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2 Two-dimensional lattices Let x and y form a basis of R2 . The lattice L ⊂ R2 generated by x and y is the set of all integral linear combinations of x and y: L = { ax + by | a, b ∈ Z }. 2 The two-dimensional lattice L generated by x = [2, 0] and y = [3, 2]. The vectors v = [0, −4] and w = [−1, 2] form another basis of the same lattice. 2 shows a region near the origin of the lattice in the plane R2 generated by the vectors x = [2, 0] and y = [3, 2]. ) If we change basis by unimodular matrix multiplication, 3 −2 −2 1 2 0 3 2 = 0 −4 −1 2 , we see that the vectors v = [0, −4] and w = [−1, 2] form another basis.

5. Hadamard’s Inequality. Let X = (xij ) be an n × n matrix over R, and let B = maxi,j |xij | be the maximum of the absolute values of its entries. Then | det(X)| ≤ nn/2 B n . Proof. Let xi = (xi1 , . . , xin ) for 1 ≤ i ≤ n be the row vectors of X. If the rows are linearly dependent, then det(X) = 0 and the result is clear. If the row vectors are linearly independent, then let X ∗ be the matrix whose rows are the Gram-Schmidt orthogonal basis vectors x∗1 , x∗2 , . . , x∗n . 4(e), | det(X)| = | det(X ∗ )|.

2012 by Taylor & Francis Group, LLC 32 Lattice Basis Reduction • Input : A basis x, y of a lattice L in Z2 such that |x| ≤ |y|. • Output : A minimal basis v1 , v2 of the lattice L. (1) Set v1 ← x and v2 ← y. Set finished ← false. (2) While not finished do: v2 · v1 (a) Set µ ← . Set m ← ⌈µ⌋. Set ǫ ← sign(µ − m). v1 · v1 (b) Set v2 ← ǫ(v2 − mv1 ). (c) If |v1 | ≤ |v2 | then (i) set finished ← true else (ii) set u ← v1 , v1 ← v2 , v2 ← u (interchange v1 and v2 ). (3) Return v1 and v2 . 4 guarantees that the new v2 makes an acute angle with the old v1 .

### An Explicit Approach To Elementary Number Theory by stein

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