# Variant of orthogonal Procrustes problem

I want to find an anti-symmetric matrix $T$ which minimizes $\|A-e^TBe^{-T}\|^2 + \mu\|T\|^2$, where $A$ and $B$ are symmetric positive definite matrices and the norm is the Frobenius matrix norm. The original orthogonal Procrustes problem is to find an orthogonal matrix $\Omega$ that minimizes $\|A - \Omega B\|$. That problem can be solved by doing an SVD of the matrix $B^TA$, as explained in for example wikipedia, http://en.wikipedia.org/wiki/Orthogonal_Procrustes_problem. My problem is different in that I want to penalize large rotations that don't really do much to improve the fit between $A$ and the rotated $B$ matrix. Is there any simple way to solve my penalized problem (the non-simple way being direct minimization of the expression, but that is much less straight forward than doing a standard SVD).

Edit: The non-penalized version of my problem is called the Two sided orthogonal Procrustes problem, and can be solved by doing an eigendecomposition of $A$ and $B$ separately. Then apparently $e^T = V_B D V_A^T$ is a solution for any matrix diagonal matrix $D$ with +1 and -1 on the diagonal.

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Is there any reason why you've chosen the penalty term to be the norm of $\mathbf T$ instead of its exponential (something like $\mu\|\exp(\mathbf T)^\top\exp(\mathbf T)-\mathbf I\|^2$ perhaps)? –  Ｊ. Ｍ. Nov 8 '11 at 10:05
$\exp(T)$ is an orthogonal matrix, so that expression would always be zero(?). But in general I don't have any objections to other forms of the penalty function if they make the problem easier to solve. –  uekstrom Nov 8 '11 at 10:11
Oh, your starting $\mathbf T$ is antisymmetric? I thought you just started with something else and was hoping $\mathbf T$ converges to an antisymmetric matrix. Still, I'm not sure your current choice of penalty makes for an easy problem... –  Ｊ. Ｍ. Nov 8 '11 at 10:15
I could use the penalty $\|1 - e^T\|^2$ instead, that would be the same for small $T$. Any ideas how to use that in practice? –  uekstrom Nov 8 '11 at 11:44