The Upper Semi-Weylness and Positive Nullity for Operator Matrices

Abstract

Let H and K be infinite dimensional separable complex Hilbert spaces and B(K, H) the algebra of all bounded linear operators from K into H. Let $A\in B(H)$ and $B\in B(K)$. We denote by $M_C$ the operator acting on $H\oplus K$ of the form $M_C=\left( \begin{array}{cc}A&{}C\\ 0&{}B\\ \end{array}\right)$. In this paper, we give necessary and sufficient conditions for $M_C$ to be an upper semi-Fredholm operator with $n(M_C)>0$ and $\hbox {ind}(M_C)<0$ for some left invertible operator $C\in B(K,H)$. Meanwhile, we discover the relationship between $n(M_C)$ and n(A) during the exploration. And we also describe all left invertible operators $C\in B(K,H)$ such that $M_C$ is an upper semi-Fredholm operator with $n(M_C)>0$ and $\hbox {ind}(M_C)<0$.

1 Introduction

Throughout this paper, $\mathbb {C}$ and $\mathbb {N}$ denote the set of complex numbers and the set of nonnegative integers, respectively. Let H and K be complex separable infinite dimensional Hilbert spaces and B(K, H) the algebra of all bounded linear operators from K into H. If $K=H$, then we denote B(H, H) by B(H). Let $T\in B(H)$. We denote by n(T) the nullity of T, which is the dimension of the kernel N(T), and by d(T) the codimension of the range R(T). If R(T) is closed and $n(T)<\infty$, T is called an upper semi-Fredholm operator (or left semi-Fredholm operator). T is said to be a lower semi-Fredholm operator (or right semi-Fredholm operator) if $d(T)<\infty$[1, Corollary 1.15]. An operator T is said to be a Fredholm operator if it is both lower and upper semi-Fredholm. Especially, if T is an upper semi-Fredholm operator and $n(T)=0$, then T is called a left invertible operator. Put $B_{l}^{-1}(K, H)=\{T\in B(K, H): T\ \hbox {is\ a\ left\ invertible\ operator}\}$. If T is a lower or upper semi-Fredholm operator, the index of T is defined by $\hbox {ind}(T)=n(T)-d(T)$. An operator T is said to be an upper semi-Weyl operator if it is an upper semi-Fredholm operator with $\hbox {ind}(T)\le 0$. If T is a Fredholm operator and $\hbox {ind}(T)=0$, then T is called a Weyl operator. Let $T^*$ denote the conjugate of $T\in B(H)$.

Let $A\in B(H)$, $B\in B(K)$ and $C\in B(K,H)$. We define a $2\times 2$ upper triangular operator matrix $M_C$ acting on $H\oplus K$ of the form

$$\begin{aligned} M_C=\left( \begin{array}{cc}A&{}C\\ 0&{}B\\ \end{array}\right) :\left( \begin{array}{c} H \\ K \\ \end{array}\right) \rightarrow \left( \begin{array}{c} H \\ K \\ \end{array}\right) . \end{aligned}$$

The completion problems of operator matrices were considered for many types of matrices, for example, $M_C$, $M_{X}$ and $M_{(X,Y)}$. In the past 20 years, the study of completion problems for upper triangular operator matrices has been an attractive subject for many scholars and many valuable results have been obtained [2,3,4,5,6]. In [7], the authors gave the necessary and sufficient conditions for $M_C$ to be an upper semi-Weyl operator for some operator $C\in B(K,H)$. Later, Wu and Huang et al. studied the self-adjoint perturbations of spectra for operator matrices in [8, 9]. In recent years, Dong and Cao gave the necessary and sufficient conditions which make $M_C$ be a CFI operator for some operator $C\in B(K,H)$ in [10].

The idea of our article comes from the following papers:

In [11], the authors have shown that there exists some operator $C\in B(K,H)$ such that $M_C$ is upper semi-Fredholm and $\hbox {ind}(M_C)\le 0$ if and only if there exists some left invertible operator $C\in B(K,H)$ such that $M_C$ is upper semi-Fredholm and $\hbox {ind}(M_C)\le 0$.

In [2], the authors briefly discuss the deficiency and the nullity of upper triangular operator matrices in the process of proving the left and right invertibility of $M_X$.

In this paper, we further give necessary and sufficient conditions for $M_C$ to be an upper semi-Fredholm operator, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$ for some left invertible operator $C\in B(K,H)$ by different methods. During the discussion, we also found a new relationship between $n(M_C)$ and n(A), which further reflects the fixed nullity problem of operator matrices. The main results in this paper are the following theorems:

Theorem 1

Suppose that $A\in B(H)$ and $B\in B(K)$ are given. Then $M_C$ is an upper semi-Fredholm operator, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$ for some left invertible operator $C\in B(K,H)$ if and only if A is an upper semi-Fredholm operator, $n(A)+n(B)>0$ and one of the following statements holds:

• (1) B is an upper semi-Fredholm operator and $n(A)+n(B)<d(A)+d(B)$;

• (2) $d(A)=\infty$.

Theorem 2

Let $A\in B(H)$ and $B\in B(K)$. Then, $M_C$ is an upper semi-Fredholm operator, $n(M_C)=\hbox {max}\{n(A), 1\}$ and $\hbox {ind}(M_C)<0$ for some left invertible operator $C\in B(K,H)$ if and only if A is an upper semi-Fredholm operator, $n(A)+n(B)>0$ and one of the following statements holds:

• (1) B is upper semi-Fredholm, $n(A)+n(B)<d(A)+d(B)$ and $n(A)>0$, $n(B)\le d(A)$;

• (2) B is upper semi-Fredholm, $n(A)+n(B)<d(A)+d(B)$ and $n(A)=0$, $n(B)\le d(A)+1$;

• (3) $d(A)=\infty$.

Furthermore, we will discuss all $C\in B_{l}^{-1}(K, H)$ such that $M_C$ is upper semi-Fredholm, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$ by Theorem 2. We get the following result:

Theorem 3

Let $A\in B(H)$ and $B\in B(K)$. Then, $M_C$ is an upper semi-Fredholm operator, $n(M_C)=\hbox {max}\{n(A), 1\}$ and $\hbox {ind}(M_C)<0$ for all left invertible operators $C\in B(K,H)$ if and only if one of the following statements holds:

• (1) A is upper semi-Fredholm with $n(A)>0$, B is left invertible and $n(A)+n(B)<d(A)+d(B)$;

• (2) A is invertible, B is upper semi-Fredholm with $n(B)=1$ and $n(A)+n(B)<d(A)+d(B)$.

2 Auxiliary Results

In order to prove Theorem 1, we give some useful lemmas.

Lemma 1

Let $A\in B(H)$ be a left invertible operator with $d(A)=\infty$ and $B\in B(K)$ with $n(B)>0$, then there exists some left invertible operator $C\in B(K,H)$ such that $M_C$ is an upper semi-Fredholm operator, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$.

Proof

We take $e_1\in N(B)$ with $\Vert e_1\Vert =1$ and extend it into an orthonormal basis of K, written as $\{e_{n}\}_{n=1}^{\infty }$. Put $M=\hbox {span}\{e_1\}$. Then $\hbox {dim}M^{\perp }=\infty =d(A)=\hbox {dim} R(A)^{\perp }$. So, there is an isometric invertible linear operator $T_1:M^{\perp }\rightarrow R(A)^{\perp }$. Take $u_1=Aw_1\in R(A)$ with $\Vert u_1\Vert =1$, and let $N=\hbox {span}\{u_1\}$. Let $T_2$ be an isometric invertible linear operator as $T_2: M\rightarrow N$ satisfying $T_2e_1=u_1=Aw_1$. Define $C: K\rightarrow H$ by

$$\begin{aligned} C=\left( \begin{array}{cc}T_1 &{} 0 \\ 0 &{} T_2 \\ \end{array}\right) :\left( \begin{array}{c} M^{\perp } \\ M \\ \end{array}\right) \rightarrow \left( \begin{array}{c} R(A)^{\perp } \\ R(A) \\ \end{array}\right) . \end{aligned}$$

It is clear that $C\in B(K,H)$ is left invertible. We claim that $M_C$ is an upper semi-Fredholm operator with $n(M_C)>0$ and $\hbox {ind}(M_C)<0$.

• (i) $0<n(M_C)=1<\infty$.

Let $\left( \begin{array}{c}x \\ y \\ \end{array}\right) \in N(M_C)$. Then $Ax+Cy=0$ and $By=0$. Let $y=\alpha _1+k_1e_1$, where $k_1\in \mathbb {C}$ and $\alpha _1\in M^{\perp }$. It follows from $Ax+Cy=Ax+T_1\alpha _1+T_2(k_1e_1)=A(x+k_1w_1)+T_1\alpha _1$ that $A(x+k_1w_1)=-T_1\alpha _1\in R(A)\cap R(A)^{\perp }=\{0\}$. From $T_1$ is invertible and A is left invertible, we can get that $\alpha _1=0$, $x=-k_1w_1$ and $y=k_1e_1$. Hence, $N(M_C)\subseteq \hbox {span}\left\{ \left( \begin{array}{ccc} w_1 \\ -e_1 \\ \end{array}\right) \right\}$. Since the inclusion “$\supseteq$” is obvious, we get that $N(M_C)=\hbox {span}\left\{ \left( \begin{array}{ccc} w_1 \\ -e_1 \\ \end{array}\right) \right\}$. Therefore, $0<n(M_C)=1<\infty$.

• (ii) $R(M_C)$ is closed.

Suppose that $M_C\left( \begin{array}{c}x_{n} \\ y_{n} \\ \end{array}\right) \rightarrow \left( \begin{array}{c}u_{0} \\ v_{0} \\ \end{array}\right)$ ($n\rightarrow \infty$). Then, $Ax_{n}+Cy_{n}\rightarrow u_{0}$ and $By_{n}\rightarrow v_{0}$ ($n\rightarrow \infty$). Let $y_n=\alpha _n+k_ne_1$, where $k_n\in \mathbb {C}$ and $\alpha _n\in M^{\perp }$. So $Ax_n+Cy_n=Ax_n+T_1\alpha _n+T_2(k_ne_1)=A(x_n+k_nw_1)+T_1\alpha _n\rightarrow u_{0}$ ($n\rightarrow \infty$). Since $A(x_n+k_nw_1)\in R(A)$ and $T_1\alpha _n\in R(A)^{\perp }$, we get that both $\{A(x_n+k_nw_1)\}_{n=1}^{\infty }$ and $\{T_1\alpha _n\}_{n=1}^{\infty }$ are Cauchy sequences. From the fact that $T_1$ is invertible, we can get that $\{\alpha _n\}_{n=1}^{\infty }$ is a Cauchy sequence. From R(A) and $M^{\perp }$ are closed, there exist some $x_{0}\in H$ and $\alpha _0\in M^{\perp }$ such that $A(x_n+k_nw_1)\rightarrow Ax_{0}$, $\alpha _n\rightarrow \alpha _0$ ($n\rightarrow \infty$). Now, $Ax_n+Cy_n=A(x_n+k_nw_1)+C\alpha _n\rightarrow Ax_{0}+C\alpha _0$ and $By_n=B\alpha _n\rightarrow B\alpha _0$ as $n\rightarrow \infty$. This implies that $M_C\left( \begin{array}{c}x_{n} \\ y_{n} \\ \end{array}\right) \rightarrow M_C\left( \begin{array}{c}x_0 \\ \alpha _0 \\ \end{array}\right)$. Therefore $\left( \begin{array}{c}u_{0} \\ v_{0} \\ \end{array}\right) =M_{C}\left( \begin{array}{c}x_{0} \\ \alpha _{0} \\ \end{array}\right) \in R(M_{C})$, which means that $R(M_C)$ is closed.

• (iii) $\hbox {ind}(M_C)<0$.

Claim 1: $\left( \begin{array}{c}0 \\ e_{i} \\ \end{array}\right) \notin R(M_{C})$ for any $i\ge 2$.

In fact, if there exists some $k\ge 2$ such that $\left( \begin{array}{c}0 \\ e_{k} \\ \end{array} \right) \in R(M_{C})$, we may suppose that $\left( \begin{array}{c}0 \\ e_{k} \\ \end{array}\right) =M_{C}\left( \begin{array}{c}x_0 \\ y_0 \\ \end{array}\right)$, then $Ax_0+Cy_0=0$ and $By_0=e_{k}$. Let $y_0=\alpha _0+k_1e_1$, where $k_1\in \mathbb {C}$ and $\alpha _0\in M^{\perp }$. From $Ax_0+Cy_0=Ax_0+T_1\alpha _0+T_2(k_1e_1)=A(x_0+k_1w_1)+T_1\alpha _0=0$ we can get that $A(x_0+k_1w_1)=-T_1\alpha _0\in R(A)\cap R(A)^{\perp }=\{0\}$. Since $T_1$ is invertible and A is left invertible, it follows that $\alpha _0=0$ and $x_0=-k_1w_1$. Then, $y_0=k_1e_1$ and $By_0=B(k_1e_1)=0$, a contradiction.

Claim 2: The nonzero linear combinations of $\left\{ \left( \begin{array}{ccc} 0 \\ e_i \\ \end{array}\right) \right\} _{i=2}^\infty$ are not in $R(M_{C})$.

In fact, if there exists $\left( \begin{array}{c}x_1 \\ y_1 \\ \end{array} \right) \in H\oplus K$ such that $\sum \limits _{i=2}^na_i\left( \begin{array}{c} 0 \\ e_{i} \\ \end{array}\right) =M_{C}\left( \begin{array}{c}x_1 \\ y_1 \\ \end{array} \right)$, where $a_i\ne 0$ for some $2\le i\le n$. Then $Ax_1+Cy_1=0$ and $By_1=a_2e_2+\cdots +a_ne_n$. Let $y_1=\alpha _1+k_2e_1$, where $k_2\in \mathbb {C}$ and $\alpha _1\in M^{\perp }$. Similar to the proof of Claim 1, we get that $y_1=k_2e_1$. Then, $By_1=B(k_2e_1)=0$, a contradiction.

Claim 3: $d(M_C)=\infty$.

We assume that $\left( \begin{array}{c}0 \\ e_{i} \\ \end{array}\right) =g_i+h_i$ for any $i\ge 2$, where $g_i\in R(M_C)$ and $h_i\in R(M_C)^{\perp }$. For any $n\ge 2$, if $\sum \limits _{i=2}^na_ih_i=0$, then $a_2\left( \begin{array}{c}0 \\ e_{2} \\ \end{array}\right)$+$\cdots$+$a_n\left( \begin{array}{c}0 \\ e_{n} \\ \end{array}\right) =a_2g_2+\cdots +a_ng_n\in R(M_C)$. From Claim 2 we know that $a_2\left( \begin{array}{c}0 \\ e_{2} \\ \end{array}\right)$+$\cdots$+$a_n\left( \begin{array}{c}0 \\ e_{n} \\ \end{array}\right) =0$. Since $\{e_i\}_{i=2}^{\infty }$ is linearly independent, we get that $a_i=0$, $2\le i\le n$ and so $\{h_i\}_{i\ge 2}$ is linearly independent. It follows that $\hbox {dim}R(M_C)^{\perp }=d(M_C)=\infty$, which implies $\hbox {ind}(M_C)<0$.

From Lemma 1, if $A\in B(H)$ is a left invertible operator with $d(A)=\infty$ and $B\in B(K)$ with $n(B)>0$, then there exists some left invertible operator $C\in B(K,H)$ such that $M_C$ is an upper semi-Fredholm operator, $n(M_C)=1$ and $\hbox {ind}(M_C)<0$. If $A\in B(H)$ is upper semi-Fredholm with $n(A)>0$, we can get that:

Lemma 2

Let $A\in B(H)$ and $B\in B(K)$. If A is an upper semi-Fredholm operator with $d(A)=\infty$ and $n(A)>0$, then there exists $C\in B_{l}^{-1}(K, H)$ such that $M_C$ is an upper semi-Fredholm operator, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$.

Proof

We will complete the proof by the following two cases.

Case 1. R(B) is not closed.

There exists an isometric invertible linear operator $T:\ K\rightarrow R(A)^{\perp }$ since $\dim K=\dim R(A)^{\perp }=\infty$.

Define an operator $C:\ K\rightarrow H$ by $Cy=Ty$ for any $y\in K$. It is clear that C is left invertible. Next we will show $M_C$ has the desired properties.

• (i) $0<n(M_C)=n(A)<\infty$.

Let $\left( \begin{array}{c}x \\ y \\ \end{array}\right) \in N(M_C)$. Then $Ax+Cy=0$ and $By=0$. It follows that $Ax=-Cy=-Ty\in R(A)\cap R(A)^{\perp }=\{0\}$. From T is invertible, we can get that $y=0$ and $x\in N(A)$. Hence, $N(M_C)\subseteq N(A)\oplus \{0\}$. The inclusion “$\supseteq$" is obvious, then $N(M_C)=N(A)\oplus \{0\}$. Therefore, $0<n(M_C)=n(A)<\infty$.

• (ii) $R(M_C)$ is closed.

Suppose that $M_C\left( \begin{array}{c} x_{n} \\ y_{n} \\ \end{array} \right) \rightarrow \left( \begin{array}{c} u_{0} \\ v_{0} \\ \end{array} \right)$ ($n\rightarrow \infty$). Then $Ax_{n}+Cy_{n}=Ax_n+Ty_n\rightarrow u_{0}$ and $By_{n}\rightarrow v_{0}$ ($n\rightarrow \infty$). Since $Cy_n=Ty_n\in R(A)^{\perp }$, we get that both $\{Ax_n\}_{n=1}^{\infty }$ and $\{Ty_n\}_{n=1}^{\infty }$ are Cauchy sequences. It follows that $\{y_n\}_{n=1}^{\infty }$ is a Cauchy sequence. We suppose that $y_n\rightarrow y_0$ ($n\rightarrow \infty$). Since R(A) is closed, there exists some $x_{0}\in H$ such that $Ax_n\rightarrow Ax_{0}$($n\rightarrow \infty$). Now, $Ax_n+Cy_n\rightarrow Ax_{0}+Cy_0$ and $By_n\rightarrow By_0$ ($n\rightarrow \infty$). Therefore, $\left( \begin{array}{c}u_{0} \\ v_{0} \\ \end{array}\right) =M_{C}\left( \begin{array}{c}x_{0} \\ y_{0} \\ \end{array}\right) \in R(M_{C})$. It means that $R(M_C)$ is closed.

• (iii) From R(B) is not closed we get that $d(M_C)\ge d(B)=\infty$ and so $\hbox {ind}(M_C)<0$.

Case 2. R(B) is closed.

Now, we divide it according to the nullity of B.

• (1) Suppose $n(B)<\infty$.

In this case, B is an upper semi-Fredholm operator. It is obvious that $M_C$ is an upper semi-Fredholm operator and $\hbox {ind}(M_C)=\hbox {ind}(A)+\hbox {ind}(B)<0$ for all left invertible operators $C\in B(K,H)$. Since $\hbox {dim}R(A)^{\perp }=\infty$, there is an isometric invertible linear operator $T:K\rightarrow R(A)^{\perp }$. Define an operator $C:K\rightarrow H$ by $Cy=Ty$ for any $y\in K$. It is clear that C is left invertible. Similar to the proof of Case 1, $n(M_C)=n(A)$ is trivial.

• (2) Suppose $n(B)=\infty$ and $n(A)<d(B)$.

Since $\hbox {dim}N(B)=\hbox {dim}R(A)^{\perp }=\infty$, there exists an isometric invertible linear operator $T_2:N(B)\rightarrow R(A)^{\perp }$. It is clear that $\hbox {dim}R(A)=\infty$. Therefore, there is a closed subspace $M\subseteq R(A)$ such that $\hbox {dim}N(B)^{\perp }=\hbox {dim}M$. So, there exists an isometric invertible linear operator $T_1:N(B)^{\perp }\rightarrow M$. Define $C:K\rightarrow H$ by

$$\begin{aligned} C=\left( \begin{array}{cc}T_1 &{} 0 \\ 0 &{} T_2 \\ \end{array}\right) :\left( \begin{array}{c}N(B)^{\perp } \\ N(B) \\ \end{array}\right) \rightarrow \left( \begin{array}{c} R(A) \\ R(A)^{\perp } \\ \end{array}\right) . \end{aligned}$$

It is clear that $C\in B(K,H)$ is left invertible. In the following, we prove that $M_C$ satisfies the above conditions.

• (i) $0<n(M_C)=n(A)<\infty$.

Let $\left( \begin{array}{c}x \\ y \\ \end{array}\right) \in N(M_C)$. Then $Ax+Cy=0$ and $By=0$. Then, $y\in N(B)$ and $Ax=-Cy=-T_2y\in R(A)\cap R(A)^{\perp }=\{0\}$. From $T_2$ is invertible, we can get that $y=0$ and $x\in N(A)$. Hence, $N(M_C)\subseteq N(A)\oplus \{0\}$. The inclusion “$\supseteq$" is obvious, then $N(M_C)=N(A)\oplus \{0\}$. Therefore, $0<n(M_C)=n(A)<\infty$.

• (ii) $R(M_C)$ is closed.

Suppose that $M_C\left( \begin{array}{c}x_{n} \\ y_{n} \\ \end{array}\right) \rightarrow \left( \begin{array}{c}u_{0} \\ v_{0} \\ \end{array}\right)$ ($n\rightarrow \infty$). Then $Ax_{n}+Cy_{n}\rightarrow u_{0}$ and $By_{n}\rightarrow v_{0}$ ($n\rightarrow \infty$). Let $y_n=\alpha _n+\beta _n$, where $\alpha _n\in N(B)^{\perp }$ and $\beta _n\in N(B)$. So $Ax_n+Cy_n=(Ax_n+T_1\alpha _n)+T_2\beta _n\rightarrow u_{0}$ ($n\rightarrow \infty$). Since $Ax_n+T_1\alpha _n\in R(A)$ and $T_2\beta _n\in R(A)^{\perp }$, we get that both $\{Ax_n+T_1\alpha _n\}_{n=1}^{\infty }$ and $\{T_2\beta _n\}_{n=1}^{\infty }$ are Cauchy sequences. It follows that $\{\beta _n\}_{n=1}^{\infty }$ is a Cauchy sequence. Note the fact that $B|_{N(B)^{\perp }}$ is left invertible and $By_n=B\alpha _n$, then we can obtain that $\{\alpha _n\}_{n=1}^{\infty }$ is a Cauchy sequence. Since R(A) is closed, there exist some $x_{0}\in H$ and $y_0\in K$ such that $Ax_n\rightarrow Ax_{0}$, $y_n\rightarrow y_0$ ($n\rightarrow \infty$). Now, $Ax_n+Cy_n\rightarrow Ax_{0}+Cy_0$ and $By_n\rightarrow By_0$ ($n\rightarrow \infty$). Therefore, $\left( \begin{array}{c}u_{0} \\ v_{0} \\ \end{array}\right) =M_{C}\left( \begin{array}{c}x_{0} \\ y_{0} \\ \end{array}\right) \in R(M_{C})$.

• (iii) From $d(M_C)\ge d(B)>n(A)=n(M_C)$ we get that $\hbox {ind}(M_C)<0$.

• (3) Suppose $n(B)=\infty$ and $d(B)\le n(A)$.

Now, B is a lower semi-Fredholm operator and then B is not compact. So $\hbox {dim}N(B)^{\perp }=\hbox {dim}R(B^*)=\infty$. Since $d(B)\le n(A)<\infty$, there exists a positive integer m such that $n(A)<d(B)+m$. Put $N(B)^{\perp }=M\oplus M^{\perp }$, where $\hbox {dim}M=m$. There are isometric invertible linear operators $T_1:M^{\perp }\rightarrow R(A)$ and $T_2:N(B)\oplus M\rightarrow R(A)^{\perp }$. Define $C:K\rightarrow H$ by

$$\begin{aligned} C=\left( \begin{array}{cc}T_1 &{} 0 \\ 0 &{} T_2 \\ \end{array} \right) :\left( \begin{array}{c}M^{\perp } \\ N(B)\oplus M \\ \end{array} \right) \rightarrow \left( \begin{array}{c}R(A) \\ R(A)^{\perp } \\ \end{array}\right) . \end{aligned}$$

It is clear that $C\in B(K,H)$ is invertible and so is left invertible.

• (i) $0<n(M_C)=n(A)<\infty$.

It is trivial to get this result.

• (ii) $R(M_C)$ is closed.

Suppose that $M_C\left( \begin{array}{c} x_{n} \\ y_{n} \\ \end{array} \right) \rightarrow \left( \begin{array}{c} u_{0} \\ v_{0} \\ \end{array} \right)$ ($n\rightarrow \infty$). Then, $Ax_{n}+Cy_{n}\rightarrow u_{0}$ and $By_{n}\rightarrow v_{0}$ ($n\rightarrow \infty$). Let $y_n=\alpha _n+\beta _n$, where $\alpha _n\in M^{\perp }$ and $\beta _n\in N(B)\oplus M$. So $Ax_n+Cy_n=(Ax_n+T_1\alpha _n)+T_2\beta _n\rightarrow u_{0}$ ($n\rightarrow \infty$). Since $Ax_n+T_1\alpha _n\in R(A)$ and $T_2\beta _n\in R(A)^{\perp }$, we get that both $\{Ax_n+T_1\alpha _n\}_{n=1}^{\infty }$ and $\{T_2\beta _n\}_{n=1}^{\infty }$ are Cauchy sequences. It follows that $\{\beta _n\}_{n=1}^{\infty }$ is a Cauchy sequence. From $By_n=B\alpha _n+B\beta _n$ we can get that $\{B\alpha _n\}_{n=1}^{\infty }$ is a Cauchy sequence. Note the fact that $B|_{N(B)^{\perp }}$ is left invertible and $\{\alpha _n\}_{n=1}^{\infty }\subseteq N(B)^{\perp }$. We can obtain that $\{\alpha _n\}_{n=1}^{\infty }$ is a Cauchy sequence, then $\{y_n\}_{n=1}^{\infty }$ is a Cauchy sequence. Since R(A) is closed, there exist some $x_{0}\in H$ and $y_0\in K$ such that $Ax_n\rightarrow Ax_{0}$, $y_n\rightarrow y_0$ ($n\rightarrow \infty$). Now, $Ax_n+Cy_n\rightarrow Ax_{0}+Cy_0$ and $By_n\rightarrow By_0$ ($n\rightarrow \infty$). Therefore, $\left( \begin{array}{c} u_{0} \\ v_{0} \\ \end{array} \right) =M_{C}\left( \begin{array}{c} x_{0} \\ y_{0} \\ \end{array} \right) \in R(M_{C})$.

• (iii) $\hbox {ind}(M_C)<0$.

Let $\{e_i\}_{i=1}^{m}$ be an orthonormal basis of M. From $\{e_i\}_{i=1}^{m}\subseteq N(B)^{\perp }=R(B^*)$ we can suppose that $e_i=B^*u_i$ for some $u_i$, $1\le i\le m$. Since $Ce_i\in R(A)^{\perp }=N(A^*)$, we have $A^*(Ce_i)=0$. Moreover, $C^*(Ce_i)=C^*(T_2e_i)=T_2^*T_2e_i=e_i$, so $\left\{ \left( \begin{array}{ccc} Ce_i \\ -u_i \\ \end{array}\right) \right\} _{i=1}^{m}\subseteq N(M_C^*)$. From the fact that $T_2$ is invertible and $Ce_i=T_2e_i$, we get that $\left\{ \left( \begin{array}{ccc} Ce_i \\ -u_i \\ \end{array}\right) \right\} _{i=1}^{m}$ is linearly independent. Since $n(B^*)=d(B)<\infty$, we assume that $h_1, h_2, \cdots , h_{d(B)}\in N(B^*)$ are linearly independent. Then, $\left\{ \left( \begin{array}{ccc} 0 \\ h_i \\ \end{array}\right) \right\} _{i=1}^{d(B)}\subseteq N(M_C^*)$.

We claim that: $\left( \begin{array}{c} 0 \\ h_1 \\ \end{array} \right) , \cdots , \left( \begin{array}{c} 0 \\ h_{d(B)} \\ \end{array} \right) , \left( \begin{array}{c} Ce_1 \\ -u_1 \\ \end{array} \right) , \cdots , \left( \begin{array}{c} Ce_m \\ -u_m \\ \end{array} \right)$ are linearly independent.

In fact, we assume that

$$\begin{aligned} a_1\left( \begin{array}{c} 0 \\ h_1 \\ \end{array} \right) +\cdots +a_{d(B)}\left( \begin{array}{c} 0 \\ h_{d(B)} \\ \end{array} \right) +\lambda _1\left( \begin{array}{c} Ce_1 \\ -u_1 \\ \end{array} \right) +\cdots +\lambda _m\left( \begin{array}{c} Ce_m \\ -u_m \\ \end{array} \right) =0, \end{aligned}$$

then

$$\begin{aligned} \left\{ \begin{aligned}&C(\lambda _1e_1+\cdots +\lambda _me_m)=0,\\&a_1h_1+\cdots +a_{d(B)}h_{d(B)}-(\lambda _1u_1+\cdots +\lambda _mu_m)=0.\\ \end{aligned} \right. \end{aligned}$$

Since C is invertible and $\{e_i\}_{i=1}^{m}$ is linearly independent, we obtain that $\lambda _1=\cdots =\lambda _m=0$. Hence, $a_1h_1+\cdots +a_{d(B)}h_{d(B)}=0$ and then $a_1=a_2=\cdots =a_{d(B)}=0$. So this claim is proved.

Therefore, $n(M_C^*)\ge d(B)+m$ and $\hbox {ind}(M_C)\le n(A)-(d(B)+m)<0$.

From the proof of Lemma 2, we can see: If A is an upper semi-Fredholm operator with $d(A)=\infty$ and $n(A)>0$, then there exists a left invertible operator $C\in B(K,H)$ such that $M_C$ is upper semi-Fredholm, $n(M_C)=n(A)>0$ and $\hbox {ind}(M_C)<0$.

By Lemma 1 and Lemma 2, we obtain the following corollary.

Corollary 1

Suppose that $B\in B(K)$ and $A\in B(H)$ are given. If A is an upper semi-Fredholm operator with $d(A)=\infty$ and $n(A)+n(B)>0$, then there exists a left invertible operator $C\in B(K,H)$ such that $M_C$ is an upper semi-Fredholm operator, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$.

In Lemma 1 and Lemma 2, $A\in B(H)$ is upper semi-Fredholm with $d(A)=\infty$. If $A\in B(H)$ is a Fredholm operator, we can get that:

Lemma 3

Suppose that $A\in B(H)$ is a Fredholm operator and $B\in B(K)$ is an upper semi-Fredholm operator. If $n(A)+n(B)>0$ and $\hbox {ind}(A)+\hbox {ind}(B)<0$, then there exists $C\in B_{l}^{-1}(K, H)$ such that $M_C$ is an upper semi-Fredholm operator, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$.

Proof

It is obvious that $M_C$ is an upper semi-Fredholm operator and $\hbox {ind}(M_C)=\hbox {ind}(A)+\hbox {ind}(B)<0$ for any $C\in B(K,H)$.

Case 1. $n(A)>0$. By $n(M_C)\ge n(A)>0$ for any $C\in B(K,H)$, the result is obvious.

Case 2. $n(A)=0$.

Now, $n(B)>0$ and $\hbox {dim}N(B)^{\perp }=\hbox {dim}R(B^*)=\infty$. Take $e_1\in N(B)$ with $\Vert e_1\Vert =1$ and then we can extend it into an orthonormal basis of K, written as $\{e_{n}\}_{n=1}^{\infty }$. Put $M=\hbox {span}\{e_1\}$. Take $u_1=Aw_1\in R(A)$ with $\Vert u_1\Vert =1$, and let $N=\hbox {span}\{u_1\}$. Then, there are isometric invertible linear operators $T_1:M^{\perp }\rightarrow H\ominus N$ and $T_2:M\rightarrow N$ such that $T_2e_1=u_1=Aw_1$. Define $C:K\rightarrow H$ by

$$\begin{aligned} C=\left( \begin{array}{cc} T_1 &{} 0 \\ 0 &{} T_2 \\ \end{array} \right) :\left( \begin{array}{c} M^{\perp } \\ M \\ \end{array} \right) \rightarrow \left( \begin{array}{c} H\ominus N \\ N \\ \end{array} \right) . \end{aligned}$$

It is clear that $C\in B(K,H)$ is invertible and so is left invertible. It follows from $\hbox {span}\left\{ \left( \begin{array}{ccc} -w_1 \\ e_1 \\ \end{array}\right) \right\} \subseteq N(M_C)$ that $n(M_C)\ge 1>0$.

According to above lemmas, we will give the necessary and sufficient condition for which $M_C$ is an upper semi-Fredholm operator, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$ for some $C\in B_{l}^{-1}(K, H)$ and the relationship between $n(M_C)$ and n(A) will be described in section 3. However, we naturally propose such a question:

Given $A\in B(H)$ and $B\in B(K)$, what is the necessary and sufficient condition for which $M_C$ is an upper semi-Fredholm operator, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$ for all left invertible operators $C\in B(K, H)$?

In order to answer the above question, we also need to introduce some lemmas as follows.

Lemma 4

Let $A\in B(H)$ be a left invertible operator with $d(A)=\infty$ and $B\in B(K)$. Then, there exists a left invertible operator $C\in B(K,H)$ such that $M_C$ is left invertible.

Proof

Since $\hbox {dim}R(A)^{\perp }=\hbox {dim}K=\infty$, there exists an isometric invertible linear operator $T:K\rightarrow R(A)^{\perp }$. Define an operator $C:K\rightarrow H$ by $Cy=Ty$ for any $y\in K$. It is clear that C is left invertible. Similar to the Case 1 of Lemma 2, the desired properties are obvious.

Similar to the condition (1) of sufficiency in the proof of Theorem 2, we have the following lemma.

Lemma 5

[12, Corollary 2.3] Let $B\in B(K)$ be an upper semi-Fredholm operator and $A\in B(H)$ be a left invertible operator with $n(B)\le d(A)$, then there exists a left invertible operator $C\in B(K,H)$ such that $M_C$ is left invertible.

Let $T\in B(K)$. If R(T) is closed, then $T|_M$ is left invertible for all infinite dimensional closed subspaces $M\subseteq N(T)^{\perp }$. If R(T) is not closed, we can get:

Lemma 6

Suppose that $T\in B(K)$ and R(T) is not closed. Then, there exists an infinite dimensional closed subspace $M\subseteq N(T)^{\perp }$ such that $T|_M$ is left invertible.

Proof

Since R(T) is not closed, there exists an infinite dimensional closed subspace $E\subseteq R(T)$. Let $M=\{x\in N(T)^{\perp }: Tx\in E\}$. It is obvious that $M\subseteq N(T)^{\perp }$. Then, we prove the following results in turn.

• (1) $\hbox {dim}M=\infty$.

Suppose that $\{y_n\}_{n=1}^{\infty }$ is an orthonormal basis of E and let $y_i=Te_i$, where $e_i\in N(T)^{\perp }$, $i=1, 2, \ldots$. So, $\{e_n\}_{n=1}^{\infty }\subseteq M$. From $T|_{N(T)^{\perp }}$ is injective, we can get that $\{e_n\}_{n=1}^{\infty }$ is linearly independent. Therefore, $\hbox {dim}M=\infty$.

• (2) M is a closed subspace.

It is clear that M is a subspace. Suppose that $\{x_n\}\subseteq M$ and $x_n\rightarrow x_0$, so $Tx_n\rightarrow Tx_0$ ($n\rightarrow \infty$). From $x_n\in M$ and E is closed, we can get that $Tx_0\in E$. Since $N(T)^{\perp }$ is closed, we have $x_0\in N(T)^{\perp }$. Therefore, $x_0\in M$ and M is closed.

• (3) $T|_M$ is left invertible.

From $T|_M$ is injective, we only need $R(T|_M)$ is closed. Suppose that $Tx_n\rightarrow y_0$ ($n\rightarrow \infty$), where $\{x_n\}\subseteq M$. It follows that $\{x_n\}\subseteq N(T)^{\perp }$ and $Tx_n\in E$. So, $y_0\in E\subseteq R(T)$. We suppose that $y_0=Tx_0$, where $x_0\in N(T)^{\perp }$. So $x_0\in M$ and $Tx_n\rightarrow y_0=Tx_0\in R(T|_M)$ as $n\rightarrow \infty$. Therefore, $R(T|_M)$ is closed and $T|_M$ is left invertible.

Lemma 7

Assume $A\in B(H)$ is an upper semi-Fredholm operator with $d(A)=\infty$ and $B\in B(K)$ is not an upper semi-Fredholm operator. Then, there exists a left invertible operator $C\in B(K,H)$ such that $M_C$ is not an upper semi-Fredholm operator.

Proof

We will use two cases to complete the proof.

Case 1. $n(B)=\infty$.

Now, there exists a closed subspace $M\subseteq R(A)^{\perp }$ such that $\hbox {dim}N(B)^{\perp }=\hbox {dim}M$. Then, there are two isometric invertible linear operators $T_1:N(B)^{\perp }\rightarrow M$ and $T_2:N(B)\rightarrow R(A)$. Define $C:K\rightarrow H$ by

$$\begin{aligned} C=\left( \begin{array}{cc}T_1 &{} 0 \\ 0 &{} T_2 \\ \end{array}\right) :\left( \begin{array}{c}N(B)^{\perp } \\ N(B) \\ \end{array}\right) \rightarrow \left( \begin{array}{c}R(A)^{\perp } \\ R(A) \\ \end{array}\right) . \end{aligned}$$

It is clear that C is left invertible. We suppose that $\{y_n\}_{n=1}^{\infty }$ is an orthonormal basis of N(B), then there exists $\{x_n\}_{n=1}^{\infty }$ such that $T_2y_n=Ax_n$. Hence, $\left\{ \left( \begin{array}{ccc} -x_n \\ y_n \\ \end{array}\right) \right\} _{n=1}^{\infty }\subseteq N(M_C)$ is linearly independent. So, $n(M_C)=\infty$ and $M_C$ is not an upper semi-Fredholm operator.

Case 2. $n(B)<\infty$ and R(B) is not closed.

Now, $\hbox {dim}N(B)^{\perp }=\hbox {dim}R(B^*)=\infty$, it follows from Lemma 6 that there exists an infinite dimensional closed subspace $M\in N(B)^{\perp }$ such that $B|_M$ is left invertible. Put $N(B)^{\perp }=M\oplus M^{\perp }$. Since $R(B)=R(B|_M)+R(B|_{M^{\perp }})$, $R(B|_M)$ is closed and R(B) is not closed, we can get that $B|_{M^{\perp }}$ is not left invertible and $\hbox {dim}M^{\perp }=\infty$. So, there exists $\{y_n\}_{n=1}^{\infty }\subseteq M^{\perp }$ with $\Vert y_n\Vert =1$ such that $By_n\rightarrow 0$ ($n\rightarrow \infty$). Let $T_1:M^{\perp }\rightarrow R(A)$ and $T_2:M\oplus N(B)\rightarrow R(A)^{\perp }$ be isometric invertible. Define $C\in B(K, H): K\rightarrow H$ by

$$\begin{aligned} C=\left( \begin{array}{cc}T_1 &{} 0 \\ 0 &{} T_2 \\ \end{array}\right) :\left( \begin{array}{c}M^{\perp } \\ M\oplus N(B) \\ \end{array} \right) \rightarrow \left( \begin{array}{c}R(A) \\ R(A)^{\perp } \\ \end{array}\right) . \end{aligned}$$

It is clear that $C\in B(K,H)$ is invertible and so is left invertible. From $C(N(B))\subseteq R(A)^{\perp }$ we obtain that $N(M_C)=N(A)\oplus \{0\}$. We suppose that $T_1y_n=Ax_n$, where $x_n\in N(A)^{\perp }$. Then, $\left\{ \left( \begin{array}{ccc} -x_n \\ y_n \\ \end{array}\right) \right\} \subseteq N(M_C)^{\perp }$ and $M_C\left( \begin{array}{c} -x_n \\ y_n \\ \end{array} \right) =\left( \begin{array}{c} 0 \\ By_n \\ \end{array} \right) \rightarrow 0$ ($n\rightarrow \infty$). From the fact that $\Vert Ax_n\Vert =\Vert T_1y_n\Vert =\Vert y_n\Vert =1$, $x_n\in N(A)^{\perp }$ and $A|_{N(A)^{\perp }}$ is left invertible, we can get that there is some $k>0$ such that $\Vert Ax_n\Vert =1\ge k\Vert x_n\Vert$ for any n, which implies $\Vert x_n\Vert \le \frac{1}{k}$. So $1\le \Vert w_n\Vert =\sqrt{\Vert x_n\Vert ^2+\Vert y_n\Vert ^2}\le \sqrt{1+\frac{1}{k^2}}$. Let $w_n=\left( \begin{array}{c} -x_n \\ y_n \\ \end{array} \right)$ and $u_n=\frac{w_n}{\Vert w_n\Vert }$, then $\Vert u_n\Vert =1$ and $u_n\in N(M_C)^{\perp }$. It follows from $M_C\left( \begin{array}{c} -x_n \\ y_n \\ \end{array} \right) \rightarrow 0$ that $M_C(u_n)\rightarrow 0$ ($n\rightarrow \infty$) and hence $M_C|_{N(M_C)^{\perp }}$ is not left invertible. From $M_C|_{N(M_C)^{\perp }}$ is injective, we get that $R(M_C|_{N(M_C)^{\perp }})$ is not closed. It follows that $M_C$ is not an upper semi-Fredholm operator since $R(M_C)=R(M_C|_{N(M_C)^{\perp }})$.

3 The Proof of the Main Results

Proof of Theorem 1

$``\Leftarrow "$ From Lemma 3 and Corollary 1, the result is obvious.

$``\Rightarrow "$ If $M_C$ is an upper semi-Fredholm operator, then A is an upper semi-Fredholm operator and $n(A)+n(B)\ge n(M_C)>0$. If $d(A)=\infty$, then the condition (2) holds. If $d(A)<\infty$, then A is a Fredholm operator and hence B is an upper semi-Fredholm operator ([13, Theorem 2.2]), $\hbox {ind}(A)+\hbox {ind}(B)=\hbox {ind}(M_C)<0$. Therefore, the condition (1) holds.

In the proof of Lemma 1 and Lemma 2, we found that there is a connection between $n(M_C)$ and n(A). Therefore, we give the following corollary.

Corollary 2

Let $A\in B(H)$ and $B\in B(K)$. If A is an upper semi-Fredholm operator with $d(A)=\infty$ and $n(A)+n(B)>0$, then there exists a left invertible operator $C\in B(K,H)$ such that $M_C$ is an upper semi-Fredholm operator, $\hbox {ind}(M_C)<0$, and $n(M_C)=\hbox {max}\{n(A), 1\}$.

In Corollary 2, the opposite is not true. For example, let $A, B\in B(\ell ^2)$ be defined by

$$\begin{aligned} A(x_{1}, x_{2}, x_{3}, \cdots )=(x_{2}, x_{3}, x_{4}, \cdots ), B(x_{1}, x_{2}, x_{3}, \cdots )=(0, 0, x_{1}, x_{2}, \cdots ). \end{aligned}$$

It is clear that $M_C$ is upper semi-Fredholm and $\hbox {ind}(M_C)=-1<0$ for all $C\in B(\ell ^2)$. From $n(A)\le n(M_C)\le n(A)+n(B)$, we can get that $n(M_C)=n(A)=\hbox {max}\{n(A), 1\}$. But $d(A)<\infty$. Therefore, Corollary 2 is just a sufficient condition. Theorem 2, as a necessary and sufficient condition for this connection, plays an important role in the proof of Theorem 3. Next, we will give the proof of Theorem 2.

Proof of Theorem 2

$``\Rightarrow "$ It is obvious that A is an upper semi-Fredholm operator, $n(A)+n(B)>0$. If $d(A)=\infty$, then the condition (3) holds. Without loss of generality, we suppose that $d(A)<\infty$, then A is a Fredholm operator and hence B is an upper semi-Fredholm operator, $\hbox {ind}(A)+\hbox {ind}(B)=\hbox {ind}(M_C)<0$. We have two cases to complete this proof.

Case 1. $n(A)>0$.

From $n(M_C)=\hbox {max}\{n(A), 1\}$, we obtain that $n(M_C)=n(A)$. By using [13, Corollary 2.6], we can get that $n(B)\le d(A)$. Therefore, (1) holds.

Case 2. $n(A)=0$.

From $n(M_C)=\hbox {max}\{n(A), 1\}$, we obtain that $n(M_C)=1$. If $n(B)>d(A)+1$, we suppose that $n(B)=m$, $d(A)=k$ and hence $m-1>k$. Let $\{e_i\}_{i=1}^{m}$ be an orthonormal basis of N(B) and put $Ce_i=\alpha _i+\beta _i$, where $\alpha _i\in R(A)$ and $\beta _i\in R(A)^{\perp }$, $i=1,2,\cdots , m$. So, $\{Ce_1-\alpha _1, \cdots , Ce_{m-1}-\alpha _{m-1}\}\subseteq R(A)^{\perp }$. It follows from $m-1>k$ that $\{Ce_1-\alpha _1, \cdots , Ce_{m-1}-\alpha _{m-1}\}$ is a linearly dependent set. Therefore, there exists $\{a_i\}_{i=1}^{m-1}\subseteq \mathbb {C}$ such that $a_j\ne 0$ for some j and $\sum \limits _{i=1}^{m-1}a_iCe_i=\sum \limits _{i=1}^{m-1}a_i\alpha _i\in R(A)$. Put $Ax_1=\sum \limits _{i=1}^{m-1}a_i\alpha _i$, then we get that $\left( \begin{array}{c} -x_1 \\ a_1e_1+\cdots +a_{m-1}e_{m-1} \\ \end{array} \right) \in N(M_C)$ and $\left( \begin{array}{c} -x_1 \\ a_1e_1+\cdots +a_{m-1}e_{m-1} \\ \end{array} \right) \ne 0.$ We may directly assume that $a_1\ne 0$ for convenience. Similar to the discussion above, we could see that $\{Ce_2-\alpha _2, \cdots , Ce_{m}-\alpha _{m}\}$ is linearly dependent, so there exists $\{b_i\}_{i=2}^{m}\subseteq \mathbb {C}$ such that $b_j\ne 0$ for some j and $\sum \limits _{i=2}^{m}b_iCe_i=\sum \limits _{i=2}^{m}b_i\alpha _i\in R(A)$. Put $Ax_2=\sum \limits _{i=2}^{m}b_i\alpha _i$, then $\left( \begin{array}{c} -x_2 \\ b_2e_2+\cdots +b_{m}e_{m} \\ \end{array} \right) \in N(M_C)$ and $\left( \begin{array}{c} -x_2 \\ b_2e_2+\cdots +b_{m}e_{m} \\ \end{array} \right) \ne 0$.

We claim that $\left( \begin{array}{c} -x_1 \\ a_1e_1+\cdots +a_{m-1}e_{m-1} \\ \end{array} \right)$ and $\left( \begin{array}{c} -x_2 \\ b_2e_2+\cdots +b_{m}e_{m} \\ \end{array} \right)$ are linearly independent.

In fact, we suppose that $k_1\left( \begin{array}{c} -x_1 \\ a_1e_1+\cdots +a_{m-1}e_{m-1} \\ \end{array} \right) +k_2\left( \begin{array}{c} -x_2 \\ b_2e_2+\cdots +b_{m}e_{m} \\ \end{array} \right)$ $=0,$ then

$$\begin{aligned} k_1a_1e_1+(k_1a_2+k_2b_2)e_2+\cdots +(k_1a_{m-1}+k_2b_{m-1})e_{m-1}+k_2b_me_m=0. \end{aligned}$$

Since $\{e_i\}_{i=1}^{m}$ is linearly independent and $a_1\ne 0$, we get that $k_1=0$ and $k_2b_2=\cdots =k_2b_m=0$. From $b_j\ne 0$ we can get $k_2=0$, then the claim is proved. It is in contradiction with $n(M_C)=1$. Therefore, $n(B)\le d(A)+1$.

$``\Leftarrow "$ If the condition (3) holds, from Corollary 2 the result is obvious.

If the condition (1) holds, then $M_C$ is an upper semi-Fredholm operator with $\hbox {ind}(M_C)<0$ for all left invertible operators $C\in B(K,H)$. Since $n(B)\le d(A)$, there is a closed subspace $M\subseteq R(A)^{\perp }$ such that $\hbox {dim}N(B)=\hbox {dim}M$. Therefore, there exists an isometric invertible linear operator $T_2:N(B)\rightarrow M$. Since A and B are not compact, $\hbox {dim}N(B)^{\perp }=\hbox {dim}R(B^*)=\infty =\hbox {dim}R(A)$. Then, there exists an isometric invertible linear operator $T_1:N(B)^{\perp }\rightarrow R(A)$. Define $C:K\rightarrow H$ by

$$\begin{aligned} C=\left( \begin{array}{cc} T_1 &{} 0 \\ 0 &{} T_2 \\ \end{array} \right) :\left( \begin{array}{c} N(B)^{\perp } \\ N(B) \\ \end{array} \right) \rightarrow \left( \begin{array}{c} R(A) \\ R(A)^{\perp } \\ \end{array} \right) . \end{aligned}$$

It is clear that C is left invertible. Similar to the (2) of Case 2 in Lemma 2, $n(M_C)=n(A)$ is trivial.

If the condition (2) holds, then $n(B)>0$. We take $e_1\in N(B)$ with $\Vert e_1\Vert =1$ and then we can extend it into an orthonormal basis of K, written as $\{e_{n}\}_{n=1}^{\infty }$. Put $M=\hbox {span}\{e_1\}$ and take $u_1=Aw_1\in R(A)$ with $\Vert u_1\Vert =1$. Let $N=\hbox {span}\{u_1\}$, then there is an isometric invertible linear operator $T_2:M\rightarrow N$ such that $T_2e_1=u_1=Aw_1$. Since $n(B)\le d(A)+1$, there is a closed subspace $F\subseteq R(A)^{\perp }$ such that $\hbox {dim}(N(B)\ominus M)=\hbox {dim}F$. Then there is an isometric invertible linear operator $T_3:N(B)\ominus M\rightarrow F$. Since $\hbox {dim}N(B)^{\perp }=\infty =\hbox {dim}(R(A)\ominus N)$, there is an isometric invertible linear operator $T_1:N(B)^{\perp }\rightarrow R(A)\ominus N$. Define a left invertible operator $C:K\rightarrow H$ by

$$\begin{aligned} C=\left( \begin{array}{ccc} T_1 &{} 0 &{} 0 \\ 0 &{} T_2 &{} 0 \\ 0 &{} 0 &{} T_3 \\ \end{array} \right) :\left( \begin{array}{c} N(B)^{\perp } \\ M \\ N(B)\ominus M \\ \end{array} \right) \rightarrow \left( \begin{array}{c} R(A)\ominus N \\ N \\ R(A)^{\perp } \\ \end{array} \right) . \end{aligned}$$

Let $\left( \begin{array}{c} x \\ y \\ \end{array} \right) \in N(M_C)$. Then, $Ax+Cy=0$ and $By=0$. Let $y=k_1e_1+\alpha _1$, where $k_1\in \mathbb {C}$ and $\alpha _1\in N(B)\ominus M$. It follows from $Ax+Cy=Ax+T_2(k_1e_1)+T_3\alpha _1=A(x+k_1w_1)+T_3\alpha _1$ that $A(x+k_1w_1)=-T_3\alpha _1\in R(A)\cap R(A)^{\perp }=\{0\}$. From $T_3$ is invertible and $n(A)=0$, we can get that $\alpha _1=0$ and $x=-k_1w_1$. Hence, $N(M_C)\subseteq \hbox {span}\left\{ \left( \begin{array}{ccc} -w_1 \\ e_1 \\ \end{array}\right) \right\}$. The inclusion “$\supseteq$" is obvious. Then $N(M_C)=\hbox {span}\left\{ \left( \begin{array}{ccc} -w_1 \\ e_1 \\ \end{array}\right) \right\}$. Therefore, $n(M_C)=1$ and $n(M_C)=\hbox {max}\{n(A), 1\}$. $\square$

Proof of Theorem 3

$``\Rightarrow "$ It is clear that A is upper semi-Fredholm, then we complete this proof from two steps.

Step 1. $d(A)<\infty$.

Now, A is a Fredholm operator, B is an upper semi-Fredholm operator, and $n(A)+n(B)<d(A)+d(B)$. Then, we have two cases:

Case 1. $n(A)>0$.

It is clear that $n(M_C)=\hbox {max}\{n(A), 1\}=n(A)$. We claim that $n(B)=0$. If not, we suppose that $n(B)=n>0$ and $\{e_i\}_{i=1}^{n}$ is an orthonormal basis of N(B). From $\hbox {dim}R(A)=\infty$ we can get that there exists a closed subspace $M\subseteq R(A)$ such that $\hbox {dim}N(B)=\hbox {dim}M$. It is easy to get $\hbox {dim}N(B)^{\perp }=\infty =\hbox {dim}(H\ominus M)$. Therefore, there exist two isometric invertible linear operators $T_1:N(B)\rightarrow M$ and $T_2:N(B)^{\perp }\rightarrow H\ominus M$. Define $C:K\rightarrow H$ by

$$\begin{aligned} C=\left( \begin{array}{cc} T_1 &{} 0 \\ 0 &{} T_2 \\ \end{array} \right) :\left( \begin{array}{c} N(B) \\ N(B)^{\perp } \\ \end{array} \right) \rightarrow \left( \begin{array}{c} M \\ H\ominus M \\ \end{array} \right) . \end{aligned}$$

It is clear that $C\in B(K,H)$ is invertible and so is left invertible. Let $T_1e_i=Au_i$, $i=1, 2, \cdots , m$, then $\left\{ \left( \begin{array}{ccc} -u_i \\ e_i \\ \end{array}\right) \right\} _{i=1}^{n}\subseteq N(M_C)$ is linearly independent. Supposed that $\{x_i\}_{i=1}^m$ is an orthonormal basis of N(A) and it follows that $\left\{ \left( \begin{array}{ccc} x_i \\ 0 \\ \end{array}\right) \right\} _{i=1}^{m}\subseteq N(M_C)$. It is easy to find that $\left\{ \left( \begin{array}{ccc} x_i \\ 0 \\ \end{array}\right) \right\} _{i=1}^{m}$, $\left\{ \left( \begin{array}{ccc} -u_i \\ e_i \\ \end{array}\right) \right\} _{i=1}^{n}$ are linearly independent. Therefore, $n(M_C)\ge n(A)+n(B)>n(A)$, a contradiction with $n(M_C)=n(A)$. Then, $n(B)=0$ and the condition (1) holds.

Case 2. $n(A)=0$.

Now, A is left invertible, $n(M_C)=1$ and $n(B)>0$. We claim that $n(B)=1$. If not, then $n(B)\ge 2$. Suppose that $n(B)=k$ and $\{e_i\}_{i=1}^k$ is an orthonormal basis of N(B). Similar to the proof of Case 1, we can get that $n(M_C)\ge n(B)\ge 2$, a contradiction. From Lemma 5, we obtain that $n(B)>d(A)$. Hence, $d(A)=0$ and A is invertible. The condition (2) holds.

Step 2. $d(A)=\infty$.

From Lemma 4 we can get that $n(A)>0$. It follows from $n(M_C)=\hbox {max}\{n(A), 1\}$ that $n(M_C)=n(A)$. By Lemma 7 we know that B is an upper semi-Fredholm operator. Similar to the proof of Case 1, we can get that $n(B)=0$. Therefore, the condition (1) holds.

$``\Leftarrow "$ If the condition (1) holds, it is obvious that $M_C$ is an upper semi-Fredholm operator and $\hbox {ind}(M_C)<0$ for all left invertible operators $C\in B(K,H)$. From $0<n(A)\le n(M_C)\le n(A)+n(B)$ we can get that $n(M_C)=\hbox {max}\{n(A), 1\}$.

If the condition (2) holds, it is clear that $M_C$ is an upper semi-Fredholm operator, $n(M_C)\le n(B)=1$ and $\hbox {ind}(M_C)<0$ for all left invertible operators $C\in B(K,H)$. If there exists a left invertible operator $C_0\in B(K,H)$ such that $n(M_{C_0})=0$, then B is left invertible, since $M_{C_0}$ is left invertible and A is invertible, which contradicts with $n(B)=1$. Therefore, $n(M_C)=1$ for all left invertible operators $C\in B(K,H)$, which implies $n(M_C)=\hbox {max}\{n(A), 1\}$. $\square$

Based on the above auxiliary results and the proof process of Theorem 3, we can easily obtain the following corollary:

Corollary 3

Let $A\in B(H)$ and $B\in B(K)$. Then, $M_C$ is an upper semi-Fredholm operator, $n(M_C)>0$ and $\hbox {ind}(M_C)<0$ for all left invertible operators $C\in B(K,H)$ if and only if one of the following statements holds:

• (1) A and B are both upper semi-Fredholm with $n(A)>0$ and $n(A)+n(B)<d(A)+d(B)$;

• (2) A is left invertible, B is upper semi-Fredholm with $n(B)>d(A)$ and $n(A)+n(B)<d(A)+d(B)$.