I have a slightly weird linear algebra problem that I am unable to solve. Let's consider a given product $UC$ of matrices, where $U$ is diagonalizable and $C$ is symmetric and positive definite. We assume that $UC$ is symmetric, i.e. $UC = (UC)^\top = CU^\top$.
Now let $V$ be another matrix diagonalizable in the same basis as $U$, so in particular $U$ and $V$ commute. (It is OK to assume that we are given the diagonalizations $U = SD_US^{-1}$ and $V = SD_VS^{-1}$, and we can even assume that $U$ and $V$ are both invertible).
Now I would like to prove that $VC$ is also symmetric, i.e. $VC = CV^T$.
I tried using commutativity $UV=VU$ (for example I tried $UVC = VUC = VCU^\top$, leading nowhere), but I haven't been able to come up with a way to make it work
I have been able to show this for the special case of $V = U^{-1}$: $C^{-1}U = C^{-1}UCC^{-1} = C^{-1}CU^TC^{-1} = U^TC^{-1}$, and thus $U^{-1}C = (C^{-1}U)^{-1} = (U^TC^{-1})^{-1} = CU^{-T}$.
Let $p(x)=a_mx^m+\ldots +a_1x+a_0$ and $V=p(U).$ Then $p(U)=p(SD_US^{-1})=Sp(D_U)S^{-1},$ because $U^m=(SD_US^{-1})^m=SD_U^mS^{-1}.$ Hence $p(U)$ depends only on the values of $p$ on the set $\Lambda$ of eigenvalues of $U.$ Next $$U^mC=U^{m-1}CU^T=\ldots =UC(U^{m-1})^T=C(U^m)^T$$ hence $VC=p(U)C=Cp(U)^T=CV^T.$
Let $f:\Lambda \to \mathbb{R}.$ Define $f(U)=Sf(D_U)S^{-1},$ where $f(D_U)$ is the diagonal matrix with entries $f(\lambda_k).$ There exists a polynomial $p$ of degree $\#\Lambda$ such that $f(\lambda_k)=p(\lambda_k).$ Then $p(U)=f(U),$ i.e. it suffices to restrict to polynomials, when considering $f(U).$