Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to solve the orthogonal procrustes problem with the additional constraint the rotation matrix is orientation-preserving: $$\min_R \|AR-B\|\quad \textrm{s.t.}\quad R^TR=I,\ \det R = 1$$ where the norm is the Frobenius norm. The standard approach is to compute the SVD of $A^TB$: $$A^TB = UDV^T.$$

If $\det A^TB > 0$, the solution is simply $R=UV^T$. Otherwise, it is $U\Sigma V^T$, where $\Sigma$ is formed by replacing the smallest singular value of $D$ by $-1$, and the rest of them by $1$.

I want to find $R$ without using SVD, by using the polar decomposition $A^TB = UP$ instead. If $\det A^TB>0$, there is no problem: $R$ and $U$ coincide.

If $\det A^TB <0$, is there any way of finding $R$ by using (a modification of) the polar decomposition?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.