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Suppose I have Givens matrices (as defined here), $G_1,...G_k$, and some permutation $P$ which mapping $\{1,..., k\}$ to $\{1,..., k\}$. Is it true that:

$$G_1G_2... G_k = G_{p(1)}G_{p(2)}...G_{p(k)}$$

The reason I think this true is that I heard they represent rotations and rotations can be applied in any order. If this is not true then what is the flaw in my logic here?

JDoe2
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  • I do not understand the equation you gave, there is a comma separated list on the left hand side and a multiplication on the right? – Vinyl_cape_jawa Feb 23 '19 at 12:05
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    Here is the link for such matrices. Let us consider $k=2$. Why do we have $G_1G_2=G_2G_1$? Did you see this post? – Dietrich Burde Feb 23 '19 at 12:09
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    You might want to put the link to Givens matrices in your question (or add some basic background). This really confused me when I read it as "given matrices $G_1,\dots,G_k$, ..." the first time. – YiFan Tey Feb 23 '19 at 12:24
  • @Bernard your edit is wrong, I'm afraid. It appears that Givens matrices are a special type of matrices; the statement is obviously not true for any "given" matrices. (See the link posted by Dietrich Burde.) – YiFan Tey Feb 23 '19 at 12:34
  • @YiFan: Sorry for that misinterpretation. I've reverted to the original state. Thank you for informing me! – Bernard Feb 23 '19 at 12:39
  • Here is a link to the Givens matrices English Wikipedia page (the previous link is the German page). – Minus One-Twelfth Feb 23 '19 at 12:49
  • Apologies, there was a typo in the question and I have added the link. – JDoe2 Feb 23 '19 at 13:04

1 Answers1

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Doubt should be eliminated through experimentation. Here we examine two Given's rotations given by $$G_{13} = \begin{bmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}, \quad G_{23} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}. $$ Let $x = \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}^T$. It is clear that $$G_{13} x = \begin{bmatrix} -x_3 \\ x_2 \\ x_1 \end{bmatrix}, \quad G_{23} G_{13} x = \begin{bmatrix} -x_3 \\ -x_1 \\ x_2 \end{bmatrix},$$ whereas $$G_{23} x = \begin{bmatrix} x_1\\ -x_3 \\ x_2 \end{bmatrix}, \quad G_{13} G_{23} x = \begin{bmatrix} -x_2 \\ -x_3 \\ x_1 \end{bmatrix}.$$ We conclude that $G_{23}G_{13}x = G_{13} G_{23} x$ if and only $x_1 = x_2 = x_3$ or equivalently $x_j = \lambda \in \mathbb{R}$ for $j=1,2,3$. It follows that Given's rotations do not commute in general.

Rotations around a common axis commute. In the example given above $G_{13}$ rotates around the $x_2$ axis, whereas $G_{23}$ rotates around the $x_1$ axis.
Carl Christian
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    Shortcut to the same answer: If rotations in $\mathbb{R}^3$ would always commute, then gimbal lock would appear at every set of Euler angles rather than at some :) – darij grinberg Feb 24 '19 at 04:47