On the existence of equiangular tight frames
... Lemma 8 Let A be an Hermitian matrix whose entries are algebraic integers. Then the eigenvalues of A are real algebraic integers. In addition, assume that the entries of A belong to a subfield F of the complex numbers. If A has an eigenvalue ω whose multiplicity differs from the multiplicity of eve ...
... Lemma 8 Let A be an Hermitian matrix whose entries are algebraic integers. Then the eigenvalues of A are real algebraic integers. In addition, assume that the entries of A belong to a subfield F of the complex numbers. If A has an eigenvalue ω whose multiplicity differs from the multiplicity of eve ...
Extended Affine Root Systems II (Flat Invariants)
... on ^9 N is a semisimple endomorphism of ^ (defining the grading on ^), and V * is an integrable torsion free connection on ^ such that V * J = 0 and V#N = 0. Actually {J, N, F#} is defined as the "leading term" of the triplet {Iw, Ol9 ¥} := {Killing form, multiplication of the coordinate Bl of the h ...
... on ^9 N is a semisimple endomorphism of ^ (defining the grading on ^), and V * is an integrable torsion free connection on ^ such that V * J = 0 and V#N = 0. Actually {J, N, F#} is defined as the "leading term" of the triplet {Iw, Ol9 ¥} := {Killing form, multiplication of the coordinate Bl of the h ...
topics in discrete mathematics - HMC Math
... relations are simply generalizations of equality. An important example illustrating this is congruence modulo m: if m is any non-negative integer we say a is congruent to b modulo m (and express this by a ≡ b (mod m)) provided a − b is divisible by m, i.e.: provided m|(a − b). Equivalently, a ≡ b (m ...
... relations are simply generalizations of equality. An important example illustrating this is congruence modulo m: if m is any non-negative integer we say a is congruent to b modulo m (and express this by a ≡ b (mod m)) provided a − b is divisible by m, i.e.: provided m|(a − b). Equivalently, a ≡ b (m ...
IDEAL FACTORIZATION 1. Introduction
... Prime factorization in Z can be proved by contradiction: if some integer greater than 1 has no prime factorization then let n > 1 be minimal without a prime factorization. Of course n is not prime, so n = ab with a, b > 1. Then a, b < n, so a and b are products of primes. Hence n = ab is a product o ...
... Prime factorization in Z can be proved by contradiction: if some integer greater than 1 has no prime factorization then let n > 1 be minimal without a prime factorization. Of course n is not prime, so n = ab with a, b > 1. Then a, b < n, so a and b are products of primes. Hence n = ab is a product o ...