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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
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