An interlacing property of eigenvalues strictly totally positive
... Gantmacher and Krein [3] is a somewhat expanded version of this Paper.) Among the results obtained in that Paper is that an n x n STP matrix A has n positive, simple eigenvalues, and that the n - 1 positive, simple eigenvalues of the principal submatrices obtained by deleting the first (or last) row ...
... Gantmacher and Krein [3] is a somewhat expanded version of this Paper.) Among the results obtained in that Paper is that an n x n STP matrix A has n positive, simple eigenvalues, and that the n - 1 positive, simple eigenvalues of the principal submatrices obtained by deleting the first (or last) row ...
Theorem 2.9. Any finite-dimensional complex Lie algebra g has a
... members of t∗ is to be as in the corresponding example of §II.1 (with h = t). ...
... members of t∗ is to be as in the corresponding example of §II.1 (with h = t). ...
Math 5A: Homework #10 Solution
... Any vector in the first quadrant can be expressed (x, y) where x > 0 and y > 0. Adding any two vectors of this form, we get (x, y) + (z, t) = (x + z, y + t) and x + z > 0 since x, z > 0 and y + t > 0, since y, t > 0. So the first quadrant is closed under addition. However −1(1, 1) = (−1, −1) is not ...
... Any vector in the first quadrant can be expressed (x, y) where x > 0 and y > 0. Adding any two vectors of this form, we get (x, y) + (z, t) = (x + z, y + t) and x + z > 0 since x, z > 0 and y + t > 0, since y, t > 0. So the first quadrant is closed under addition. However −1(1, 1) = (−1, −1) is not ...
Homework 2
... 4. (6pts) Let T1 , T2 : Rn → Rm be functions that satisfy linear conditions. Suppose further that T1 (ei ) = T2 (ei ) for all standard vectors ei in Rn . Show that T1 (v) = T2 (v) for all vectors v ∈ Rn . 5. (20pts) Do 2.2.13, 2.2.17, 2.2.15, 2.2.19, 2.2.23 (5pts each). 6. (8pts) The trace of an n × ...
... 4. (6pts) Let T1 , T2 : Rn → Rm be functions that satisfy linear conditions. Suppose further that T1 (ei ) = T2 (ei ) for all standard vectors ei in Rn . Show that T1 (v) = T2 (v) for all vectors v ∈ Rn . 5. (20pts) Do 2.2.13, 2.2.17, 2.2.15, 2.2.19, 2.2.23 (5pts each). 6. (8pts) The trace of an n × ...
Semidefinite and Second Order Cone Programming Seminar Fall 2001 Lecture 10
... maps J[x] back to itself; therefore, its restriction to the r-dimensional subspace J[x] may be considered as an r × r matrix, denoted by L0 (x). The matrix L0 (x) actually can be written explicitly with respect to the basis {e, x, . . . , xr−1 }: This is the matrix that maps the αi to the βi : ...
... maps J[x] back to itself; therefore, its restriction to the r-dimensional subspace J[x] may be considered as an r × r matrix, denoted by L0 (x). The matrix L0 (x) actually can be written explicitly with respect to the basis {e, x, . . . , xr−1 }: This is the matrix that maps the αi to the βi : ...
1300Y Geometry and Topology, Assignment 1 Exercise 1. Let Γ be a
... Exercise 1. Let Γ be a discrete group (a group with a countable number of elements, each one of which is an open set). Show (easy) that Γ is a zerodimensional Lie group. Suppose that Γ acts smoothly on a manifold M̃ , meaning that the action map θ :Γ × M̃ −→ M̃ (h, x) 7→ h · x is C ∞ . Suppose also ...
... Exercise 1. Let Γ be a discrete group (a group with a countable number of elements, each one of which is an open set). Show (easy) that Γ is a zerodimensional Lie group. Suppose that Γ acts smoothly on a manifold M̃ , meaning that the action map θ :Γ × M̃ −→ M̃ (h, x) 7→ h · x is C ∞ . Suppose also ...
A minimal route to the classification of simple compact Lie groups. 1
... Σf (g). If F : G → C is a function taking values in a convex set we can forming |G| also average F over G. Compact case. In the case of a compact group the role of averaging by counting is replaced by averaging by integrating. Theorem 4. Every compact topological group admits a unique Borel measure ...
... Σf (g). If F : G → C is a function taking values in a convex set we can forming |G| also average F over G. Compact case. In the case of a compact group the role of averaging by counting is replaced by averaging by integrating. Theorem 4. Every compact topological group admits a unique Borel measure ...
Groups, Representations, and Particle Physics
... Representations (homomorphisms to matrices) that are to one-to-one. Notice that the kernel of a homomorphism is always a normal subgroup of the first group (the domain of the mapping). Unfaithful representations “blur” structure by “dividing” out a normal subgroup. This “lumps” elements of G togethe ...
... Representations (homomorphisms to matrices) that are to one-to-one. Notice that the kernel of a homomorphism is always a normal subgroup of the first group (the domain of the mapping). Unfaithful representations “blur” structure by “dividing” out a normal subgroup. This “lumps” elements of G togethe ...
The Mathematics Department of Piscataway High School is proud to
... Piscataway High School Common Mathematics Course Sequences The sequences below do not represent all possible course options available at each level. Please refer to the Course Offerings Booklet for complete advisement information. *Students are encouraged to move to more challenging sequences when a ...
... Piscataway High School Common Mathematics Course Sequences The sequences below do not represent all possible course options available at each level. Please refer to the Course Offerings Booklet for complete advisement information. *Students are encouraged to move to more challenging sequences when a ...
TUTORIAL SHEET 13 Let p be a prime and F q the finite field with q
... Let p be a prime and Fq the finite field with q = pr elements. We outline a proof that there are q n−1 (q − 1) p-regular conjugacy classes in the group G = GLn (Fq ) of invertible n × n matrices with entries in Fq . The relevance for representation theory of the above is that as a consequence there ...
... Let p be a prime and Fq the finite field with q = pr elements. We outline a proof that there are q n−1 (q − 1) p-regular conjugacy classes in the group G = GLn (Fq ) of invertible n × n matrices with entries in Fq . The relevance for representation theory of the above is that as a consequence there ...
Proposition 7.3 If α : V → V is self-adjoint, then 1) Every eigenvalue
... Proof The corresponding linear map α : Cn → Cn , given by α(v) = Av, is self-adjoint and it follows from the last proposition that all the eigenvalues are real. 2. Remark. (Version for self-adjoint maps). If we state last corollary in terms of linear maps, we get the following: if α : V → is self-ad ...
... Proof The corresponding linear map α : Cn → Cn , given by α(v) = Av, is self-adjoint and it follows from the last proposition that all the eigenvalues are real. 2. Remark. (Version for self-adjoint maps). If we state last corollary in terms of linear maps, we get the following: if α : V → is self-ad ...
Problem Set 5 Solutions MATH 110: Linear Algebra
... k x − s k≤k x − t k with equality if and only if s − t. b) Find the linear polynomial nearest to sin πt on the interval [−1, 1] (Here nearest is based on defining norm using the inner product defined in problem 3). By part (a) we simply have to calculate the projection of f = sin πt onto the subspac ...
... k x − s k≤k x − t k with equality if and only if s − t. b) Find the linear polynomial nearest to sin πt on the interval [−1, 1] (Here nearest is based on defining norm using the inner product defined in problem 3). By part (a) we simply have to calculate the projection of f = sin πt onto the subspac ...
aa9pdf
... the same time, it is not invertible. (Thus, as opposed to the case of finite dimensional algebras, ‘most’ of noninvertible elements of A are not zero-divisors!) 2. Let A be a central simple k-algebra. Prove that: (i) Any two A-modules of the same dimension over k are isomorphic (as A-modules). (ii) ...
... the same time, it is not invertible. (Thus, as opposed to the case of finite dimensional algebras, ‘most’ of noninvertible elements of A are not zero-divisors!) 2. Let A be a central simple k-algebra. Prove that: (i) Any two A-modules of the same dimension over k are isomorphic (as A-modules). (ii) ...
LECTURE 17 AND 18 - University of Chicago Math Department
... these are called linear transformations (or linear maps), and their study forms the basis of linear algebra. Definition. A function T : Rn → Rm is a linear transformation2 if it satisfies the following properties: i) T (λx) = λT (x) for all λ ∈ R and x ∈ Rn ; ii) T (x + y) = T (x) + T (y) for all x, ...
... these are called linear transformations (or linear maps), and their study forms the basis of linear algebra. Definition. A function T : Rn → Rm is a linear transformation2 if it satisfies the following properties: i) T (λx) = λT (x) for all λ ∈ R and x ∈ Rn ; ii) T (x + y) = T (x) + T (y) for all x, ...
33-759 Introduction to Mathematical Physics Fall Semester, 2005 Assignment No. 8.
... 1. Each of the following statements is almost, but not quite, correct. In each case find the (or at leaat a) correct statement by making a (small) change in the original statement, or perhaps adding a qualification. While doing this, or in addition, indicate what was wrong with the original statemen ...
... 1. Each of the following statements is almost, but not quite, correct. In each case find the (or at leaat a) correct statement by making a (small) change in the original statement, or perhaps adding a qualification. While doing this, or in addition, indicate what was wrong with the original statemen ...
Problem Set 2 - Massachusetts Institute of Technology
... 1. Density matrices. A density matrix (also sometimes known as a density operator) is a representation of statistical mixtures of quantum states. This exercise introduces some examples of density matrices, and explores some of their properties. (a) Let |ψi = a|0i + b|1i be a qubit state. Give the ma ...
... 1. Density matrices. A density matrix (also sometimes known as a density operator) is a representation of statistical mixtures of quantum states. This exercise introduces some examples of density matrices, and explores some of their properties. (a) Let |ψi = a|0i + b|1i be a qubit state. Give the ma ...
Math 110, Fall 2012, Sections 109-110 Worksheet 121 1. Let V be a
... gives 2(A−A0 ) = 0. Thus A = A0 . Now substituting above yields 0 = i(B −B 0 ), so B = B 0 . It follows that the A and B given above are unique. ...
... gives 2(A−A0 ) = 0. Thus A = A0 . Now substituting above yields 0 = i(B −B 0 ), so B = B 0 . It follows that the A and B given above are unique. ...
We assume all Lie algebras and vector spaces are finite
... diagonalizable. We thus only need to show that all eigenvalues of ad X|t are zero. If not, let λ 6= 0 be an eigenvalue, with eigenvector Y ∈ t. Thus ad X(Y ) = λY . This implies that X and Y are linearly independent; we also have ad Y (X) = −λY . Thus ad Y leaves the span of X, Y invariant, and must ...
... diagonalizable. We thus only need to show that all eigenvalues of ad X|t are zero. If not, let λ 6= 0 be an eigenvalue, with eigenvector Y ∈ t. Thus ad X(Y ) = λY . This implies that X and Y are linearly independent; we also have ad Y (X) = −λY . Thus ad Y leaves the span of X, Y invariant, and must ...
CLASSICAL GROUPS 1. Orthogonal groups These notes are about
... which is the definition of the special orthogonal group SO(n). It is the identity component of O(n), and therefore has the same dimension and the same Lie algebra. Because there are lots of nice theorems about connected compact Lie groups, some people prefer SO(n) to O(n), and like to call SO(n) a c ...
... which is the definition of the special orthogonal group SO(n). It is the identity component of O(n), and therefore has the same dimension and the same Lie algebra. Because there are lots of nice theorems about connected compact Lie groups, some people prefer SO(n) to O(n), and like to call SO(n) a c ...
3 The positive semidefinite cone
... which implies, since A 0, u ∈ ker(A) (see Exercise 3.2). Since im(A) = ker(A)⊥ for any symmetric matrix A we get im(A) = ker(A)⊥ ⊆ span(x). This means that A is of the form A = λxxT . One can show in a similar way that B is a nonnegative multiple of xxT . • We now show that these are the only extr ...
... which implies, since A 0, u ∈ ker(A) (see Exercise 3.2). Since im(A) = ker(A)⊥ for any symmetric matrix A we get im(A) = ker(A)⊥ ⊆ span(x). This means that A is of the form A = λxxT . One can show in a similar way that B is a nonnegative multiple of xxT . • We now show that these are the only extr ...
Kevin M. Reynolds
... Singular Value Decomposition - Decomposing a matrix into two orthogonal matrices, one being a change of basis in the domain and the other a change of basis for the range such that the matrix becomes diagonal. As an example the method was applied to image compression using MATLAB. ...
... Singular Value Decomposition - Decomposing a matrix into two orthogonal matrices, one being a change of basis in the domain and the other a change of basis for the range such that the matrix becomes diagonal. As an example the method was applied to image compression using MATLAB. ...
Orbital measures and spline functions Jacques Faraut
... determinantal formula due to Olshanski. In particular (for k = 1) the entry X11 is distributed according to a probability measure on R, the density of which is a spline function, as observed by Okounkov. In other words these results describe, for the action of the unitary group U (n) on the space Hn ...
... determinantal formula due to Olshanski. In particular (for k = 1) the entry X11 is distributed according to a probability measure on R, the density of which is a spline function, as observed by Okounkov. In other words these results describe, for the action of the unitary group U (n) on the space Hn ...
Physics 3730/6720 – Maple 1b – 1 Linear algebra, Eigenvalues and Eigenvectors
... Note the square brackets. ...
... Note the square brackets. ...
6301 (Discrete Mathematics for Computer Scientists)
... A-level Mathematics or equivalent Prof Y Kurylev Prof A Sokal ...
... A-level Mathematics or equivalent Prof Y Kurylev Prof A Sokal ...
aa6.pdf
... 7. Let U+ be an associative C-algebra with two generators E, H, and one defining relation HE − EH = 2E. Let M be an U+ -module. (i) Show that if v ∈ M is a nonzero eigenvector of the operator H : M → M , then E(v) is either zero or is an eigenvector of H again. (ii) Show that if M is a finite dimen ...
... 7. Let U+ be an associative C-algebra with two generators E, H, and one defining relation HE − EH = 2E. Let M be an U+ -module. (i) Show that if v ∈ M is a nonzero eigenvector of the operator H : M → M , then E(v) is either zero or is an eigenvector of H again. (ii) Show that if M is a finite dimen ...