# Minimize Frobenius norm with unitary constraint

I am trying to find a unitary tramsformation, $M$, that minimizes $\Vert MA-B \Vert_F^2$ where $A$ and $B$ are $N\times L,\;L\ge N$.

I know how to solve it without the unitary constraint. I thought using Lagrange multipliers with the constraint $\Vert M^HM-I \Vert_F^2 = tr\left\{ \left( M^HM-I \right)^H \left( M^HM-I \right)\right\} = 0$ but the it is quite difficult to solve. Is there any simpler way?

Thanks.

• Why wouldn't the constraint $M^HM=I$ be appropriate? (I may not grasp the context of the question fully -- it looks to me like you are talking about unitary matrices $M$.) Feb 25, 2014 at 16:29
• @JeffSnider I am indeed talking about a unitary matrix $M$. However, $M^HM=I$ imposes $N^2$ constraints to my understanding.
– ThP
Feb 25, 2014 at 20:10
• Since $M^HM$ is Hermitian and the non-diagonals are constrained to be zero it imposes only $N(N+1)/2$ constraints. There may be a special way to ensure $M$ is unitary other than that constraint, but I don't know what it would be. Feb 25, 2014 at 20:36
• This is the Orthogonal Procrustes Problem ( en.wikipedia.org/wiki/Orthogonal_Procrustes_problem ) Aug 8, 2016 at 10:32

Since $M$ is unitary you can write $$\Vert MA-B\Vert_F^2=\Vert MA\Vert_F^2-2\langle MA,B\rangle_F+\Vert B\Vert_F^2\\ = \Vert A\Vert_F^2-2\mathrm{Re}\langle M,BA^*\rangle_F+\Vert B\Vert_F^2.$$ Therefore, your optimization reduces to maximizing $\mathrm{Re}\langle M,BA^*\rangle_F$. Using H$\ddot{\text{o}}$lder inequality and SVD of $BA^*=USV^*$, it's straightforward to show that the maximum is the Schatten 1-norm of $BA^*$ which is attained at $M=UV^*$.
• Can you explain how to show that the maximum of $\mathrm{Re}\langle M,BA^*\rangle_F$ is obtained at $M=UV^*$? Feb 26, 2014 at 9:17
• First, use the fact that $\Vert M\Vert = 1$ and H$\ddot{\text{o}}$lder inequality to show that $\mathrm{Re}\langle M,BA^*\rangle_F$ is upper-bounded by the Schatten 1-norm of $BA^*$. Then, simply verify that this bound is attained at $M=UV^*$.