Let $A, B$ be self-adjoint positive operators, $C, D$ be orthogonal operators on an Euclidean space, and $AC = DB$.
Prove that $A=B \Leftrightarrow AC$ is normal.
I know many properties of self-adjoint and orthogonal operators and tried to apply them but with no success. Could you please help me?
Recall that $AC$ is normal iff $(AC)^\ast (AC) = (AC)(AC)^\ast$. Since $AC = DB$ holds, this is equivalent to $(DB)^\ast (DB) = (AC)(AC)^\ast$. This holds iff $B^\ast D^\ast DB = ACC^\ast A$. But $C$ and $D$ are orthogonal, hence it follows that $B^\ast B = A A^\ast$. Since $A$ and $B$ are self-adjoint, this simplifies to $A^2 = B^2$. Finally, this implies $A = B$ since $A^2$ and $B^2$ are positive and therefore have a unique square root.