2
$\begingroup$

I am almost sure this question has been asked before but I had a long look and its possible I lack the language to describe my question to the search box properly.

Assume we are working with real numbers. If we call a "simple" rotation one represented by a matrix "$R$" that is an identity matrix except for 4 entries defined by indices x and y. Where these changed entries $r_{ij}$ can be represented by: $$ r_{xx}=r_{yy}=cos(\theta) $$ $$ r_{xy}=-sin(\theta) $$ $$ r_{yx}=-r_{xy}=sin(\theta) $$

For example this matrix: $$ \begin{bmatrix} 1& 0& 0& 0& 0\\ 0& cos(\theta)& 0& -sin(\theta)& 0\\ 0& 0& 1& 0& 0\\ 0& sin(\theta)& 0& cos(\theta)& 0\\ 0& 0& 0& 0& 1 \end{bmatrix} $$

Which rotates the plane spanned by $e_2$ and $e_4$ by theta.

I have two questions:

  1. Does this notion of "simple rotations" have a proper name?

  2. My main question, if one has a rotation in 1 arbitrary plane in n-dimensions spanned by non-basis vectors is it possible, and more importantly always possible, to decompose that as a combination of these simple rotations? If so is there an algorithmic way to do this and does it have a name?

For bonus points, if there's anything I should know about how Complex co-ordinates or Complex theta behave in this context I would be happy to hear about it.

$\endgroup$
3
  • 2
    $\begingroup$ They're called Givens rotations (en.wikipedia.org/wiki/Givens_rotation) and yes, I believe arbitrary rotations can be decomposed into Givens rotations. $\endgroup$ – Qiaochu Yuan Sep 1 '20 at 6:31
  • 1
    $\begingroup$ What Qiaochu said. IIRC a proof for the fact that every rotation can be written as a composition of Givens rotations would go by induction on $n$. Prove that a sequence of Givens rotations turns the first column into $(1,0,\ldots,0)^T$ and go from there. $\endgroup$ – Jyrki Lahtonen Sep 1 '20 at 6:45
  • 2
    $\begingroup$ This will be a special case of the QR decomposition applied to a rotation matrix. One of the algorithms does it by performing a series of Givens rotations, a.k.a. elementary or basic rotations, thus giving a desired decomposition into them. $\endgroup$ – Conifold Sep 1 '20 at 6:45
1
$\begingroup$

These rotations are called Givens rotations, and every rotation can be decomposed into Givens rotations. Think of an $n \times n$ orthogonal matrix in terms of its columns $v_1, \dots v_n$, which form an orthonormal basis. Multiplying such an orthogonal matrix by a Givens rotation on the left has the effect of applying that rotation to each of the vectors $v_i$. Our goal will be to "straighten out" this basis by repeatedly applying Givens rotations until it's the standard basis $e_1, \dots e_n$ of $\mathbb{R}^n$.

A Givens rotation allows us to rotate in any coordinate plane, so we can argue as follows. Write $v_1 = (v_{11}, v_{12}, ...)$. First, by rotating $90^{\circ}$ in a coordinate plane we can swap any two entries up to sign, $(x, y) \mapsto (-y, x)$. So swap any nonzero entry into the first coordinate, so that $v_{11} \neq 0$. Next, by an appropriate rotation in the $e_i, e_j$-coordinate plane, if $v_{1i}, v_{1j}$ are both nonzero we can rotate so that $v_{1j} = 0$. So rotate in the $e_1, e_j$-coordinate plane for any $j$ such that $v_{1j}$ is nonzero until all entries other than $v_{11}$ are equal to zero. At the end of this process we have $v_1 = \pm e_1$ (and if $v_1 = -e_1$ we can arrange $v_1 = e_1$ by a final $180^{\circ}$ rotation), and $v_2, \dots v_n$ must be orthogonal to it so are contained in the copy of $\mathbb{R}^{n-1}$ spanned by $e_2, \dots e_{n-1}$ (in matrix terms, our original orthogonal matrix is now a block matrix). Now we can induct on $n$.

At the very last step we may get $v_n = -e_n$ rather than $v_n = e_n$ but this could only happen if our original matrix was a reflection rather than a rotation.

$\endgroup$
2
  • $\begingroup$ This is fantastic thanks. Is there any notion of a minimum number of Givens rotations required to emulate an arbitrary rotation? $\endgroup$ – Disgusting Sep 1 '20 at 7:30
  • $\begingroup$ Probably? The above argument (slightly modified to be a bit trickier about signs) does it in ${n+1 \choose 2}$ Givens rotations, and if we're even trickier about zero entries we should be able to get it down to ${n \choose 2}$ and this should be best possible generically by dimension considerations, I think? $\endgroup$ – Qiaochu Yuan Sep 1 '20 at 7:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.