Prove there exists a self-adjoint transformation $C$ s.t. $CA=B$ if $A$ and $AB$ are self adjoint If $A$ and $B$ are linear transformations such that $A$ and $AB$ are self adjoint and such that $\ker (A)\subset \ker (B)$, then prove there exists a self-adjoint transformation $C$ s.t. $CA=B$.
 A: *

*If $\text{ker}(A)\subset\text{ker}(B)$ then $\text{im}(B^*)\subset\text{im}(A^*)$, that is 
$$
\forall x\ \exists y_x\colon \ B^*x=A^*y_x.
$$
Taking $x=e_k$ and introducing $C^*=[y_{e_1}\ y_{e_2}\ldots y_{e_n}]$ we get $B^*=A^*C^*$ $\Leftrightarrow$  $CA=B$.

*Since $A$ is self-adjoint, we can diagonalize it by a unitary $U$ for simplicity.
$$
D=U^*AU.
$$
Introduce also $\tilde B=U^*BU$ and $\tilde C=U^*CU$. Then 
$$
CA=B\qquad\Leftrightarrow\qquad
\left[\matrix{\tilde C_1 & \tilde C_2\\\tilde C_3 & \tilde C_4}\right]
\left[\matrix{D_1 & 0\\0 & 0}\right]=\left[\matrix{\tilde B_1 & \tilde B_2\\\tilde B_3 & \tilde B_4}\right]
$$
where $D_1$ is invertible (collecting non-zero eigenvalues of $A$).

*Nothing depends on the blocks $\tilde C_2$ and $\tilde C_4$ (multiplied by zero blocks), so we can change them to $\tilde C_2=\tilde C_3^*$ and $\tilde C_4=0$. Further, 
$$
\tilde C_1 D_1=\tilde B_1\qquad\Rightarrow \qquad
\tilde C_1 =\tilde B_1D_1^{-1}=D_1^{-1}\underbrace{(D_1\tilde B_1)}_{\text{self-adjoint}}D_1^{-1}=\tilde C_1^*.
$$
Finally, pick
$$
C=U\left[\matrix{\tilde C_1 & \tilde C_3^*\\\tilde C_3 & 0}\right]U^*.
$$

