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**Example text**

Then E is called a solution to the extended matrix gcd problem. If E1 is the first m and E2 the last m colums of E then E1 A1 + E2 A2 = G. We will use “extended matrix GCD” to mean, more generally, the problem of recovering the first r rows of a unimodular 80 CHAPTER 5. ECHELON FORMS OVER PIDS 81 transforming matrix which transforms to echelon form an A ∈ Rn×m with rank r. Consider the scalar case of the extended matrix GCD problem, when m = 1. 7. Let M ∈ R(n−r)×n have rank n − r and satisfy M A = 0.

13, the principal (n − h − 1) × m submatrix of A is rank deficient. Now apply the argument used to prove statement 3). 5) By definition of an index echelon transform, we have Im V I PA ∗ ∗ ∗ ∗ ∗ = ∗ (1/d)B2 . 8) is equal to the last n − m rows of the triangularizing adjoint of d10 P A. Now fix some i with n − h − 1 ≤ i ≤ n, and let A¯ and P¯ be the principal i × i submatrices of A and P respectively. Since P = diag(∗, Ih ) we have that P¯ A¯ is equal to principal ith submatrix of P A. 5. THE TRIANGULARIZING ADJOINT 51 first i entries in row i of the triangularizing adjoint of P A is given by ¯ adj .

Because of the preconditioning with C we have h∗k+1 equal to the gcd of all entries in the first column of B. Using (b) we conclude that h∗k+1 is the product of the first k + 1 entries in an echelon form of A. Let T2 [k + 1, k + 1] = ck+1 h∗k+1 . Then (d2 /h∗k+1 , ck+1 ) = (1). Since h∗k+1 | d we may conclude that ck+1 ⊥ d. This gives the inductive step for (a). Using (a) we also have that T2 [k, k] = ck h∗k with ck ⊥ d. Now we are going to combine fraction free Gaussian elimination of B with the decreasing modulus approach of the previous section.