An Approximate Equilibrium Theorem of the Generalized Game
... i.e., F has a continuous approximate selection. This completes the proof.□ Lemma 2. Suppose that X and Y are as in Lemma 1. Let F : X → 2 is a multivalued mapping with nonempty convex values. If F is a.l.s.c. then F has a continuous approximate selection. Proof. Taking F = G in Lemma 1.□ Remark 2. W ...
... i.e., F has a continuous approximate selection. This completes the proof.□ Lemma 2. Suppose that X and Y are as in Lemma 1. Let F : X → 2 is a multivalued mapping with nonempty convex values. If F is a.l.s.c. then F has a continuous approximate selection. Proof. Taking F = G in Lemma 1.□ Remark 2. W ...
Math 215 HW #7 Solutions
... 6. Problem 3.4.4. If Q1 and Q2 are orthogonal matrices, so that QT Q = I, show that Q1 Q2 is also orthogonal. If Q1 is rotation through θ and Q2 is rotation through φ, what is Q1 Q2 ? Can you find the trigonometric identities for sin(θ + φ) and cos(θ + φ) in the matrix multiplication Q1 Q2 ? Answer: ...
... 6. Problem 3.4.4. If Q1 and Q2 are orthogonal matrices, so that QT Q = I, show that Q1 Q2 is also orthogonal. If Q1 is rotation through θ and Q2 is rotation through φ, what is Q1 Q2 ? Can you find the trigonometric identities for sin(θ + φ) and cos(θ + φ) in the matrix multiplication Q1 Q2 ? Answer: ...
Solutions - UBC Math
... where g ′ is a polynomial with integer coefficients. Now, if the gcd of all the coefficients of g ′ is some integer m > 1, we get: m|D, since g was a monic polynomial, and so the leading coefficient of g ′ is D. Then we can cancel m on the both sides of equation (1) (and replace D with D/m). Thus, w ...
... where g ′ is a polynomial with integer coefficients. Now, if the gcd of all the coefficients of g ′ is some integer m > 1, we get: m|D, since g was a monic polynomial, and so the leading coefficient of g ′ is D. Then we can cancel m on the both sides of equation (1) (and replace D with D/m). Thus, w ...
Relatives of the quotient of the complex projective plane by complex
... under some representation (of group U (1) in the Hermitian case and of group SU (2) in the the hyperhermitian case), namely under a representation which is a multiple of an irreducible one. The corresponding generalized von Neuman-Wigner theorems (see [6], [3]) claim in our present terminology that ...
... under some representation (of group U (1) in the Hermitian case and of group SU (2) in the the hyperhermitian case), namely under a representation which is a multiple of an irreducible one. The corresponding generalized von Neuman-Wigner theorems (see [6], [3]) claim in our present terminology that ...
Solutions to Math 51 Final Exam — June 8, 2012
... • The main issue was forgetting to give a basis, and instead giving the span of the basis. • A few students computed N (A) instead of N (A − 3I). (b) Determine the definiteness of Q. Justify your answer. (4 points) Since Q(e1 ) = −1 and Q(e3 ) = 2, Q assumes both positive and negative values and hen ...
... • The main issue was forgetting to give a basis, and instead giving the span of the basis. • A few students computed N (A) instead of N (A − 3I). (b) Determine the definiteness of Q. Justify your answer. (4 points) Since Q(e1 ) = −1 and Q(e3 ) = 2, Q assumes both positive and negative values and hen ...
Linear recursions over all fields
... It’s more difficult to show nk λn for a k ≥ 1 satisfies (1.1) if λ is a reciprocal root of 1 − c1 x − · · · − cd xd with multiplicity greater than k. To do this, we will rely on the following theorem characterizing linearly recursive sequences in terms of their generating functions. Theorem 2.1. If ...
... It’s more difficult to show nk λn for a k ≥ 1 satisfies (1.1) if λ is a reciprocal root of 1 − c1 x − · · · − cd xd with multiplicity greater than k. To do this, we will rely on the following theorem characterizing linearly recursive sequences in terms of their generating functions. Theorem 2.1. If ...
Algebras. Derivations. Definition of Lie algebra
... is not commutative, λ 6= 0. Thus change variables once more setting x := x/λ. We finally get ...
... is not commutative, λ 6= 0. Thus change variables once more setting x := x/λ. We finally get ...
CLASS NOTES ON LINEAR ALGEBRA 1. Matrices Suppose that F is
... dot product on F n . Lemma 1.1. The following are equivalent for two matrices A, B ∈ Mm,n (F ). 1. A = B. 2. Ax = Bx for all x ∈ F n . 3. Aei = Bei for 1 ≤ i ≤ n. N will denote the natural numbers {0, 1, 2 . . .}. ...
... dot product on F n . Lemma 1.1. The following are equivalent for two matrices A, B ∈ Mm,n (F ). 1. A = B. 2. Ax = Bx for all x ∈ F n . 3. Aei = Bei for 1 ≤ i ≤ n. N will denote the natural numbers {0, 1, 2 . . .}. ...
Division algebras
... In this case, we can give a full classification. Theorem 2 (Frobenius). Every finite-dimensional associative division algebra D over R is isomorphic to R, C or H. The following proof is a combination of [1, Theorem 2B.5] and [3]. It uses techniques in algebraic topology to dispose of the commutative ...
... In this case, we can give a full classification. Theorem 2 (Frobenius). Every finite-dimensional associative division algebra D over R is isomorphic to R, C or H. The following proof is a combination of [1, Theorem 2B.5] and [3]. It uses techniques in algebraic topology to dispose of the commutative ...
BALANCING UNIT VECTORS
... would be best possible, as the standard unit vectors in the d-dimensional space with the L1 norm show. Regarding Theorem 4, it is not even clear what the best upper bound in Theorem 3 should be. Bárány and Grinberg [1] claim that they can replace 2d by 2d − 1. On the other hand, the upper bound ca ...
... would be best possible, as the standard unit vectors in the d-dimensional space with the L1 norm show. Regarding Theorem 4, it is not even clear what the best upper bound in Theorem 3 should be. Bárány and Grinberg [1] claim that they can replace 2d by 2d − 1. On the other hand, the upper bound ca ...
Basis (linear algebra)
Basis vector redirects here. For basis vector in the context of crystals, see crystal structure. For a more general concept in physics, see frame of reference.A set of vectors in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every vector in the vector space is a linear combination of this set. In more general terms, a basis is a linearly independent spanning set.Given a basis of a vector space V, every element of V can be expressed uniquely as a linear combination of basis vectors, whose coefficients are referred to as vector coordinates or components. A vector space can have several distinct sets of basis vectors; however each such set has the same number of elements, with this number being the dimension of the vector space.