Trace norm is invariant under conjugation by an isometry

Question

How would I go about showing that for the trace norm of a matrix, given by $||M||_1=Tr(\sqrt{M^{\dagger}M})$, is invariant under conjugation by an isometry? That is, $||UMU^{\dagger}||_1=||M||_1$. The text I am reading says this is the case because the eigenvalues of the two are the same, but I fail to see even why this is the case. For an isometry I know we have that $U^{\dagger}U=I$, but I don't know how to proceed. (If it helps, the book is discussing Hermitian matrices.

Answer 1

We can start with \begin{align} \|UMU^T\|_1 &= Tr\sqrt{\Big((UMU^T)^T(UMU^T)\Big)} \\ & = Tr\sqrt{\Big(UM^TU^TUMU^T\Big)}\\ &= Tr\sqrt{\Big(UM^TMU^T\Big)}\\ \end{align}

Now, we want to use the following facts. First, $Tr AB = Tr BA$ for any matrices $A$ and $B$. But we have a nasty square root in the way. So let's fix that. We know that $M^TM$ is positive semi-definite. So $$ M^TM = SDS^{-1} $$ for some invertible $S$ and diagonal positive $D$. So we can compute

$$ (US\sqrt{D}S^{-1}U^T)^2 = (US\sqrt{D}S^{-1}U^T)(US\sqrt{D}S^{-1}U^T)=US\sqrt{D}\sqrt{D}S^{-1}U^T=UM^TMU^T $$ So we can finish with \begin{align} Tr\sqrt{\Big(UM^TMU^T\Big)} &= Tr\Big(U\sqrt{M^TM}U^T\Big)\\ &=Tr\Big(U^TU\sqrt{M^TM}\Big) \end{align}