to M , this proves the existence
part of the theorem.
6.2.1 Comments
In the list of invariant factors of a matrix some d j ’s may equal zero. In
that case, d j = 0 implies d j+1 = ··· = d r = 0. Moreover, some invariant