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Let $A$ be a fixed 3-by-3 matrix and $Q$ be a rotation matrix whose yaw, pitch, and roll angles are $\phi\in[0,\pi]$, $\theta\in[0,\pi]$, and $\psi\in[0,\pi/2]$, respectively: \begin{equation} Q= \begin{bmatrix} \cos\phi \cos\theta \cos\psi - \sin\phi \sin\psi & \sin\phi \cos\theta \cos\psi + \cos\phi \sin\psi & -\sin\theta \cos\psi \\ -\cos\phi \cos\theta \sin\psi - \sin\phi \cos\psi & -\sin\phi \cos\theta \sin\psi + \cos\phi \cos\psi & \sin\theta \sin\psi \\ \cos\phi \sin\theta & \sin\phi \sin\theta & \cos\theta \end{bmatrix} \end{equation}

We define the rotational transformation$B=QAQ^\top$ and a scalar \begin{equation} F(\phi,\theta,\psi)=\left|B_{12}^2-B_{21}^2\right|+\left|B_{13}^2-B_{31}^2\right|+\left|B_{23}^2-B_{23}^2\right| \end{equation}

The entries $B_{ij}$ depend on $(\phi,\theta,\psi)$.

Is is possible to prove the concavity of $F$ for a given $A$? or otherwise find an example of $A$ for which $F$ is not concave ?

I can find numerically the maximum of $F$ for a set of random matrix $A$, but I am enable to find a rigorous way that this maximum is indeed a global or local one.

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  • $\begingroup$ Welcome to MO. You question is unclear and needs more explanations: Do you fix $A$ and look at $F=F(Q)$ as a function of $Q$? Or on the contrary do you fix $Q$ and look at $F=F(A)$ as a function defined on the whole vector space of matrices? In the first case your question makes no sense, because the set of rotation matrices is not convex. In the second case the answer is no, there is no concavity (take $Q=Id$ and look matrices with zero coefficients except $A_{12}$, in which case $F(A)=\frac 12 |A_{12}|^2$ is clearly not concave). I vote as off-topics, not research level. $\endgroup$
    – leo monsaingeon
    Commented Apr 17, 2020 at 6:28
  • $\begingroup$ Thanks @leomonsaingeon for you answer. I will try to make my question clearer. I fix $A$ (which is in my case the tensor of velocity gradient at point $(x,y,z)$ in the space), then I will look for a new frame, that consists of rotating the laboratory frame (to get a new frame $x*=Qx$), for which the new velocity gradient $B=QAQ^\top$ maximizes F. Practically, $F$ depends on $\phi$, $\theta$ and $\psi$ as variable and $A$ as a fixed matrix. $\endgroup$
    – Benjamin Techer
    Commented Apr 17, 2020 at 7:07

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Multiplying by $S = \operatorname{diag}(\pm 1, \pm 1, \pm 1)$ (any combination of signs) leaves the objective function unchanged. So this function has multiple maxima: if $Q$ is one, then $SQ$ is another one. If you construct a curve that joins two of these maxima, then either the function is constant on the curve (unlikely) or you'll find out that the function is not concave (more likely).

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    $\begingroup$ Thanks @Federico for your answer. I think I didn't correctly formulate my question. My question is to find $Q$ that maximizes the objectif function $F$. What. I do practically is that for a fixed A, I write $F(\phi,\theta,\psi)$ and I look for the maximum of $F$. My concern is that I don't know if this maximum obtained numerically is unique or not. $\endgroup$
    – Benjamin Techer
    Commented Apr 17, 2020 at 7:30
  • $\begingroup$ I think what I wrote answers your question as intended. If $Q = Q(\phi_1,\theta_1,\psi_1)$ is a maximum, then $SQ = Q(\phi_2,\theta_2,\psi_2)$ is another maximum. (You probably will have to take $S$ with exactly two minus signs because what you have is a parametrization of $SO_n$, not of $O_n$) Just look at your function on a path from $(\phi_1,\theta_1,\psi_1)$ to $(\phi_2,\theta_2,\psi_2)$ and you will probably see a non-concave section. $\endgroup$
    – Federico Poloni
    Commented Apr 17, 2020 at 7:45

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