Let $U$ and $V$ be two $n\times n$ unitary matrices such that $UV=\lambda VU$.Then prove that $\lambda^n=1$ Let $U$ and $V$ be two $n\times n$ unitary matrices such that $UV=\lambda VU$.Then prove that $\lambda^n=1$  
Totally stuck on it.Can I get some help?

Unitary is not needed, just non-singular. Eigenvalues are not needed either. Just use the multi-linearity and multiplicativity of the determinant:
$$
  \det(UV)=
  \det(\lambda VU)=\lambda^n\det(VU)=\lambda^n\det(V)\det(U)=\lambda^n\det(UV)
$$
so $\lambda^n=1$ since $\det(UV)\neq0$.
