论文标题
Hilbert太空运营商的Drazin逆和M-Selfadjointness的伴随左M的可逆性
Left m-invertibility by the adjoint of Drazin inverse and m-selfadjointness of Hilbert space operators
论文作者
论文摘要
hilbert太空运营商$ a \ in \ b $留下$（x，m）$ - $ b \ in \ b $可逆（resp。，$ b \ in \ b $是$（x，m）$ - $ a \ in \ in \ b $的$ a \ in \ b $）用于某些操作员$ x \ in \ b $ $ \ triangle_ {b，a}^m（x）= \ sum_ {j = 0}^m（-1）^j \ left（\ begin {array} {array} {clcr} m \\ j \ end end {array} {array} \ right） $Δ_{b，a}^m（x）= \ sum_ {j = 0}^m（-1）^j \ left（\ begin {array} {clcr} {clcr} m \\ j \ end {array} \ right） 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.对于drazin nirverrtible运算符$ a $，可以看出，$ x $的存在是含义$ \ triangle_ {b，a}（x）= 0 \longrightArrowΔ^m_ { $（a^*，a^*_ d）$或$（a^*_ d，a^*）$。反向含义失败。假设有一定的通勤条件，可以看出$ \ triangle_ {a^*_ d，a}^m（x）= 0 = \ triangle^n_ {b^*_ d，b} $δ^{m+n-1} _ {a^*b^*，ab}（xy）= 0 =δ^{m+n-1} _ {a^*+b^*，a+b}（xy）$。
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 $\triangle_{B,A}^m(X)=\sum_{j=0}^m(-1)^j\left(\begin{array}{clcr}m\\j\end{array}\right)B^{m-j}XA^{m-j}=0$ (resp., $δ_{B,A}^m(X)=\sum_{j=0}^m(-1)^j\left(\begin{array}{clcr}m\\j\end{array}\right)B^{(m-j)}XA^j=0$). 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 $\triangle_{B,A}(X)=0\Longrightarrow δ^m_{C,A}(X)=0$, where the pair $(B,C)=$ either $(A,A_d)$ or $(A_d,A)$ or $(A^*,A^*_d)$ or $(A^*_d,A^*)$. Reverse implications fail. Assuming certain commutativity conditions, it is seen that $\triangle_{A^*_d,A}^m(X)=0=\triangle^n_{B^*_d,B}(Y)$ implies $δ^{m+n-1}_{A^*B^*,AB}(XY)=0=δ^{m+n-1}_{A^*+B^*,A+B}(XY)$.
