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... gauss divisible random variable could be taken to be the sum of a gaussian variable and a discrete variable, and in particular is thus exponentially decaying. The arguments in this paper will be a nonsymmetric version of those in . Thus, for instance, everywhere eigenvectors are used in , left and r ...
... gauss divisible random variable could be taken to be the sum of a gaussian variable and a discrete variable, and in particular is thus exponentially decaying. The arguments in this paper will be a nonsymmetric version of those in . Thus, for instance, everywhere eigenvectors are used in , left and r ...
Towers of Free Divisors
... E = V \U, the “exceptional orbit variety”, is a hypersurface formed from the positive codimension orbits. We introduce the condition that the representation is a “block representation”, which is a refinement of the decomposition arising from the Lie–Kolchin theorem for solvable linear algebraic grou ...
... E = V \U, the “exceptional orbit variety”, is a hypersurface formed from the positive codimension orbits. We introduce the condition that the representation is a “block representation”, which is a refinement of the decomposition arising from the Lie–Kolchin theorem for solvable linear algebraic grou ...
Solvable Groups, Free Divisors and Nonisolated
... E = V \U, the “exceptional orbit variety”, is a hypersurface formed from the positive codimension orbits. We introduce the condition that the representation is a “block representation”, which is a refinement of the decomposition arising from the Lie-Kolchin theorem for solvable linear algebraic grou ...
... E = V \U, the “exceptional orbit variety”, is a hypersurface formed from the positive codimension orbits. We introduce the condition that the representation is a “block representation”, which is a refinement of the decomposition arising from the Lie-Kolchin theorem for solvable linear algebraic grou ...
Document
... Let V = R2, the set of all ordered pairs of real number, with the standard addition and the following nonstandard definition of scalar multiplication: c(x1, x2) = (cx1, 0). Show that V is not a vector space. pf: This example satisfies the first nine axioms of the definition of a vector space. For ex ...
... Let V = R2, the set of all ordered pairs of real number, with the standard addition and the following nonstandard definition of scalar multiplication: c(x1, x2) = (cx1, 0). Show that V is not a vector space. pf: This example satisfies the first nine axioms of the definition of a vector space. For ex ...
Matrices and Linear Algebra with SCILAB
... write the summation symbol, Σ, with its associated indices, if he used the convention that, whenever two indices were repeated in an expression, the summation over all possible values of the repeating index was implicitly expressed. Thus, the equation for the generic term of a matrix multiplication, ...
... write the summation symbol, Σ, with its associated indices, if he used the convention that, whenever two indices were repeated in an expression, the summation over all possible values of the repeating index was implicitly expressed. Thus, the equation for the generic term of a matrix multiplication, ...
Full Text - J
... the results of [4] depend upon the assumption that the diagonal entries of A1 are not integers (this is called assumption (i) in [4]). Special systems of the form (1) appear in the analytic theory of semisimple Frobenius manifolds [8], [9], where A0 has distinct eigenvalues as in (2). The matrix A1 ...
... the results of [4] depend upon the assumption that the diagonal entries of A1 are not integers (this is called assumption (i) in [4]). Special systems of the form (1) appear in the analytic theory of semisimple Frobenius manifolds [8], [9], where A0 has distinct eigenvalues as in (2). The matrix A1 ...
GENERATING SETS 1. Introduction In R
... Since (ab) = (ba), without loss of generality a < b. We will argue by induction on b − a that (ab) is a product of transpositions (i i + 1). This is obvious when b − a = 1, since (ab) = (a a + 1) is one of the transpositions we want in the desired generating set. Now suppose b − a = k > 1 and the th ...
... Since (ab) = (ba), without loss of generality a < b. We will argue by induction on b − a that (ab) is a product of transpositions (i i + 1). This is obvious when b − a = 1, since (ab) = (a a + 1) is one of the transpositions we want in the desired generating set. Now suppose b − a = k > 1 and the th ...
211 - SCUM – Society of Calgary Undergraduate Mathematics
... Y = (1, 1) and Z = (0, −2), in the form X = aY + bZ. We may do this either by solving a system of linear equations, or simply by observing that in order to get the “5” part of X, we must have a = 5 (because the first coordinate of Z is zero.) But 5Y = (5, 5) so in order to get the “1” part of X we m ...
... Y = (1, 1) and Z = (0, −2), in the form X = aY + bZ. We may do this either by solving a system of linear equations, or simply by observing that in order to get the “5” part of X, we must have a = 5 (because the first coordinate of Z is zero.) But 5Y = (5, 5) so in order to get the “1” part of X we m ...
Fast structured matrix computations: tensor rank and Cohn Umans method
... Except for the case of skew-symmetric matrices, we obtain algorithms with optimum bilinear complexities for all structured matrix–vector products listed above. In particular we obtain the rank and border rank of the structure tensors in all cases but the last. A reader who follows the developments i ...
... Except for the case of skew-symmetric matrices, we obtain algorithms with optimum bilinear complexities for all structured matrix–vector products listed above. In particular we obtain the rank and border rank of the structure tensors in all cases but the last. A reader who follows the developments i ...
Matrix Lie groups and their Lie algebras
... is a sequence in SL(n), det(Ak ) = 1, such that Ak → A, then by continuity of the determinant det(A) = 1 also; therefore, A ∈ SL(n). (c) The orthogonal group O(n): Recall that A ∈ O(n) if and only if AT A = I . Now let {Ak } be a sequence in O(n) such that Ak → A. Passing to limit in the equation AT ...
... is a sequence in SL(n), det(Ak ) = 1, such that Ak → A, then by continuity of the determinant det(A) = 1 also; therefore, A ∈ SL(n). (c) The orthogonal group O(n): Recall that A ∈ O(n) if and only if AT A = I . Now let {Ak } be a sequence in O(n) such that Ak → A. Passing to limit in the equation AT ...
Compressed Sensing
... Az = 0}. Equivalently, the spark (A) is the minimum number of linearly dependent columns of A. (b) If we assume that m < n, which it is for an underdetermined system, spark (A) ∈ [2, m + 1] 2. Recovering at most a unique x given a signal y Theorem 1.1. Let y ∈ Rm . There exists at most one x ∈ Σk su ...
... Az = 0}. Equivalently, the spark (A) is the minimum number of linearly dependent columns of A. (b) If we assume that m < n, which it is for an underdetermined system, spark (A) ∈ [2, m + 1] 2. Recovering at most a unique x given a signal y Theorem 1.1. Let y ∈ Rm . There exists at most one x ∈ Σk su ...
Jordan normal form
In linear algebra, a Jordan normal form (often called Jordan canonical form)of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called a Jordan matrix, representing the operator with respect to some basis. Such matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal (on the superdiagonal), and with identical diagonal entries to the left and below them. If the vector space is over a field K, then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix lie in K, or equivalently if the characteristic polynomial of the operator splits into linear factors over K. This condition is always satisfied if K is the field of complex numbers. The diagonal entries of the normal form are the eigenvalues of the operator, with the number of times each one occurs being given by its algebraic multiplicity.If the operator is originally given by a square matrix M, then its Jordan normal form is also called the Jordan normal form of M. Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given M is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed; it is conventional to group blocks for the same eigenvalue together, but no ordering is imposed among the eigenvalues, nor among the blocks for a given eigenvalue, although the latter could for instance be ordered by weakly decreasing size.The Jordan–Chevalley decomposition is particularly simple with respect to a basis for which the operator takes its Jordan normal form. The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.The Jordan normal form is named after Camille Jordan.