I'm looking into unital *-algebras and I'm trying to understand if the following is true:
If $A, B$ are elements of the algebra such that $AA^* = BB^*$ then there exists a unitary element $U$ such that
$$A U^* = B U$$
I know that this is true for non-necessarily-commutative C*-algebra, but it seems to me that the proofs use spectral theory and singular value decomposition, which do not seem to arise without the C*-condition.
The *-algebra may not be even commutative (think about the algebra of polynomials with complex matrix coefficients equipped with transpose conjugation).
Do you have any proof/counterexample for this?
Without closure, there could be many things missing from those that make C$^*$-algebras useful.
Consider $\mathbb C[t]$ (complex polynomials of a real variable) as your algebra. Since unitaries are invertible, it is immediate that the only unitaries are the scalars, that is the constant polynomials of absolute value $1$. Let $$ p=1+it,\qquad\qquad q=1-it. $$ Then $$ pp^*=|1+it|^2=1-t^2=|1-it|=qq^*. $$ But $p$ and $q$ are not scalar multiples of each other, so there cannot be any unitaries $u,v$ with $pu=qv$.