Physlib.Mathematics.CrossProductMatrix

Given a vector $\omega = (\omega_0, \omega_1, \omega_2) \in \mathbb{R}^3$, the cross product matrix (also known as the hat map and denoted by $[\omega]_{\times}$) is the $3 \times 3$ skew-symmetric matrix defined as: $$[\omega]_{\times} = \begin{pmatrix} 0 & -\omega_2 & \omega_1 \\ \omega_2 & 0 & -\omega_0 \\ -\omega_1 & \omega_0 & 0 \end{pmatrix}$$ This matrix is characterized by the property that for any vector $v \in \mathbb{R}^3$, the matrix-vector product $[\omega]_{\times} v$ is equal to the cross product $\omega \times v$.

For any vectors $\omega, v \in \mathbb{R}^3$, the product of the cross product matrix $[\omega]_{\times}$ and the vector $v$ is equal to the cross product $\omega \times v$. Here, $[\omega]_{\times}$ (the hat map) is the $3 \times 3$ skew-symmetric matrix associated with $\omega$.

For any vector $\omega \in \mathbb{R}^3$, let $[\omega]_\times$ denote the cross product matrix (also known as the hat map). The transpose of this matrix is equal to its negative, i.e., $$[\omega]_\times^\top = -[\omega]_\times$$ This property characterizes the cross product matrix as a skew-symmetric matrix.

The vee map takes a $3 \times 3$ real matrix $A$ and returns a vector in $\mathbb{R}^3$ defined by $(A_{2,1}, A_{0,2}, A_{1,0})$. This operation is the left inverse of the hat map (cross-product matrix map); specifically, it satisfies $(\omega_\times)^\vee = \omega$ for any vector $\omega \in \mathbb{R}^3$, where $\omega_\times$ is the skew-symmetric cross-product matrix.

For any vector $\omega \in \mathbb{R}^3$, applying the vee map $(\cdot)^\vee$ to the cross product matrix (the hat map) $[\omega]_\times$ returns the original vector $\omega$. That is, $$([\omega]_\times)^\vee = \omega$$ where $[\omega]_\times$ is the skew-symmetric matrix such that $[\omega]_\times v = \omega \times v$ for any $v \in \mathbb{R}^3$. This demonstrates that the vee map is a left inverse of the hat map.

For any $3 \times 3$ real matrix $A$ that is skew-symmetric (i.e., $A^T = -A$), applying the vee map $(\cdot)^\vee$ followed by the hat map $[\cdot]_\times$ returns the original matrix. That is, $$[A^\vee]_\times = A$$ where the vee map is defined as $A^\vee = (A_{2,1}, A_{0,2}, A_{1,0})$ and the hat map (cross product matrix) is defined as $$[\omega]_{\times} = \begin{pmatrix} 0 & -\omega_2 & \omega_1 \\ \omega_2 & 0 & -\omega_0 \\ -\omega_1 & \omega_0 & 0 \end{pmatrix}$$ This result establishes that the hat map is a right inverse of the vee map on the space of skew-symmetric matrices $\mathfrak{so}(3)$.