Some Rules of Elementary Transformation in $B^{+}(E,F)$ and their applications

TL;DR

This paper introduces rules for elementary transformations in the space of bounded operators between Banach spaces and explores their applications in topology and analysis.

Contribution

It presents new rules for elementary transformations in $B^{+}(E,F)$ and applies them to problems in topology and analysis.

Findings

New rules for elementary transformations in $B^{+}(E,F)$

Applications to topology of operators

Applications to analysis of operators

Abstract

In this paper, some rules of elementary transformation in $B^{+}(E,F)$ are presented. Then we give some applications to topology and analysis for operators.

Equations

$$E_1=\left\{x+\alpha x:\forall x\in E_{\ast }\right\};$$

$$P_{E_1}^R=P_{E_{\ast }}^R+\alpha P_{E_{\ast }}^R\mathrm{andso}{P}_R^{E_1}=P_R^{E_{\ast }}-\alpha P_{E_{\ast }}^R.$$

$$P_{E_{\ast }}^{R}x+\alpha P_{E_{\ast }}^{R}x=0\mathrm{for}x\in E\iff P_{E_{\ast }}^{R}x=0\iff x\in R$$

$$P_{E_1}^R=P_{E_{\ast }}^R+\alpha P_{E_{\ast }}^R,\mathrm{and}\mathrm{so},P_R^{E_1}=P_R^{E_{\ast }}-\alpha P_{E_{\ast }}^R.$$

$$P_{E_1}^R=P_{E_{\ast }}^R+\alpha P_{E_{\ast }}^R.$$

$$P(\lambda )=(1-2\lambda )P_{E_{\ast }}^R+(1-\lambda )P_{E_{\ast }}^{R}0\le \lambda \le 1.$$

$$R(P(\lambda ))=R(P_{E_{\ast }}^R+\frac{1-\lambda }{1-2\lambda }\alpha P_{E_{\ast }}^R),0\le \lambda \le 1.$$

$$P(\lambda )=P_{E_{\ast }}^R+\lambda \alpha P_{E_{\ast }}^{R}0\le \lambda \le 1.$$

$$R(T_0)=\left\{y+\alpha y:\forall y\in F_{\ast }\right\},P_{R(T_0)}^N=P_{F_{\ast }}^N+\alpha P_{F_{\ast }}^N,$$

$$T_0=\left\{P_{F_{\ast }}^N+\alpha P_{F_{\ast }}^N\right\}{T}_0.$$

$$F_{\lambda }=\left\{y+\lambda \alpha y:\forall y\in F_{\ast }\right\}\mathrm{for}\mathrm{all}\lambda \in [0,1].$$

$$P_{F_{\lambda }}^N=P_{F_{\ast }}^N+\lambda \alpha P_{F_{\ast }}^N\text{mbox}and{P}_N^{F_{\lambda }}=P_N^{F_{\ast }}-\lambda \alpha P_{F_{\ast }}^N,\forall \lambda \in [0,1].$$

$$P(\lambda )=P_{F_{\lambda }}^N{T}_0,\forall \lambda \in [0,1].$$

$$R(P(\lambda ))=F_{\lambda },\forall \lambda \in [0,1],$$

$$R(P(\lambda ))\oplus N=F,i.e.,P(\lambda )\in C_r(N).$$

$$y\in N(P(\lambda ))\iff P_{F_{\lambda }}^N{T}_{0}y=0\iff T_{0}y\in N\iff y\in N(T_0)$$

$$P(\lambda )\in \{T\in C_r(N):N(T)=N(T_0)\},\lambda \in [0,1].$$

$$P_{N(T_0)}^{R_0}=P_{E_{\ast }}^{R_0}+\alpha P_{E_{\ast }}^{R_0},$$

$$T_0=T_0{P}_{R_0}^{N(T_0)}=T_0(P_{R_0}^{E_{\ast }}-\alpha P_{E_{\ast }}^{R_0}).$$

$$P(\lambda )=T_0(P_{R_0}^{E_{\ast }}-\lambda \alpha P_{E_{\ast }}^{R_0}),0\le \lambda \le 1.$$

$$x\in N(P(\lambda ))⟺(P_{R_0}^{E_{\ast }}-\lambda \alpha P_{E_{\ast }}^{R_0})x=0⟺x\in R(P_{E_{\ast }}^{R_0}+\lambda \alpha P_{E_{\ast }}^{R_0})$$

$$E=H\oplus (E_1+E_2),E_1=E_1^{\ast }\oplus (E_1\cap E_2),\text{mbox}and{E}_2=E_2^{\ast }\oplus (E_1\cap E_2).$$

$$E_1+E_2=(E_1^{\ast }\oplus E_2^{\ast })\oplus (E_1\cap E_2).$$

$$H_1=\{x+\alpha x:\forall x\in E_1^{\ast }\}=\{x+{\alpha }^{-1}x:\forall x\in E_2^{\ast }\}.$$

$$E_1^{\ast }\oplus E_2^{\ast }=H_1\oplus E_2^{\ast }=H_1\oplus E_1^{\ast }$$

$$E=H\oplus (E_1+E_2)=H\oplus (E_1^{\ast }\oplus E_2^{\ast })\oplus (E_1\cap E_2)=H\oplus H_1\oplus E_2^{\ast }\oplus (E_1\cap E_2)=H\oplus H_1\oplus E_2$$

$$E=H\oplus H_1\oplus E_1^{\ast }\oplus (E_1\cap E_2)=H\oplus H_1\oplus E$$

$$E=N(T_i)\oplus R_i,i=1,2.$$

$$E=N_0\oplus R_1=N_0\oplus R_2$$

$$L_{i}x=\left\{\begin{array}[]{rcl}T_{i}x,&&{x\in R_{i}}\\ 0,&&{x\in N_{0}}\end{array}\right.$$

Full text

**Some Rules of Elementary Transformation in $B^{+}(E,F)$ and their applications

**Ma Jipu1,2

Abstract Let $E,F$ be two Banach spaces, $B(E,F)$ the set of all bounded linear operators from $E$ into $F$, and $B^{+}(E,F)$ the set of double splitting operators in $B(E,F)$. In this paper, we present some rules of elementary transformations in $B^{+}(E,F)$, consisting of five theorems. Let $\Phi_{m,n}$ be the set of all Fredholm operators $T$ in $B(E,F)$ with dim$N(T)=m$ and codim$R(T)=n$, and $F_{k}=\{T\in B(E,F):$rank$T=k<\infty\}$. Applying the rules we prove that $F_{k}(k<$dim$F)$ and $\Phi_{m,n}(n>0)$ are path connected, so that they are not only smooth submanifolds in $B(E,F)$ with tangent space $M(X)=\{T\in B(E,F):TN(X)\subset R(X)\}$ at $X$ in them, but also path connected. Hereby we obtain the following topological construction: $B(\mathbf{R}^{m},\mathbf{R}^{n})=\bigcup^{n}\limits_{k=0}F_{k}(m\geq n),B(\mathbf{R}^{m},\mathbf{R}^{n})=\bigcup^{m}\limits_{k=0}F_{k}(m<n),F_{k}$ is path connected and smooth sub-hypersurface in $B(\mathbf{R}^{m},\mathbf{R}^{n})(k<n),$ and especially dim$F_{k}=(m+n-k)k$ for $k=0,1,\cdots,n$. Finally we introduce an equivalent relation in $B^{+}(E,F)$ and prove that the equivalent class $\tilde{T}$ generated by $T$ is path connected for any $T\in B^{+}(E,F)$ with $R(T)\varsubsetneq F$.

Key words Elementary Transformation Path Connected Set of Operators Dimension Sub-hypersurface Smooth Submanifold Equivalent Relation.

2000 Mathematics Subject Classification: 47B38, 46T20, 58A05, 15A09.

1 Introduction

Let $E,F$ be two Banach spaces, $B(E,F)$ the set of all linear bounded operators from $E$ into $F$, and $B^{+}(E,F)$ the set of all double splitting operators in $B(E,F)$. It is well known that the elementary transformation of a matrix is a power tool in matrix theory. For example, by the elementary transformation one can show that $F_{k}=\{T\in B(\mathbf{R}^{n}):$rank$T=k\}(k<n)$ is path connected. When dim$E=$dim$F=\infty$ there is no such elementary transformation in $B(E,F)$, so it is difficult to prove that $F_{k}(k<$dim$F)$ in $B(E,F)$ is path connected. In this paper, we present some rules of elementary transformation in $B^{+}(E,F)$, which consist of five theorems, see Section 2. Using the rules we prove that $F_{k}(k<$dim$F)$ and $\Phi_{m,n}(n>0)$ are path connected, where $\Phi_{m,n}$ denotes the set of all Fredholm operators $T$ in $B(E,F)$ with dim$N(T)=m$ and codim$R(T)=n$. Then by Theorem 4.2 in [Ma4] we obtain the following result $F_{k}(k<$dim$F)$ and $\Phi_{m,n}(n>0)$ are not only smooth submanifolds in $B(E,F)$ with tangent space $M(X)=\{T\in B(E,F):TN(X)\subset R(V)\}$ at any $X$ in then, but also path connected. As its application we have that $B(\mathbf{R}^{m},\mathbf{R}^{n})=\bigcup^{n}\limits_{k=0}F_{k}(m\geq n),B(\mathbf{R}^{m},\mathbf{R}^{n})=\bigcup^{m}\limits_{k=0}F_{k}(m<n),F_{k}(k<n)$, is path connected and smooth sub-hpersurface in $B(\mathbf{R}^{m},\mathbf{R}^{n})$ and especially, dim$F_{k}=(m+n-k)k$ for $k=0,1,\cdots,n$. Finally we introduce an equivalent relation in $B^{+}(E,F)$ and prove that the equivalent class $\tilde{T}$ generated by $T$ is path connected for $T\in B^{+}(E,F)$ with $R(T)\varsubsetneq F.$

2 Some Rules of Elementary Transformation in $B^{+}(E,F)$

In this section, we will introduce some rules of elementary transformation in $B^{+}(E,F)$, which consist of five theorems. It is useful to imagene the trace of these rules of elementary Transformations from Euclidean space to Banach space.

Theorem 2.1 If $E=E_{1}\oplus R=E_{*}\oplus R$ , then the following conclusions hold:

$(i)$* there exists a unique $\alpha\in B(E_{*},R)$ such that*

[TABLE]

conversely, for any $\alpha\in B(E_{*},R)$ the subspace $E_{1}$ defined by $(2.1)$ satisfies $E=E_{1}\oplus R$

$(ii)$

[TABLE]

Proof For the proof of $(i)$ see [Ma3] and [Abr].

Obviously,

$\left(P^{R}_{E_{*}}+\alpha P^{R}_{E_{*}}\right)^{2}=P^{R}_{E_{*}}+\alpha P^{R}_{E_{*}},$ and

[TABLE]

Then by (2.1), one concludes

[TABLE]

The proof ends.  $\Box$

Let $B^{+}(E)$ be the set of all double splitting operators in $B(E)$ and $C_{r}(R)=\left\{T\in B^{+}(E):E=R(T)\oplus R\right\}$

Theorem 2.2 Suppose that $E=E_{*}\oplus R$ and $\dim R>0.$ Then $P^{R}_{E_{*}}$ and $(-P^{R}_{E_{*}})$ are path connected in the set $\left\{T\in C_{r}(R):N(T)=R\right\}.$

Proof Due to $\dim R>0,$ one can assume that $B(E_{*},R)$ contains a non-zero operator $\alpha$, otherwise $E_{*}=\{0\}$ the theorem is trivial. Let $E_{1}=\left\{x+\alpha x:\forall x\in E_{*}\right\}.$ Then by Theorem 2.1, $E=E_{1}\oplus R$ and

[TABLE]

Consider the path

[TABLE]

Clearly,

[TABLE]

Then by Theorem 2.1, $R(P(\lambda))\oplus R=E$, i.e.$P(\lambda)\in C_{r}(R)$ for all $\lambda\in[0,1]$. In addition, $P(0)=P^{R}_{E_{1}},P(1)=-P^{R}_{E_{*}}$ and $N(P(\lambda))=R,\forall\lambda\in[0,1]$. This shows that $P^{R}_{E_{1}}$ and $-P^{R}_{E_{*}}$ are path connected in $\left\{T\in C_{r}(R):N(T)=R\right\}.$

Next go to show that $P^{R}_{E_{1}}$ and $P^{R}_{E_{*}}$ are path connected in $\left\{T\in C_{r}(R):N(T)=R\right\}.$ Consider the path

[TABLE]

By Theorem 2.1, $P(1)=P^{R}_{E_{*}}+\alpha P^{R}_{E_{*}}=P^{R}_{E_{1}},$ and $R(P(\lambda))\oplus R=E\,\,\,\,\forall\lambda\in[0,1],$ where $E_{1}=\left\{x+\alpha x:\forall x\in E_{*}\right\}.$ Obviously, $N(P(\lambda))=R$ and $P(0)=P^{R}_{E_{*}}$. Therefor $P^{R}_{E_{1}}$ and $P^{R}_{E_{*}}$ are path connected in $\left\{T\in C_{r}(R):N(T)=R\right\}.$ Thus the theorem is proved.  $\Box$

For simplicity, still write $C_{r}(N)=\left\{T\in B^{+}(E,F):R(T)\oplus N=F\right\}$ in the sequal.

Theorem 2.3 Suppose $T_{0}\in C_{r}(N)$ and $F=F_{*}\oplus N$. Then $T_{0}$ and $P^{N}_{F_{*}}T_{0}$ are path connected in the set $\left\{T\in C_{r}(N):N(T)=N(T_{0})\right\}.$

Proof One can assume $R(T_{0})\neq F_{*}$, otherwise the theorem is trivial. Then by Theorem 2.1, there exists a non-zero operator $\alpha\in B(F_{*},N)$ such that

[TABLE]

and $P_{N}^{R(T_{0})}=P_{N}^{F_{*}}-\alpha P^{N}_{F_{*}}.$ So

[TABLE]

Let

[TABLE]

Note $\lambda\alpha\in B(F_{*},N).$ According to Theorem 2.1 we also have

[TABLE]

Consider the path

[TABLE]

Because $F=R(T_{0})\oplus N=F_{\lambda}+N$ one observers

[TABLE]

and so

[TABLE]

Note

[TABLE]

i.e., $N(P(\lambda))=N(T_{0}),\forall\lambda\in[0,1].$ Thus

[TABLE]

In addition, $P(1)=T_{0}$ by (2.3), and $P(0)=P^{N}_{F_{*}}T_{0}$.

Finally we conclude that $T_{0}$ and $P^{N}_{F_{*}}T_{0}$ are path connected in the set $\{T\in C_{r}(N):N(T)=N(T_{0})\}$. The proof ends. $\Box$

Let $C_{d}(R)=\{T\in B^{+}(E,F):E=N(T)\oplus R\}$.

Theorem 2.4 Suppose that $T_{0}\in C_{d}(R_{0})$ and $E=E_{*}\oplus R_{0}.$ Then $T_{0}$ and $T_{0}P^{E_{*}}_{R_{0}}$ are path connected in the set $\{T\in C_{d}(R_{0}):R(T)=R(T_{0})\}$.

Proof One can assume $E_{*}\neq N(T_{0}),$ otherwise the theorem is trivial. Then by Theorem 2.1, there exists a non-zero operator $\alpha\in B(E_{*},R_{0})$ such that

[TABLE]

and so,

[TABLE]

Consider the path as follows,

[TABLE]

Since $(P^{E_{*}}_{R_{0}}-\lambda\alpha P_{E_{*}}^{R_{0}})x=x\;\forall x\in R_{0},$ we conclude $R(P(\lambda))=R(T_{0})$. We also have $N(P(\lambda))=R(P^{E_{*}}_{R_{0}})+\lambda\alpha P^{E_{*}}_{R_{0}})$.Indeed,

[TABLE]

since $\lambda\alpha\in B(E_{*},R_{0})$ for all $\lambda\in[0,1].$

Thus $P(\lambda)\in\{T\in C_{d}(R):R(T)=R_{0}\},\forall\lambda\in[0,1].$ In addition, $P(1)=T_{0}$ by (2.4), and $P(0)=T_{0}P^{E_{*}}_{R_{0}}.$ Then the theorem is proved. $\Box$

Theorem 2.5 Suppose that the subspaces $E_{1}$ and $E_{2}$ in $E$ satisfy $dimE_{1}=dimE_{2}<\infty.$ Then $E_{1}$ and $E_{2}$ possess a common complement R, i.e.,$E=E_{1}\oplus R=E_{2}\oplus R$.

Proof According to the assumption $dimE_{1}=dimE_{2}<\infty$, we have the following decompositions:

[TABLE]

It is easy to observe that $(E^{*}_{1}\oplus E^{*}_{2})\cap(E_{1}\cap E_{2})=\{0\}$ and $dimE^{*}_{1}=dimE^{*}_{2}<\infty.$ Indeed, if $e^{*}_{1}+e^{*}_{2}$ belongs to $E_{1}\cap E_{2}$, for $e^{*}_{i}\in E^{*}_{i},\;i=1,2,$ then $e_{2}^{*}=(e^{*}_{1}+e^{*}_{2})-e^{*}_{1}\in E_{1}$ and $e_{1}^{*}=(e^{*}_{1}+e^{*}_{2})-e^{*}_{2}\in E_{2}$, so that $e^{*}_{1}=e^{*}_{2}=0$ because of $E^{*}_{i}\cap(E_{1}\cap E_{2})=\{0\},\,i=1,2.$ Hereby, one can see

[TABLE]

we now are in the position to determine $R$. We may assume $dimE^{*}_{1}=dimE^{*}_{2}>0,$ otherwise the theorem is trivial. Then $B^{X}(E^{*}_{1},E^{*}_{2})$ contains an operator $\alpha$, which bears the subspace $H_{1}$ as follows,

[TABLE]

By Theorem 2.1,

[TABLE]

Finally, according to (2.5) we have

[TABLE]

and

[TABLE]

This says $R=H\oplus H_{1}.$ The proof ends. $\Box$

3 Some Applications

In this section we will give some application of the rules in Section 2.

Theorem 3.1 For $k<dimF$, $F_{k}$ is path connected.

Proof In what follows, we may assume $k>0$, otherwise, the theorem is trivial. Let $T_{1}$ and $T_{2}$ be arbitrary two operators in $F_{k}$, and

[TABLE]

Then $0<$ dim $R_{1}=$dim $R_{2}=k<\infty$, and by Theorem 2.5, there exists a subspace $N_{0}$ in $E$ such that

[TABLE]

Let

[TABLE]

$i=1,2.$

We claim that $T_{i}$ and $L_{i}$ are path connected in $F_{k},\,i=1,2.$ Due to $(3.1)$ the theorem 2.1 shows that there exists an operator $\alpha_{i}\in B(N_{0},R_{i})$ such that

[TABLE]

Consider the following paths:

[TABLE]

Obviously, $P_{i}(0)=T_{i}P^{N_{0}}_{R_{i}}=L_{i},$ and $P(1)=T_{i}(P^{N_{0}}_{R_{i}}-\alpha_{i}P^{R_{i}}_{N_{0}})=T_{i}P^{N(T_{i})}_{R_{i}}=T_{i},\;i=1,2.$ Next go to show $R(P_{i}(\lambda))=T_{i}R_{i}=R(T_{i}).$

Evidently, $R(P^{N_{0}}_{R_{i}}-\lambda\alpha_{i}P^{R_{i}}_{N_{0}})\subset R_{i},$ and the converse rlation follows from $(P^{N_{0}}_{R_{i}}-\lambda\alpha_{i}P^{R_{i}}_{N_{0}})r=r,\forall r\in R_{i}.$ This says $P_{i}(\lambda)\in F_{k}$ for $0\leq\lambda\leq 1\,\mbox{and}\,i=1,2,$ so that $L_{i}$ and $T_{i}$ are path connected in $F_{k},\;i=1,2.$ Hence, in what follows, we can assume $N(T_{1})=N(T_{2})=N_{0}.$ In the other hand, according to Theorem 2.5 we have

[TABLE]

where $N_{*}$ is a closed subspace in $F$. Then by Theorem 2.3, the proof of the theorem turns to that of the following conclusion: $P^{N^{*}}_{R(T_{1})}T_{2}$ and $T_{1}$ are path connected in $F_{k}$. Let

[TABLE]

It is easy to observe

[TABLE]

In fact, note $N(T_{1})=N(T_{2})=N_{0}$ and $T^{+}_{1}T_{1}=P^{N_{0}}_{R_{1}},$ then

[TABLE]

We claim $P^{N_{*}}_{R_{(T_{1})}}T_{2}T^{+}_{1}|_{R(T_{1})}\in B^{X}(R(T_{1})).$

Evidently,

[TABLE]

i.e.,$P^{N_{*}}_{R_{(T_{1})}}T_{2}T^{+}_{1}|_{R(T_{1})}$ is injective.

In the other hand, since the decomposition (3.2) implies $P^{N_{*}}_{R_{(T_{1})}}|_{R(T_{2})}\in B^{X}\left(R(T_{2}),R(T_{1})\right)$, there is a unique $r_{2}\in R_{2}$ for any $y\in R(T_{1})$, such that

[TABLE]

Then $y_{0}=T_{1}P^{N_{0}}_{R_{1}}r_{2}$ fulfills

[TABLE]

This says that $P^{N_{*}}_{R(T_{1})}T_{2}T^{+}_{1}|_{R(T_{1})}$ is surjective. So $P^{N_{*}}_{R(T_{1})}T_{2}T^{+}_{1}|_{R(T_{1})}\in B^{X}(R(T_{1}))$ of all invertible operators in $B(R(T_{1})$.

It is well known that $P^{N_{*}}_{R(T_{1})}T_{2}T^{+}_{1}|_{R(T_{1})}$ is path connected with some one of $-I_{R(T_{1})}$ and $I_{R(T_{1})}$ is $B^{X}(R(T_{1}))$, where $I_{R(T_{1})}$ denotes the identity on $R(T_{1})$. Let $Q(t)$ be a path in $B^{X}(R(T_{1}))$ with $Q(0)=P^{N_{*}}_{R(T_{1})}T_{2}T_{1}^{+}|_{R(T_{1})}$ and $Q(1)=-I_{R(T_{1})}$ (or $I_{R(T_{1})})$. Then $Q_{1}(t)=Q(t)T_{1}$ is a path in the set $S=\{T\in B(E,F):R(T)=R(T_{1})$ and $N(T)=N(T_{1})\}$ satisfying

[TABLE]

Since $S\subset F_{k}$, it follows that $P^{N_{*}}_{R(T_{1})}T_{2}T^{+}_{1}T_{1}$ and $-T_{1}($or $T_{1})$ are path connected in $F_{k}$. By (3.3) we merely need to show the following conclusion: if $P^{N_{*}}_{R(T_{1})}T_{2}T^{+}_{1}T_{1}$ is path connected with $-T_{1}$ in $F_{2}$ then $P^{N_{*}}_{R(T_{1})}T_{2}T^{+}_{1}T_{1}$ and $T_{1}$ are path connected in $F_{k}$. Note $F=R(T_{1})\oplus N_{*}$ and dim$N_{*}>0$. By Theorem 2.2, $P^{N_{*}}_{R(T_{1})}$ and $-P^{N_{*}}_{R(T_{1})}$ are path connected in the set $\{T\in G_{r}(N_{*}):N(T)=N_{*}\}=\{T\in B(F):F=R(T)\oplus N_{*}$ and $N(T)=N_{*}\}$. Let $Q(t)$ be such a path in the set with $Q(0)=-^{N_{*}}_{R(T_{1})}$ and $Q(1)=P^{N_{*}}_{R(T_{1})}.$ Since $F=R(Q(t))\oplus N_{*}$ and $N(Q(t))=N_{*}\forall t\in[0,1],R(Q(t)T_{1})=Q(t)R(T_{1})=Q(t)F,$ $Q(t)$ for $t\in[0,1]$ belongs to $F_{k}$, and satisfies $Q(0)T_{1}=-P^{N_{*}}_{R(T_{1})}T_{1}=-T_{1}$ and $Q(1)T_{1}=T_{1}.$ This says that $-T_{1}$ and $T_{1}$ are path connected in $F_{k}$, so that the conclusion is proved.  $\Box$

It is obvious that if either dim$E$ or dim$F$ is finite, then $B(E,F)$ consists of all finite rank operators. Hence we assume dim$E=$dim$F=\infty$ in the sequel.

Let $\Phi_{m,n}=\left\{T\in B^{+}(E,F):\dim N(T)=m<\infty\;\mbox{and}\;\mathrm{codim}R(T)=n<\infty\right\}$, we have

Theorem 3.2 $\Phi_{m,n},(n>0)$ is path connected.**

Proof Let $T_{1}$ and $T_{2}$ be arbitrary two operators in $\Phi_{m,n}$ and

[TABLE]

i.e. $T_{1}\in C_{r}(N_{1})$ and $T_{2}\in C_{r}(N_{2})$.

Clearly, $\dim N_{1}=\dim N_{2}=n<\infty$. Then by Theorem 2.5, there exists a subspace $F_{*}$ in $F$ such that

[TABLE]

So, by Theorem 2.3 the proof of the theorem turns to that of the operators $P^{N_{1}}_{F_{*}}T_{1}$ and $P^{N_{2}}_{F_{2}}T_{2}$ being path connected in $\Phi_{m,n}.$ For simplicity, still write $P^{N_{1}}_{F_{1}}T_{1}$ and $P^{N_{2}}_{F_{*}}T_{2}$ as $T_{1}$ and $T_{2}$, respectively. However, here $R(T_{1})=R(T_{2})=F_{*}$ while $N(T_{1}),N(T_{2})$ keep invariant. Due to dim$N(T_{1})=$dim$N(T_{2})=m<\infty.$ Theorem 2.5 shows that there exists a subspace $R$ in $E$ such that

[TABLE]

Then by Theorem 2.4, one can conclude that $T_{2}P^{N(T_{1})}_{R}$ and $T_{2}$ are path connected in $\Phi_{m,n}.$ Thus the proof of the theorem turns once more to that of $T_{2}P^{N(T_{1})}_{R}$ and $T_{1}$ being path connected in $\Phi_{m,n}.$ In what follows, we do this.

Let

[TABLE]

Then

[TABLE]

It is easy to observe $T_{2}T^{+}_{1}\mid_{F_{*}}\in B^{X}(F_{*}).$ Indeed,

[TABLE]

i.e., $T_{2}T^{+}_{1}|_{F_{*}}$ is injective; let; $r\in R$ satisfy $T_{2}r=y$ for any $y\in F_{*}$, and $T_{1}r-y_{0}\in F_{*}$, then

[TABLE]

and so $T_{2}T^{+}_{1}|_{F_{*}}$ is also surjective. Thus $T_{2}T^{+}_{1}|_{F_{*}}\in B^{X}(F_{*})$ and is path connected with some one of $-I_{F_{*}}$ and $I_{F_{*}}$ is in $B^{X}(F_{*})$. Similar to the way in the proof of Theorem 3.1 one can prove that $T_{2}T^{+}_{1}T_{1}$ is path connected with some one of $-T_{1}$ and $T_{2}$ is in $\Phi_{m,n}$. Note the equality $F=F_{*}\oplus N_{1}$ in (3.4) and dim$N_{1}>0.$ Due to dim $N_{1}>0$ Theorem 2.2 shows that $-P^{N_{1}}_{F_{*}}$ and $P^{N_{1}}_{F_{*}}$ are path connected in the set $S=\{T\in C_{F}(N_{1})\subset(F);N(T)=N_{1}\}$, say that $Q(X)\in S$ for $\lambda\in[0,1]$ fulfills $Q(0)=-P^{N_{1}}_{F_{*}}$ and $Q(1)=P^{N_{1}}_{F_{*}}.$ Obviously,

[TABLE]

and

[TABLE]

for all $\lambda\in[0,1].$ So $Q(\lambda)T_{1}$ for any $\lambda\in[0,1]$ belongs to $\Phi_{m,n}$. This says that $-T_{1}$ and $T_{1}$ are path connected in $\Phi_{m,n}$. Therefore by (3.6) $T_{2}P^{N(T_{1})}_{R}$ is also path connected with $T_{1}$ in $\Phi_{m,n}$.$\Box$

By Theorem 4.2 in [Ma4] we futher have

Theorem 3.3 $F_{k}(k<$dim$F$) and $\Phi_{m,n}(n>0)$ are not only path connected set but also smooth submanifolds in $B(E,F)$ with tangent space $M(X)=\{T\in B(E,T):TN(X)\subset R(X)\}$ at any $X$ in them.

Applying the theorem to $B(\mathbf{R}^{m},\mathbf{R}^{n})$ we obtain the geomatrical and topologial construction of $B(\mathbf{R}^{m},\mathbf{R}^{n})$.

Theorem 3.4 $B(\mathbf{R}^{m},\mathbf{R}^{n})=\bigcup^{n}\limits_{k=0}F_{k}(m\geq n),B(\mathbf{R}^{m},\mathbf{R}^{n})=\bigcup^{n}\limits_{k=0}F_{k}(m<n),F_{k}$ is a smooth and path connected submanifold in $B(\mathbf{R}^{m},\mathbf{R}^{n})$, and especially, dim$F_{k}=(m+n-k)k$ for $k=0,1,\cdots,n.$

Proof We need only to prove the formula dim$F_{k}=(m+n-k)k$ for $k=0,1,\cdots,n$ since otherwise the theorem is immediate from Theorem 3.3. Let $T=\{T_{i,j}\}^{m,n}_{i,j=1}$ for any $T\in B(\mathbf{R}^{m},\mathbf{R}^{n}),I_{k}=\{T\in B(\mathbf{R}^{m},\mathbf{R}^{n}):t_{i,j}=0$ except $t_{i,i}=1,1\leq i\leq k\}$, and $I^{+}_{k}=\{(s_{i,j}\}^{m,n}_{i,j=1}\in B(\mathbf{R}^{n},\mathbf{R}^{m}):s_{i,j}=0$ except $s_{i,i}=1,1\leq i\leq k\}$.

Obviously, $I_{k}I^{+}_{k}I_{k}=I_{k}$ and $I^{+}_{k}T_{k}T^{+}_{k}=T^{+}_{k}.$ So

[TABLE]

and

[TABLE]

where $I^{*}_{n},I^{*}_{m}$ denote the identity on $\mathbf{R}^{n}$ and $\mathbf{R}^{m}$, respectively. Let

[TABLE]

By direct computing

[TABLE]

so that dim$M^{+}=(n-k)(m-k).$ By Lemma 4.1 in [Ma4], $B(\mathbf{R}^{m},\mathbf{R}^{n})=M(I_{k})\oplus M^{+}$ and so, dim$M(I_{k})=m\times n-(m-k)(n-k)=(m+n)-k)k.$ Due to $F_{k}$ being path connected, by Theorem 3.1, one can conclude ${\rm dim}F_{k}={\rm dim}M(I_{k})=(m+n-k)k.$  $\Box$

Let $G(\cdot)$ denote the set of all splitling subspaces in the Banach space in the parentheses, $U_{E}(R)=\left\{H\in G(E):E=R\oplus H\right\}$ for any $R\in G(E)$, and $U_{F}(S)=\{L\in G(E):$ $F=S\oplus L\}$ for any $S\in G(E)$. In order to consider of more general results then that of the previous theorems 3.1 and 3.2, we introduce the equivalent relation as follows.

Definition 3.1 $T_{0}$ and $T_{*}$ in $B^{+}(E,F)$ are said to be equivalent provided there exist finite number of subspaces $N_{1},\cdots,N_{m}$ in $G(E)$, and $F_{1},\cdots,F_{n}$ in $G(F)$ such that all**

[TABLE]

and

[TABLE]

are non-empty. For abbreviation, write it as $T_{0}\sim T_{*},$ and let $\widetilde{T}$ denote the equivalent class generated by $T$ in $B(E,F).$

Theorem 3.5 $\widetilde{T_{0}}$ for any $T_{0}\in B^{+}(E,F)$ with $\mathrm{codim}R(T_{0})>0$ is path connected.**

Proof Assume that $T_{*}$ is any operator in $\widetilde{T_{0}}$, and

[TABLE]

We define by induction:

[TABLE]

It is easy to observe

[TABLE]

Indeed, $R(T_{1})=T_{0}R(P^{N_{1}}_{R_{1}})=T_{0}R_{1}=R(T_{0})$ and$N(T_{1})=\{x\in E:P^{N_{1}}_{R_{1}}x\in N(T_{0})\}=\{x\in E:P^{N_{1}}_{R_{1}}x=0\}=N_{1}$ because of $E=R_{1}\oplus N_{1}$; similarly, by induction one can conclude $R(T_{k})=R(T_{0})$ and $N(T_{k})=N_{k}$ for $k=1,2,\cdots,m.$ So $T_{k}\in\tilde{T}_{0}$ for $k=1,2,\cdots,m.$

Let $S_{k}=\{T\in G_{d}(R_{k}):R(T)=R(T_{0})\}=\{T\in B(E,F):E=N(T)\oplus R_{k}$ and $R(T)=R(T_{0})\}$ for $k=1,2,\cdots,m.$ Evidently $S_{k}\subset\tilde{T}_{0},k=1,2,\cdots,m.$ In fact, by (3.7) $R_{1}\in U_{E}(N(T_{0}))\cap U_{E}(N_{1}),\cdots,R_{k}\in U_{E}(N_{k-1})\cap U_{E}(N_{k});$ while $R_{k}\in U_{E}(N_{k})\cap U_{E}(N(T))$ and $R(T)=R(T_{0});$ so that $S_{k}\subset\tilde{T}_{0},k=1,2,\cdots,m.$ Next go to show that $T_{0}$ and $T_{m}$ are path connected in $\tilde{T}_{0}$. Due to $N_{k-1}\oplus R_{k})=N(T_{k-1})\oplus R_{k}=N_{k}\oplus R_{k}.$ Theorem 2.4 shows that $T_{k+}$ and $T_{k}$ are path connected in $S_{k}$, and so in $\tilde{T}_{0}$ for $k=1,2,\cdots,m$. Therefore $T_{0}$ and $T_{m}$ are path connected in $\tilde{T}_{0}$.

Let $M_{k}=\{T\in C_{d}(R_{k}):\mathbf{R}(T)=R(T_{k})\}.$ By (3.8)

[TABLE]

so that $M_{k}\subset\tilde{T}_{0}$. Note $E=N(T_{k-1})\oplus R_{k}=N_{k}\oplus R_{k}.$ Then by Theorem 2.4, $T_{k}=T_{k-1}P^{N_{k}}_{R_{k}}$ and $T_{k-1}$ are path connected in $M_{k-1}=\{T\in C_{d}(R_{k}):R(T)=R(T_{k-1})\}\subset\tilde{T}_{0},k=1,2,\cdots,m.$ This shows that $T_{0}$ and $T_{m}$ are poth connected in $\tilde{T}_{0}$. Finally go to prove that $T_{m}$ and $T_{*}$ are path connected in $\tilde{T}_{0}.$

In other hand, assume

[TABLE]

Write $T_{m,0}=T_{m}.$ We define by induction,

[TABLE]

According to the equality $R(T_{m})=R(T_{0})$ in (3.8), we infer

[TABLE]

In fact wsmite $F_{0}=\mathbf{R}(T_{m})$, then by (3.9)

[TABLE]

and so

[TABLE]

Thus, take $T_{m,i-1},F_{i}$ and $S_{i}$ for $i=1,2,n$ in the places of $T_{0},F_{*}$ and $N$, in Theorem 2.3, respectively, then the theorem shows that $T_{m,k}$ and $T_{m,k-1}$ are path connected in $\left\{T\in C_{r}(S_{k}):N(T)=N_{m}\right\}\subset\widetilde{T_{0}},k=1,2,\cdots,n,$ so that $T_{m}$ and $T_{m,n}$ are path connected in $\widetilde{T_{0}}$. Since $T_{0}$ and $T_{m}$ are path connected in $\widetilde{T_{0}},$ $T_{0}$ and $T_{m,n}$ are path connected in $\widetilde{T_{0}},$ so the proof of the theorem reduces to that of $T_{m,n}$ being path connected with $T_{*}$ in $\widetilde{T_{0}},$ where $T_{m,n}$ and $T_{*}$ satisfy

[TABLE]

and

[TABLE]

[TABLE]

The first equality in (3.11) shows $T_{*}\in C_{\alpha}(R_{m+1})$ and $E=N_{m}\oplus R_{m+1}.$ Then by Theorem 2.4, $T_{*}P^{N_{m}}_{R_{m+1}}$ and $T_{*}$ are path connected in $\left\{T\in C_{\alpha}(R_{m+1}):R(T)=R(T_{*})\right\}\subset\widetilde{T_{0}}.$

The second equality in (3.11) shows $T_{*}\in C_{r}(S_{n+1})$ an d $F=F_{n}\oplus S_{n+1}.$ Then by Theorem 2.3, $P^{S_{n+1}}_{F_{n}}T_{*}$ and $T_{*}$ are path connected in $\left\{T\in C_{r}(S_{n+1}):N(T)=N(T_{*})\right\}\subset\widetilde{T_{0}}.$ Combining the preceding results, we conclude that $P^{S_{n+1}}_{F_{n}}T_{*}P^{N_{m}}_{R_{n+1}}$ and $T_{*}$ are path connected in $\widetilde{T_{0}}.$ So far, the proof of the theorem reduces to that of $P^{S_{n+1}}_{F_{n}}T_{*}P^{N_{m}}_{R_{m+1}}$ being path connected with $T_{m,n}$ in $\widetilde{T_{0}}.$

According to (3.10) we have

[TABLE]

where $N_{m}=N(T_{m,n})$ and $R(T_{m,n})=F_{n}.$

Let

[TABLE]

Obviously,

[TABLE]

We claim

[TABLE]

$P^{S_{n+1}}_{F_{n}}T_{*}T^{+}_{m,n}y=0\,\,\mathrm{for}\,\,y\in F_{n}\Leftrightarrow T_{*}{T^{+}_{m,n}}y=0$ because of (3.11) $\Leftrightarrow T^{+}_{m,n}y=0$ since $T^{+}_{m,n}y\in R_{m+1}\Leftrightarrow y=0;$ while from the assumption in (3.7)and (3.8), $R_{m+1}\in U_{E}(N_{m})\cap U_{E}(N(T_{*}))$ and $S_{n+1}\in U_{F}(F_{n})\cap U_{F}(R(T_{*})),$ it follows that there exist $r\in R_{m+1}$ and $y_{0}\in F_{n},$ for any $y\in F_{n}$ such that $P^{S_{n+1}}_{F_{n}}T_{*}r=y\,\,\mathrm{and}\,\,T^{+}_{m,n}y_{0}=r,\,i.e.,P^{S_{n+1}}_{F_{n}}T_{*}T^{+}_{m,n}$ is surjective.

It is well known that $P^{S_{n+1}}_{F_{n}}T_{*}T^{+}_{m,n}|_{F_{n}}$ is path connected with some one of $I_{F_{n}}$ and $(-I_{F_{n}})$ in $B^{X}(F_{n}).$ Hence $P^{S_{n+1}}_{F_{n}}T_{*}T^{+}_{m,n}P^{S_{n+1}}_{F_{n}}$ is path connected with some one $P^{S_{n+1}}_{F_{n}}$ and $(-P^{S_{n+1}}_{F_{n}})$ in the set $S=$ $\left\{F\in B(F_{n}):R(T)=F_{n}\,\,\mathrm{and}\,\,N(T)=S_{n+1}\right\}$ $\subset\left\{T\in C_{r}(S_{n+1}):N(T)=S_{n+1}\right\}.$ If $P^{S_{n+1}}_{F_{n}}T_{*}T^{+}_{m,n}P^{S_{n+1}}_{F}$ is path connected with $(-P^{S_{n+1}}_{F_{n}})$, then by Theorem 2.2, it is path connected with $P^{S_{n+1}}_{F_{n}}$ in $S$. Thus $P^{S_{n+1}}_{F_{n}}T_{*}T^{+}_{m,n}P^{S_{n+1}}_{F_{n}}$ is also path connected with $P^{S_{n+1}}_{F_{n}}$ in $\left\{T\in C_{r}(S_{n+1}):N(T)=S_{n+1}\right\}$. Note the equality (3.10). Finally, by the simular way to that in the proof of Theorem 3.2, one can infer that $P^{S_{n+1}}_{F_{n}}T_{*}P^{N_{m}}_{R_{m+1}}$ and $T_{m}$ are path connected in $\tilde{T}_{0}.$ The proof ends. $\Box$

It is easy to see that $\widetilde{T}=F_{k}$ for any $T\in F_{k}(k<\infty)$ as well as $\widetilde{T}=\Phi_{m,n}$ for any $T\in\Phi_{m,n},\,n>0.$

References

[Abr]  R. Abraham, J. E. Marsden, and T. Ratin, Manifolds, tensor analysis and applications, 2nd ed., Applied Mathematical Sciences 75, Springer, New York, 1988.

[An]  V. I. Arnol’d, Geometrical methods in the theory of ordinary differential equations, 2nd ed., Grundlehren der Mathematischen Wissenschaften 250, Springer, New York, 1988.

[Bo]  B. Booss,D.D.Bleecker, Topology and Analysis: the Atyah-Singer Index Formula and Gauge-Theoretic physics, New York: Springer-Verlag 1985.

[Caf]  V. Cafagra, Global invertibility and finite solvability, pp. 1-30 in Nonlinear functional analysis (New York, NJ, 1987), edited by P. S. Milojevic, Lecture Notes in Pure and Appl. Math. 121, Dekker, New York, 1990.

[Ma1]  Jipu Ma, Dimensions of Subspaces in a Hilbert Space and Index of Semi-Fredholm Operators, Sci, China, Ser. A 39:12(1986).

[Ma2]  Jipu Ma, Generalized Indices of Operators in B(H), Sci, China, Ser. A , 40:12(1987).

[Ma3]  Jipu Ma, Complete Rank Theorem in Advanced Calculus and Frobenius Theorem in Banach Space, arxiv:1407.5198v5[math.FA]23 Jan. 2015.

[Ma4]  Jipu Ma Frobenius Theorem in Banach Space, submited to Ivent. Math..(Also see) arXiv: submit/1512157[math.FA] 19 Mar 2016).

1. Department of Mathematics, Nanjing University, Nanjing, 210093, P. R. China

2. Tseng Yuanrong Functional Analysis Research Center, Harbin Normal University, Harbin, 150080, P. R. China

E-mail address: jipuma@126.com; jipuma1935@sina.com