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Given a $3\times3$ matrix $M$, if we would like to get the closest $\mathrm{SO}(3)$ matrix $R$ that minimizes \begin{equation} \|R-M\|_F \end{equation}

then $R$ = $UV^{T}$ where $U$ and $V^{T}$ are the orthogonal matrices from the singular value decompositon of $M$. i.e. $M = U\Sigma V^{T}$ as explained in this answer. When $UV$ is not a "special" orthogonal matrix i.e. $det(UV^{T}) = -1$ we replace the singular vector $u_3$ by $-u_3$.

My question is why only $u_3$ and why not any other column of $U$ or row of $V^{T}$?

(Posting this as a question instead of a comment since I don't have enough reputation yet, sorry!)

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  • $\begingroup$ You must mean $\det(UV^{T}) = -1$ $\endgroup$ – Jean Marie Becker Aug 14 at 16:03
  • $\begingroup$ Oops, yes thanks! $\endgroup$ – Karnik Ram Aug 14 at 17:09
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By right-multiplying both $R$ and $M$ by $V$ and left-multiplying by $U^T$ you are leaving the objective invariant. Now the new $M$ is diagonal (actually, it's $\Sigma$), and it's not hard to convince yourself that so should the new $R$ be. If $\det(UV^T) = -1$, then the constraint means $R$ is a diagonal matrix with $\pm 1$ as its diagonal entries, and an odd number of $-1$s. Since $\Sigma$ is nonnegative, the choice that minimizes the distance is to make the entry corresponding to the smallest diagonal entry in $\Sigma$ the sole $-1$.

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  • $\begingroup$ Thanks for your answer! I am able to convince myself that the new rotation $R'$ (after left-multiplying and right-multiplying $R$) should be diagonal, but I am unable to understand why it should have a determinant of -1. I think the old rotation, $R$, should have a determinant of -1 (to make $det(U^{T}RV) = 1$ when det(UV^{T} = -1), but then it need not be diagonal? $\endgroup$ – Karnik Ram 2 days ago
  • $\begingroup$ The old rotation R is what you constrain to be in SO(3), so it needs to be determinant 1. If you let $R' = U^T R V$, then you need $\det R' = -1$. $\endgroup$ – Yoav Kallus 2 days ago
  • $\begingroup$ Thanks! I get it now. And negating the last element of $\Sigma$ is equivalent to negating the smallest singular vector of $U$. And we know $R = UV^{T}$. But what seems a little unintuitive to me here is that we're modifying our input $M$ itself to minimize the objective. $\endgroup$ – Karnik Ram 2 days ago

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