The reversibility of the adjoint operator Let $H$ - a Hilbert space, operator $A \in \mathcal{B}$ and $A$ is reversible. How to prove the fact that $(A^*)^{-1} = (A^{-1})^*$? 
 A: Simply notice
$\langle Ax,y\rangle=\langle x,A^{*}y\rangle=\langle A^{-1}Ax,A^{*}y\rangle=\langle Ax,(A^{-1})^{*}A^{*}y\rangle$.
Thus $(A^{-1})^{*}A^{*}=I\Rightarrow (A^{-1})^{*}=(A^{*})^{-1}$
A: For any $v,w \in H$ you get:
$(v,w) = (T T^{-1}v,w) = (T^{-1}v,T^*w) = (v,(T^{-1})^*T^*w)$,
yielding $(T^{-1})^*T^* = \mathrm{Id}$.
A: Well for any operator $T$, $T^*$ is defined by the equation $(T^*u,v)=(u,Tv)$ for all $u,v$. So, in particular, $(A^{-1})^*$ will be the operator satisfying $((A^{-1})^*u,v)=(u,A^{-1}v).$ Putting $v=Aw$, we have $$(u,w)=((A^{-1})^*u,Aw)=(A^*(A^{-1})^*u,w).$$
Thus $A^*(A^{-1})^*=1$, so $(A^{-1})^*=(A^*)^{-1}.$
A: ziggurism shows that $(A^{-1})^*$ is a right inverse of $A^*$, while Konstantin and Basanta R. Pahari show that it is a left inverse. As the following (well known) example shows, but both of these are needed in an infinite dimensional Hilbert space. Take $H$ to be the Hilbert space of square-summable real sequences. For $x=(x_1,x_2,\ldots)\in H$ define $Lx = (x_2,x_3,\ldots)$ and $Rx:=(0,x_1,x_2,\ldots)$.  Both $L$ and $R$ are bounded linear operators on $H$, $L$ is onto but not one-to-one, while $R$ is one-to-one but not onto. Finally, $LRx =x$ for all $x\in H$. That is, $L$ is a left inverse of $R$. But $RLx=(0,x_2,x_3,\ldots)$ which is not the same as $x$ when $x_1\not=0$. 
