$T^{\alpha}\left({\mathcal{M}}_{\mu}^{s}\right)$ as $t_{s}(x)$. Then we extend
it to other leafs as:

  * •

Step 1. Denote $Q:=\left\\{r\;\big{|}\;r|s\right\\}$. Let $\tau\in Q$ and
$k=\frac{s}{\tau}\in N$. Then

$\displaystyle t_{\tau}(y)=\left(bd_{k}\right)^{*}(t_{s})(bd_{k}(y))$ (186)

can be calculated by constructing $M_{\tau}(y)$ ($y\in{\bf
T}^{\alpha}\left({\mathcal{M}}_{\tau p\times\tau q}\right)$) as follows:
First, split $M_{s}(bd(y))$ into $(\tau p)^{\alpha}\times(\tau q)^{\alpha}$
blocks

$\begin{array}[]{l}M_{s}(bd_{r}(y))=\\\
\begin{bmatrix}T^{1,\cdots,1}_{1,\cdots,1}(y)&T^{1,\cdots,1}_{1,\cdots,2}(y)&\cdots&T^{1,\cdots,1}_{rq,\cdots,\tau
q}(y)\\\
T^{1,\cdots,2}_{1,\cdots,1}(y)&T^{1,\cdots,2}_{1,\cdots,2}(y)&\cdots&T^{1,\cdots,2}_{rq,\cdots,\tau
q}(y)\\\ \vdots&~{}&~{}&~{}\\\ T^{\tau p,\cdots,\tau
p}_{1,\cdots,1}(y)&T^{\tau p,\cdots,\tau p}_{1,\cdots,2}(y)&\cdots&t^{\tau
p,\cdots,\tau p}_{\tau q,\cdots,\tau q}(y)\\\ \end{bmatrix},\end{array}$

where each block
$T^{i_{1},\cdots,i_{\alpha}}_{j_{1},\cdots,j_{\alpha}}\in{\mathcal{M}}_{k\times
k}$. Then we set

$\displaystyle
M_{r}(y):=k^{\alpha}[\xi^{i_{1},\cdots,i_{\alpha}}_{j_{1},\cdots,j_{\alpha}}(y)]\in{\mathcal{M}}_{(\tau
p)^{\alpha}\times(\tau q)^{\alpha}},$ (187)

where

$\xi^{i_{1},\cdots,i_{\alpha}}_{j_{1},\cdots,j_{\alpha}}(y)=\operatorname{Tr}\left(T^{i_{1},\cdots,i_{\alpha}}_{j_{1},\cdots,j_{\alpha}}\right).$

  * •

Step 2. Assume $gcd(\ell,s)=\tau$. If $\ell=\tau$, $t_{\ell}=t_{\tau}$ has
already been defined in Step 1. So we assume $\ell=k\tau$, where $k\geq 2$.
Then

$\displaystyle t_{\ell}(z)=\left(pr_{k}\right)^{*}(t_{\tau})(pr_{k}(y))$ (188)

can be calculated by constructing

$\displaystyle
M_{\ell}(z):=\frac{1}{k^{\alpha}}\left[M_{\tau}(y=pr_{k}(z))\otimes
I_{k^{\alpha}}\right],\quad z\in{\mathcal{M}}_{\ell p\times\ell q}.$ (189)

######  Example 130.

Consider a covariant tensor field
$\left<t\right>\in{\mathcal{T}}^{2}\left(\Sigma_{\mu}\right)$, where
$\mu=\frac{2}{3}$. Moreover, $\left<t\right>$ is firstly defined at
$x=(x_{i,j})\in{\mathcal{M}}_{2\times 3}$ with its structure matrix as

$\displaystyle M_{1}(x)=\begin{bmatrix}1&x_{12}&0&0&0&1&0&0&x_{22}\\\
0&1&0&-1&0&0&0&0&0\\\ 1&0&-1&0&0&0&1&-1&0\\\
0&1&0&0&0&x_{22}&0&0&x_{23}\end{bmatrix}.$ (190)

  1. 1.

Let $X(\left<x\right>),Y(\left<x\right>)\in V\left(\Sigma_{\mu}\right)$ be
defined at $x\in{\mathcal{M}}_{2\times 3}$ as

$X(x)=\begin{bmatrix}x_{13}&0&0\\\ 0&0&x_{21}\end{bmatrix},\quad
Y(x)=\begin{bmatrix}1&0&1\\\ 0&x_{11}^{2}&0\end{bmatrix}.$

Evaluate $\left<t\right>(X,Y)$.

First,

$\begin{array}[]{ccl}t_{1}(X,\cdot)&=&\left(M_{1}(x)\;\big{|}\;X\right)_{W}\\\
{}\hfil&=&\begin{bmatrix}x_{13}+x_{21}&x_{12}x_{13}-x_{21}&0\\\
0&x_{13}&x_{21}x_{23}\end{bmatrix}.\end{array}$

Then

$t_{1}(X,Y)=x_{13}\left(1+x_{11}^{2}\right)+x_{21}.$

  2. 2.

Expressing $\left<t\right>$ on leaf ${\mathcal{M}}_{4\times 6}$: Note that

$x=pr_{2}(y)=\begin{bmatrix}\frac{y_{11}+y_{22}}{2}&\frac{y_{13}+y_{24}}{2}&\frac{y_{15}+y_{26}}{2}\\\
\frac{y_{31}+y_{42}}{2}&\frac{y_{33}+y_{44}}{2}&\frac{y_{35}+y_{46}}{2}\end{bmatrix},$

where $y=\left[y_{i,j}\right]\in{\mathcal{M}}_{4\times 6}$.

Using (189), we have

$M_{2}(y)=\frac{1}{2^{2}}\begin{bmatrix}t^{11}&t^{12}&t^{13}\\\
t^{21}&t^{22}&t^{23}\\\ \end{bmatrix},$

where

$\begin{array}[]{ccl}t^{11}&=&\begin{bmatrix}1&\frac{y_{13}+y_{24}}{2}&0\\\
0&1&0\\\ \end{bmatrix}\otimes I_{2};\\\ t^{12}&=&\begin{bmatrix}0&0&1\\\
-1&0&0\\\ \end{bmatrix}\otimes I_{2};\\\
t^{13}&=&\begin{bmatrix}0&0&\frac{y_{33}+y_{44}}{2}\\\ 0&0&0\\\
\end{bmatrix}\otimes I_{2};\\\ t^{21}&=&\begin{bmatrix}1&0&-1\\\ 0&1&0\\\
\end{bmatrix}\otimes I_{2};\\\ t^{22}&=&\begin{bmatrix}0&0&0\\\
0&0&\frac{y_{33}+y_{44}}{2}\\\ \end{bmatrix}\otimes I_{2};\\\
t^{23}&=&\begin{bmatrix}1&-1&0\\\ 0&0&\frac{y_{35}+y_{46}}{2}\\\
\end{bmatrix}\otimes I_{2}.\end{array}$

##  5 Lie Algebra on Square Equivalence Space

###  5.1 Ring Structure on $\Sigma$

Consider the vector space of the equivalent classes of square matrices
$\Sigma:=\Sigma_{1}$. Since $\Sigma$ is closed under the STP, more algebraic
structures may be posed on it. First, polynomials; Second, Lie algebra
structure.

Then we extend some fundamental concepts of matrices to their equivalent
classes.

######  Definition 131.

  1. 1.

$\left<A\right>$ is nonsingular (symmetric, skew symmetric, positive/negative
(semi-)definite, upper/lower (strictly) triangular, diagonal, etc.) if its
irreducible element $A_{1}$ is (equivalently, every $A_{i}$ is).

  2. 2.

$\left<A\right>$ and $\left<B\right>$ are similar, denoted by
$\left<A\right>\sim\left<B\right>$, if there exists a nonsingular
$\left<P\right>$ such that

$\displaystyle\left<P^{-1}\right>\left<A\right>\left<P\right>=\left<B\right>.$
(191)

  3. 3.

$\left<A\right>$ and $\left<B\right>$ are congruent, denoted by
$\left<A\right>\simeq\left<B\right>$, if there exists a nonsingular
$\left<P\right>$ such that

$\displaystyle\left<P^{T}\right>\left<A\right>\left<P\right>=\left<B\right>.$
(192)

  4. 4.

$\left<J\right>$ is called the Jordan normal form of $\left<A\right>$, if the
irreducible element $J_{1}\in\left<J\right>$ is the Jordan normal form of
$A_{1}$.

######  Definition 132.

[29] A set $R$ with two operators $+,~{}\times$ is a ring. If the followings
hold:

  1. 1.

$(R,+)$ is an Abelian group;

  2. 2.

$(R,\times)$ is a monoid;

  3. 3.

(Distributive Rule)

$\begin{array}[]{l}(a+b)\times c=a\times c+b\times c\\\ c\times(a+b)=c\times
a+c\times b,\quad a,b,c\in R.\end{array}$

Observing ${\mathcal{M}}_{1}$, which consists of all square matrices, both
$\pm$ (including $\vdash$) and $\ltimes$ are well defined. Unfortunately,
$\left({\mathcal{M}}_{1},\rotatebox{-90.0}{$\pm$}\right)$ is not a group
because there is no unit element. Since both  $\pm$ and $\ltimes$ are
consistent with the equivalence $\sim$, we consider
$\Sigma:={\mathcal{M}}_{1}/\sim$. Then it is easy to verify the following:

######  Proposition 133.

$\left(\Sigma,\rotatebox{-90.0}{$\pm$},\ltimes\right)$ is a ring.

Consider a polynomial on a ring $R$, $p:R\rightarrow R$, as

$\displaystyle\begin{array}[]{ccl}p(x)&=&a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0},\\\
{}\hfil&{}\hfil&~{}~{}~{}a_{i}\in R,\;i=0,1,\cdots,n.\end{array}$ (195)

It is obvious that this is well defined. Set
$R=(\Sigma,\rotatebox{-90.0}{$\pm$},\ltimes)$, then $p(x)$ defined in (195) is
also well defined on $R$. Particularly, the coefficients $a_{i}$ can be chosen
as $\left<a_{i}\right>$ for $a_{i}\in{\mathbb{F}}$, $i=1,\cdots,n$. Then
$p(x)$ is as a “standard” polynomial. Unless elsewhere is stated, in this
paper only such standard polynomials are considered.

For any $\left<A\right>\in\Sigma$, the polynomial $p(\left<A\right>)$ is well
defined, and it is clear that

$\displaystyle p(\left<A\right>)=\left<p(A)\right>,\quad\mbox{for
any}~{}A\in\left<A\right>.$ (196)

Using Taylor expansion, we can consider general matrix functions. For
instance, we have the following result:

######  Theorem 134.

Let $f(x)$ be an analytic function. Then $f(\left<A\right>)$ is well defined
provided $f(A)$ is well defined. Moreover,

$\displaystyle f(\left<A\right>)=\left<f(A)\right>.$ (197)

In fact, the above result can be extended to multi-variable case.

######  Definition 135.

Let $F(x_{1},\cdots,x_{k})$ be a $k$-variable analytic function. Then
$F(\left<A\right>_{1},\cdots,\left<A\right>_{k})$ is a well posed expression,
where $\left<A\right>_{i}\in\Sigma$, $i=1,\cdots,k$. Assume
$A_{i}\in\left<A\right>_{i}$, $i=1,\cdots,k$, then $F(A_{1},\cdots,A_{k})$ is
a realization of $F(\left<A\right>_{1},\cdots,\left<A\right>_{k})$.
Particularly, if $A_{i}\in{\mathcal{M}}_{r\times r}$, $\forall i$, we call
$F(A_{1},\cdots,A_{k})$ a realization on $r$-th leaf.

Similar to (197), we also have

$\displaystyle
F(\left<A\right>_{1},\cdots,\left<A\right>_{k})=\left<F(A_{1},\cdots,A_{k})\right>,$
(198)

provided $F(A_{1},\cdots,A_{k})$ is well defined.

In the following we consider some fundamental matrix functions for $\Sigma$.
We refer to [13] for the definitions and basic properties of some fundamental
matrix functions. Using these acknowledges, the following result is obvious:

######  Theorem 136.

Let $\left<A\right>,~{}\left<B\right>$ in $\Sigma$ (i.e., $A$, $B$ are square
matrices). Then the following hold:

  1. 1.

$\displaystyle e^{\left<A\right>}\ltimes e^{\left<B\right>}=e^{\left<A\ltimes
B\right>}.$ (199)

  2. 2.

If $\left<A\right>$ is real skew symmetric, then $e^{\left<A\right>}$ is
orthogonal.

  3. 3.

Assume $B$ is invertible, we denote $\left<B\right>^{-1}=\left<B^{-1}\right>$.
Then

$\displaystyle
e^{\left<B\right>^{-1}\left<A\right>\left<B\right>}=\left<B\right>^{-1}\ltimes
e^{\left<A\right>}\ltimes\left<B\right>.$ (200)

  4. 4.

Let $A,~{}B$ be closed enough to identity so that $\log(A)$ and $\log(B)$ are
defined. Then

$\displaystyle\log(\left<A\right>\ltimes\left<B\right>)=\log(\left<A\right>)\rotatebox{-90.0}{$\pm$}\log(\left<B\right>).$
(201)

Many known results for matrix functions can be extended to $\Sigma$. For
instance, it is easy to prove the following Euler fromula:

######  Proposition 137.

Consider ${\mathbb{F}}={\mathbb{R}}$ and let $\left<A\right>\in\Sigma$. Then
the Euler formula holds. That is,

$\displaystyle
e^{i\left<A\right>}=\cos(\left<A\right>)~{}\rotatebox{-90.0}{$\pm$}~{}i\sin(\left<A\right>).$
(202)

Recall the modification of trace and determinant in Definitions 19 and 21. The
following proposition shows that for the modifications the relationship
between $\operatorname{trace}(A)$ and $\det(A)$ [13] remain available.

######  Proposition 138.

Assume ${\mathbb{F}}={\mathbb{R}}$ (or ${\mathbb{F}}={\mathbb{C}}$),
$\left<A\right>\in\Sigma$, then

$\displaystyle
e^{\operatorname{Tr}(\left<A\right>)}=\operatorname{Dt}\left(e^{\left<A\right>}\right).$
(203)

Proof. Let $A_{0}\in\left<A\right>$ and $A_{0}\in{\mathcal{M}}_{n\times n}$.
Then

$\begin{array}[]{ccl}e^{\operatorname{Tr}(\left<A\right>)}&=&\left<e^{\frac{1}{n}\operatorname{trace}(A_{0})}\right>\\\
{}\hfil&=&\left<\left(e^{\operatorname{trace}(A_{0})}\right)^{\frac{1}{n}}\right>\\\
{}\hfil&=&\left<\operatorname{Dt}(e^{A_{0}})\right>\\\
{}\hfil&=&\operatorname{Dt}\left(e^{\left<A\right>}\right).\end{array}$

$\Box$

Next, we consider the characteristic polynomial of an equivalent class, they
comes from standard matrix theory [23].

######  Definition 139.

Let $\left<A\right>\in\Sigma$. $A_{1}\in\left<A\right>$ is its irreducible
element. Then

$\displaystyle p_{\left<A\right>}(\lambda):=\det(\lambda\vdash A_{1})$ (204)

is called the characteristic polynomial of $\left<A\right>$.

The following result is an immediate consequence of the definition.

######  Theorem 140.

(Cayley-Hamilton) Let $p_{\left<A\right>}$ be the characteristic polynomial of
$\left<A\right>$. Then

$\displaystyle p_{\left<A\right>}(\left<A\right>)=0.$ (205)

######  Remark 141.

  1. 1.

If we choose $A_{k}=A_{1}\otimes I_{k}$ and calculate the characteristic
polynomial of $p_{A^{k}}$, then
$p_{A_{k}}(\lambda)=\left(p_{A_{1}}(\lambda)\right)^{k}$. So
$p_{A_{k}}(\left<A\right>)=0$ is equivalent to $p_{A_{1}}(\left<A\right>)=0$.

  2. 2.

Choosing any $A_{i}\in\left<A\right>$, the corresponding minimal polynomials
$q_{A_{i}}(\lambda)$ are the same. So we have unique minimal polynomial as
$q_{\left<A\right>}(\lambda)=q_{A_{i}}(\lambda)$.

###  5.2 Bundled Lie Algebra

Consider the vector space of the equivalent classes of square matrices
$\Sigma:=\Sigma_{1}$, this section gives a Lie algebraic structure to it.

######  Definition 142.

[21] A Lie algebra is a vector space $g$ over some field ${\mathbb{F}}$ with a
binary operation $[\cdot,\cdot]:~{}g\times g\rightarrow g$, satisfying

  1. 1.

(bi-linearity)

$\displaystyle\begin{array}[]{ccl}[\alpha A+\beta
B,C]&=&\alpha[A,C]+\beta[B,C];\cr[C,\alpha A+\beta
B]&=&\alpha[C,A]+\beta[C,B],\end{array}$ (208)

where $\alpha,~{}\beta\in{\mathbb{F}}$.

  2. 2.

(skew-symmetry)

$\displaystyle[A,B]=-[B,A];$ (209)

  3. 3.

(Jacobi Identity)

$\displaystyle\begin{array}[]{r}[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0,\\\ \forall
A,B,C\in g.\end{array}$ (212)

######  Definition 143.

Let $(E,Pr,B)$ be a discrete bundle. If

  1. 1.

$(B,\oplus,\otimes)$ is a Lie algebra;

  2. 2.

$(E_{i},+,\times)$ is a Lie algebra, $i=1,2,\cdots$;

  3. 3.

The restriction $Pr|_{E_{i}}:~{}(E_{i},+,\times)\rightarrow(B,\oplus,\otimes)$
is a Lie algebra homomorphism, $i=1,2,\cdots$,

then $(B,\oplus,\otimes)$ is called a bundled Lie algebra.

(We refer to Definition 61 and Remark 64 for the concept of discrete bundle.)

On vector space $\Sigma$ we define an operation
$[\cdot,\cdot]:~{}\Sigma\times\Sigma\rightarrow\Sigma$ as

$\displaystyle[\left<A\right>,\left<B\right>]:=\left<A\right>\ltimes\left<B\right>\vdash\left<B\right>\ltimes\left<A\right>.$
(213)

Then we have the following Lie algebra:

######  Theorem 144.

The vector space ${\Sigma}$ with Lie bracket $[\cdot,\cdot]$ defined in (213),
is a bundled Lie algebra, denoted by $gl({\mathbb{F}})$.

Proof. Let ${\mathcal{M}}=\cup_{i=1}^{\infty}{\mathcal{M}}^{i}$, where
${\mathcal{M}}^{i}={\mathcal{M}}_{1}^{i}$. Then it is clear that
$\left({\mathcal{M}},Pr,\Sigma\right)$ is a discrete bundle.

Next, we prove $\left(\Sigma,\rotatebox{-90.0}{$\pm$},[\cdot,\cdot]\right)$ is
a Lie algebra. Equations (208) and (209) are obvious. We prove (212) only.

Assume $A_{1}\in\left<A\right>$, $B_{1}\in\left<B\right>$ and
$C_{1}\in\left<C\right>$ are irreducible, and $A_{1}\in{\mathcal{M}}_{m\times
m}$, $B_{1}\in{\mathcal{M}}_{n\times n}$, and $C_{1}\in{\mathcal{M}}_{r\times
r}$. Let $t=\operatorname{lcm}(n,m,r)$. Then it is easy to verify that

$\displaystyle\begin{array}[]{l}[\left<A\right>,[\left<B\right>,\left<C\right>]]=\\\
~{}\left<\left[(A_{1}\otimes I_{t/m}),[(B_{1}\otimes I_{t/n}),(C_{1}\otimes
I_{t/r})]\right]\right>.\end{array}$ (216)

Similarly, we have

$\displaystyle\begin{array}[]{l}[\left<B\right>,[\left<C\right>,\left<A\right>]]=\\\
~{}\left<\left[(B_{1}\otimes I_{t/n}),[(C_{1}\otimes I_{t/r}),(A_{1}\otimes
I_{t/m})]\right]\right>.\end{array}$ (219)
$\displaystyle\begin{array}[]{l}[\left<C\right>,[\left<A\right>,\left<B\right>]]=\\\
~{}\left<\left[(C_{1}\otimes I_{t/r}),[(A_{1}\otimes I_{t/m}),(B_{1}\otimes
I_{t/n})]\right]\right>.\end{array}$ (222)

Since (212) is true for any $A,B,C\in gl(t,{\mathbb{F}})$, it is true for
$A=A_{1}\otimes I_{t/m}$, $B=B_{1}\otimes I_{t/n}$, and $C=C_{1}\otimes
I_{t/r}$. Using this fact and equations (216)-(222), we have

$\begin{array}[]{l}[\left<A\right>,[\left<B\right>,\left<C\right>]]\rotatebox{-90.0}{$\pm$}[\left<B\right>,[\left<C\right>,\left<A\right>]]\rotatebox{-90.0}{$\pm$}[\left<C\right>,[\left<A\right>,\left<B\right>]]\\\
=\left<\left[(A_{1}\otimes I_{t/m}),[(B_{1}\otimes I_{t/n}),(C_{1}\otimes
I_{t/r})]\right]\right.\\\ +~{}~{}\left[(B_{1}\otimes I_{t/n}),[(C_{1}\otimes
I_{t/r}),(A_{1}\otimes I_{t/m})]\right]{}{}\\\ +\left.\left[(C_{1}\otimes
I_{t/r}),[(A_{1}\otimes I_{t/m}),(B_{1}\otimes I_{t/n})]\right]\right>\\\
=\left<0\right>=0.\end{array}$

Let the Lie algebraic structure on $M^{n}$ be $gl(n,{\mathbb{F}})$ (where
${\mathbb{F}}={\mathbb{R}}$ or ${\mathbb{F}}={\mathbb{C}}$). It follows from
the consistence of  $\pm$ and $\ltimes$ with the equivalence that
$Pr:gl(n,{\mathbb{F}})\rightarrow gl({\mathbb{F}})$ is a Lie algebra
homomorphism. $\Box$

###  5.3 Bundled Lie Sub-algebra

This section considers some useful Lie sub-algebras of Lie algebra
$gl({\mathbb{F}})$. Assume $g$ is a Lie algebra, $h\subset g$ is a vector
subspace. Then $h$ is called a Lie sub-algebra if and only if $[h,h]\subset
h$.

######  Definition 145.

Let $(E,Pr,B)$ be a bundled Lie algebra. if

  1. 1.

$(H,\oplus,\otimes)$ is a Lie sub-algebra of $(B,\oplus,\otimes)$;

  2. 2.

$(F_{i},+,\times)$ is a Lie sub-algebra of $(E_{i},+,\times)$, $i=1,2,\cdots$;

  3. 3.

The restriction $Pr|_{F_{i}}:~{}(F_{i},+,\times)\rightarrow(H,\oplus,\otimes)$
is a Lie algebra homomorphism, $i=1,2,\cdots$,

then $(F,Pr,H)$ is called a bundled Lie sub-algebra of $(E,Pr,B)$.

It is well known that there are some useful Lie sub-algebras of Lie algebra
$gl(n,{\mathbb{F}})$. When $gl(n,{\mathbb{F}})$, $\forall n$, are generalized
to the bundled Lie algebra $gl({\mathbb{F}})$ (over $\Sigma$), the
corresponding Lie sub-algebras are investigated one-by-one in this section.

  * •

Bundled orthogonal Lie sub-algebra

######  Definition 146.

$\left<A\right>\in\Sigma$ is said to be symmetric (skew symmetric) if
$A^{T}=A$ ($A^{T}=-A$), $\forall A\in\left<A\right>$.

The symmetric (skew symmetric) $\left<A\right>$ is well defined because if
$A\sim B$ and $A^{T}=A$ (or $A^{T}=-A$), then so is $B$. It is also easy to
verify the following:

######  Proposition 147.

Assume $\left<A\right>$ and $\left<B\right>$ are skew symmetric, then so is
$[\left<A\right>,\left<B\right>]$.

We, therefore, can define the following bundled Lie sub-algebra.

######  Definition 148.

$o({\mathbb{F}}):=\left\\{\left<A\right>\in
gl({\mathbb{F}})\;|\;\left<A\right>^{T}=-\left<A\right>\right\\}$

is called the bundled orthogonal algebra.

  * •

Bundled special linear algebra

######  Definition 149.

$sl({\mathbb{F}}):=\left\\{\left<A\right>\in
gl({\mathbb{F}})\;|\;\operatorname{Tr}(\left<A\right>)=0\right\\}$

is called the bundled special linear algebra.

Similar to the case of orthogonal algebra, it is easy to verify that
$\sl({\mathbb{F}})$ is a Lie sub-algebra of $gl({\mathbb{F}})$.

  * •

Bundled upper triangular algebra

######  Definition 150.

$t({\mathbb{F}}):=\left\\{\left<A\right>\in
gl({\mathbb{F}})\;|\;\left<A\right>~{}\mbox{is upper triangular}\right\\}$

is called the bundled upper triangular algebra.

Similarly, we can define bundled lower triangular algebras.

  * •

Bundled strictly upper triangular algebra

######  Definition 151.

$n({\mathbb{F}}):=\left\\{\left<A\right>\in
gl({\mathbb{F}})\;|\;\left<A\right>~{}\mbox{is strictly upper
triangular}\right\\}$

is called the bundled strictly upper triangular algebra.

  * •

Bundled diagonal algebra

######  Definition 152.

$d({\mathbb{F}}):=\left\\{\left<A\right>\in
gl({\mathbb{F}})\;|\;\left<A\right>~{}\mbox{is diagonal}\right\\}$

is called the bundled diagonal algebra.

  * •

Bundled symplectic algebra

######  Definition 153.

$\begin{array}[]{ccl}sp({\mathbb{F}})&:=&\left\\{\left<A\right>\in
gl({\mathbb{F}})\;|\;\left<A\right>~{}\mbox{satisfies
(\ref{4.4.2.0})}\right.\\\
{}\hfil&{}\hfil&~{}\left.\mbox{and}~{}A_{1}\in{\mathcal{M}}_{2n\times
2n},~{}n\in N\right\\},\end{array}$

is called the bundled symplectic algebra.

$\displaystyle\left<J\right>\ltimes\left<A\right>\rotatebox{-90.0}{$\pm$}\left<A\right>^{T}\ltimes\left<J\right>=0,$
(223)

where

$J=\begin{bmatrix}0&1\\\ -1&0\end{bmatrix}.$

######  Definition 154.

A Lie sub-algebra ${\mathcal{J}}\subset{\mathcal{G}}$ is called an idea, if

$\displaystyle\left[g,{\mathcal{J}}\right]\in{\mathcal{J}}.$ (224)

######  Example 155.

$sl({\mathbb{F}})$ is an idea of $gl({\mathbb{F}})$. Because

$\operatorname{Tr}[g,h]=\operatorname{Tr}(g\ltimes h\vdash h\ltimes
g)=0,\quad\forall g\in gl({\mathbb{F}}),\forall h\in\sl({\mathbb{F}}).$

Hence,
$\left[gl({\mathbb{F}}),\sl({\mathbb{F}})\right]\subset\sl({\mathbb{F}})$.

Many properties of the sub-algebra of $gl(n,{\mathbb{F}})$ can be extended to
the sub-algebra of $gl({\mathbb{F}})$. The following is an example.

######  Proposition 156.

$\displaystyle
gl({\mathbb{F}})=sl({\mathbb{F}})\rotatebox{-90.0}{$\pm$}r\left<1\right>,\quad
r\in{\mathbb{F}}.$ (225)

Proof. It is obvious that

$\left<A\right>=\left(\left<A\right>\vdash\operatorname{Tr}(\left<A\right>)\right)\rotatebox{-90.0}{$\pm$}\operatorname{Tr}(\left<A\right>)\left<1\right>.$

Since
$\operatorname{Tr}\left(\left<A\right>\vdash\operatorname{Tr}(\left<A\right>)\right)=0$,
which means

$\left(\left<A\right>\vdash\operatorname{Tr}(\left<A\right>)\right)\in\sl({\mathbb{F}}).$

The conclusion follows. $\Box$

######  Example 157.

  1. 1.

Denote by

$\displaystyle
gl(\left<A\right>,{\mathbb{F}}):=\left\\{\left<X\right>\;\big{|}\;\left<X\right>\left<A\right>\rotatebox{-90.0}{$\pm$}\left<A\right>\left<X\right>=0\right\\}.$
(226)

It is obvious that $gl(\left<A\right>,{\mathbb{F}})$ is a vector sub-space of
$gl({\mathbb{F}})$. Let $\left<X\right>,~{}\left<Y\right>\in
gl(\left<A\right>,{\mathbb{F}})$. Then

$\begin{array}[]{l}\left[\left<X\right>,\left<Y\right>\right]\left<A\right>\rotatebox{-90.0}{$\pm$}\left<A\right>\left[\left<X\right>,\left<Y\right>\right]^{T}\\\
=\left<X\right>\left<Y\right>\left<A\right>\vdash\left<Y\right>\left<X\right>\left<A\right>\\\
~{}+\left<A\right>\left<Y^{T}\right>\left<X^{T}\right>\vdash\left<X^{T}\right>\left<Y^{T}\right>\left<A\right>\\\
=\left<X\right>\left<Y\right>\left<A\right>\vdash\left<Y\right>\left<X\right>\left<A\right>\\\
~{}+\left<A\right>\left<Y\right>\left<X\right>\left<A\right>\vdash\left<X\right>\left<Y\right>\left<A\right>\\\
=0.\end{array}$

Hence, $gl(\left<A\right>,{\mathbb{F}})\subset gl({\mathbb{F}})$ is a bundled
Lie sub-algebra.

  2. 2.

Assume $\left<A\right>$ and $\left<B\right>$ are congruent. That is, there
exists a non-singular $\left<P\right>$ such that
$\left<A\right>=\left<P^{T}\right>\left<B\right>\left<P\right>$. Then it is
easy to verify that
$\pi:~{}gl(\left<A\right>,{\mathbb{F}})\rightarrow~{}gl(\left<B\right>,{\mathbb{F}})$
is an isomorphism, where

$\pi(\left<X\right>)=\left<P^{-T}\right>\left<X\right>\left<P\right>.$

###  5.4 Further Properties of $gl({\mathbb{F}})$

######  Definition 158.

Let $p\in{\mathbb{N}}$.

  1. 1.

The $p$-truncated equivalent class is defined as

$\displaystyle\left<A\right>^{[\cdot,p]}:=\left\\{A_{i}\in\left<A\right>\;\big{|}\;i|p\right\\}.$
(227)

  2. 2.

The $p$-truncated square matrices is defined as

$\displaystyle{\mathcal{M}}^{[\cdot,p]}:=\left\\{A\in{\mathcal{M}}_{n\times
n}\;\big{|}\;n|p\right\\}.$ (228)

  3. 3.

The $p$-truncated equivalence space is defined as

$\displaystyle\Sigma^{[\cdot,p]}:={\mathcal{M}}^{[\cdot,p]}/\sim.$ (229)

######  Remark 159.

  1. 1.

If $A_{1}\in\left<A\right>$ is irreducible, $A_{1}\in{\mathcal{M}}_{n\times
n}$ and $n$ is not a divisor of $p$, then
$\left<A\right>^{[\cdot,p]}=\emptyset$.

  2. 2.

It is obvious that
$\left({\mathcal{M}}^{[\cdot,p]},Pr,\Sigma^{[\cdot,p]}\right)$ is a discrete
bundle, which is a sub-bundle of $\left({\mathcal{M}},Pr,\Sigma\right)$. That
is, the following (233) is commutative:

$\displaystyle\begin{array}[]{ccc}{\mathcal{M}}^{[\cdot,k]}&\xrightarrow{~{}~{}~{}\pi~{}~{}~{}}&{\mathcal{M}}\\\
Pr\downarrow&{}\hfil&Pr\downarrow\\\
\Sigma^{[\cdot,k]}&\xrightarrow{~{}~{}~{}\pi^{\prime}~{}~{}~{}}&\Sigma\end{array}$
(233)

Where $\pi$ and $\pi^{\prime}$ are including mappings.

  3. 3.

It is ready to verify that $\Sigma^{[\cdot,p]}$ is closed with  $\pm$ and
$[\cdot,\cdot]$, defined in (213), hence the including mapping
$\pi^{\prime}:\Sigma^{[\cdot,p]}\rightarrow\Sigma$ is a Lie algebra
homomorphism. Identifying $\left<A\right>^{[\cdot,p]}$ with its image
$\left<A\right>=\pi^{\prime}\left(\left<A\right>^{[\cdot,p]}\right)$, then
$\Sigma^{[\cdot,p]}$ becomes a Lie sub-algebra of $gl({\mathbb{F}})$. We
denote this as

$\displaystyle
gl^{[\cdot,p]}({\mathbb{F}}):=\left(\Sigma^{[\cdot,p]},~{}\rotatebox{-90.0}{$\pm$},~{}[\cdot,\cdot]\right),$
(234)

and call $gl^{[\cdot,p]}({\mathbb{F}})$ the $p$-truncated Lie sub-algebra of
$gl({\mathbb{F}})$.

  4. 4.

Let $\Gamma\subset gl({\mathbb{F}})$ be a Lie sub-algebra. Then its
$p$-truncated sub-algebra $\Gamma^{[\cdot,p]}$ is defined in a similar way as
for $gl^{[\cdot,p]}$. Alternatively, it can be considered as

$\displaystyle\Gamma^{[\cdot,p]}:=\Gamma\bigcap\Sigma^{[\cdot,p]}.$ (235)

######  Definition 160.

[25] Let $g$ be a Lie algebra.

  1. 1.

Denote the derived serious as ${\mathcal{D}}(g):=[g,g]$, and

${\mathcal{D}}^{(k+1)}(g):={\mathcal{D}}\left({\mathcal{D}}^{k}(g)\right),\quad
k=1,2,\cdots.$

Then $g$ is solvable, if there exists an $n\in{\mathbb{N}}$ such that
${\mathcal{D}}^{(n)}(g)=\\{0\\}$.

  2. 2.

Denote the descending central series as ${\mathcal{C}}(g):=[g,g]$, and

${\mathcal{C}}^{(k+1)}(g):=\left[g,{\mathcal{C}}^{(k)}(g)\right],\quad
k=1,2,\cdots.$

The $g$ is nilpotent, if there exists an $n\in{\mathbb{N}}$ such that
${\mathcal{C}}^{(n)}(g)=\\{0\\}$.

######  Definition 161.

Let $\Gamma\subset gl({\mathbb{F}})$ be a sub-algebra of $gl({\mathbb{F}})$.

  1. 1.

$\Gamma$ is solvable, if for any $p\in{\mathbb{N}}$, the truncated sub-algebra
$\Gamma^{[\cdot,p]}$ is solvable.

  2. 2.

$\Gamma$ is nilpotent, if for any $p\in{\mathbb{N}}$, the truncated sub-
algebra $\Gamma^{[\cdot,p]}$ is nilpotent.

######  Definition 162.

[25] Let $g$ be a Lie algebra.

  1. 1.

$g$ is simple if it has no non-trivial idea (the only ideas are $\\{0\\}$ and
$g$ itself).

  2. 2.

$g$ is semi-simple if it has no solvable idea except $\\{0\\}$.

Though the following proposition is simple, it is also fundamental.

######  Proposition 163.

The Lie algebra $gl^{[\cdot,p]}({\mathbb{F}})$ is isomorphic to the classical
linear algebra $gl(p,{\mathbb{F}})$.

Proof. First we construct a mapping
$\pi:gl^{[\cdot,p]}({\mathbb{F}})\rightarrow gl(p,{\mathbb{F}})$ as follows:
Assume $\left<A\right>\in gl^{[\cdot,p]}({\mathbb{F}})$ and
$A_{1}\in\left<A\right>$ is irreducible. Say, $A_{1}\in{\mathcal{M}}_{n\times
n}$, then by definition, $n|p$. Denote by $s=p/n$, then define

$\pi(\left<A\right>):=A_{1}\otimes I_{s}\in gl(p,{\mathbb{F}}).$

Set $\pi^{\prime}:gl(p,{\mathbb{F}})\rightarrow gl^{[\cdot,p]}({\mathbb{F}})$
as

$\pi^{\prime}(B):=\left<B\right>\in gl^{[\cdot,p]}({\mathbb{F}}),$

then it is ready to verify that $\pi$ is a bijective mapping and
$\pi^{-1}=\pi^{\prime}$.

By the definitions of  $\pm$ and $[\cdot,\cdot]$, it is obvious that $\pi$ is
a Lie algebra isomorphism. $\Box$

The following properties are available for classical $gl(n,{\mathbb{F}})$
[25], [41]. Using Proposition 163, it is easy to verify that they are also
available for $gl({\mathbb{F}})$.

######  Proposition 164.

Let $g\subset gl({\mathbb{F}})$ be a Lie sub-algebra.

  1. 1.

If $g$ is nilpotent then it is solvable.

  2. 2.

If $g$ is solvable (or nilpotent) then so is its sub-algebra, its homomorphic
image.

  3. 3.

If $h\subset g$ is an idea of $g$ and $h$ and $g/h$ are solvable, then $g$ is
also solvable.

######  Definition 165.

Let $\left<A\right>\in gl({\mathbb{F}})$. The adjoint representation
$\operatorname{ad}_{\left<A\right>}:~{}gl({\mathbb{F}})\rightarrow
gl({\mathbb{F}})$ is defined as

$\displaystyle\operatorname{ad}_{\left<A\right>}\left<B\right>=[\left<A\right>,\left<B\right>].$
(236)

To see (236) is well defined, we have to prove that

$\displaystyle\operatorname{ad}_{\left<A\right>}\left<B\right>=\left<\operatorname{ad}_{A}B\right>,\quad
A\in\left<A\right>,~{}\left<B\right>\in\left<B\right>.$ (237)

It follows from the consistence of $\ltimes$ and  $\pm$ ($\vdash$) with
$\sim_{\ell}$ immediately.

######  Example 166.

Consider $\left<A\right>\in\Sigma$. Assume $\left<A\right>$ is nilpotent, that
is, there is a $k>0$ such that $\left<A\right>^{k}=0$. Then
$\operatorname{ad}_{\left<A\right>}$ is also nilpotent.

Note that $A^{k}=0$ if and only if $(A\otimes I_{s})^{k}=0$. Similarly,
$\operatorname{ad}_{A}^{k}=0$ if and only if $\operatorname{ad}^{k}_{A\otimes
I_{s}}=0$. Hence, we need only to show that $\operatorname{ad}_{A}$ is
nilpotent, where $A\in\left<A\right>$ and $A\in{\mathcal{M}}_{n\times n}$. It
is easy to verify that

$\operatorname{ad}^{m}_{A}B=\mathop{\sum}\limits_{i=0}^{m}(-1)^{i}\binom{m}{i}A^{m-i}BA^{i}.$

As $m=2k-1$, it is clear that $\operatorname{ad}^{m}_{A}B=0$, $\forall
B\in{\mathcal{M}}_{n\times n}$. It follows that

$\operatorname{ad}^{2k-1}_{A}=0.$

######  Definition 167.

  1. 1.

Let $A,~{}B\in gl(n,{\mathbb{F}})$. Then the Killing form
$(\cdot,\cdot)_{K}:gl(n,{\mathbb{F}})\times
gl(n,{\mathbb{F}})\rightarrow{\mathbb{F}}$ is defined as (We refer to [25] for
original definition. The following definition is with a mild modification.)

$\displaystyle(A,B)_{K}:=\operatorname{Tr}(\operatorname{ad}_{A}\operatorname{ad}_{B}).$
(238)

  2. 2.

Assume $\left<A\right>,\left<B\right>\in gl({\mathbb{F}})$. The killing form
$(\cdot,\cdot):gl({\mathbb{F}})\times gl({\mathbb{F}})\rightarrow{\mathbb{F}}$
is defined as

$\displaystyle(\left<A\right>,\left<B\right>)_{K}:=\operatorname{Tr}\left(\operatorname{ad}_{\left<A\right>}\ltimes\operatorname{ad}_{\left<B\right>}\right).$
(239)

To see the killing form is well defined, we also need to prove

$\displaystyle(\left<A\right>,\left<B\right>)_{K}=(A,B)_{K},\quad
A\in\left<A\right>,~{}B\in\left<B\right>.$ (240)

Similar to (237), it can be verified by a straightforward calculation.

Because of the equations (237) and (240), the following properties of finite
dimensional Lie algebras [41] can easily be extended to $gl({\mathbb{F}})$:

######  Proposition 168.

Consider $g=gl({\mathbb{F}})$. Let $\left<A\right>$, $\left<A\right>_{1}$,
$\left<A\right>_{2}$, $\left<B\right>$, $\left<E\right>\in g$,
$c_{1},~{}c_{2}\in{\mathbb{F}}$. Then

  1. 1.

$\displaystyle(\left<A\right>,\left<B\right>)_{K}=(\left<B\right>,\left<A\right>)_{K}.$
(241)

  2. 2.

$\displaystyle\begin{array}[]{l}\left(c_{1}\left<A\right>_{1}\rotatebox{-90.0}{$\pm$}c_{2}\left<A\right>_{2},\left<B\right>\right)_{K}\\\
=c_{1}\left(\left<A\right>_{1},\left<B\right>\right)_{K}\rotatebox{-90.0}{$\pm$}c_{2}\left(\left<A\right>_{2},B\right)_{K},\\\
~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}c_{1},\;c_{2}\in{\mathbb{F}}.\end{array}$ (245)

  3. 3.

$\displaystyle\left(\operatorname{ad}_{\left<A\right>}\left<B\right>,~{}\left<E\right>\right)_{K}\rotatebox{-90.0}{$\pm$}\left(\left<B\right>,~{}\operatorname{ad}_{\left<A\right>}\left<E\right>\right)_{K}=0.$
(246)

  4. 4.

Let $h\subset g$ be an idea of $g$, and $\left<A\right>,~{}\left<B\right>\in
h$. Then

$\displaystyle\left(\left<A\right>,~{}\left<B\right>\right)_{K}=\left(\left<A\right>,~{}\left<B\right>\right)_{K}^{h}.$
(247)

The right hand side means the Killing form on the ideal $h$.

  5. 5.

A sub-algebra $\xi\subset g$ is semi-simple, if and only if, its Killing form
is non-degenerated.

The Engel theorem can easily be extended to $gl({\mathbb{F}})$:

######  Theorem 169.

Let $\\{0\\}\neq g\subset gl({\mathbb{F}})$ be a bundled Lie sub-algebra.
Assume each $\left<A\right>\in g$ is nilpotent, (i.e., for each
$\left<A\right>\in g$ there exists a $k>0$ such that
$\left<A\right>^{k}=\left<A^{k}\right>=0$).

  1. 1.

If $g$ is finitely generated, then there exists a vector $X\neq 0$ (of
suitable dimension) such that

$G\ltimes X=0,\quad\forall G\in g.$

  2. 2.

$g$ is nilpotent.

######  Definition 170.

[25] Let $V$ be an $n$ dimensional vector space. A flag in $V$ is a chain of
subspaces

$0=V_{0}\subset V_{1}\subset V_{2}\subset\cdots\subset V_{n}=V,$

with $\dim(V_{i})=i$. Let $A\in End(V)$ be an endomorphism of $V$. $A$ is said
to stabilize this flag if

$AV_{i}\subset V_{i},\quad i=1,\cdots,n.$

Lie theorem can be extended to $gl({\mathbb{F}})$ as follows.

######  Theorem 171.

Assume $g\subset gl({\mathbb{F}})$ is a solvable Lie sub-algebra. Then for any
$p>0$ there is a flag of ideals
$0={\mathcal{I}}_{0}\subset{\mathcal{I}}_{1}\subset\cdots\subset{\mathcal{I}}_{p}$,
such that the truncated $g^{[\cdot,p]}$ stabilizes the flag.

######  Corollary 172.

Assume $g\subset gl({\mathbb{F}})$ is a Lie sub-algebra. $g$ is solvable if
and only if, $D(g)$ is nilpotent.

######  Example 173.

Consider the bundled Lie sub-algebras $t({\mathbb{F}})$ and $n({\mathbb{F}})$.
It is easy to verify the following:

  1. 1.

$t({\mathbb{F}})$ is solvable;

  2. 2.

$n({\mathbb{F}})$ is nilpotent.

Even though $gl({\mathbb{F}})$ is an infinite dimensional Lie algebra, it has
almost all the properties of finite dimensional Lie algebras. This claim can
be verified one by one easily. The reason is: $gl({\mathbb{F}})$ is a union of
finite dimensional Lie algebras.

##  6 Lie Group on Nonsingular Equivalence Space

###  6.1 Bundled Lie Group

Consider $\Sigma:=\Sigma_{1}$, we define a subset

$\displaystyle
GL({\mathbb{F}}):=\left\\{\left<A\right>\in\Sigma\;|\;\operatorname{Dt}(\left<A\right>)\neq
0\right\\}.$ (248)

We emphasize the fact that $GL({\mathbb{F}})$ is an open subset of $\Sigma$.
For an open subset of bundled manifolds we have the following result.

######  Proposition 174.

Let $M$ be a bundled manifold, and $N$ an open subset of $M$. Then $N$ is also
a bundled manifold.

Proof. It is enough to construct an open cover of $N$. Starting from the open
cover of $M$, which is denoted as

${\mathcal{C}}=\\{U_{\lambda}\;|\;\lambda\in\Lambda\\},$

we construct

${\mathcal{C}}_{N}:=\\{U_{\lambda}\cap
N\;|\;U_{\lambda}\in{\mathcal{C}},~{}\lambda\in\Lambda\\}$

Then we can prove that it is $C^{r}$ ($C^{\infty}$ or $C^{\omega}$) comparable
as long as ${\mathcal{C}}$ is. Verifying other conditions is trivial. $\Box$

######  Corollary 175.

$GL({\mathbb{F}})$ is a bundled manifold.

Note that, Proposition 45 shows that $\ltimes:GL({\mathbb{F}})\times
GL({\mathbb{F}})\rightarrow GL({\mathbb{F}})$ is well define.

######  Definition 176.

A topological space $G$ is a bundled Lie group, if

  1. 1.

it is a bundled analytic manifold;

  2. 2.

it is a group;

  3. 3.

the product $A\times B\rightarrow AB$ and the inverse mapping $A\rightarrow
A^{-1}$ are analytic.

The following result is an immediate consequence of the definition.

######  Theorem 177.

$GL({\mathbb{F}})$ is a bundled Lie group.

Proof. We already known then $GL({\mathbb{F}})$ is a bundled analytic
manifold. We first prove $GL({\mathbb{F}})$ is a group. It is ready to verify
that $\left<1\right>$ is the identity. Moreover,
$\left<A\right>^{-1}=\left<A^{-1}\right>$. The conclusion follows.

Using a simple coordinate chart, it is obvious that the inverse and product
are two analytic mappings. $\Box$

###  6.2 Relationship with $gl({\mathbb{F}})$

Denote by

${\mathcal{W}}:=\left\\{A\in{\mathcal{M}}_{1}\;|\;\det(A)\neq 0\right\\},$

and

${\mathcal{W}}_{s}:={\mathcal{W}}\bigcap{\mathcal{M}}_{s\times s},\quad
s=1,2,\cdots.$

Consider the bundle: $\left({\mathcal{M}}_{1},Pr,\Sigma\right)$, where the map
$Pr:A\mapsto\left<A\right>$ is the national projection. It has a natural sub-
bundle: $\left({\mathcal{W}},Pr,GL({\mathbb{F}})\right)$ via the following
bundle morphism as

$\displaystyle\begin{array}[]{ccc}{\mathcal{W}}&\xrightarrow{~{}~{}~{}\pi~{}~{}~{}}&{\mathcal{M}}_{1}\\\
Pr\downarrow&{}\hfil&Pr\downarrow\\\
GL({\mathbb{F}})&\xrightarrow{~{}~{}~{}\pi^{\prime}~{}~{}~{}}&\Sigma\end{array}$
(252)

In fact, the projection leads to Lie group homomorphism.

######  Theorem 178.

  1. 1.

With natural group and differential structures,
${\mathcal{W}}_{s}=GL(s,{\mathbb{F}})$ is a Lie group.

  2. 2.

Consider the projection $Pr$. Restrict it to each leaf yields

$\displaystyle Pr\big{|}_{{\mathcal{W}}_{s}}:GL(s,{\mathbb{F}})\rightarrow
GL({\mathbb{F}}).$ (253)

Then $Pr\big{|}_{{\mathcal{W}}_{s}}$ is a Lie group homomorphism.

  3. 3.

Set the image set as $Pr(GL(s,{\mathbb{F}})):=\Psi_{s}$. Then
$\Psi_{s}<GL({\mathbb{F}})$ is a Lie sub-group. Moreover,

$\displaystyle
Pr\big{|}_{{\mathcal{W}}_{s}}:GL(s,{\mathbb{F}})\rightarrow\Psi_{s}$ (254)

is a Lie group isomorphism.

######  Definition 179.

A vector field $\left<\xi\right>\in V\left(GL({\mathbb{F}})\right)$, (where
for each $P\in GL({\mathbb{F}})$, $\left<\xi(P)\right>\in
T_{P}\left(GL({\mathbb{F}})\right)$), is called a left-invariant vector field,
if for any $\left<A\right>\in GL({\mathbb{F}})$

$\begin{array}[]{ccl}\left(L_{\left<A\right>}\right)_{*}\left(\left<\xi\right>(P)\right)&:=&\left<\left(L_{A}\right)_{*}(\xi(P))\right>\\\
{}\hfil&=&\left<\xi(AP)\right>=\left<\xi\right>(AP).\end{array}$

Then it is easy to verify the following relationship between
$GL({\mathbb{F}})$ and $gl({\mathbb{F}})$:

######  Theorem 180.

The corresponding Lie algebra of the bundled Lie group $GL({\mathbb{F}})$ is
$gl({\mathbb{F}})$ in the following natural sense:

$gl({\mathbb{F}})\simeq
T_{\left<1\right>}\left(GL({\mathbb{F}})\right)\xrightarrow{\left(L_{\left<A\right>}\right)_{*}}T_{\left<A\right>}\left(GL({\mathbb{F}})\right).$

That is, $gl({\mathbb{F}})$ is a Lie algebra isomorphic to the Lie algebra
consists of the vectors on the tangent space of $GL({\mathbb{F}})$ at
identity. Then these vectors generate the left-invariant vector fields which
form the tangent space at any $\left<A\right>\in GL({\mathbb{F}})$.

Let $Pr:{\mathcal{M}}_{n\times n}\rightarrow\Sigma$ be the natural mapping
$A\mapsto\left<A\right>$. Then we have the following commutative picture:

$\displaystyle\begin{array}[]{ccc}gl(n,{\mathbb{F}})&\xrightarrow{~{}~{}~{}\exp~{}~{}~{}}&GL(n,{\mathbb{F}})\\\
Pr\downarrow&{}\hfil&Pr\downarrow\\\
gl({\mathbb{F}})&\xrightarrow{~{}~{}~{}\exp~{}~{}~{}}&GL({\mathbb{F}})\end{array}$
(258)

where $n=1,2,\cdots$. Recall (18), we know that the exponential mapping $\exp$
is well defined $\forall\left<X\right>\in gl({\mathbb{F}})$.

Graph (258) also shows the relationship between $gl({\mathbb{F}})$ and
$GL({\mathbb{F}})$, which is a generalization of the relationship between
$gl(n,{\mathbb{F}})$ and $GL(n,{\mathbb{F}})$.

###  6.3 Lie Subgroups of $GL({\mathbb{F}})$

It has been discussed that $gl({\mathbb{F}})$ has some useful Lie sub-
algebras. It is obvious that $GL({\mathbb{F}})$ has some Lie sub-groups,
corresponding to those sub-algebras of $gl({\mathbb{F}})$. They are briefly
discussed as follows.

  * •

Bundled orthogonal Lie sub-group

######  Definition 181.

$\left<A\right>\in GL({\mathbb{F}})$ is said to be orthogonal, if
$A^{T}=A^{-1}$.

It is also easy to verify the following:

######  Proposition 182.

Assume $\left<A\right>$ and $\left<B\right>$ are orthogonal, then so is
$\left<A\right>\ltimes\left<B\right>$.

We, therefore, can define the following bundled Lie sub-group of
$GL({\mathbb{F}})$ as follows.

######  Definition 183.

$O({\mathbb{F}}):=\left\\{\left<A\right>\in
GL({\mathbb{F}})\;\big{|}\;\left<A\right>^{T}=\left<A\right>^{-1}\right\\}$

is called the bundled orthogonal group.

It is easy to verify the following proposition:

######  Proposition 184.

Consider the bundled orthogonal group.

    1. 1.

$O({\mathbb{F}})$ is a Lie sub-group of $GL({\mathbb{F}})$, i.e.,
$O({\mathbb{F}})<GL({\mathbb{F}})$.

    2. 2.

$SO({\mathbb{F}}):=\left\\{\left<A\right>\in
O({\mathbb{F}})\;\big{|}\;\operatorname{Dt}(\left<A\right>)=1\right\\}.$

Then $SO({\mathbb{F}})<O({\mathbb{F}})<GL({\mathbb{F}})$.

    3. 3.

The Lie algebra for both $O({\mathbb{F}})$ and $SO({\mathbb{F}})$ is
$o({\mathbb{F}})$.

  * •

Bundled special linear group

######  Definition 185.

$SL({\mathbb{F}}):=\left\\{\left<A\right>\in
GL({\mathbb{F}})\;\big{|}\;\operatorname{Dt}(\left<A\right>)=1\right\\}$

is called the bundled special linear group.

Similar to the case of orthogonal algebra, it is easy to verify the following:

######  Proposition 186.

Consider the bundled special linear group.

    1. 1.

$SL({\mathbb{F}})$ is a Lie sub-group of $GL({\mathbb{F}})$, i.e.,
$SL({\mathbb{F}})<GL({\mathbb{F}})$.

    2. 2.

The Lie algebra of $SL({\mathbb{F}})$ is $sl({\mathbb{F}})$.

  * •

Bundled upper triangular group

######  Definition 187.

$T({\mathbb{F}}):=\left\\{\left<A\right>\in
GL({\mathbb{F}})\;\big{|}\;\left<A\right>~{}\mbox{is upper
triangular}\right\\}$

is called the bundled upper triangular group.

######  Proposition 188.

Consider the bundled upper triangular group.

    1. 1.

$T({\mathbb{F}})$ is a Lie sub-group of $GL({\mathbb{F}})$, i.e.,
$T({\mathbb{F}})<GL({\mathbb{F}})$.

    2. 2.

The Lie algebra of $T({\mathbb{F}})$ is $t({\mathbb{F}})$.

  * •

Bundled special upper triangular group

######  Definition 189.

$N({\mathbb{F}}):=\left\\{\left<A\right>\in
T({\mathbb{F}})\;\big{|}\;Dt(\left<A\right>)=1\right\\}$

is called the bundled special upper triangular group.

######  Proposition 190.

Consider the bundled special upper triangular group.

    1. 1.

$N({\mathbb{F}})$ is a Lie sub-group of $T({\mathbb{F}})$, i.e.,
$N({\mathbb{F}})<T({\mathbb{F}})<GL({\mathbb{F}})$.

    2. 2.

The Lie algebra of $N({\mathbb{F}})$ is $n({\mathbb{F}})$.

  * •

Bundled symplectic group

######  Definition 191.

$\begin{array}[]{ccl}SP({\mathbb{F}})&:=&\left\\{\left<A\right>\in
GL({\mathbb{F}})\;\big{|}\;A_{1}\in{\mathcal{M}}_{2n}\right.\\\
{}\hfil&{}\hfil&~{}~{}~{}~{}~{}\left.\mbox{satisfies
(\ref{5.7.3.01})},\;n\in{\mathbb{N}}\right\\},\end{array}$

is called the bundled symplectic group.

$\displaystyle\left<A\right>^{T}\left<J\right>\left<A\right>=\left<J\right>,$
(259)

where $J$ is defined in (31).

######  Proposition 192.

Consider the bundled symplectic group.

  1. 1.

$SP({\mathbb{F}})$ is a Lie sub-group of $GL({\mathbb{F}})$, i.e.,
$SP({\mathbb{F}})<GL({\mathbb{F}})$.

  2. 2.

The Lie algebra of $SP({\mathbb{F}})$ is $sp({\mathbb{F}})$.

###  6.4 Symmetric Group

Let ${\bf S}_{k}$ be the $k$-th order symmetric group. Denote

${\bf S}:=\bigcup_{k=1}^{\infty}{\bf S}_{k}.$

######  Definition 193.

A matrix $A\in{\mathcal{M}}_{k}$ is called a permutation matrix, if
$\operatorname{Col}(A)\subset\Delta_{k}$ and
$\operatorname{Col}(A^{T})\subset\Delta_{k}$. The set of $k\times k$
permutation matrices is denoted by ${\mathcal{P}}_{k}$.

######  Proposition 194.

Consider the set of permutation matrix.

  1. 1.

If $P\in{\mathcal{P}}_{k}$, then

$\displaystyle P^{T}=P^{-1}.$ (260)

  2. 2.

${\mathcal{P}}_{k}<O(k,{\mathbb{R}})<GL(k,{\mathbb{R}}).$

Let $\sigma\in{\bf S}_{k}$. Define a permutation matrix
$M_{\sigma}\in{\mathcal{P}}_{k}$ as $M_{\sigma}:=\left[m_{i,j}\right]$, where

$\displaystyle m_{i,j}=\begin{cases}1,\quad\sigma(j)=i\\\
0,\quad\mbox{otherwise}.\end{cases}$ (261)

The following proposition is easily verifiable.

######  Proposition 195.

Define $\pi:{\bf S}_{k}\rightarrow{\mathcal{P}}_{k}$, where
$\pi(\sigma):=M_{\sigma}\in{\mathcal{P}}_{k}$ is constructed by (261). Then
$\pi$ is an isomorphism.

Assume $\sigma,~{}\lambda\in{\bf S}_{k}$. Then Proposition 195 leads to

$\displaystyle M_{\sigma\circ\lambda}=M_{\sigma}M_{\lambda}.$ (262)

Next, assume $\sigma\in{\bf S}_{m}$, $\lambda\in{\bf S}_{n}$, we try to
generalize (262).

######  Definition 196.

Assume $\sigma\in{\bf S}_{m}$, $\lambda\in{\bf S}_{n}$. The (left) STP of
$\sigma$ and $\lambda$ is defined by

$\displaystyle M_{\sigma\ltimes\lambda}=M_{\sigma}\ltimes
M_{\lambda}\in{\mathcal{P}}_{t},$ (263)

where $t=\operatorname{lcm}(m,n)$. That is,

$\displaystyle\sigma\ltimes\lambda:=\pi^{-1}\left(M_{\sigma}\ltimes
M_{\lambda}\right)\in{\bf S}_{t}.$ (264)

Similarly, we can define the right STP of $\sigma$ and $\lambda$.

Now, it is clear that $(S,\ltimes)<({\mathcal{M}}_{1},\ltimes)$ is a sub-
monoid. To get a bundled Lie subgroup structure, we consider the quotient
space

${\mathcal{D}}:=\left({\bf S},\ltimes\right)/\sim_{\ell}.$

Then we have the following:

######  Theorem 197.

${\mathcal{D}}$ is a discrete bundled sub-Lie group of $GL({\mathbb{F}})$.

${\mathcal{D}}$ may be used to investigate the permutation of uncertain number
of elements.

##  7 Vector Equivalence

###  7.1 Equivalence of Vectors of Different Dimensions

Consider the set of vectors on field ${\mathbb{F}}$. Denote it as

$\displaystyle{\mathcal{V}}:=\bigcup_{i=1}^{\infty}{\mathcal{V}}_{i},$ (265)

where ${\mathcal{V}}_{i}$ is the $i$ dimensional vector subspace of
${\mathcal{V}}$.

Our purpose is to build a vector space structure on ${\mathcal{V}}$. To this
end, we first propose an equivalence relation.

######  Definition 198.

  1. 1.

Let $X,~{}Y\in{\mathcal{V}}$. $X$ and $Y$ are said to be vector equivalent,
denoted by $X\leftrightarrow Y$, if there exist two one-vectors ${\bf 1}_{s}$
and ${\bf 1}_{t}$ such that

$\displaystyle X\otimes{\bf 1}_{s}=Y\otimes{\bf 1}_{t}.$ (266)

  2. 2.

The equivalent class is denoted as

$\lceil X\rceil:=\left\\{Y\;\big{|}\;Y\leftrightarrow X\right\\}.$

  3. 3.

In an equivalent class $\lceil X\rceil$ a partial order ($\leqslant$) is
defined as: $X\leqslant Y$, if there exists a one-vector ${\bf 1}_{s}$ such
that $X\otimes{\bf 1}_{s}=Y$. $X_{1}\in\lceil X\rceil$ is irreducible, if
there are no $Y$ and ${\bf 1}_{s}$, $s>1$, such that $X_{1}=Y\otimes{\bf
1}_{s}$.

######  Remark 199.

  1. 1.

The equivalence defined above can be seen as the left equivalence. Formally,
we set $\leftrightarrow:=\leftrightarrow_{l}$.

  2. 2.

The right equivalence can be defined as follows: Let $X,~{}Y\in{\mathcal{V}}$.
$X$ and $Y$ are said to be right equivalent, denoted by
$X\leftrightarrow_{r}Y$, if there exist two one-vectors ${\bf 1}_{s}$ and
${\bf 1}_{t}$ such that

$\displaystyle{\bf 1}_{s}\otimes X={\bf 1}_{t}\otimes Y.$ (267)

  3. 3.

The right equivalent class of $X$ is denoted as $\lceil X\rceil_{r}$. Of
course, we have $\lceil X\rceil_{l}=\lceil X\rceil$.

  4. 4.

Hereafter, the left equivalence is considered as the defaulted one. That is,
we always assume $\leftrightarrow=\leftrightarrow_{l}$. But with some obvious
modifications one sees easily that the arguments/results in the sequel are
also valid for right equivalence.

The following properties of the matrix equivalence are also true for vector
equivalence. The proofs are also similar to the matrix equivalence. Therefore,
they are omitted.

######  Theorem 200.

  1. 1.

If $X\leftrightarrow Y$, then there exists a vector $\Gamma$ such that

$\displaystyle X=\Gamma\otimes\operatorname{{\bf 1}}_{\beta},\quad
Y=\Gamma\otimes\operatorname{{\bf 1}}_{\alpha}.$ (268)

  2. 2.

In each class $\lceil X\rceil$ there exists a unique $X_{1}\in\lceil X\rceil$,
such that $X_{1}$ is irreducible.

######  Remark 201.

  1. 1.

If $X=Y\otimes{\bf 1}_{s}$, then $Y$ is called a divisor of $X$ and $X$ is
called a multiple of $Y$. This relation determined the order $Y\leqslant X$.

  2. 2.

If (268) holds and $\alpha,~{}\beta$ are co-prime, then the $\Gamma$
satisfying (268) is called the greatest common divisor of $X$ and $Y$.
Moreover, $\Sigma=gcd(X,Y)$ is unique.

  3. 3.

If (266) holds and $s,~{}t$ are co-prime, then

$\displaystyle\Xi:=X\otimes\operatorname{{\bf
1}}_{s}=Y\otimes\operatorname{{\bf 1}}_{t}$ (269)

is called the least common multiple of $X$ and $Y$. Moreover,
$\Xi=\operatorname{lcm}(X,Y)$ is unique.

  4. 4.

Consider an equivalent class $\lceil X\rceil$, denote the unique irreducible
element by $X_{1}$, which is called the root element. All the elements in
$\lceil X\rceil$ can be expressed as

$\displaystyle X_{i}=X_{1}\otimes\operatorname{{\bf 1}}_{i},\quad
i=1,2,\cdots.$ (270)

$X_{i}$ is called the $i$-th element of $\lceil X\rceil$. Hence, an equivalent
class $\lceil X\rceil$ is a well ordered sequence as:

$\lceil X\rceil=\left\\{X_{1},~{}X_{2},~{}X_{3},\cdots\right\\}.$

We also have a lattice structure on $\lceil X\rceil=\\{X_{1},X_{2},\cdots\\}$:

######  Proposition 202.

$(\lceil X\rceil,~{}\leqslant)$ is a lattice.

Proof. It is easy to verify that for $U,~{}V\in\lceil X\rceil$,

$\sup(U,~{}V)=lcm(U,~{}V);\quad\inf(U,~{}V)=gcd(U,~{}V).$

The conclusion follows. $\Box$

######  Proposition 203.

Let $A_{1}\in{\mathcal{M}}$ and $X_{1}\in{\mathcal{V}}$ be both irreducible.
Then the two lattices $\left<A\right>$ and $\lceil X\rceil$, generated by
$A_{1}$ and $X_{1}$ are isomorphic. Precisely,

$\displaystyle\left(\left<A\right>,~{}\prec\right)\approxeq\left(\lceil
X\rceil,~{}\leqslant\right).$ (271)

The isomorphism is: $\varphi:A_{s}=A_{1}\otimes I_{s}\mapsto
X_{s}=X_{1}\otimes\operatorname{{\bf 1}}_{s}$.

Next, we consider the lattice structure of

${\mathcal{V}}=\bigcup_{i=1}^{\infty}{\mathcal{V}}_{i},$

which is obviously a partition. Observing that if $q|p$, say $p=qs$, Then
${\mathcal{V}}_{q}\otimes\operatorname{{\bf 1}}_{s}$ is a subspace of
${\mathcal{V}}_{p}$. We give an order as
${\mathcal{V}}_{q}\vartriangleleft{\mathcal{V}}_{p}$. Under this partial order
we have a lattice structure on ${\mathcal{V}}$ as follows:

######  Proposition 204.

$\left({\mathcal{V}},~{}\vartriangleleft\right)$ is a lattice with

$\displaystyle\begin{array}[]{ccl}\sup\left({\mathcal{V}}_{\alpha},{\mathcal{V}}_{\beta}\right)&=&{\mathcal{V}}_{\alpha\vee\beta}\\\
\inf\left({\mathcal{V}}_{\alpha},{\mathcal{V}}_{\beta}\right)&=&{\mathcal{V}}_{\alpha\wedge\beta}\\\
\end{array}$ (274)

We also have the following isomorphic relation:

######  Proposition 205.

Define $\varphi:{\mathcal{V}}\rightarrow{\mathcal{M}}_{\mu}$ as:

$\displaystyle\varphi\left({\mathcal{V}}_{i}\right):={\mathcal{M}}_{\mu}^{i}.$
(275)

Then $\varphi$ is a lattice isomorphism.

######  Definition 206.

  1. 1.

Let $p\in{\mathbb{N}}$. The $p$-lower truncated vector space is defined as

$\displaystyle{\mathcal{V}}^{[p,\cdot]}:=\bigcup_{\\{s\;\big{|}\;p|s\\}}{\mathcal{V}}_{s}.$
(276)

  2. 2.

The quotient space of ${\mathcal{V}}/\leftrightarrow$ is denoted as

$\displaystyle\Omega_{{\mathcal{V}}}:=\left\\{\lceil
X\rceil\;\big{|}\;X\in{\mathcal{V}}\right\\}.$ (277)

  3. 3.

The subspace

$\displaystyle\begin{array}[]{ccl}\Omega_{i}&:=&{\mathcal{V}}^{[i,\cdot]}/\leftrightarrow\\\
{}\hfil&=&\left\\{\lceil
X\rceil\;\big{|}\;X_{1}\in{\mathcal{V}}_{i}\right\\}.\end{array}$ (280)

###  7.2 Vector Space Structure on Vector Equivalence Space

Next, we define an addition between vectors of different dimensions.

######  Definition 207.

Let $X\in{\mathcal{V}}_{p}$ and $Y\in{\mathcal{V}}_{q}$.
$t=\operatorname{lcm}(p,~{}q)$. Then

  1. 1.

the vector addition
$<+:{\mathcal{V}}\times{\mathcal{V}}\rightarrow{\mathcal{V}}$ of $X$ and $Y$
is defined as

$\displaystyle X<+Y:=\left(X\otimes\operatorname{{\bf
1}}_{t/p}\right)+\left(Y\otimes\operatorname{{\bf 1}}_{t/q}\right);$ (281)

  2. 2.

the subtraction $<-$ is defined as

$\displaystyle X<-Y:=X<+(-Y).$ (282)

######  Remark 208.

Let $X\in{\mathcal{V}}^{T}_{p}$ and $Y\in{\mathcal{V}}^{T}_{q}$ be tow row
vectors. Then we define

  1. 1.

the vector addition as

$\displaystyle X<+Y:=\left(X\otimes\operatorname{{\bf
1}}^{T}_{t/p}\right)+\left(Y\otimes\operatorname{{\bf 1}}^{T}_{t/q}\right);$
(283)

  2. 2.

the subtraction as

$\displaystyle X<-Y:=X<+(-Y).$ (284)

######  Proposition 209.

The vector addition $<+$ is consistent with the equivalence $\leftrightarrow$.
That is, if $X\leftrightarrow\tilde{X}$ and $Y\leftrightarrow\tilde{Y}$, then
$X<+Y\leftrightarrow\tilde{X}<+\tilde{Y}$.

Proof. Since $X\leftrightarrow\tilde{X}$, according to Theorem 200, there
exists $\Gamma$, say, $\Gamma\in{\mathcal{V}}_{p}$, such that

$X=\Gamma\otimes\operatorname{{\bf
1}}_{\alpha},\quad\tilde{X}=\Gamma\otimes\operatorname{{\bf 1}}_{\beta}.$

Similarly, there exists $\Pi$, say, $\Pi\in{\mathcal{V}}_{q}$, such that

$Y=\Pi\otimes\operatorname{{\bf
1}}_{s},\quad\tilde{Y}=\Pi\otimes\operatorname{{\bf 1}}_{t}.$

Let $\xi=lcm(p,q)$, $\eta=lcm(p\alpha,sq)$, and $\eta=\xi\ell$. Then

$\begin{array}[]{l}X<+Y=\left(\Gamma\otimes\operatorname{{\bf
1}}_{\alpha}\right)<+\left(\Pi\otimes\operatorname{{\bf 1}}_{s}\right)\\\
~{}=\left[\Gamma\otimes\operatorname{{\bf
1}}_{\alpha}\otimes\operatorname{{\bf 1}}_{\eta/(\alpha
p)}\right]+\left[\Pi\otimes\operatorname{{\bf 1}}_{s}\otimes\operatorname{{\bf
1}}_{\eta/(sq)}\right]\\\ ~{}=\left[\Gamma\otimes\operatorname{{\bf
1}}_{\eta/p}\right]+\left[\Pi\otimes\operatorname{{\bf 1}}_{\eta/q}\right]\\\
~{}=\left(\left[\Gamma\otimes\operatorname{{\bf
1}}_{\xi/p}\right]+\left[\Pi\otimes\operatorname{{\bf
1}}_{\xi/q}\right]\right)\otimes\operatorname{{\bf 1}}_{\ell}\\\
~{}=(\Gamma<+\Pi)\otimes\operatorname{{\bf 1}}_{\ell}.\end{array}$

Hence $X<+Y\leftrightarrow\Gamma<+\Pi$. Similarly, we can show that
$\tilde{X}<+\tilde{Y}\leftrightarrow\Gamma<+\Pi$. The conclusion follows.
$\Box$

######  Corollary 210.

The vector addition $<+$ (or subtraction $<-$) is well defined on the quotient
space $\Omega$, as well as $\Omega_{i}$, $i=1,2,\cdots$. That is,

$\displaystyle\begin{array}[]{l}\lceil X\rceil<+\lceil Y\rceil:=\lceil
X<+Y\rceil,\\\ ~{}~{}~{}~{}X,~{}Y\in{\mathcal{V}},\left(\mbox{or}~{}\lceil
X\rceil,~{}\lceil Y\rceil\in\Omega_{{\mathcal{V}}}\right).\end{array}$ (287)

Let $\lceil X\rceil\in\Omega$ (or $\lceil X\rceil\in\Omega_{i}$). Then we
define a scale product

$\displaystyle a\lceil X\rceil:=\lceil aX\rceil,\quad a\in{\mathbb{F}}.$ (288)

Using (287) and (288), one sees easily that $\Omega$ becomes a vector space:

######  Theorem 211.

Using the vector addition defined in (287) and the scale product defined in
(288), we have

  1. 1.

$\Omega$ is a vector space over ${\mathbb{F}}$;

  2. 2.

$\Omega_{i}$, $i=1,2,\cdots$ are the subspaces of $\Omega$;

  3. 3.

if $i|j$, then $\Omega_{i}$ is a subspace of $\Omega_{j}$.

Let $E\subset{\mathcal{V}}$ be a set of vectors. Then

$\displaystyle\lceil E\rceil:=\left\\{\lceil X\rceil\;\big{|}\;X\in
E\right\\}.$ (289)

The following proposition shows that the vector equivalence keeps the space-
subspace relationship unchanged.

######  Proposition 212.

Assume $E\subset{\mathcal{V}}_{i}$ is a subspace of ${\mathcal{V}}_{i}$. Then
$\lceil E\rceil\subset\Omega_{i}$ is a subspace of $\Omega_{i}$.

Definition 207 can be translated to its corresponding right one as follows:

######  Definition 213.

Let $X\in{\mathcal{V}}_{p}$ and $Y\in{\mathcal{V}}_{q}$.
$t=\operatorname{lcm}(p,~{}q)$. Then

  1. 1.

the (right) vector addition
$+>:{\mathcal{V}}\times{\mathcal{V}}\rightarrow{\mathcal{V}}$ of $X$ and $Y$
is defined as

$\displaystyle X+>Y:=\left(\operatorname{{\bf 1}}_{t/p}\otimes
X\right)+\left(\operatorname{{\bf 1}}_{t/q}\otimes Y\right);$ (290)

  2. 2.

the subtraction $->$ is defined as

$\displaystyle X->Y:=X+>(-Y).$ (291)

Then all the arguments in this subsection and the following several
subsections can be stated in a parallel way for right addition and
corresponding linear spaces.

###  7.3 Inner Product and Linear Mappings

######  Definition 214.

Let $X\in{\mathcal{V}}_{m}\subset{\mathcal{V}}$,
$Y\in{\mathcal{V}}_{n}\subset{\mathcal{V}}$, and $t=\operatorname{lcm}(m,n)$.
Then the weighted inner product is defied as

$\displaystyle\left(X\;\big{|}\;Y\right)_{W}:=\frac{1}{t}\left(X\otimes{\bf
1}_{t/m}\;\big{|}\;Y\otimes{\bf 1}_{t/n}\right),$ (292)

where $\left(X\;\big{|}\;Y\right)$ is the conventional inner product. Say, if
${\mathbb{F}}={\mathbb{R}}$, $\left(X\;\big{|}\;Y\right)=X^{T}Y$, and if
${\mathbb{F}}={\mathbb{C}}$, $\left(X\;\big{|}\;Y\right)=\bar{X}^{T}Y$.

######  Definition 215.

Let $\lceil X\rceil,\lceil Y\rceil\in\Omega$. Their inner product is defined
as

$\displaystyle\left(\lceil X\rceil\;\big{|}\;\lceil
Y\rceil\right):=\left(X\;\big{|}\;Y\right)_{W},\quad X\in\lceil
X\rceil,Y\in\lceil Y\rceil.$ (293)

It is easy to verify the following proposition, which assures the Definition
215 is reasonable.

######  Proposition 216.

Equation (293) is well defined. That is, it is independent of the choice of
$X$ and $Y$.

Since $\Omega$ is a vector space, using (293) as an inner product on $\Omega$,
then we have the follows:

######  Proposition 217.

The $\Omega$ with inner product defined by (293) is an inner product space. It
is not a Hilbert space.

Proof. The first part is obvious. As for the second part, we construct a
sequence as

$\begin{cases}X_{1}=a\in{\mathbb{F}}\\\ X_{i+1}=X_{i}\otimes\operatorname{{\bf
1}}_{2}+\frac{1}{2^{i+1}}\left(\delta_{2^{i+1}}^{1}-\delta_{2^{i+1}}^{2}\right),\quad
i=1,2,\cdots.\end{cases}$

Then we can prove that $\\{X_{i}\\}$ is a Cauchy sequence and it does not
converge to any $X\in{\mathcal{V}}$. $\Box$

Given a vector $X\in{\mathcal{V}}$, then it determined a linear mapping
$\varphi_{X}:~{}{\mathcal{V}}\rightarrow{\mathbb{F}}$ via inner product as

$\varphi_{X}:Y\mapsto(X\;\big{|}\;Y)_{W}.$

Similarly, $\lceil X\rceil\in\Sigma$ can also determine a linear mapping
$\varphi_{\lceil X\rceil}:~{}\Sigma\rightarrow{\mathbb{F}}$ as

$\varphi_{\lceil X\rceil}:\lceil Y\rceil\mapsto(\lceil
X\rceil\;\big{|}\;\lceil Y\rceil).$

Unfortunately, the inverse is not true. Because $\Sigma$ is an infinite
dimensional vector space but any vector in $\Sigma$ has only finite
dimensional representatives.

Next, we consider the linear mappings on ${\mathcal{V}}$. It is well known
that assume $A\in{\mathcal{M}}_{m\times n}$ and $X\in{\mathcal{V}}_{n}$. Then
the product $\times:~{}{\mathcal{M}}_{m\times
n}\times{\mathcal{V}}_{n}\rightarrow{\mathcal{V}}_{m}$, defined as
$(A,~{}X)\mapsto AX$, for given $A$ is a linear mapping. We intend to
generalize such a linear mapping to arbitrary matrix and arbitrary vector.

######  Definition 218.

Let $A\in{\mathcal{M}}_{m\times n}$, $X\in{\mathcal{V}}_{p}$, and
$t=\operatorname{lcm}(n,p)$. Then the vector product, denoted by $<\times$, is
defined as

$\displaystyle A<\times X:=\left(A\otimes
I_{t/n}\right)\left(X\otimes\operatorname{{\bf 1}}_{t/p}\right).$ (294)

######  Remark 219.

  1. 1.

The vector product defined in (294) is the left vector product. Of course, we
can define right vector product, denoted by $\times>$, defined as

$\displaystyle A\times>X:=\left(I_{t/n}\otimes
A\right)\left(\operatorname{{\bf 1}}_{t/p}\otimes X\right).$ (295)

  2. 2.

In fact, $<\times$ is a combination of $\sim_{\ell}$ of matrices and
$\leftrightarrow_{\ell}$ of vectors, and $\times>$ is a combination of
$\sim_{r}$ of matrices and $\leftrightarrow_{r}$ of vectors. Of course, we may
define two more vector products by combinations of $\sim_{l}$ with
$\leftrightarrow_{r}$ and $\sim_{r}$ with $\leftrightarrow_{l}$ respectively.

  3. 3.

Note that when $n=P$, $A<\times X=AX$. That is, the linear mapping defined in
(294) is a generalization of conventional linear mapping. It is also true for
other vector products.

  4. 4.

To avoid similar but tedious arguments, hereafter the default vector product
is $<\times$.

The following proposition is easily verifiable.

######  Proposition 220.

Consider the vector product

$<\times:~{}{\mathcal{M}}\times{\mathcal{V}}\rightarrow{\mathcal{V}}.$

  1. 1.

It is linear with respect to the second variable, precisely,

$\displaystyle A<\times(aX<+bY)=A<\times X<+B<\times Y,\quad
a,b\in{\mathbb{F}}.$ (296)

  2. 2.

Assume both $A,~{}B\in{\mathcal{M}}_{\mu}$, then the vector product is also
linear with respect to the first variable, precisely,

$\displaystyle(A\rotatebox{-90.0}{$\pm$}B)<\times X=A<\times X<+B<\times X.$
(297)

The following proposition shows that the vector product is consistent with
both matrix equivalence and vector equivalence.

######  Proposition 221.

Assume $A\sim B$ and $X\leftrightarrow Y$. Then

$\displaystyle A<\times X\leftrightarrow B<\times Y.$ (298)

Proof. Assume $A=\Lambda\otimes I_{s}$, $B=\Lambda\otimes I_{\alpha}$;
$X=\Gamma\otimes\operatorname{{\bf 1}}_{t}$,
$Y=\Gamma\otimes\operatorname{{\bf 1}}_{\beta}$, where
$\Lambda\in{\mathcal{M}}_{n\times p}$ and $\Gamma\in{\mathcal{V}}_{q}$. Denote
$\xi=\operatorname{lcm}(p,q)$, $\eta=\operatorname{lcm}(ps,qt)$, and
$\eta=k\xi$. Then we have

$\begin{array}[]{l}A<\times X=\left(\Lambda\otimes
I_{s}\right)\left(\Gamma\otimes\operatorname{{\bf 1}}_{t}\right)\\\
~{}=\left(\Lambda\otimes I_{s}\otimes
I_{\eta/ps}\right)\left(\Gamma\otimes\operatorname{{\bf
1}}_{t}\otimes\operatorname{{\bf 1}}_{\eta/qt}\right)\\\
~{}=\left(\Lambda\otimes I_{\xi/p}\otimes
I_{k}\right)\left(\Gamma\otimes\operatorname{{\bf
1}}_{\xi/q}\otimes\operatorname{{\bf 1}}_{k}\right)\\\
~{}=\left[\left(\Lambda\otimes
I_{\xi/p}\right)\left(\Gamma\otimes\operatorname{{\bf
1}}_{\xi/q}\right)\right]\otimes\left[I_{k}\operatorname{{\bf
1}}_{k}\right]\\\
~{}=\left(\Lambda<\times\Gamma\right)\otimes\operatorname{{\bf
1}}_{k}.\end{array}$

Hence

$A<\times X\leftrightarrow\Lambda<\times\Gamma.$

Similarly, we have

$B<\times Y\leftrightarrow\Lambda<\times\Gamma.$

Equation (298) follows. $\Box$

The above propositions have an immediate consequence as follows:

######  Corollary 222.

The vector product
$<\times:{\mathcal{M}}\times{\mathcal{V}}\rightarrow{\mathcal{V}}$ can be
extended to $<\times:\Sigma_{{\mathcal{M}}}\times\Omega\rightarrow\Omega$.
Particularly, each $\left<A\right>\in\Sigma_{{\mathcal{M}}}$ determines a
linear mapping on the vector space $\Omega$.

Next, we extend the vector equivalence to matrices.

######  Definition 223.

  1. 1.

Assume $V,~{}W\in{\mathcal{M}}_{\cdot\times n}$. $V$ and $W$ are said to be
vector equivalent, denoted by $V\leftrightarrow W$, if there exist
$\operatorname{{\bf 1}}_{s}$ and $\operatorname{{\bf 1}}_{t}$ such that

$\displaystyle V\otimes\operatorname{{\bf 1}}_{s}=W\otimes\operatorname{{\bf
1}}_{t}.$ (299)

  2. 2.

The equivalence class of $V$ is denoted as

$\displaystyle\lceil V\rceil:=\left\\{W\;\big{|}\;W\leftrightarrow
V\right\\}.$ (300)

######  Remark 224.

  1. 1.

It is clear the equivalence defined by (299) is left vector equivalence
$\leftrightarrow_{\ell}$. The right vector equivalence $\leftrightarrow_{r}$
can be defined similarly. Moreover, the right vector equivalence class is
denoted by $\lceil\cdot\rceil_{r}$.

  2. 2.

A matrix can be considered as a linear mapping, and correspondingly the matrix
equivalence is considered. A matrix can also be considered as a vector
subspace generated by its columns. In this way its vector equivalence, defined
by (299)-(300), makes sense. Denote the equivalence class as

$\Omega_{{\mathcal{M}}}:={\mathcal{M}}/\leftrightarrow.$

All the results about vector space ${\mathcal{V}}$ can be extended to
${\mathcal{M}}_{\cdot\times n}$ for $n\in{\mathbb{N}}$. They are briefly
summarized as follows:

######  Proposition 225.

  1. Given $n\in{\mathbb{N}}$.

  2. 1.

Assume $V\in{\mathcal{M}}_{r\times n}$, $W\in{\mathcal{M}}_{s\times n}$,
$V\leftrightarrow W$ and $r|s$. Then we said $V$ is a divisor of $W$ or $W$ is
a multiple of $V$. Moreover, an order is determined by this as $V\leqslant W$.

  3. 2.

$(\lceil V\rceil,\leqslant)$ is a lattice.

  4. 3.

Assume $r|s$, then an order is given in ${\mathcal{M}}_{\cdot\times n}$ as

${\mathcal{M}}_{r\times n}\vartriangleleft{\mathcal{M}}_{s\times n}.$

  5. 4.

$\left({\mathcal{M}}_{\cdot\times n},\vartriangleleft\right)$ is a lattice.

  6. 5.

Assume $V\in{\mathcal{M}}_{r\times n}$ is irreducible. then $\varphi:\lceil
V\rceil\rightarrow{\mathcal{M}}_{\cdot\times n}$ defined as
$V_{i}\mapsto{\mathcal{M}}_{ir\otimes n}$ is a lattice homomorphism.

The operators $<+$ and $<\times$ can also be defined in a similar way as for
vectors:

######  Definition 226.

  1. 1.

Assume $V,~{}\in{\mathcal{M}}_{p\times n}$ and $W\in{\mathcal{M}}_{q\times
n}$, $t=\operatorname{lcm}(p,~{}q)$. The vector addition is defined as

$V<+W:=\left(V\otimes\operatorname{{\bf
1}}_{t/p}\right)+\left(W\otimes\operatorname{{\bf 1}}_{t/q}\right).$

  2. 2.

Let $A\in{\mathcal{M}}_{m\times n}$, $V\in{\mathcal{M}}_{p\times q}$, and
$t=\operatorname{lcm}(n,p)$. Then the vector product of $A$ and $V$ is defined
as

$\displaystyle A<\times V:=\left(A\otimes
I_{t/n}\right)\left(V\otimes\operatorname{{\bf 1}}_{t/p}\right).$ (301)

We can denote the equivalence class of ${\mathcal{M}}_{\cdot\times n}$ as

$\Sigma_{{\mathcal{M}}}^{n}:={\mathcal{M}}_{\cdot\times n}/\leftrightarrow.$

For vector product and vector addition we also have the distributive law with
respect to these two operators for matrix case.

######  Proposition 227.

$<\times:{\mathcal{M}}\times{\mathcal{M}}\rightarrow{\mathcal{M}}$ is
distributive. Precisely, Let $A,~{}B\in{\mathcal{M}}_{\mu}$ and
$C\in{\mathcal{M}}_{n\times\beta}$, $D\in{\mathcal{M}}_{p\times\beta}$. Then

$\displaystyle(aA\rotatebox{-90.0}{$\pm$}bB)<\times C=aA<\times C<+bB<\times
C,\quad a,b\in{\mathbb{F}}.$ (302) $\displaystyle A<\times(aC+bD)=aA<\times
C<+bA<\times D,\quad a,b\in{\mathbb{F}}.$ (303)

Proof. We prove (302). The proof of (303) is similar. Denote
$\operatorname{lcm}(m,n)=s$, $\operatorname{lcm}(m,p)=r$,
$\operatorname{lcm}(n,p)=t$, and $\operatorname{lcm}(m,n,p)=\xi$. Then for
(302) we have

$\begin{array}[]{l}LHS=\left(aA\otimes I_{s/m}+bB\otimes I_{s/n}\right)<\times
C\\\ ~{}=\left[\left(aA\otimes I_{s/m}+bB\otimes I_{s/n}\right)\otimes
I_{\xi/s}\right]\left[C\otimes{\bf 1}_{\xi/p}\right]\\\ ~{}=a\left(A\otimes
I_{\xi/m}\right)\left(C\otimes{\bf 1}_{\xi/p}\right)+b\left(B\otimes
I_{\xi/n}\right)\left(C\otimes{\bf 1}_{\xi/p}\right)\\\
~{}=a\left[\left(A\otimes I_{s/m}\right)\left(C\otimes{\bf
1}_{s/p}\right)\right]\otimes{\bf 1}_{\xi/s}\\\ ~{}+b\left[\left(B\otimes
I_{t/n}\right)(C\otimes{\bf 1}_{t/p})\right]\otimes{\bf 1}_{\xi/t}\\\
~{}=a\left[\left(A\otimes I_{s/m}\right)\left(C\otimes{\bf
1}_{s/p}\right)\right]\\\ ~{}<+b\left[\left(B\otimes
I_{t/n}\right)(C\otimes{\bf 1}_{t/p})\right]\\\ ~{}=RHS.\end{array}$

$\Box$

Finally, we can extend the distributive law to equivalence spaces as follows.

######  Corollary 228.

$<\times:\Sigma_{{\mathcal{M}}}\times\Sigma_{{\mathcal{M}}}^{n}\rightarrow\Sigma_{{\mathcal{M}}}^{n}$
is distributive. Precisely, Let
$\left<A\right>,~{}\left<B\right>\in\Sigma_{\mu}$ and $\lceil C\rceil,\lceil
D\rceil\in\Sigma_{{\mathcal{M}}}^{\beta}$. Then

$\displaystyle(a\left<A\right>\rotatebox{-90.0}{$\pm$}b\left<B\right>)<\times\lceil
C\rceil=a\left<A\right><\times\lceil C\rceil<+b\left<B\right><\times\lceil
C\rceil,\quad a,b\in{\mathbb{F}}.$ (304)
$\displaystyle\left<A\right><\times(a\lceil C\rceil+b\lceil
D\rceil)=a\left<A\right><\times\lceil C\rceil<+b\left<A\right><\times\lceil
D\rceil,\quad a,b\in{\mathbb{F}}.$ (305)

###  7.4 Type-1 Invariant Subspace

Given $A\in{\mathcal{M}}_{\mu}^{i}$, we seek a subspace
$S\subset{\mathcal{V}}$ which is $A$ invariant.

######  Definition 229.

Let $S\subset{\mathcal{V}}$ be a vector subspace. If $A<\times S\subset S$,
$S$ is called an $A$-invariant subspace. Moreover, if
$S\subset{\mathcal{V}}_{t}$, it is called the type 1 invariant subspace;
Otherwise, it is called the type 2 invariant subspace.

This subsection considers the type 1 invariant subspace only.

######  Proposition 230.

Let $A\in{\mathcal{M}}_{\mu}$ and $S={\mathcal{V}}_{t}$. Then $S$ is
$A$-invariant, if and only if,

  * (i)

$\displaystyle\mu_{y}=1;$ (306)

  * (ii)

$A\in{\mathcal{M}}^{i}_{\mu}$, where $i$ satisfies

$\displaystyle\operatorname{lcm}(i\mu_{x},t)=t\mu_{x}.$ (307)

Proof. (Necessity) Assume $\xi=\operatorname{lcm}(i\mu_{x},t)$, by definition
we have

$\begin{array}[]{l}A<\times X=\left(A\otimes
I_{\xi/i\mu_{x}}\right)\left(X\otimes\operatorname{{\bf 1}}_{\xi/t}\right)\\\
~{}\in{\mathcal{V}}_{t},\quad X\in{\mathcal{V}}_{t}.\end{array}$

Hence we have

$\displaystyle i\mu_{y}\left(\frac{\xi}{i\mu_{x}}\right)=t.$ (308)

It follows from (308) that

$\displaystyle\frac{\xi}{t}=\frac{\mu_{x}}{\mu_{y}}.$ (309)

Since $\mu_{x}$ and $\mu_{y}$ are co-prime and the left hand side of (309) is
an integer, we have $\mu_{y}=1$. It follows that

$\xi=t\mu_{x}.$

(Sufficiency) Assume (307) holds, then (308) holds. It follows that

$A<\times X\in{\mathcal{V}}_{t},\quad\mbox{when}~{}X\in{\mathcal{V}}_{t}.$

$\Box$

Assume $A\in{\mathcal{M}}_{\mu}$ and $S={\mathcal{V}}_{t}$, a natural question
is: Can we find $A$ such that $S$ is $A$-invariant? According to Proposition
230, we know that it is necessary that $\mu_{y}=1$. Let
$k_{1},\cdots,k_{\ell}$ be the prime divisors of $\gcd(\mu_{x},~{}t)$, then we
have

$\displaystyle\mu_{x}=k_{1}^{\alpha_{1}}\cdots
k_{\ell}^{\alpha_{\ell}}p;~{}t=k_{1}^{\beta_{1}}\cdots
k_{\ell}^{\beta_{\ell}}q,$ (310)

where $p,~{}q$ are co-prime and $k_{i}\nmid p$, $k_{i}\nmid q$, $\forall i$.

Now it is obvious that to meet (307) it is necessary and sufficient that

$\displaystyle i=k_{1}^{\beta_{1}}\cdots k_{\ell}^{\beta_{\ell}}\lambda,$
(311)

where $\lambda|q$.

Summarizing the above argument we have the following result.

######  Proposition 231.

Assume $A\in{\mathcal{M}}_{\mu}$, $S={\mathcal{V}}_{t}$. $S$ is $A$-invariant,
if and only if, (i) $\mu_{y}=1$, and (ii) $A\in{\mathcal{M}}_{\mu}^{i}$, where
$i$ satisfies (311).

######  Remark 232.

Assume $A\in{\mathcal{M}}_{1}$, $S={\mathcal{V}}_{t}$. Using Proposition 231,
it is clear that $S$ is $A$-invariant, if and only if,
$A\in{\mathcal{M}}_{1}^{i}$ with

$i\in\\{\ell\;\big{|}\;\ell|t\\}.$

Particularly, when $i=1$ then $A=a$ is a number. So it is a scale product of
$S$, i.e, for $V\in S$ we have $a<\times V=aV\in S$. when $i=t$ then
$A\in{\mathcal{M}}_{t\times t}$. So we have $A<\times V=AV\in S$. This is the
classical linear mapping on ${\mathcal{V}}_{t}$. We call these two linear
mapping the standard linear mapping. The following example shows that there
are lots of non-standard linear mappings.

######  Example 233.

  1. 1.

Assume $\mu=0.5$, $t=6$. Then $\mu_{y}=1$, $\mu_{x}=2$. $\gcd(\mu_{x},t)=2$.
Using (310), $\mu_{x}=2\times 1$ and $t=2\times 3$. According to (311)
$i=2^{\beta_{1}}\lambda$, where $\beta_{1}=1$, $\lambda=1$ or $\lambda=3$.
According to Proposition 231, ${\mathcal{V}}_{6}$ is $A$-invariant, if and
only if, $A\in{\mathcal{M}}_{0.5}^{2}$ or $A\in{\mathcal{M}}_{0.5}^{6}$.

  2. 2.

Next, we give a numerical example: Assume

$\displaystyle A=\begin{bmatrix}1&-1&0&0\\\
0&0&1&0\end{bmatrix}\in{\mathcal{M}}_{0.5}^{2}.$ (312)

Then ${\mathbb{R}}^{6}$ is $A$-invariant space. For instance, let

$\displaystyle
X=\begin{bmatrix}1+i&2&1-i&0&0&0\end{bmatrix}^{T}\in{\mathbb{C}}^{6}.$ (313)

Then

$\displaystyle A<\times X=iX.$ (314)

Motivated by (314), we give the following definition.

######  Definition 234.

Assume $A\in{\mathcal{M}}$ and $X\in{\mathcal{V}}$. If

$\displaystyle A<\times X=\alpha X,\quad\alpha\in{\mathbb{F}},X\neq 0,$ (315)

then $\alpha$ is called an eigenvalue of $A$, and $X$ is called an eigenvector
of $A$ with respect to $\alpha$.

######  Example 235.

Recall Example 233. In fact it is easy to verify that matrix $A$ in (312), as
a linear mapping on ${\mathbb{R}}^{6}$, has $6$ eigenvalues: Precisely,

$\sigma(A)=\left\\{i,-i,0,0,0,1\right\\}.$

Correspondingly, the first eigenvector is $X$, defined in (313). The other $5$
eigenvectors are $(1-i,2,1+i,0,0,0)^{T}$, $(1,1,1,0,0,0)^{T}$,
$(0,0,0,0,1,0)^{T}$ (this is a root vector), $(0,0,0,0,0,1)^{T}$, and
$(0,0,0,1,1,1)^{T}$, respectively.

Assume $S={\mathcal{V}}_{t}$ is $A$-invariant with respect to $<\times$, that
is,

$\displaystyle A<\times S\subset S.$ (316)

Then the restriction $A|_{S}$ is a linear mapping on $S$. It follows that
there exists a matrix, denoted by $A|_{t}\in{\mathcal{M}}_{t\times t}$, such
that $A|_{S}$ is equivalent to $A|_{t}$. We state it as a proposition.

######  Proposition 236.

Assume $A\in{\mathcal{M}}$ and $S={\mathcal{V}}_{t}$ is $A$-invariant. Then
there exists a unique matrix $A|_{t}\in{\mathcal{M}}_{t\times t}$, such that
$A|_{S}$ is equivalent to $A|_{t}$. Precisely,

$\displaystyle A<\times X=A|_{t}X,\quad\forall X\in S.$ (317)

$A|_{t}$ is called the realization of $A$ on $S={\mathcal{V}}_{t}$.

######  Remark 237.

  1. 1.

To calculate $A|_{t}$ from $A$ is easy. In fact, it is clear that

$\displaystyle\operatorname{Col}_{i}\left(A_{[t]}\right)=A<\times\delta_{t}^{i},\quad
i=1,\cdots,t.$ (318)

  2. 2.

Consider Example 233 again.

$\begin{array}[]{l}\operatorname{Col}_{1}(A|_{6})=A<\times\delta_{6}^{1}\\\
~{}=\left[\begin{bmatrix}1&-1&0&0\\\ 0&0&1&0\end{bmatrix}\otimes
I_{3}\right]\left[\delta_{6}^{1}\otimes\operatorname{{\bf 1}}_{2}\right]\\\
~{}=(1,1,0,0,0,0)^{T}.\end{array}$

Similarly, we can calculate all other columns. Finally, we have

$A|_{6}=\begin{bmatrix}1&-1&0&0&0&0\\\ 1&0&-1&0&0&0\\\ 0&1&-1&0&0&0\\\
0&0&0&1&0&0\\\ 0&0&0&1&0&0\\\ 0&0&0&0&1&0\end{bmatrix}$

Finally, given a matrix $A\in{\mathcal{M}}_{\mu}^{i}$, we would like to know
whether it has (at least one) type-1 invariant subspace $S={\mathcal{V}}_{t}$?

Then as a corollary of Proposition 230, we can prove the following result.

######  Corollary 238.

Assume $A\in{\mathcal{M}}_{\mu}^{i}$, then $A$ has (at least one) type 1
invariant subspace $S={\mathcal{V}}_{t}$, if and only if,

$\displaystyle\mu_{y}=1;$ (319)

Proof. According to Proposition 230, $\mu_{y}=1$ is obvious necessary. We
prove it is also sufficient. Assume $i$ is factorized into its prime factors
as

$\displaystyle i=\prod_{j=1}^{n}i_{j}^{k_{j}},$ (320)

and correspondingly, $\mu_{x}$ is factorized as

$\displaystyle\mu_{x}=\prod_{j=1}^{n}i_{j}^{t_{j}}p,$ (321)

where $p$ is co-prime with $i$; $t$ is factorized as

$\displaystyle t=\prod_{j=1}^{n}i_{j}^{r_{j}}q,$ (322)

where $q$ is co-prime with $i$.

Using Proposition 230 again, we have only to prove that there exists at least
one $t$ satisfying (307). Calculate that

$\begin{array}[]{ccl}(i\mu_{x},~{}t)&=&\prod_{j=1}^{n}i_{j}^{\max(t_{j}+k_{j},~{}r_{j})}\operatorname{lcm}(p,~{}q);\\\
\mu_{x}t&=&\prod_{j=1}^{n}i_{j}^{t_{j}+r_{j}}pq.\end{array}$

To meet (307) a necessary condition is: $p$ and $q$ are co-prime. Next, fix
$j$, we consider two cases: (i) $r_{j}>k_{j}+t_{j}$: Then on the LHS (left
hand side) of (307) we have factor $i_{j}^{r_{j}}$ and on the RHS of (307) we
have factor $i_{j}^{r_{j}+t_{j}}$. Hence, as long as $t_{j}=0$, we can choose
$r_{j}>k_{j}$ to meet (307). (ii) $r_{j}<k_{j}+t_{j}$: Then on the LHS we have
factor $i_{j}^{k_{j}+t_{j}}$ and on the RHS we have factor
$i_{j}^{r_{j}+t_{j}}$. Hence, as long as $r_{j}=k_{j}$, (307) is satisfied.
$\Box$

Using above notations, we also have the following result

######  Corollary 239.

Assume $\mu_{y}=1$. Then ${\mathcal{V}}_{t}$ is $A$-invariant, if and only if,
(i) for $t_{j}=0$, the corresponding $r_{j}\geq k_{j}$; (ii) for $t_{j}>0$,
the corresponding $r_{j}=k_{j}$.

######  Example 240.

Recall Example 233 again.

  1. 1.

Since the matrix $A$ defined in (312) is in ${\mathcal{M}}_{0.5}^{2}$. We have
$i=2$, ${\mu}_{y}=1$, ${\mu}_{x}=2=ip$ and hence $p=1$. According to Corollary
239, $S={\mathcal{V}}_{t}$ is $A$-invariant, if and only if, $t=iq=2q$ and $q$
is co-prime with $i=2$. Hence

${\mathcal{V}}_{2(2n+1)},\quad n=0,1,2,\cdots,$

are type-1 invariant subspaces of $A$.

  2. 2.

Assume $q=5$. Then the restriction is

$A|_{{\mathbb{R}}^{10}}=A_{[10]}=\begin{bmatrix}1&0&-1&0&0&0&0&0&0&0\\\
1&0&0&-1&0&0&0&0&0&0\\\ 0&1&0&-1&0&0&0&0&0&0\\\ 0&1&0&0&-1&0&0&0&0&0\\\
0&0&1&0&-1&0&0&0&0&0\\\ 0&0&0&0&0&1&0&0&0&0\\\ 0&0&0&0&0&1&0&0&0&0\\\
0&0&0&0&0&0&1&0&0&0\\\ 0&0&0&0&0&0&1&0&0&0\\\
0&0&0&0&0&0&0&1&0&0\end{bmatrix}$

The eigenvalues are

$\sigma(A|_{{\mathbb{R}}^{10}}=\\{-1,i,-i,0,1,0,0,0,0,1\\}.$

The corresponding eigenvectors are:

$\begin{array}[]{l}E_{1}=(0,1,0,1,2,0,0,0,0,0)^{T}\\\
E_{2}=(0.3162+0.1054i,0.5270,0.4216-0.2108i,\\\
~{}0.3162-0.4216i,0.1054-0.3162i,0,0,0,0,0)^{T}\\\
E_{3}=(0.3162-0.1054i,0.5270,0.4216+0.2108i,\\\
~{}0.3162+0.4216i,0.1054+0.3162i,0,0,0,0,0)^{T}\\\
E_{4}=(1,1,1,1,1,0,0,0,0,0)^{T}\\\ E_{5}=(2,1,0,1,0,0,0,0,0,0)^{T}\\\
R_{6}=(0,0,0,0,0,0,1,0,0,0)^{T}\\\ R_{7}=(0,0,0,0,0,0,1,1,0,0)^{T}\\\
E_{8}=(0,0,0,0,0,0,0,0,0,1)^{T}\\\ E_{9}=(0,0,0,0,0,0,0,0,1,0)^{T}\\\
E_{10}=(0,0,0,0,0,1,1,1,1,1)^{T}.\end{array}$

Note that $R_{6}$, $R_{7}$ are two root vectors. That is,

$A<\times R_{6}=R_{7};\quad A<\times R_{7}=E_{8}.$

######  Remark 241.

In fact, the set of type-1 $A$-invariant subspace depends only on the shape of
$A$. Hence we define

${\mathcal{I}}_{\mu}^{i}:=\left\\{{\mathcal{V}}_{t}\;|\;{\mathcal{V}}_{t}~{}\mbox{is}~{}A\in{\mathcal{M}}_{\mu}^{i}~{}\mbox{invariant}\right\\}.$

We can also briefly call ${\mathcal{I}}_{\mu}^{i}$ the set of
${\mathcal{M}}_{\mu}^{i}$-invariant subspaces.

Before ending this subsection we consider a general type-1 invariant subspace
$S\subset{\mathcal{V}}_{t}$. The following proposition is obvious.

######  Proposition 242.

If $S\subset{\mathcal{V}}_{t}$ is an $A$-invariant subspace, then
${\mathcal{V}}_{t}$ is also an $A$-invariant subspace.

Because of Proposition 242 searching $S$ becomes a classical problem. Because
we can first find a matrix $P\in{\mathcal{M}}_{t\times t}$, which is
equivalent to $A|_{{\mathcal{V}}_{t}}$. Then $S$ must be a classical invariant
subspace of $P$.

###  7.5 Type-2 Invariant Subspace

Denote the set of Type-1 $A$-invariant subspaces as

${\mathcal{I}}_{A}:=\left\\{{\mathcal{V}}_{s}\;\big{|}\;{\mathcal{V}}_{s}~{}\mbox{is}~{}A-\mbox{invariant}\right\\}.$

Assume $A\in{\mathcal{M}}_{\mu}^{i}$. Then
${\mathcal{I}}_{A}={\mathcal{I}}_{\mu}^{i}$.

To assure ${\mathcal{I}}_{A}\neq\emptyset$, through this subsection we assume
$\mu_{y}=1$. We give a definition for this.

######  Definition 243.

  1. 1.

Assume $A\in{\mathcal{M}}_{\mu}$. $A$ is said to be a bounded operator (or
briefly, $A$ is bounded,) if $\mu_{y}=1$.

  2. 2.

A sequence $\left\\{{\mathcal{V}}_{i}\;\big{|}\;i=1,2,\cdots\right\\}$, is
called an $A$ generated sequence if

$A<\times{\mathcal{V}}_{i}\subset{\mathcal{V}}_{i+1},\quad i=1,2,\cdots.$

  3. 3.

A finite sequence
$\left\\{{\mathcal{V}}_{i}\;\big{|}\;i=1,2,\cdots,p\right\\}$, is called an
$A$ generated loop if ${\mathcal{V}}_{p}={\mathcal{V}}_{1}$.

######  Lemma 244.

Assume there is an $A$ generated loop ${\mathcal{V}}_{p}$,
${\mathcal{V}}_{q_{1}}$, $\cdots$, ${\mathcal{V}}_{q_{r}}$,
${\mathcal{V}}_{p}$, as depicted in (323).

$\displaystyle{\mathcal{V}}_{p}\xrightarrow{\mbox{A}}{\mathcal{V}}_{q_{1}}\xrightarrow{\mbox{A}}\cdots\xrightarrow{\mbox{A}}{\mathcal{V}}_{q_{r}}\xrightarrow{\mbox{A}}{\mathcal{V}}_{p}$
(323)

Then

$\displaystyle q_{j}=p,\quad j=1,\cdots,r.$ (324)

Proof. Assume $A\in{\mathcal{M}}_{\mu}^{i}$. According to Definition 218, we
have the following dimension relationship:

$\displaystyle\begin{array}[]{lll}s_{0}:=\operatorname{lcm}(i\mu_{x},p)&\Rightarrow&q_{1}=\mu
s_{0};\\\ s_{1}:=\operatorname{lcm}(i\mu_{x},q_{1})&\Rightarrow&q_{2}=\mu
s_{1};\\\ {}&\vdots&{}\\\
s_{r-1}:=\operatorname{lcm}(i\mu_{x},q_{r-1})&\Rightarrow&q_{r}=\mu
s_{r-1};\\\ s_{r}:=\operatorname{lcm}(i\mu_{x},q_{r})&\Rightarrow&p=\mu
s_{r}.\\\ \end{array}$ (330)

Next, we

$\begin{array}[]{llll}\mbox{set}&s_{0}:=t_{0}p&\mbox{then}&q_{1}=\mu
t_{0}p;\\\
\mbox{set}&s_{1}:=t_{1}q_{1}&\mbox{then}&q_{2}=\mu^{2}t_{1}t_{0}p;\\\
{}&\vdots&{}&{}\\\
\mbox{set}&s_{r-1}:=t_{r-1}q_{r-1}&\mbox{then}&q_{r}=\mu^{r}t_{r-1}\cdots
t_{1}t_{0}p;\\\
\mbox{set}&s_{r}:=t_{r}q_{r}&\mbox{then}&p=\mu^{r+1}t_{r}\cdots
t_{1}t_{0}p.\\\ \end{array}$

We conclude that

$\displaystyle\mu^{r+1}t_{r}t_{r-1}\cdots t_{0}=1.$ (331)

Equivalently, we have

$\frac{\mu_{x}^{r+1}}{\mu_{y}^{r+1}}=t_{r}t_{r-1}\cdots t_{0}.$

It follows that

$\mu_{y}=1.$

Define

$s_{r}=\operatorname{lcm}(i\mu_{x},q_{r}):=i\mu_{x}\xi,$

where $\xi\in{\mathbb{N}}$. Then from the last equation of (330) we have

$\displaystyle p=i\xi.$ (332)

That is,

$\mu\operatorname{lcm}(i\mu_{x},q_{r})=i\xi.$

Using (331), and the expression

$q_{r}=\mu^{r}t_{r-1}\cdots t_{1}t_{0}p,$

we have

$\operatorname{lcm}\left(\mu_{x},\frac{\mu_{x}\xi}{t_{r}}\right)=\xi\mu_{x}.$

From above it is clear that

$\displaystyle\begin{array}[]{l}t_{r}\big{|}\mu_{x}\\\
\gcd(\mu_{x},\xi)=1.\end{array}$ (335)

Next, using last two equation in (330), we have

$\mu\operatorname{lcm}\left(i\mu_{x},\mu\operatorname{lcm}(i\mu_{x},q_{r-1})\right)=p=i\xi.$

Similar to the above argument, we have

$\xi\mu_{x}\big{|}\operatorname{lcm}\left(\mu_{x},\mu\operatorname{lcm}(\mu,\frac{q_{r-1}}{i})\right).$

Hence

$\xi\mu_{x}\big{|}\operatorname{lcm}\left(\mu_{x},\mu\operatorname{lcm}(\mu,\frac{\mu_{x}^{2}}{t_{r}t_{r-1}}\xi)\right).$

To meet this requirement, it is necessary that

$t_{r}t_{r-1}\big{|}\mu_{x}^{2}.$

Continuing this process, finally we have

$\displaystyle t_{r}t_{r-1}\cdots t_{s}\big{|}\mu_{x}^{r-s+1},\quad
s=r-1,r-2,\cdots,0.$ (336)

Combining (331) with (336) yields that

$t_{s}=\mu_{x},\quad s=0,1,\cdots,r.$

That is, $q_{1}=q_{2}=\cdots=q_{r}=p$. $\Box$

######  Theorem 245.

A finite dimensional subspace $S\subset{\mathcal{V}}$ is $A$-invariant, if and
only if, $S$ has the following structure:

$\displaystyle S=\oplus_{i=1}^{\ell}S^{i},$ (337)

where

$A<\times S^{i}\subset S^{i+1},\quad i=1,\cdots,\ell-1.$

Proof. Since $S$ is of finite dimension, there are only finite
${\mathcal{V}}_{t_{i}}$ such that

$S^{j}:=S\cap{\mathcal{V}}_{t_{j}}\neq\\{0\\}.$

Now for each $0\neq X_{0}\in S^{j}\subset{\mathcal{V}}_{t_{j}}$ we construct
$X_{1}:=A<\times X\in V_{t_{r}}$ for certain $t_{r}$. Note that $S$ is
$A$-invariant, if for all $X_{0}\in S^{j}$, we have $t_{r}=t_{j}$, then this
$S^{j}=S^{\ell}$ is the end element in the sequence. Otherwise, we can find a
successor $S^{r}=S\cap{\mathcal{V}}_{t_{r}}$. Note that since there are only
finite $S^{j}$, according to Lemma 244, starting from $X_{0}\in S$ there are
only finite sequence of differen $S^{j}$ till it reach an $A$-invariant
$S^{\ell}$ (equivalently, $A$-invariant ${\mathcal{V}}_{t_{\ell}}$). The claim
follows. $\Box$

###  7.6 Higher Order Linear Mapping

######  Definition 246.

Let $A\in{\mathcal{M}}_{\mu}^{i}$, ${\mathcal{V}}_{t}$ is $A$-invariant
subspace of ${\mathcal{V}}$. That is,
$A:{\mathcal{V}}_{t}\rightarrow{\mathcal{V}}_{t}$ is a linear mapping. The
higher order linear mapping of $A$, is defined as

$\displaystyle\begin{cases}A^{[1]}<\times X:=A<\times X,\quad
X\in{\mathcal{V}}_{t}\\\ A^{[k+1]}<\times X:=A<\times\left(A^{[k]}<\times
X\right),\quad k\geq 1.\end{cases}$ (338)

######  Definition 247.

Let $X\in{\mathcal{V}}$. The $A$-sequence of $X$ is the sequence
$\left\\{X_{i}\right\\}$, where

$\begin{cases}X_{0}=X\\\ X_{i+1}=A<\times X_{i},\quad
i=0,1,2,\cdots.\end{cases}$

Using notations (320)-(322), we have the following result.

######  Lemma 248.

Assume $A$ is bounded. Then $A<\times
X\in{\mathcal{V}}_{s}\in{\mathcal{I}}_{A}$, if and only if, for each $0<j<n$,
one of the following is true:

$\displaystyle r_{j}=0;$ (339)

or

$\displaystyle t_{j}\leq k_{j}+r_{j}.$ (340)

Proof.

$\begin{array}[]{l}\operatorname{lcm}(i\mu_{x},t)=\operatorname{lcm}\left(\prod_{j=1}^{n}i_{j}^{\max(k_{j}+r_{j},t_{j})}q,\prod_{j=1}^{n}i_{j}^{t_{j}}q\right)\\\
~{}=\prod_{j=1}^{n}i_{j}^{\max(k_{j}+r_{j},t_{j})}\operatorname{lcm}(p,q),\end{array}$

and

$\begin{array}[]{l}\tilde{t}=\operatorname{lcm}(i\mu_{x},t)\mu\\\
~{}=\prod_{j=1}^{n}i_{j}^{\max(k_{j}+r_{j},t_{j})-r_{j}}\operatorname{lcm}(p,q)/p.\end{array}$

Then

$\operatorname{lcm}(i\mu_{x},\tilde{t})=\prod_{j=1}^{n}i_{j}^{\max\left(k_{j}+r_{j},max(k_{j}+r_{j},t_{j})-r_{j}\right)}\operatorname{lcm}(p,q),$

and

$\mu_{x}\tilde{t}=\prod_{j=1}^{n}i_{j}^{\max\left(k_{j}+r_{j},t_{j}\right)}\operatorname{lcm}(p,q).$

Now $A<\times X\in{\mathcal{V}}_{s}\in{\mathcal{I}}_{A}$ is assured by (307),
which leads to

$\displaystyle\max\left(k_{j}+r_{j},\max(k_{j}+r_{j},t_{j})-r_{j}\right)=\max\left(k_{j}+r_{j},t_{j}\right).$
(341)

  1. Case 1

: $t_{j}>k_{j}+2r_{j}$, (341) leads to $r_{j}=0$. Hence, we have

$\displaystyle r_{j}=0~{}\mbox{and}~{}t_{j}>k_{j}.$ (342)

  2. Case 2

: $k_{j}+r_{j}\leq t_{j}\leq k_{j}+2r_{j}$, which leads to
$k_{j}+r_{j}=t_{j}$.

  3. Case 3

: $t_{j}<k_{j}+r_{j}$, which assures (341). Combining Case 2 and Case 3 yields

$\displaystyle t_{j}\leq k_{j}+r_{j}.$ (343)

Note that when $r_{j}=0$, if $t_{j}\leq k_{j}$ we have (343). Hence $t_{j}\leq
k_{j}$ is not necessary. The conclusion follows. $\Box$

The following result is important.

######  Theorem 249.

Let $A\in{\mathcal{M}}_{\mu}^{i}$ be bounded. Then for any $X\in{\mathcal{V}}$
the $A$-sequence of $X$ will enter a ${\mathcal{V}}_{t}\in{\mathcal{I}}_{A}$
at finite steps.

Proof. Assume $A<\times X\in{\mathcal{V}}_{\tilde{t}}$. Using notations
(320)-(322), it is easy to calculate that after one step the $j$-th index of
$t$ becomes

$t^{1}_{j}=\max\left(k_{j}+r_{j},t^{0}_{j})\right)-r_{j}.$

Assume $r_{j}=0$, this component already meets the requirement of Lemma 248.
Assume for some $j$, $r_{j}>0$ and $t^{0}_{j}>k_{j}+r_{j}$, then we have

$t^{1}_{j}=t^{0}_{j}-r_{j}<t^{0}_{j}.$

Hence after finite times, say $k$, the $j$-th index of $X_{k}$, denoted by
$t^{k}_{j}$, satisfies

$\displaystyle t^{k}_{j}=t^{0}_{j}-kr_{j},$ (344)

will satisfy (340), and as long as (340) holds, $t^{s}_{j}=t^{k}_{j}$ $\forall
s>k$. Hence, after finite steps either (339) or (340) (or both) is satisfied.
Then at the next step the sequence enters into
${\mathcal{V}}_{t}\in{\mathcal{I}}_{A}$. $\Box$

######  Definition 250.

Given a polynomial

$\displaystyle p(x)=x^{n}+c_{n-1}x^{n-1}+\cdots+c_{1}x+c_{0},$ (345)

a matrix $A\in{\mathcal{M}}$ and a vector $X\in{\mathcal{V}}$.

  1. 1.

$p(x)$ is called an $A$-annihilator of $X$, if

$\displaystyle\begin{array}[]{ccl}p(A)X&:=&A^{[n]}<\times
X<+c_{n-1}A^{[n-1]}<\times X\\\ {}\hfil&{}\hfil&<+\cdots<+c_{1}A<\times
X<+c_{0}=0.\end{array}$ (348)

  2. 2.

Assume $q(x)$ is the $A$-annihilator of $X$ with minimum degree, then $q(x)$
is called the minimum $A$-annihilator of $X$.

The following result is obvious:

######  Proposition 251.

The minimum $A$-annihilator of $X$ divides any $A$-annihilator of $X$.

The following result is an immediate consequence of Theorem 249.

######  Corollary 252.

Assume $A$ is bounded, then for any $X\in{\mathcal{V}}$ there exists at least
one $A$-annihilator of $X$.

Proof. According to Theorem 249, there is a finite $k$ such that
$A^{[k]}X\in{\mathcal{V}}_{s}$ with ${\mathcal{V}}_{s}$ being $A$-invariant.
Now in ${\mathcal{V}}_{s}$ assume the minimum annihilator polynomial for
$A^{[k]}X$ is $q(x)$, then $p(x)=x^{k}q(x)$ is an $A$-annihilator of $X$.
$\Box$

######  Example 253.

  1. 1.

Assume $A\in{\mathcal{M}}_{2/3}^{1}$. Since $\mu_{y}=2\neq 1$, we know any $X$
does not have its $A$-annihilator.

Now assume $X_{0}\in{\mathcal{V}}_{k}$, where $k=3^{s}p$, and $3,~{}p$ are co-
prime. Then it is easy to see that the $A$ sequence of $X_{0}$ has the
dimensions, $\dim(X_{i}):=d_{i}$, which are: $d_{1}=2\times 3^{s-1}p$,
$d_{2}=2^{2}\times 3^{s-2}p$ $\cdots$ $d_{s}=2^{s}p$, $d_{s+1}=2^{s+1}p$,
$d_{s+2}=2^{s+2}p$, $\cdots$. It can not reach a
${\mathcal{V}}_{t}\in{\mathcal{I}}_{A}$.

  2. 2.

Given

$A=\begin{bmatrix}1&0&1&1\\\ 0&1&0&1\end{bmatrix};\quad X=\begin{bmatrix}1\\\
0\\\ 0\end{bmatrix}.$

We try to find the minimum $A$-annihilator of $X$. Set $X_{0}=X$. It is easy
to see that

$X_{1}=A<\times X_{0}\in{\mathcal{V}}_{6}\in{\mathcal{I}}_{A}.$

Hence, we can find the annihilator of $X$ in the space of ${\mathbb{R}}^{6}$.
Calculating

$\begin{array}[]{lll}X_{1}=\begin{bmatrix}1\\\ 1\\\ 0\\\ 0\\\ 0\\\
0\end{bmatrix};&X_{2}=\begin{bmatrix}1\\\ 1\\\ 1\\\ 1\\\ 0\\\
0\end{bmatrix};&X_{3}=\begin{bmatrix}2\\\ 2\\\ 1\\\ 0\\\ 0\\\
1\end{bmatrix};\\\ X_{4}=\begin{bmatrix}2\\\ 1\\\ 1\\\ 2\\\ 1\\\
1\end{bmatrix};&X_{5}=\begin{bmatrix}3\\\ 3\\\ 1\\\ -1\\\ -1\\\
0\end{bmatrix};&X_{6}=\begin{bmatrix}3\\\ 2\\\ 2\\\ 4\\\ 2\\\
2\end{bmatrix},\end{array}$

it is easy to verify that $X_{1},~{}X_{2},~{}X_{3},~{}X_{4},~{}X_{5}$ are
linearly independent. Moreover,

$X_{6}=X_{1}+X_{2}+X_{3}+X_{4}-X_{5}.$

The minimum $A$-annihilator of $X=X_{0}$ follows as

$p(x)=x^{6}+x^{5}-x^{4}-x^{3}-x^{2}-x.$

###  7.7 Invariant Subspace on Equivalence Space

Denote

$\Omega_{{\mathcal{M}}}:={\mathcal{M}}/\leftrightarrow.$

We extend the vector product to the equivalence spaces.

######  Definition 254.

Let $A\in{\mathcal{M}}$ and $B\in{\mathcal{M}}_{\cdot\times q}$. Then we
define
$<\times:\Sigma_{{\mathcal{M}}}\times\Omega_{{\mathcal{M}}}^{q}\rightarrow\Omega_{{\mathcal{M}}}^{q}$
as

$\displaystyle\left<A\right><\times\lceil B\rceil:=\lceil A<\times B\rceil.$
(349)

The following proposition shows that (349) is well defined.

######  Proposition 255.

(349) is independent of the choice of $A$ and $B$.

Proof. Assume $A_{1}\in\left<A\right>$ is irreducible and
$A_{1}\in{\mathcal{M}}_{m\times n}$; $B_{1}\in\left<B\right>$ is also
irreducible and $B_{1}\in{\mathcal{M}}_{p\times q}$. $A_{i}=A_{1}\otimes
I_{i}$ and $B_{j}=B_{1}\otimes\operatorname{{\bf 1}}_{j}$. Set
$s=\operatorname{lcm}(n,p)$, $t=\operatorname{lcm}(ni,pj)$, and $s\xi=t$. Then

$\begin{array}[]{l}\left(A_{i}<\times B_{j}\right)\\\ =\left(A_{i}\otimes
I_{t/ni}\right)\left(B_{j}\otimes\operatorname{{\bf 1}}_{t/pj}\right)\\\
=\left(A_{1}\otimes I_{i}\otimes
I_{t/ni}\right)\left(B_{1}\otimes\operatorname{{\bf
1}}_{t}\otimes\operatorname{{\bf 1}}_{t/pj}\right)\\\
=\left(A_{1}\otimes\otimes I_{t/n}\right)\left(B_{1}\otimes\operatorname{{\bf
1}}_{t/p}\right)\\\ =(A_{1}<\times B_{1})\otimes\operatorname{{\bf
1}}_{\xi}\leftrightarrow(A_{1}<\times B_{1}).\end{array}$

$\Box$

Precisely speaking, because ${\mathcal{V}}$ is not a vector space, “invariant
subspace” is not a rigorous subspace. But
$\Omega_{{\mathcal{V}}}:={\mathcal{V}}/\leftrightarrow$ is a vector space. It
is easy to see that the results about ${\mathcal{V}}$, and
${\mathcal{M}}_{\cdot\times n}$ can be extended to $\Omega_{{\mathcal{V}}}$
and $\Omega_{{\mathcal{M}}}$. For instance, $\left<A\right>\in\Sigma_{\mu}$ is
a bounded operator on $\Omega_{{\mathcal{V}}}$ if and only if, $\mu_{y}=1$.

##  8 Conclusion

Matrix theory is one of the most fundamental and useful tools in modern
science and technology. But one of the major weaknesses is its dimension
restriction. To overcome this barrier, the purpose of this paper is to set up
a framework for an almost dimension-free matrix theory.

First, we review the STP ($\ltimes$), which extends the conventional matrix
product to overall matrices ${\mathcal{M}}$. The related monoid structure for
$\left({\mathcal{M}},~{}\ltimes\right)$ is obtained. The matrix equivalence
$\sim$ is proposed. A lattice structure over each equivalence class is
obtained. The equivalence space, as the quotient space ${\mathcal{M}}/\sim$ is
discussed.

Second, the set of overall matrices is partitioned into subspaces as
${\mathcal{M}}=\bigcup_{\mu\in{\mathbb{Q}}_{+}}{\mathcal{M}}_{\mu}$. The STA (
$\pm$ ) is proposed. Under this addition the quotient spaces
$\Sigma_{\mu}={\mathcal{M}}_{\mu}/\sim$ become vector spaces. Certain
geometric and algebraic structures are revealed. Including topological
structure, inner product structure. differential manifold structure, etc.

Particularly, when $\mu=1$, (corresponding to square matrices) we have
extended the Lie algebra and Lie group theory to ${\mathcal{M}}_{1}$. A fiber
bundle structure, called the discrete bundle, is proposed for
${\mathcal{M}}_{\mu}$ and the extended Lie group and Lie algebra.

Finally, the set of overall vectors ${\mathcal{V}}$ are considered as a
universal vector space, based on the vector equivalence $\leftrightarrow$. A
matrix $A$ of any dimension can be considered either a linear mapping on
${\mathcal{V}}$, or a subspace generated by its columns. The $A$-invariant
subspace is discussed in details. Many key concepts such as
eigenvalue/eigenvector, characteristic polynomial of a matrix have been
extended from square matrices to no-square matrices.

It was said by Asimov that “Only in mathematics is there no significant
correction - only existence. Once the Greeks had developed the deductive
method, they were correct in what they did, correct for all the time.” [38]
All extensions we did in this paper consist with the classical ones. That is,
when the dimension restrictions required by the classical matrix theory are
satisfied the new operators proposed in this paper coincide with the classical
ones.

There are many questions remain for further discussion. For instance, is it
possible to construct an equivalence over ${\mathcal{M}}$, which is consistent
with certain matrix product, such that the quotient space becomes a vector
space?

The followings are some possible equivalences on ${\mathcal{M}}$.

  1. 1.

Equivalence 1:

######  Definition 256.

Let $A,~{}B\in{\mathcal{M}}$. $A$ is said to be equivalent to $B$, if there
exist $I_{i}$, $I_{j}$, $\operatorname{{\bf 1}}_{s}$, $\operatorname{{\bf
1}}_{t}$ such that

$\displaystyle\operatorname{{\bf 1}}_{\alpha}^{T}\otimes A\otimes
I_{i}=\operatorname{{\bf 1}}_{\beta}^{T}\otimes B\otimes I_{j}$ (350)

It is easy to verify that $\simeq$ is an equivalence relation. Moreover,
similar to matrix equivalence or vector equivalence, we have the following
result:

######  Theorem 257.

Assume $A\simeq B$, then there exists a $\Lambda$, such that

$\displaystyle\begin{array}[]{l}A=\operatorname{{\bf
1}}^{T}_{p}\otimes\Lambda\otimes I_{s}\\\ B=\operatorname{{\bf
1}}^{T}_{q}\otimes\Lambda\otimes I_{t}.\end{array}$ (353)

Hence the lattice structure similar to matrix equivalence exists.

It may be considered as a combination of matrix and vector equivalences.
Unfortunately, (i) it is not consistent with STP ($\ltimes$); (ii) the
quotient space is not a vector space.

  2. 2.

Equivalence 2:

######  Definition 258.

Let $A,~{}B\in{\mathcal{M}}$. $A$ is said to be equivalent to $B$, if there
exist $\operatorname{{\bf 1}}_{i}$, $\operatorname{{\bf 1}}_{j}$,
$\operatorname{{\bf 1}}_{s}$, $\operatorname{{\bf 1}}_{t}$ such that

$\displaystyle\operatorname{{\bf 1}}_{\alpha}^{T}\otimes
A\otimes\operatorname{{\bf 1}}_{i}=\operatorname{{\bf 1}}_{\beta}^{T}\otimes
B\otimes\operatorname{{\bf 1}}_{j}$ (354)

The lattice structure can also be determined in a similar way. Moreover, the
quotient space is a vector space. Unfortunately, a proper product, which is
consistent with the equivalence, is unknown.

Further geometric/algebraic structures may be investigated.

  1. 1.

More geometric structure on equivalence space could be interesting. For
instance, a Riemannian geometric structure or a Symplectic geometric structure
may be posed on the equivalence space.

  2. 2.

Under the STP and  $\pm$ ($\vdash$), for any $A\in{\mathcal{M}}_{\mu}$ and
$B\in{\mathcal{M}}_{\lambda}$,

$\displaystyle[A,B]=A\ltimes B\vdash B\ltimes A$ (355)

is well defined. Moreover, the three requirements (208)-(212) in Definition
142 can also be satisfied (under obvious modification). Exploring the
properties of this generalized Lie algebra is challenging and interesting.

One may be more interested in its applications. For instance, can we use the
extended structure proposed in this paper to the analysis and control of
certain dynamic systems? Particularly, we may consider the following special
cases:

  1. 1.

Consider a dynamic system

$\dot{x}=A(t)x,\quad x\in{\mathbb{R}}^{n},$

where $A(t)$ satisfies

$\dot{A}_{t}=V(x),$

where $V(x)$ is a vector field on ${\mathcal{M}}_{n\times n}$. What can we say
about this system? Is it possible to extend this system to the equivalent
space $\Sigma$?

  2. 2.

A dimension-varying dynamic control system as

$\begin{cases}x(t+1)=A<\times x(t)<+B<\times u(t)\\\ y(t)=C<\times
x(t),\end{cases}$

where $x(t)\in{\mathcal{V}}$.

What can we say about this, say, controllability? observability etc.?

In one word, this paper could be the beginning of investigating dimension-free
matrix theory and its applications.

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