The maximal rotation matrix

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 $3$).

$\newcommand{\Basis}{\mathbf{e}}\newcommand{\Brak}{\langle #1\rangle}$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.
• 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. – Widawensen Jun 28 '16 at 21:36