$$% Arcus cosine. \def\acos{\cos^{-1}} % Vector projection. \def\projection#1#2{{proj_{#1}\left(#2\right)}} % Vector rejection. \def\rejection#1#2{{rej_{#1}\left(#2\right)}} % Norm. \def\norm#1{{\left\|#1\right\|}} % Cross product. \def\cross#1#2{\mathit{cross}\left(#1,#2\right)} % Dot product. \def\dot#1#2{{#1 \cdot #2}} % Magnitude. \def\mag#1{{\left|#1\right}} \def\group#1{\left(#1\right)}} \def\sbgrp#1{\left\{#1\right\}}$$

## Second power of the cross product matrix

The 2nd power of the cross product matrix $\widetilde{V} = \left[\begin{matrix} 0 & -v_z & v_y \\ v_z & 0 & -v_x \\ -v_y & v_x & 0 \\ \end{matrix}\right]$ is $\widetilde{V}^2 = \left[\begin{matrix} -v_z^2 - v_y^2 & v_x v_y & v_x v_z \\ v_x v_y & -v_x^2 - v_z^2 & v_y v_z \\ v_x v_z & v_y v_z & -v_x^2 - v_y^2 \end{matrix}\right]$ Furthemore if $$v$$ is unit then $\widetilde{V}^2 = \left[\begin{matrix} v_x^2 - 1 & v_x v_y & v_x v_z \\ v_x v_y & v_y^2 - 1 & v_y v_z \\ v_x v_z & v_y v_z & v_z^2 - 1 \end{matrix}\right]$

### Proof

\begin{aligned} &\widetilde{V}^2 = \widetilde{V}\widetilde{V} \end{aligned}
\begin{aligned} &= \left[\begin{matrix} 0 & -v_z & v_y \\ v_z & 0 & -v_x \\ -v_y & v_x & 0 \\ \end{matrix}\right] \left[\begin{matrix} 0 & -v_z & v_y \\ v_z & 0 & -v_x \\ -v_y & v_x & 0 \\ \end{matrix}\right] \end{aligned}
Definition of $$\widetilde{V}$$.
\begin{aligned} &= \left[\begin{matrix} 0 \cdot 0 + -v_z \cdot v_z + v_y \cdot -v_y & 0 \cdot -v_z + -v_z \cdot 0 + v_y \cdot v_x & 0 \cdot v_y + -v_z \cdot -v_x + v_y \cdot 0 \\ v_z \cdot 0 + 0 \cdot v_z + -v_x \cdot -v_y & v_z \cdot -v_z + 0 \cdot 0 + -v_x \cdot v_x & v_z \cdot v_y + 0 \cdot -v_x + -v_x \cdot 0 \\ -v_y \cdot 0 + v_x \cdot v_z + 0 \cdot -v_y & -v_y \cdot -v_z + v_x \cdot 0 + 0 \cdot v_x & -v_y \cdot v_y + v_x \cdot -v_x + 0 \cdot 0 \end{matrix}\right] \end{aligned}
\begin{aligned} &= \left[\begin{matrix} -v_z^2 - v_y^2 & v_x v_y & v_x v_z \\ v_x v_y & -v_x^2 - v_z^2 & v_y v_z \\ v_x v_z & v_y v_z & -v_x^2 - v_y^2 \end{matrix}\right] \end{aligned}
If $$v$$ is unit then
\begin{aligned} &= \left[\begin{matrix} v_x^2 - 1 & v_x v_y & v_x v_z \\ v_x v_y & v_y^2 - 1 & v_y v_z \\ v_x v_z & v_y v_z & v_z^2 - 1 \end{matrix}\right] \end{aligned}
$$\vec{u}$$ is unit. Hence $$-v_z^2 - v_y^2 = v_x^2 - 1$$, $$-v_x^2 - v_z^2 = v_y^2 - 1$$, and $$-v_x^2 - v_y^2 = v_z^2 - 1$$.