Procrustes Problem with Maximization (Instead of Minimization)

Question

The classical (orthogonal) Procrustes problem is to solve the optimization problem $$ \begin{array}{rl} \min&\|\Omega{A}-B\|_F\\ \text{s.t.}&\Omega^\mathrm{T}\Omega=I \end{array} $$ The solution to this problem is well-known. I'm curious about the solution to $$ \begin{array}{rl} \max&\|\Omega{A}-B\|_F\\ \text{s.t.}&\Omega^\mathrm{T}\Omega=I \end{array} $$ (max instead of min). Given $A$ and $B$, we could solve the minimization problem to get $\Omega_0$. My intuition tells me that the optimal solution to the maximization problem is then $-\Omega_0$. We rotate $A$ to be as close to $B$ as possible, then flip it "180 degrees" in the opposite direction. Is there a simple proof or counterexample to this claim?

Answer 1

I think that my suggested solution is correct. Following the Wikipedia article, notice that $$ \|\Omega{A}-B\|_F^2=\langle\Omega{A}-B,\Omega{A}-B\rangle=\|A\|_F^2+\|B\|_F^2-2\langle\Omega{A},B\rangle $$ and thus maximizing $\|\Omega{A}-B\|_F^2$ is equivalent to minimizing $\langle\Omega{A},B\rangle$. But, using the SVD $AB^\mathrm{T}=U\Sigma{V^\mathrm{T}}$, $$ \langle\Omega{A},B\rangle=\langle\Omega,AB^\mathrm{T}\rangle=\langle\Omega,U\Sigma{V^\mathrm{T}}\rangle=\langle{U^\mathrm{T}}\Omega{V},\Sigma\rangle. $$ The matrix ${U^\mathrm{T}}\Omega{V}$ is orthogonal as a product of orthogonal matrices, and thus the expression is minimized when ${U^\mathrm{T}}\Omega{V}=-I$ (since $\Sigma$ is diagonal with non-negative entries). Hence, the solution is $\Omega^*=-UV^\mathrm{T}$, the negative of the solution to the minimization problem.