A Hilbert space operator $A\in B\left(\mathcal{H}\right)$ is left $(X,m)$-invertible by $B\in B\left(\mathcal{H}\right)$ (resp., $B\in B\left(\mathcal{H}\right)$ is an $(X,m)$-adjoint of $A\in B\left(\mathcal{H}\right)$) for some operator $X\in B\left(\mathcal{H}\right)$ if ${\mathrm{△}}_{B,A}^m\left(X\right)=\sum _{j=0}^m(-1{)}^j(\begin{array}{c}m\\ j\end{array})B^{m-j}X{A}^{m-j}=0$ (resp., ${\delta }_{B,A}^m\left(X\right)=\sum _{j=0}^m(-1{)}^j(\begin{array}{c}m\\ j\end{array})B^{(m-j)}X{A}^j=0$). No Drazin invertible operator $A\in B\left(\mathcal{H}\right)$, 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 invertible operators A, it is seen that the existence of an X acts as a conduit for implications ${\mathrm{△}}_{B,A}\left(X\right)=0⟹{\delta }_{C,A}^m\left(X\right)=0$, where the pair $is$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 ${\mathrm{△}}_{A_d^{\ast },A}^m\left(X\right)=0={\mathrm{△}}_{B_d^{\ast },B}^n\left(Y\right)$ implies ${\delta }_{A^{\ast }{B}^{\ast },AB}^{m+n-1}\left(XY\right)=0={\delta }_{A^{\ast }+B^{\ast },A+B}^{m+n-1}\left(XY\right)$.

希尔伯特空间算子$\mathrm{一个}\in 乙\left(\mathcal{H}\right)$离开了$(X,米)$-可逆的$乙\in 乙\left(\mathcal{H}\right)$（分别，$乙\in 乙\left(\mathcal{H}\right)$是一个$(X,米)$- 伴随的$\mathrm{一个}\in 乙\left(\mathcal{H}\right)$) 对于某些运算符$X\in 乙\left(\mathcal{H}\right)$如果${\mathrm{△}}_{乙,\mathrm{一个}}^{米}\left(X\right)=\sum _{j=0}^{米}(-1{)}^j(\begin{array}{c}米\\ j\end{array}){乙}^{米-j}X{\mathrm{一个}}^{米-j}=0$（分别，${\delta }_{乙,\mathrm{一个}}^{米}\left(X\right)=\sum _{j=0}^{米}(-1{)}^j(\begin{array}{c}米\\ j\end{array}){乙}^{(米-j)}X{\mathrm{一个}}^j=0$）。无 Drazin 可逆算子$\mathrm{一个}\in 乙\left(\mathcal{H}\right)$, 与 Drazin 逆${\mathrm{一个}}_d$, 可以留下$(我,米)$-可逆（等效地，m -可逆）由它的伴随或它的 Drazin 逆或它的 Drazin 逆的伴随。对于 Drazin 可逆算子A，可以看出X的存在充当了蕴涵的管道${\mathrm{△}}_{乙,\mathrm{一个}}\left(X\right)=0⟹{\delta }_{C,\mathrm{一个}}^{米}\left(X\right)=0$, 其中对$\mathrm{一世}s$任何一个$(\mathrm{一个},{\mathrm{一个}}_d)$或者$({\mathrm{一个}}_d,\mathrm{一个})$或者$({\mathrm{一个}}^{*},{\mathrm{一个}}_d^{*})$或者$({\mathrm{一个}}_d^{*},{\mathrm{一个}}^{*})$. 反向暗示失败。假设一定的交换律条件，可以看出${\mathrm{△}}_{{\mathrm{一个}}_d^{*},\mathrm{一个}}^{米}\left(X\right)=0={\mathrm{△}}_{{乙}_d^{*},乙}^n\left(是\right)$暗示${\delta }_{{\mathrm{一个}}^{*}{乙}^{*},\mathrm{一个}乙}^{米+n-1}\left(X是\right)=0={\delta }_{{\mathrm{一个}}^{*}+{乙}^{*},\mathrm{一个}+乙}^{米+n-1}\left(X是\right)$.