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Let's consider two numbers calculated for a rotation matrix which are:

  • $s_e=$ the sum of all entries of a matrix
  • $s_a=$ the sum of absolute values of all entries for a given matrix.

    It would be interesting to know for what rotation matrix these numbers are maximal. Case for 2D is rather straightforward, it is known general form of a matrix and both numbers $s_e$, $s_a$ are just functions of a single variable $\theta$

    $ R= \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \ sin(\theta) & \cos(\theta) \\ \end{bmatrix} $

    Case for 3D seems much more complicated. We have 3 variables ($3$ DOF) and final form is quite complicated. Additionally it's hard to guess what form of equation we should take: Rodrigues formula? Euler rotations? How to use constraints (highly symmetrical in their structure) usually imposed on a rotation matrix ?

Examples shed some light on the problem.

Let's look at the identity matrix. Here situation is simple. $$R=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ $s_e= 3, s_a=3 $

Other example $$R=\begin{bmatrix} \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2} & 0 \\ \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$

We have here

$s_e= 1+\sqrt{2}, s_a=1+2\sqrt{2}$.

The number $s_e$ is less and $s_a$ greater comparing to the case for identity matrix.

Intuition tells that such a matrix , I call it here "the maximal" rotation matrix and denote appropriately $R_{max(s_e)}$ and $R_{max(s_a)}$, should have entries somehow evenly located in rows and columns and negative entries should be as small as possible. $0$'s should be rather absent. Good candidate for it in the second sense, it seems, is the matrix mentioned earlier by me (A certain unique rotation matrix).

$$A=\begin{bmatrix} -\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{2}{3} \\ \dfrac{2}{3} & -\dfrac{1}{3} & \dfrac{2}{3} \\ \dfrac{2}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \\ \end{bmatrix} $$
For it $s_e= 3, s_a=5 $

  • Maybe it is the maximal matrix taking into account $s_e$ but how to prove it ?
  • If not what is the maximal 3D rotation matrix in both senses mentioned above ?

  • Could we at least prove that $R_{max(s_a)}$ is from the family of rotations $R_{max(s_e)}$ ?


($App.^*$One can consider also a minimal versions of rotation matrices: $R_{min(s_e)}$ and $R_{min(s_a)}$. For them values are probably $-3$ and $3$).

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$\newcommand{\Basis}{\mathbf{e}}\newcommand{\Brak}[1]{\langle #1\rangle}$Let $A$ be an $n \times n$ orthogonal matrix, and put $\Basis = (1, 1, \dots, 1)$. The sum of the entries of $A$ is the inner product $$ \Brak{\Basis, A\Basis} = n\cos\theta, $$ with $\theta$ the angle between $\Basis$ and $A\Basis$. Since there exists an orthogonal matrix fixing $\Basis$ (i.e., with $\theta = 0$), it follows that the maximum sum-of-entries for an $n \times n$ orthogonal matrix is $n$, and this is achieved precisely by the copy of $O(n-1)$ that fixes $\Basis$.


Offhand I don't see any nice way to get at the maximum absolute value sum in general. For $n = 2$, the maximum absolute value sum is clearly $2\sqrt{2}$, so there exist (block diagonal) matrices in $O(2n)$ with absolute value sum $2n\sqrt{2}$, and matrices in $O(2n+1)$ with absolute value sum at least $1 + 2n\sqrt{2}$.

These lower bounds on the maximum absolute value sum are not optimal for $n \geq 4$: For example, the $4 \times 4$ orthogonal matrix $$ \tfrac{1}{2}\left[\begin{array}{@{}rrrr@{}} 1 & -1 & -1 & -1 \\ 1 & -1 & 1 & 1 \\ 1 & 1 & -1 & 1 \\ 1 & 1 & 1 & -1 \\ \end{array}\right] $$ has absolute value sum $8 > 4\sqrt{2}$.

For the case $n = 3$, it's plausible your matrix maximizes the absolute value sum: Naively, one expects the set of columns of an extremal matrix to be invariant under rotation about a diagonal axis of the unit cube. It's easy to check your matrix has this symmetry, and its columns are equidistant from the coordinate planes.

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    $\begingroup$ Thank you Andrew. First case is solved very fine, for the second maybe it would exist firstly some independent method of calculating the optimal axis, and then calculating the angle for 3D would be relatively simple from Rodrigues formula. $\endgroup$ – Widawensen Jun 28 '16 at 16:28
  • $\begingroup$ From the other end point. I've just noticed that if we assume that rotation would be by $180^\circ$ (case for symmetrical rotation matrix) then we have simplified Rodriques formula $R(v,\theta)=2vv^T-1$ which probably can lead to the unit axis vector $v$, especially if we assume that components of $v$ are non-negative. If so the only negative components must lie on the diagonal. We know that the sum of diagonal is $1+2cos(180^\circ)=- 1$ and from Andrew method we can assume $s_e=3$ so in this case $s_a$ must be $ = 3+1+1=5$ exactly as in the example above. $\endgroup$ – Widawensen Jun 28 '16 at 21:36

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