By Arne Storjohann
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Extra resources for Algorithms for Matrix Canonical Forms
Assume henceforth that r = m. Let F2 be the adjoint transform of (B2 , d) where B2 is the trailing (n − m) × (n − m) submatrix of U P A. Then: 4) Rows m + 1, m + 2, . . , n − h − 1 of F are zero. 5) The last min(n − m, h + 1) rows of F are given by the last min(n − m, h + 1) rows of F2 V I P det P where V is the submatrix of 1 d U comprised of the last n − m rows and first m columns. Proof. We prove each statement of the lemma in turn. 1 & 2) Obvious 3) If r < m, then for i = m+1, . . , n the principal i×(i−1) submatrix of d10 A is rank deficient; since entries in row i of d10 F are (i − 1) × (i − 1) minors of these submatrices, they must be zero.
10 becomes O(nθ (log β) + n2 (log n) B(log β)) word operations where β = nN . Proof. Computing an index k reduction transform for an n × 1 matrix requires n − k < n applications of φ∗,∗ . Let fn (k) be the number of basic operations (not counting applications of φ∗,∗ ) required to compute an index k transform for an m × m matrix A where m ≤ n. Then fn (1) = 0. The result will follow if we show that for any indices k1 , k2 with k1 + k2 = k > 1, we have fn (k) ≤ fn (k1 ) + fn (k2 ) + O(nk θ−2 ). Compute an index k1 transform U1 for the principal (n − k2 ) × (n − k2 ) submatrix of A.
To show that H is the Hermite form of A. 11 A ∼ = H d2 I . 2) Thus H is left multiple of A. 3 it will suffice to show that det L(A) = det L(H). 1. 2). All such minors which involve a row from d2 I will be a multiple of d2 . We may deduce that H has rank m and that (d2 , det L(H)) = (det L(A)). By construction each diagonal entry of H is a divisor of d2 . 12. 16 (computation of Hermite form over Z/(d2 ) with transforming matrix) we get the following. 15. Let A ∈ Zn×n be nonsingular and d be a positive multiple of det A.