Left m-invertibility by the adjoint of Drazin inverse and m-selfadjointness of Hilbert space operators

Abstract

A Hilbert space operator $A\in\B$ is left $(X,m)$-invertible by $B\in\B$ (resp., $B\in\B$ is an $(X,m)$-adjoint of $A\in\B$) for some operator $X\in\B$ if ${△}_{B,A}^m(X)={\sum }_{j=0}^m(-1{)}^j\left(\begin{array}{c}m\\ j\end{array}\right)B^{m-j}X{A}^{m-j}=0$ (resp., ${\delta }_{B,A}^m(X)={\sum }_{j=0}^m(-1{)}^j\left(\begin{array}{c}m\\ j\end{array}\right)B^{(m-j)}X{A}^j=0$). No Drazin invertible operator $A\in\B$, with Drazin inverse $A_d$, can be left $(I,m)$-invertible (equivalently, $m$-invertible) by its adjoint or its Drazin inverse or the adjoint of its Drazin inverse. For Drazin inverrtible operators $A$, it is seen that the existence of an $X$ acts as a conduit for implications ${△}_{B,A}(X)=0⟹{\delta }_{C,A}^m(X)=0$, where the pair $(B,C)=$ either $(A,A_d)$ or $(A_d,A)$ or $(A^{\ast },A_d^{\ast })$ or $(A_d^{\ast },A^{\ast })$. Reverse implications fail. Assuming certain commutativity conditions, it is seen that ${△}_{A_d^{\ast },A}^m(X)=0={△}_{B_d^{\ast },B}^n(Y)$ implies ${\delta }_{A^{\ast }{B}^{\ast },AB}^{m+n-1}(XY)=0={\delta }_{A^{\ast }+B^{\ast },A+B}^{m+n-1}(XY)$.

14 pages

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