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Lets say we have a known unit vector, $v$. Consider multiplying this vector by a known square matrix $M$ that keeps the vector in the space, but changes the direction and length of $v$ in general, $v \rightarrow v' = Mv$. If we renormalize this vector to have unit length by dividing by $\sqrt{v' v'^{*}}$, we have moved between two vectors with length-one that point in different directions, which is a rotation. Is there a way to find a unitary matrix that accomplishes this same rotation given $M$ and $v$?

This question seems similar and useful: Finding the rotation matrix in n-dimensions but the givens are the two vectors. Here we only have $M$ and $v$ to use.

Thanks!

kηives
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  • The map $v \mapsto \frac{|v|}{|Mv|}Mv$ is not generally a linear map and in particular is not necessarily a rotation – Ben Grossmann Mar 05 '20 at 23:01
  • @Omnomnomnom but isn't there a rotation which accomplishes the same result? – kηives Mar 05 '20 at 23:04
  • Yes, there exists a rotation $R$ for which $Rv$ is the normalized vector. It is incorrect to say, however, that "we have moved between two vectors with length-one that point in different directions, which is a rotation". – Ben Grossmann Mar 05 '20 at 23:24
  • With that being said, if you want such a rotation then you could use the post you linked with $v$ and the normalized $v'$ as your two vectors. – Ben Grossmann Mar 05 '20 at 23:26
  • @Omnomnomnom yes, thank you for clarifying. No, I would like to extract the rotation part of $M$ directly, if possible. – kηives Mar 05 '20 at 23:44
  • @Omnomnomnom said another way, I would like to find a form for $M$ like $s U$ where $U$ is a rotation and $s$ a rescaling. – kηives Mar 05 '20 at 23:46
  • It sounds like what you're looking for is a polar decomposition of the form $M = PU$. If this is what you're looking for, then you could take $U = M(M^M)^{-1/2} = (MM^)^{-1/2}M$. – Ben Grossmann Mar 05 '20 at 23:49
  • @Omnomnomnom ah, I meant $s$ as a multiple of the identity. – kηives Mar 05 '20 at 23:51
  • That doesn't make much sense since most matrices $M$ are not a multiple of a rotation matrix (or unitary matrix) $U$ – Ben Grossmann Mar 05 '20 at 23:55
  • @Omnomnomnom they must be right? Or $s$ is zero. Since every vector can be normalized, and two vectors of the same length can be rotated into each other. Right? – kηives Mar 05 '20 at 23:57
  • If you choose a particular $v$, then there exists a scaling $s$ and unitary $U$ such that $Mv = sUv$. This does not imply that $M = sU$; saying that $M = s U$ amounts to saying that there is a single rotation for which $Mx = sUx$ holds for every vector $x$. – Ben Grossmann Mar 06 '20 at 00:01
  • @Omnomnomnom Yes, but this is what I asked in the question. Given $M$ and $v$, you can accomplish the same thing with a $s$ and a $U$. I am wondering is there a way to find a $U$. – kηives Mar 06 '20 at 18:26
  • Ultimately, a formula for such a $U$ must depend on both $M$ and $v$. One such formula comes from applying the ideas from the linked post to $v$ and the normalized $Mv$. – Ben Grossmann Mar 06 '20 at 18:50

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