Solutions to Assignment 3
... 2.1.18 Suppose the first two columns, b1 and b2 , of B are equal. What can you say about the columns of AB (if AB is defined)? Why? Well, we can write B = [b1 b1 b2 · · · bn ], and then AB = [Ab1 Ab1 Ab2 · · · Abn ]. One can see that the first two columns of AB are equal. 2.1.24 Suppose AD = Im (the ...
... 2.1.18 Suppose the first two columns, b1 and b2 , of B are equal. What can you say about the columns of AB (if AB is defined)? Why? Well, we can write B = [b1 b1 b2 · · · bn ], and then AB = [Ab1 Ab1 Ab2 · · · Abn ]. One can see that the first two columns of AB are equal. 2.1.24 Suppose AD = Im (the ...
Factoring 2x2 Matrices with Determinant of
... The matrix has a dominant right column, therefore we multiply by . The product matrix has a dominant left column and therefore we multiply by . The product matrix of that has a dominant left column, thus we multiply by again. The product matrix again has a dominant left column, and so we multiply by ...
... The matrix has a dominant right column, therefore we multiply by . The product matrix has a dominant left column and therefore we multiply by . The product matrix of that has a dominant left column, thus we multiply by again. The product matrix again has a dominant left column, and so we multiply by ...
Notes on fast matrix multiplcation and inversion
... triangular with positive entries on the diagonal. We assume that the matrix A that we wish to invert is symmetric and positive definite. To see that we can make this assumption we use the fact that AT A is symmetric and positive definite. The first statement is an easy exercise and the second follow ...
... triangular with positive entries on the diagonal. We assume that the matrix A that we wish to invert is symmetric and positive definite. To see that we can make this assumption we use the fact that AT A is symmetric and positive definite. The first statement is an easy exercise and the second follow ...
DISCRIMINANTS AND RAMIFIED PRIMES 1. Introduction
... pe |(p) that discZ/pZ (OK /pe ) is 0 in Z/pZ if and only if e > 1. (Recall that the vanishing of a discriminant is independent of the choice of basis.) Suppose e > 1. Then any x ∈ p − pe is a nonzero nilpotent element in OK /pe . By linear algebra over fields, such an x can be used as part of a Z/pZ ...
... pe |(p) that discZ/pZ (OK /pe ) is 0 in Z/pZ if and only if e > 1. (Recall that the vanishing of a discriminant is independent of the choice of basis.) Suppose e > 1. Then any x ∈ p − pe is a nonzero nilpotent element in OK /pe . By linear algebra over fields, such an x can be used as part of a Z/pZ ...
Matrix Multiplication
... i) matrix multiplication is not commutative that is, the order in which the matrices appear in the product is important. For numbers, we know that 2 x 3 is the same as 3 x 2. However, in general, for matrices AB BA First of all, even when AB makes sense and can be calculated, the dimensions o ...
... i) matrix multiplication is not commutative that is, the order in which the matrices appear in the product is important. For numbers, we know that 2 x 3 is the same as 3 x 2. However, in general, for matrices AB BA First of all, even when AB makes sense and can be calculated, the dimensions o ...
Lecture notes
... When two vectors are added together, we sum their corresponding elements. Graphically, we put the beginning of the second vector at the end of the first one, once the second vector has been multiply by -1. ...
... When two vectors are added together, we sum their corresponding elements. Graphically, we put the beginning of the second vector at the end of the first one, once the second vector has been multiply by -1. ...
Solutions to Math 51 First Exam — January 29, 2015
... Of course, this is far from the only possible basis we could have found. Some people solved this problem by noticing right away that C(A) had to be all of R2 , because the first two columns were already linearly independent, and therefore saved themselves the trouble of row reducing (but we have to ...
... Of course, this is far from the only possible basis we could have found. Some people solved this problem by noticing right away that C(A) had to be all of R2 , because the first two columns were already linearly independent, and therefore saved themselves the trouble of row reducing (but we have to ...
118 CARL ECKART AND GALE YOUNG each two
... each two nonparallel elements of G cross each other. Obviously the conclusions of the theorem do not hold. The following example will show that the condition that no two elements of the collection G shall have a complementary domain in common is also necessary. In the cartesian plane let M be a circ ...
... each two nonparallel elements of G cross each other. Obviously the conclusions of the theorem do not hold. The following example will show that the condition that no two elements of the collection G shall have a complementary domain in common is also necessary. In the cartesian plane let M be a circ ...
18.02SC MattuckNotes: Matrices 2. Solving Square Systems of
... In the formula, Aij is the cofactor of the element aij in the matrix, i.e., its minor with its sign changed by the checkerboard rule (see section 1 on determinants). Formula (13) shows that the steps in calculating the inverse matrix are: 1. Calculate the matrix of minors. 2. Change the signs of the ...
... In the formula, Aij is the cofactor of the element aij in the matrix, i.e., its minor with its sign changed by the checkerboard rule (see section 1 on determinants). Formula (13) shows that the steps in calculating the inverse matrix are: 1. Calculate the matrix of minors. 2. Change the signs of the ...
Name: Period ______ Version A
... The Inverse Matrix: In earlier math course you also learned that every nonzero real number has a multiplicative inverse, the number you multiply it by to get the multiplicative identity, 1. For example: The multiplicative inverse of 4 is ¼ because (4)(1/4) = 1 Similarly, SOME (but not all) SQUARE ma ...
... The Inverse Matrix: In earlier math course you also learned that every nonzero real number has a multiplicative inverse, the number you multiply it by to get the multiplicative identity, 1. For example: The multiplicative inverse of 4 is ¼ because (4)(1/4) = 1 Similarly, SOME (but not all) SQUARE ma ...
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.