The image-kernel (P,Q)-inverse of an linear operator.

Ponente(s): Slavisa Djordjevic .

The idea of ``prescribing" was considered by D. Djordjevi\'c and Wei in the setting of rings where generalized inverses with certain idempotents were studied (\cite{DWei}). However, some origins of these ideas can be traced back at least to Bott-Duffin in 1953 (\cite{Duffin}). M.P. Drazin himself considered that Both-Daffins ideas has to be recognized as a direct precursor of the (b,c)-inverse {with b,c elements in a semigroup} and, consequently, of most other known types of uniquely-defined outer generalized inverses (\cite{Drazin})}.

Throughout this talk, $X$ and $Y$ denote a Banach spaces and $B(X,Y)$ denote the space of bounded linear operators from $X$ to $Y$. If $X=Y$ then $B(X,X)=B(X)$ will be a Banach algebra. For an operator $T\in B(X,Y)$, we will denote by $R(T)$ the range of $T$ and by $N(T)$ the nullspace of $T$. An operator $S\in B(Y,X)$ is called a (bounded) inverse of $T\in B(X,Y)$ if $TS=I_{Y}$ and $ST=I_{X}$, where $I_{Y},I_{X}$ are the identities in the respective spaces. If there is no confusion, we use $I$ to denote $I_ {X}$ or $I_ {Y}$.

Let $T\in B(X,Y)$. An operator $S\in B(Y,X)$ is an inner inverse for $T$ if $$TST=T$$ and, in this case, we say that $T$ is inner regular. $S$ is an outer inverse for $T$ if $$STS=S$$ and, we say that $T$ is outer regular. Recall that the inner inverse is not unique even prescribing its range and null space. On the other hand, for the outer inverse we have uniqueness if we prescribe the range and null space, although in general not be unique (for more details see \cite{Djordjevic}). {We remark, several classes of generalized inverses found in the literature are special classes of outer inverses with prescribed range and null space, hence arises the necessity of knowing how to translate the properties that "some a kind" generalized inverse could have with these characteristics in the space of bounded linear operators}.

An operator $P\in B(X)$ is a projection if $P=P^{2}$. Therefore, if $S$ is an outer or inner inverse of $T$, then $ST$ and $TS$ are projections.

If $X_{1}$ and $X_{2}$ are subspaces of $X$ such that $X_{1}\cap X_{2}=\{0\}$ and $X_{1}+X_{2}=X$, we say that $X_{1}$ is complemented with $X_{2}$. Additionally, if $X_{1}$ and $X_{2}$ are closed, then we write $X=X_{1}\oplus X_{2}$.

Recall $T\in B(X)$ is group invertible if there exist $S\in B(X)$ such that $TST=T$, $STS=S$ and $ST=TS$. If this inverse exists then it is will be unique and we write $S=T^{\#}$ for the group inverse of $T$. We have that $T$ is group invertible if and only if $X=R(T)\oplus N(T)$ \cite[Theorem 2.1.1 and Theorem 2.2.1]{WG}.

Is well-know that $T\in B(X,Y)$ is inner invertible, if and only if $N(T)$ and $R(T)$ are closed and complemented spaces of $X$ and $Y$, respectively \cite[Corollary 1.1.5]{Djordjevic}. More over, $T$ has the following matrix form: $$T=\begin{bmatrix} T_{1} & 0 \\ 0 & 0 \end{bmatrix}: \begin{bmatrix} N \\ N(T) \end{bmatrix}\to \begin{bmatrix} R(T) \\ M \end{bmatrix},$$ with $T_1\in B(N,R(T))$ invertible where $M,N$ are closed subspaces of $Y$ and $X$ respectively such that $Y=R(T)\oplus M$ and $X=N\oplus N(T)$. If $S$ is any inner inverse of $T$ such that $R(ST)=N$ and $N(TS)=M$, then $S$ has the following matrix form: $$S=\begin{bmatrix} T_{1}^{-1} & 0 \\ 0 & S_{2} \end{bmatrix}: \begin{bmatrix} R(T) \\ M \end{bmatrix}\to \begin{bmatrix} N \\ N(T) \end{bmatrix},$$ with $S_{2}\in B(M,N(T))$ is arbitrary \cite[Theorem 1.1.6]{Djordjevic}\label{base}.

In this talk we introduce a kind of (p,q)-inverse (see \cite{Kantun}) in the context of the space of bounded linear operators, with the purpose of obtaining a unique inverse following the geometric theory of generalized inverses. Also, we present several results using matrix representation techniques for bounded lineal operator on Banach spaces.