Inverting non-square matrix with cross-product

199 Views Asked by At

I want to solve for $R$ given $V$, $B$: $$V_{3x2} = R_{3x3}B_{3x2}$$

Where $R$ is a rotation (orthogonal) matrix, and the columns of $B$ are orthogonal (but not of the same length in general).

  1. Do the above facts imply the the columns of $V$ are also orthogonal?
  2. Why is it valid to solve for $R$ by creating new matrices $V'_{3x3}$ and $B'_{3x3}$ where the 3rd column is the cross product of the first two columns, and then setting $R = V'_{3x3}(B'_{3x3})^{-1}$? See section "B. Orientation Determination" of paper, where this is performed without justification.
1

There are 1 best solutions below

2
On BEST ANSWER
  1. Yes, since rotations preserve dot products, specifically if $R$ is a rotation matrix and $u,v$ are compatible-length column vectors then $u\cdot v = u^Tv = u^TR^TRv = (Ru)^T(Rv) = (Ru)\cdot(Rv)$. Consequently, rotations do not affect orthogonality, the length of a vector, and the angle between two vectors.

  2. The essential fact being asserted here is that if $v,w \in \mathbb R^3$ are orthogonal, and $R$ is a $3\times 3$ rotation matrix, then $R(v\times w) = (Rv) \times (Rw)$. In fact this holds generally even when $v,w$ are not orthogonal. Once you believe this is true then it makes sense to extract a third column as the cross product of the first two, since it too satisfies the matrix relation. One can see a couple mechanical proofs of this at this related question, and it can also be seen by tensor analysis. but you might still wonder why the authors felt it so obvious that it doesn't require proof.

    Effectively, this is saying that the cross product is somewhat intrinsic: it doesn't care how you rotate your coordinate system, it still points in the same direction (but it is sensitive to rescaling the axes so it's not entirely intrinsic).

    The property proved in 1 helps to see intuitively why this is so: the magnitude of $u \times v$ is determined by $\|u\|, \|v\|$ and the angle between them, so it is invariant under rotation. Likewise, when $u,v$ are non-parallel in $\mathbb R^3$, the direction of $u \times v$ is the basically-unique direction perpendicular to both $u$ and $v$, and so rotation commutes nicely with this as well. This direction is unique up to negation, and here the "right-hand rule" comes in to disambiguate: the cross-product is the one that makes the triple-product of $u,v,u\times v$ positive.

    If you think about this algebraically, this corresponds to the convention that a rotation matrix has $\det(R) = 1$ (when $\det(R) = -1$ it's a reflection). Hence a hint as to why the authors called the section "Orientation Determination": there are two orthogonal matrices that map the columns of $B$ to the columns of $V$, but only one is a genuine rotation (hence orientation-preserving).