Homework #9 - UC Davis Mathematics
... 1. Prove the generalization of Theorem 7.4.1 as expressed in the sentence that includes Eq. (8), for any arbitrary value of the integer k. Ans: We are proving that if x(1) , . . . , x(k) are solutions of Eq. (3), then x = c1 x(1) (t) + · · · + ck x(k) (t) is also a solution for any constants c1 , . ...
... 1. Prove the generalization of Theorem 7.4.1 as expressed in the sentence that includes Eq. (8), for any arbitrary value of the integer k. Ans: We are proving that if x(1) , . . . , x(k) are solutions of Eq. (3), then x = c1 x(1) (t) + · · · + ck x(k) (t) is also a solution for any constants c1 , . ...
Subspace sampling and relative
... • G u and E isenstat, “E fficient algorith m s for com puting a strong rank revealing Q R factorization”, S IA M J . S ci. C om puting, 19 9 6 . Main algorithm: there exist k columns of A, forming a matrix C, such that th e sm allest singular value of C is “large”. We can find such columns in O(mn2) ...
... • G u and E isenstat, “E fficient algorith m s for com puting a strong rank revealing Q R factorization”, S IA M J . S ci. C om puting, 19 9 6 . Main algorithm: there exist k columns of A, forming a matrix C, such that th e sm allest singular value of C is “large”. We can find such columns in O(mn2) ...
Solutions
... 1. The kernel always contains the identity element e, so its non-empty. Let a, b be elements in the kernel. We have that ab−1 is in the kernel since Φ(ab−1 ) = Φ(a)Φ(b)−1 = e · e−1 = e. Therefore the kernel is a subgroup. We need to check normality. Let x be any element of G. The element xax−1 is ma ...
... 1. The kernel always contains the identity element e, so its non-empty. Let a, b be elements in the kernel. We have that ab−1 is in the kernel since Φ(ab−1 ) = Φ(a)Φ(b)−1 = e · e−1 = e. Therefore the kernel is a subgroup. We need to check normality. Let x be any element of G. The element xax−1 is ma ...
General linear group
... The special linear group, SL(n, F), is the group of all matrices with determinant 1. They are special in that they lie on a subvariety – they satisfy a polynomial equation (as the determinant is a polynomial in the entries). Matrices of this type form a group as the determinant of the product of two ...
... The special linear group, SL(n, F), is the group of all matrices with determinant 1. They are special in that they lie on a subvariety – they satisfy a polynomial equation (as the determinant is a polynomial in the entries). Matrices of this type form a group as the determinant of the product of two ...
26. Determinants I
... matrix with entries in the original field k (whatever that was), but in the polynomial ring k[x], or in its field of fractions k(x). But then it is much less clear what it might mean to substitute T for x, if x has become a kind of scalar. Indeed, Cayley and Hamilton only proved the result in the 2- ...
... matrix with entries in the original field k (whatever that was), but in the polynomial ring k[x], or in its field of fractions k(x). But then it is much less clear what it might mean to substitute T for x, if x has become a kind of scalar. Indeed, Cayley and Hamilton only proved the result in the 2- ...
Algorithms for computing selected solutions of polynomial equations
... complexity for dense polynomial systems has been analyzed by Shub and Smale (1993) and recently homotopy algorithms for sparse systems have been described by Huber and Sturmfels (1992). Asymptotically speaking, the complexity of homotopy methods on well-conditioned inputs is linear in the number of ...
... complexity for dense polynomial systems has been analyzed by Shub and Smale (1993) and recently homotopy algorithms for sparse systems have been described by Huber and Sturmfels (1992). Asymptotically speaking, the complexity of homotopy methods on well-conditioned inputs is linear in the number of ...
0.1 Linear Transformations
... we shall denote the transformation by TA : Rn 7→ Rm . Thus TA (x) = Ax Since linear transformations can be identified with their standard matrices we will use [T ] as symbol for the standard matrix for T : Rn 7→ Rm . T (x) = [T ]x or [TA ] = A Geometry of linear Transformations A linear transformati ...
... we shall denote the transformation by TA : Rn 7→ Rm . Thus TA (x) = Ax Since linear transformations can be identified with their standard matrices we will use [T ] as symbol for the standard matrix for T : Rn 7→ Rm . T (x) = [T ]x or [TA ] = A Geometry of linear Transformations A linear transformati ...
570 SOME PROPERTIES OF THE DISCRIMINANT MATRICES OF A
... (i = 1, 2). Since £i is idempotent and Fis non-modular, ti(e\) =^0, (i = 1, 2). Hence by a transformation of basis of A which leaves the basis of N unchanged and the principal unit of B invariant, Ti(A) [or 7^2C4)] can be reduced to a diagonal form ||gr8r*||, where g* = 0, (i>p). A manipulation of t ...
... (i = 1, 2). Since £i is idempotent and Fis non-modular, ti(e\) =^0, (i = 1, 2). Hence by a transformation of basis of A which leaves the basis of N unchanged and the principal unit of B invariant, Ti(A) [or 7^2C4)] can be reduced to a diagonal form ||gr8r*||, where g* = 0, (i>p). A manipulation of t ...
6-2 Matrix Multiplication Inverses and Determinants page 383 17 35
... Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system. ...
... Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system. ...
The Hadamard Product
... semidefinite, then we know that all eigenvalues are positive (as ordered for the decomposition). That is, λk > 0 for all 1 ≤ k ≤ r. Clearly then Dk is positive semidefinite for all 1 ≤ k ≤ r. Now consider < Bk x, x > for any x ∈ Cn . Then < Bk x, x >=< U Dk U ∗ x, x >=< Dk U ∗ x, U ∗ x >≥ 0. This i ...
... semidefinite, then we know that all eigenvalues are positive (as ordered for the decomposition). That is, λk > 0 for all 1 ≤ k ≤ r. Clearly then Dk is positive semidefinite for all 1 ≤ k ≤ r. Now consider < Bk x, x > for any x ∈ Cn . Then < Bk x, x >=< U Dk U ∗ x, x >=< Dk U ∗ x, U ∗ x >≥ 0. This i ...
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.