Is it always possible to decompose a rotation along an arbitrary plane as a combination of “simple rotations”?

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.

• They're called Givens rotations (en.wikipedia.org/wiki/Givens_rotation) and yes, I believe arbitrary rotations can be decomposed into Givens rotations. – Qiaochu Yuan Sep 1 '20 at 6:31
• 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. – Jyrki Lahtonen Sep 1 '20 at 6:45
• 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. – Conifold Sep 1 '20 at 6:45

1 Answer

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.

• This is fantastic thanks. Is there any notion of a minimum number of Givens rotations required to emulate an arbitrary rotation? – Disgusting Sep 1 '20 at 7:30
• 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? – Qiaochu Yuan Sep 1 '20 at 7:48