Find the maximum of $\operatorname{Tr}(RZ)$ over all orthogonal matrices $R$

Question

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?

Answer 1

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.

Answer 2

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 $U=V^\dagger$.