Trace distance remaining invariant under Unitary transformations I'm working on this exercise, and I'm stuck in I believe the last line or two of the proof. If I could get some help I would be very appreciative! Note that both $\rho$ and $\sigma$ are operators and $U$ is a unitary matrix.
I have to show that $D(U\rho U^\dagger, U,\sigma U^\dagger) = D(\sigma, \rho)$, where $D(\sigma, \rho) = \frac{1}{2}$Tr $|\sigma-\rho|$.
Starting off, $D(U\rho U^\dagger, U,\sigma U^\dagger) = \frac{1}{2}$ Tr |$U\rho U^\dagger, U,\sigma U^\dagger$| = $\frac{1}{2}$ Tr |$U(\rho-\sigma)U^\dagger$|.
Now, because |$UAU^\dagger$| = $U$|$A$|$U^\dagger$ for an operator $A$, we have the distance above reduces to 
$\frac{1}{2}$ Tr |$U(\rho-\sigma)U^\dagger$| =  $\frac{1}{2}$ Tr$(U$|$\rho-\sigma$|$U^\dagger)$.
This is where I get stuck. I do not know how to conclude that this expression is equivalent to $D(\rho,\sigma)$.
Again, any clarification on the last bit (or any pointing out of any mistakes I made) would be greatly appreciated.

Using the identity $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$, which holds whenever the products $AB$ and $BA$ are defined, one gets for any square matrix $X$ and invertible matrix $U$
$$\operatorname{Tr}(UXU^{-1})=\operatorname{Tr}(XU^{-1}U)=\operatorname{Tr}(X).$$
Your case reduces to this one, as $U^{-1}=U^\dagger$ for a unitary matrix $U$.
