Linear Algebra : Modules, Linear Operators, Characteristic and Minimal Polynomials, Generators, Abelian Groups and Annihilators
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Let F be a field and . Then is an n - dimensional vector space
over F. Define a function by .
(a) Show that T is a linear operator.
(b) Find the characteristic and minimal polynomials for T, with explanation. (For the characteristic polynomial, recall that you will need to choose a basis for , find the matrix of T relative to that basis, and find the characteristic polynomial of the matrix.)
(c) By Example 47, we can use T to make into a module over the polynomial ring . Show that is cyclic by giving a generator for M, with explanation. Find the (as defined in Exercise 38).
Example 47) Let F be a field and Let be the polynomial ring over F. What does it mean for an Abelian group M to be a module over ?
First, note that the "constant polynomials" in form a copy of the
field F. Thus scalar multiplication of on M gives scalar multiplication of F on M, which makes M a vector space over F.
Second, consider the function defined by . (That is, equals the result of multiplying the polynomial x by v, using scalar multiplication of on M.) For any and , the requirements (M, where M are the properties of a module-see below) on the scalar multiplication imply and . That is, T is a linear operator on the F-vector space M.
Then given any and , the properties (M) imply
Conversely, suppose that M is any F-vector space and is a linear operator on M. We can define a scalar multiplication on on M by . This satisfies the requirements, making M an -module.
Exercise 38) Let M be an R-module. The annihilator of M in R is defined by
. Show that is an ideal of R.
Definition: Let R be a commutative ring with identity 1. A module over R (also called an
R - module) is an Abelian group M (operation written +) together with a
scalar multiplication which associates to each and an element rm
of M; and , we require:
Modules, linear operators, characteristic and minimal polynomials, generators, abelian groups and annihilators are investigated and discussed in the solution. The solution is detailed and well presented.