MHB K^n as a K[T]-module - Exercise 1.2.8 (a) - Berrick and Keating (B&K)

[Math Amateur]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

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

First, so that MHB readers will understand the relevant notation for the construction of $$K^n$$ as a $$K[T]$$-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:

$$A$$ is an $$n \times n$$ matrix over a field $$K$$

$$U$$ is a subspace of $$K^n$$

$$M$$ is the $$K[T]$$-module obtained from the vector space of column vectors $$K^n$$

So we regard $$M=K^n$$ as a right module over $$K[T]$$

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

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

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

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

We are required to show that:

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

... a subspace $$U$$ of $$K^n$$ is a submodule $$L$$ of $$M$$ ...

if and only if

... $$AU \subseteq U$$

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

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

(1) $$0 \in U$$

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

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

(1) $$0 \in L$$

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

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

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

Now assume that the subspace $$U$$ is a submodule $$L$$ of $$M$$

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

We need to show $$AU \subseteq U$$

So let $$x \in AU$$

Therefore

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

Therefore

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

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

But we are given that $$U$$ is also a submodule L

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

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

Can someone please help ... ...

Peter

 

[Euge]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

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

First, so that MHB readers will understand the relevant notation for the construction of $$K^n$$ as a $$K[T]$$-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:

$$A$$ is an $$n \times n$$ matrix over a field $$K$$

$$U$$ is a subspace of $$K^n$$

$$M$$ is the $$K[T]$$-module obtained from the vector space of column vectors $$K^n$$

So we regard $$M=K^n$$ as a right module over $$K[T]$$

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

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

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

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

We are required to show that:

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

... a subspace $$U$$ of $$K^n$$ is a submodule $$L$$ of $$M$$ ...

if and only if

... $$AU \subseteq U$$

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

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

(1) $$0 \in U$$

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

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

(1) $$0 \in L$$

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

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

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

Now assume that the subspace $$U$$ is a submodule $$L$$ of $$M$$

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

We need to show $$AU \subseteq U$$

So let $$x \in AU$$

Therefore

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

Therefore

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

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

But we are given that $$U$$ is also a submodule L

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

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

Can someone please help ... ...

Peter

Hi Peter,

I noticed that you wrote the same definition for a submodule of $K^n$ as you did for a subspace of $K^n$, but I know what you mean. Let's start from the beginning.

First, assume that $U$ is a submodule $L$ of $M$. Then for all $u\in U$, $uT\in U$, that is, $Au\in U$. Therefore, $AU\subset U$.

Conversely, suppose $AU\subset U$. Since $U$ is a subspace of $K^n$, to show that $U$ is a submodule of $M$, it suffices to show that $uf(T)\in U$ for all $u\in U$ and $f(T) \in K[T]$. Given $u\in U$, $uT = Au \in U$. Inductively, for all positive integers $j$, $uT^j = A^j u\in U$. Hence, by closure under addition and scalar multiplication in $U$, $uf(T) \in U$ for all $f(T)\in K[T]$. Since $u$ was arbitrary, the result follows.

 

[Deveno]

Let's look at a simple space, a simple subspace, and two different matrices $A$. Our field will be $\Bbb R$, our vector space will be $\Bbb R^2$, and our subspace $U$ will be:

$U = \{(x,0): x \in \Bbb R\}$.

Our first matrix will be:

$A = \begin{bmatrix}1&0\\1&1 \end{bmatrix}$

and our second matrix will be:

$A' = \begin{bmatrix}1&1\\0&1 \end{bmatrix}$.

These matrices look very similar, but we shall see that they behave very differently with respect to $U$.

Note that, for $u \in U$ we have:

$Au = \begin{bmatrix}1&0\\1&1 \end{bmatrix}\begin{bmatrix}x\\0 \end{bmatrix} = \begin{bmatrix}x\\x \end{bmatrix}$.

Since $Au = uT$, and $T$ is certainly one polynomial of $\Bbb R[T]$, we see we fail to have closure of $U$ as a (right) $\Bbb R[T]$-module, under this "scalar multiplication" (module action).

On the other hand, we have:

$A'u = \begin{bmatrix}1&1\\0&1 \end{bmatrix}\begin{bmatrix}x\\0 \end{bmatrix} = \begin{bmatrix}x\\0 \end{bmatrix}$.

It follows, then, that $uT \in U$, in fact $T$ acts as the identity on $U$, so that:

$(u)(f_0 + f_1T + \cdots + f_nT^n) = uf_0 + uf_1 + \cdots+uf_n = (f_0 + f_1 + \cdots f_n)u$ (we can put the scalar on the left since $\Bbb R$ is a commutative ring).

In fact, the reason $U$ is a $\Bbb R[T]$-submodule when $T$ acts as $A'$ and not when $T$ acts as $A$, is that $(1,0)$ is an eigenvector of $A'$. I'll say this again, because it's fairly important:

The eigenspaces of a matrix $A \in M_n(\mathcal{K})$, are $\mathcal{K}[T]$-invariant subspaces. If we have eigenspaces whose dimensions sum up to $n$, we can create a direct sum decomposition of $\mathcal{K}^n$ into $\mathcal{K}[T]$-invariant submodules.

This is a very happy occurence, because it means we can choose an eigenbasis, in which the matrix $A$ is diagonal, and so powers of $A$ can be computed by just taking the powers of the diagonal elements.

You have to realize we don't have "one" $\mathcal{K}[T]$-mdoule structure on $\mathcal{K}^n$, we have one for each matrix $A$. Different choices for $A$ will yield different module structures. So, in truth, the information we get is more about "what $A$ does", and not so much about the internal structure of $\mathcal{K}^n$.

 

[Math Amateur]

Let's look at a simple space, a simple subspace, and two different matrices $A$. Our field will be $\Bbb R$, our vector space will be $\Bbb R^2$, and our subspace $U$ will be:

$U = \{(x,0): x \in \Bbb R\}$.

Our first matrix will be:

$A = \begin{bmatrix}1&0\\1&1 \end{bmatrix}$

and our second matrix will be:

$A' = \begin{bmatrix}1&1\\0&1 \end{bmatrix}$.

These matrices look very similar, but we shall see that they behave very differently with respect to $U$.

Note that, for $u \in U$ we have:

$Au = \begin{bmatrix}1&0\\1&1 \end{bmatrix}\begin{bmatrix}x\\0 \end{bmatrix} = \begin{bmatrix}x\\x \end{bmatrix}$.

Since $Au = uT$, and $T$ is certainly one polynomial of $\Bbb R[T]$, we see we fail to have closure of $U$ as a (right) $\Bbb R[T]$-module, under this "scalar multiplication" (module action).

On the other hand, we have:

$A'u = \begin{bmatrix}1&1\\0&1 \end{bmatrix}\begin{bmatrix}x\\0 \end{bmatrix} = \begin{bmatrix}x\\0 \end{bmatrix}$.

It follows, then, that $uT \in U$, in fact $T$ acts as the identity on $U$, so that:

$(u)(f_0 + f_1T + \cdots + f_nT^n) = uf_0 + uf_1 + \cdots+uf_n = (f_0 + f_1 + \cdots f_n)u$ (we can put the scalar on the left since $\Bbb R$ is a commutative ring).

In fact, the reason $U$ is a $\Bbb R[T]$-submodule when $T$ acts as $A'$ and not when $T$ acts as $A$, is that $(1,0)$ is an eigenvector of $A'$. I'll say this again, because it's fairly important:

The eigenspaces of a matrix $A \in M_n(\mathcal{K})$, are $\mathcal{K}[T]$-invariant subspaces. If we have eigenspaces whose dimensions sum up to $n$, we can create a direct sum decomposition of $\mathcal{K}^n$ into $\mathcal{K}[T]$-invariant submodules.

This is a very happy occurence, because it means we can choose an eigenbasis, in which the matrix $A$ is diagonal, and so powers of $A$ can be computed by just taking the powers of the diagonal elements.

You have to realize we don't have "one" $\mathcal{K}[T]$-mdoule structure on $\mathcal{K}^n$, we have one for each matrix $A$. Different choices for $A$ will yield different module structures. So, in truth, the information we get is more about "what $A$ does", and not so much about the internal structure of $\mathcal{K}^n$.

Thanks to Euge and Deveno for significant help ... It is very much appreciated ...

Just working through your posts now ...

Peter