Find the maximum of $\operatorname{Tr}(RZ)$ over all orthogonal matrices $R$ Say I have the following maximization.
$$ \max_{R: R^T R=I_n} \operatorname{Tr}(RZ),$$
where $R$ is an $n\times n$ orthogonal transformational, and the SVD of $Z$ is written as $Z = USV^T$. 
I'm trying to find the optimal $R^*$ which intuitively I know is equal to $VU^T$ where
$$\operatorname{Tr}(RZ)=\operatorname{Tr}(VU^T USV^T)=\operatorname{Tr}(S).$$
I know this is the max since it is the sum of all the singular values of $Z$. However, I'm having trouble coming up with a mathematical proof justifying my intuition.
Any thoughts? 
 A: Let $A=\sqrt{S}$, and equip the space of $n\times n$ real matrices with the usual Euclidean scalar product. Then
$$\hbox{Tr}(RZ)= \hbox{Tr}(RUA^2V^T)=\hbox{Tr}((RUA)(VA)^T)=\langle RUA,VA\rangle$$
By the Cauchy-Schwarz inequality, we get
$$\hbox{Tr}(RZ)\leq \Vert RUA \Vert_2 \Vert VA \Vert_2= \Vert A \Vert_2 \Vert A \Vert_2
=\hbox{Tr}(AA^T)=\hbox{Tr}(S)$$
where we used the invariance of the $\Vert \cdot \Vert_2 $ under orthogonal transformations.
the converse inequality, is proved by choosing $R=VU^T$, and we are done.
A: I'll show how to prove the more general case with complex matrices: find the maximum of $\operatorname{Tr}(UZ)$ over all unitaries $U$:
$$\max_{U: U^\dagger U=I}|\operatorname{Tr}(UZ)|.$$
Leveraging the polar decomposition, we know that $Z$ can be always written as $Z=VP$ for some unitary $V$ and positive semi-definite $P\ge0$.

Because the product of unitaries is another unitary, this observation reduces the problem to the following:
$$\max_{U: U^\dagger U=I}\operatorname{Tr}(UP).$$
Moreover, observe that the $P$ in the polar decomposition equals $\sqrt{A^\dagger A}$. Denoting with $\{u_k\}$ the eigenvectors of $P$, and $s_k$ the eigenvalues of $P$ (i.e. the singular values of $Z$), we have
$P=\sum_k s_k u_k u_k^*,$
and therefore for every unitary $U$ there is an orthonormal basis $\{v_k\}$ such that
$$UP=\sum_k s_k v_k u_k^*.$$
It follows that the trace reads
$\operatorname{Tr}(UP)=\sum_k s_k \langle u_k,v_k\rangle$, and taking the absolute value,
$$
  \lvert\operatorname{Tr}(UP)\rvert=\left\lvert\sum_k s_k
  = |\langle u_k,v_k\rangle|\right\rvert
  \le \sum_k s_k \lvert\langle u_k,v_k\rangle\rvert
  \le\sum_k s_k
  = \operatorname{Tr}P.
\tag X
$$
therefore if there is a $U$ such that $\lvert\operatorname{Tr}(UP)\rvert=\sum_k s_k$, that ought to be maximum we are looking for.
But finding this $U$ is trivial at this point: just use $U=V^\dagger$.
