Quaternionic numerical range of a class of matrices

Question

I need to characterize the quaternionic numerical range of matrices of the form $$\begin{bmatrix} r & p \\ 0 & -r\end{bmatrix}$$ where $r$ is a positive real number and $p$ is a quaternion. Specifically I want to investigate the convexity of the quaternionic numerical range of such a matrix. I couldn'd find any counterexamples to show that it's not convex, but I can't come up with a proof either, except for when $p$ is a real number.

Answer 1

WLOG use unit vectors $[\begin{smallmatrix}\alpha \\ \beta\end{smallmatrix}]\in\mathbb{H}^2$, compute

$$ \begin{bmatrix} \alpha \\ \beta \end{bmatrix}^{\ast}\begin{bmatrix}r & p \\ 0 & -r\end{bmatrix} \begin{bmatrix}\alpha \\ \beta\end{bmatrix}=r|\alpha|^2+\overline{\alpha}p\beta-r|\beta|^2 $$

Write $|\alpha|=\cos\theta$, $|\beta|=\sin\theta$ so this is

$$ r(\cos^2\theta-\sin^2\theta)+\cos\theta\sin\theta|p|u $$

where $u$ is an arbitrary unit quaternion (do you see why?). Set $u=e^{\phi\mathbf{u}}$ so this becomes

$$ r\cos(2\theta)+\frac{1}{2}|p|\sin(2\theta)(\cos\phi+\sin\phi\,\mathbf{u}). $$

Set $q=\frac{1}{2}|p|$. This is the image of $(r\cos(2\theta),q\sin(2\theta)\cos(\phi),q\sin(2\theta)\sin(\phi)\mathbf{u})$ under the projection map $(x,y,\mathbf{z})\mapsto(x+y,\mathbf{z})$. The set of points $(x,y,\mathbf{z})\in\mathbb{R}\times\mathbb{R}\times\mathbb{R}^3$ in question is characterized by

$$ \frac{x^2}{r^2}+\frac{y^2}{q^2}+\frac{\|\mathbf{z}\|^2}{q^2}=1. $$

This is a hyperspheroid. Fixing $\mathbf{z}$ (in the ball of radius $q$ around $0$), we want to find the range of $x+y$. This is an interval of the form $(-w,w)$. Imagine drawing the ellipse $x^2/r^2+y^2/q^2=1-\|\mathbf{z}\|^2/q^2$ and overlaying lines of the form $x+y=\mathrm{const}$; the extreme values $\pm w$ occur where such a line is tangent to the ellipse. Parametrize the ellipse, implicitly differentiate to get $x\dot{x}/r^2+y\dot{y}/q^2=0$, set $\dot{y}=-\dot{x}$ (from $x+y=w$), get $x/r^2=y/q^2$; set $x=r^2c$, $y=q^2c$ and substitute to get $c=\sqrt{(1-\|\mathbf{z}\|^2/q^2)/(r^2+q^2)}$ which in turn gives $w=\sqrt{(r^2+q^2)(1-\|\mathbf{z}\|^2/q^2)}$. In conclusion, the numerical range is

$$ W=\Bigl\{w+\mathbf{z}\mid \frac{w^2}{r^2+(|p|/2)^2}+\frac{\|z\|^2}{(|p|/2)^2}=1\Bigr\}, $$

which is a solid hyperspheroid, and is convex.

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If you want an example of a quaternionic numerical range not being convex, consider the case of the $1\times1$ matrix $A=i$. The range is then the unit sphere $S^3$, which is not convex. See if you can figure out what the quaternionic numerical range of $[\begin{smallmatrix} 1 & 0 \\ 0 & i \end{smallmatrix}]$ is (it's also not convex).

It is a "cone" with apex $1\in\mathbb{R}$ and base $S^2\subset\mathbb{R}^3=\mathrm{Im}\,\mathbb{H}$.