Matrix Worksheet 7
... 1. An ice-cream stall sells both green tea and mocha ice cream. A small portion of either costs $0.75 and a large portion costs $1.25. During a short period of time, the number of ice creams sold is own in the table below. ...
... 1. An ice-cream stall sells both green tea and mocha ice cream. A small portion of either costs $0.75 and a large portion costs $1.25. During a short period of time, the number of ice creams sold is own in the table below. ...
On Some Aspects of the Differential Operator
... In chapter 3 we restrict D on Pn , polynomials of degree upto n-1. We see that D is nilpotent on Pn , and we go through the entire process to reach the Jordan ...
... In chapter 3 we restrict D on Pn , polynomials of degree upto n-1. We see that D is nilpotent on Pn , and we go through the entire process to reach the Jordan ...
Lecture 8
... (i) λ > 0 and there exists an associated eigenvector with non-negative entries. (ii) Any other eigenvalue κ of P satisfies |κ| ≤ λ. (iii) If |κ| = λ, then κ = e2πik/m λ for some k, m ∈ N with m ≤ n. Remark 2.5 Let us analyze the dominant eigenvalues of the transition matrix Π of a d-period irreduci ...
... (i) λ > 0 and there exists an associated eigenvector with non-negative entries. (ii) Any other eigenvalue κ of P satisfies |κ| ≤ λ. (iii) If |κ| = λ, then κ = e2πik/m λ for some k, m ∈ N with m ≤ n. Remark 2.5 Let us analyze the dominant eigenvalues of the transition matrix Π of a d-period irreduci ...
Latest Revision 09/21/06
... are composed of real numbers, not all properties that work for real numbers will necessarily work for matrices. For example, in multiplication of all non-zero real numbers, each number has a multiplicative inverse, where only a select set of matrices have an inverse under matrix multiplication. Also ...
... are composed of real numbers, not all properties that work for real numbers will necessarily work for matrices. For example, in multiplication of all non-zero real numbers, each number has a multiplicative inverse, where only a select set of matrices have an inverse under matrix multiplication. Also ...
Pset 9
... p , where α = 0 or 1 and all the primes p are of the form 3k + 1. Proof. By Theorem 3.19, if f = ax2 + bxy + cy 2 is a reduced positive definite binary quadratic form of discriminant −3, then a = 1. As f is reduced, either b = 0 or b = 1. But if b = 0, then the discriminant is −4c which cannot be −3 ...
... p , where α = 0 or 1 and all the primes p are of the form 3k + 1. Proof. By Theorem 3.19, if f = ax2 + bxy + cy 2 is a reduced positive definite binary quadratic form of discriminant −3, then a = 1. As f is reduced, either b = 0 or b = 1. But if b = 0, then the discriminant is −4c which cannot be −3 ...
LINEAR DEPENDENCE AND RANK
... In the example here, there are three equations and three unknowns and the rank of the coefficient matrix, A, is 3. The solution we have computed is unique. Another way to state this is: A consistent system of equations where A is of order n has a unique solution if and only if A-1 exists. This is im ...
... In the example here, there are three equations and three unknowns and the rank of the coefficient matrix, A, is 3. The solution we have computed is unique. Another way to state this is: A consistent system of equations where A is of order n has a unique solution if and only if A-1 exists. This is im ...
Solutions - UCR Math Dept.
... (a) 2(f+g)(x) = 2(f (x) + g(x)) = 2(x3 + 2x2 + (π + (−π))x + (2 + 14)) = (2 · 1)x3 + (2 · 2)x2 + (2 · 16) = 2x3 + 4x2 + 32. (b) 6(f-g)(x) = 6(f (x) − g(x)) = 6(x3 + 2x2 + (π +− (−π))x + (2 + (−14)) = (6 · 1)x3 + (6 · 2)x2 + (6 · 2π)x + (6 · (−12)) = 6x3 + 12x2 + 12πx − 72. 5. Let V be a vector space ...
... (a) 2(f+g)(x) = 2(f (x) + g(x)) = 2(x3 + 2x2 + (π + (−π))x + (2 + 14)) = (2 · 1)x3 + (2 · 2)x2 + (2 · 16) = 2x3 + 4x2 + 32. (b) 6(f-g)(x) = 6(f (x) − g(x)) = 6(x3 + 2x2 + (π +− (−π))x + (2 + (−14)) = (6 · 1)x3 + (6 · 2)x2 + (6 · 2π)x + (6 · (−12)) = 6x3 + 12x2 + 12πx − 72. 5. Let V be a vector space ...
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.