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I need help with Exercise 1.2.8 (a) (Chapter 1: Basics, page 3o) concerning \(\displaystyle K^n\) as a \(\displaystyle K[T]\)-module ... ...

First, so that MHB readers will understand the relevant notation for the construction of \(\displaystyle K^n\) as a \(\displaystyle K[T]\)-module, I am presenting the relevant text from B&K as follows:https://www.physicsforums.com/attachments/2997

Exercise 1.2.8 (a) (page 30) reads as follows:https://www.physicsforums.com/attachments/2998

https://www.physicsforums.com/attachments/2999So we are given that:

\(\displaystyle A\) is an \(\displaystyle n \times n\) matrix over a field \(\displaystyle K\)

\(\displaystyle U\) is a subspace of \(\displaystyle K^n\)

\(\displaystyle M\) is the \(\displaystyle K[T]\)-module obtained from the vector space of column vectors \(\displaystyle K^n\)

So we regard \(\displaystyle M=K^n\) as a right module over \(\displaystyle K[T]\)

Addition in \(\displaystyle M=K^n\) is normal column vector addition \(\displaystyle x+y\) where \(\displaystyle x, y \in K^n\) making \(\displaystyle M=K^n\) an abelian group as required

Given an \(\displaystyle n \times n\) matrix \(\displaystyle A\) over \(\displaystyle K\), a right action of \(\displaystyle f(T) \in K[T]\) on \(\displaystyle M\) is defined as follows:

\(\displaystyle xf(T) = xf_0 + Axf_1 + ... \ ... A^rxf_r

\)

where

\(\displaystyle f(T) = f_0 + f_1T + ... \ ... f_rT^r\)

We are required to show that:

------------------------------------------------

... a subspace \(\displaystyle U\) of \(\displaystyle K^n\) is a submodule \(\displaystyle L\) of \(\displaystyle M\) ...

if and only if

... \(\displaystyle AU \subseteq U\)

------------------------------------------------

Just reviewing the definitions of subspace and submodule in this context ...\(\displaystyle U\) is a subspace of \(\displaystyle K^n\) if \(\displaystyle U\) is a subset of \(\displaystyle K^n\) such that:

(1) \(\displaystyle 0 \in U\)

(2) \(\displaystyle x, y \in U \Longrightarrow x + y \in U \)

(3) \(\displaystyle f(T) \in k[T] , x \in U \Longrightarrow x f(T) \in U \)\(\displaystyle L\) is a submodule of \(\displaystyle M = K^n\) if \(\displaystyle L\) is a subset of \(\displaystyle K^n\) such that:

(1) \(\displaystyle 0 \in L\)

(2) \(\displaystyle x,y \in L \Longrightarrow x + y \in L

\)

(3) \(\displaystyle x \in L\) and \(\displaystyle f(T) \in K[T] \Longrightarrow xf(T) \in L\)

===========================

Now assume that the subspace \(\displaystyle U\) is a submodule \(\displaystyle L\) of \(\displaystyle M\)

Then we have that \(\displaystyle xf(T) \in U\) for all \(\displaystyle x \in L, f(T) \in K[T] \)

We need to show \(\displaystyle AU \subseteq U\)

So let \(\displaystyle x \in AU\)

Therefore

\(\displaystyle x = AU = \begin{pmatrix} a_{11} & a_{12} & a_{13} & ... & ... & a_{1n} \\ a_{21} & a_{22} & a_{23} & ... & ... & a_{2n} \\ a_{31} & a_{32} & a_{33} & ... & ... & a_{3n} \\... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... \\ a_{n1} & a_{n2} & a_{n3} & ... & ... & a_{nn}\end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \\u_3 \\.\\ .\\ u_s \\ 0 \\ 0 \\ .\\ . \\0 \end{pmatrix}\) for some integer \(\displaystyle s\) where \(\displaystyle s = dim(U)\)

Therefore

\(\displaystyle x = AU = \begin{pmatrix} a_{11}u_1 + a_{12}u_2 + a_{13}u_3 + ... \ ... + a_{1s}u_s + 0 +0 ... \ ... +0 \\ a_{21}u_1 + a_{22}u_2 + a_{23}u_3 + ... \ ... + a_{2s}u_s + 0 +0 ... \ ... +0 \\ a_{31}u_1 + a_{32}u_2 + a_{33}u_3 + ... ... + a_{3s}u_s + 0 +0 ... \ ... +0 \\... \ ... \\ ... \ ... \\ a_{n1}u_1 + a_{n2}u_2 + a_{n3}u_3 + ... ... + a_{ns}u_s +0 +0 ... \ ... +0 \end{pmatrix}\)

Thus \(\displaystyle x = AU \in U\) if \(\displaystyle a_{ij}u_j \in U\) since the terms of the matrix \(\displaystyle AU\) are sums of such elements and a subspace will contain these sums if the individual summands belong to it ...

But we are given that \(\displaystyle U\) is also a submodule L

Thus \(\displaystyle xf(T) \in L\) ...

... ... I was going to try to use this to show that each term \(\displaystyle a_{ij}u_j \in U\) ... ... BUT ... feel I have lost my way ...

Can someone please help ... ...

Peter