A variant of Ex 2(b), Sec 79, Pg 158 from PR Halmos's Finite-Dimensional Vector Spaces:
If $A$ is an arbitrary linear transformation on an $n$-dimensional complex inner product space $V$, does it follow that the transformations $A^*A$ and $AA^*$ are always unitarily equivalent? In other words, does there exist some unitary transformation $U$ on $V$ such that $A^*A = U^{-1}AA^*U$?
My approach thus far: it is clear that each of $A^*A$ and $AA^*$ are self-adjoint [because, for example, $(A^*A)^* = A^*A$]. That is, $A^*A$ is diagonalizable w.r.t. a suitable orthonormal basis in $V$, and so is $AA^*$. Also, the condition $A^*A = U^{-1}AA^*U$ means that the transformations $A^*A$ and $AA^*$ share one and the same matrix w.r.t. possibly different orthonormal bases in $V$, meaning in turn that $A^*A$ and $AA^*$ share all their eigenvalues. I'm not able to progress further however, and would appreciate guidance. Thanks.
The fact that each of the transformations $A^*A$ and $AA^*$ is self-adjoint implies that each is diagonalizable w.r.t. some orthonormal basis. Also, each transformation shares its all its non-zero Eigenvalues (if any) with the other. It follows that the diagonalized forms of $A^*A$ and $AA^*$ share their diagonal (including the zero entries if any), if the underlying orthonormal bases are suitably ordered. In other words, $A^*A$ and $AA^*$ have the same matrix w.r.t. suitable orthonormal bases. It follows (from the results around similarity of transformations and orthonormal basis change using isometry) that $A^*A = U^{-1}AA^*U$, i.e., $A^*A$ and $AA^*$ are unitarily equivalent.