This note reviews important properties of what you might call the cross product matrix. For a vector $u \in \mathbb{R}^3$ we’ll write $$[u]_\times = \begin{bmatrix} 0 & -u_z & u_y \\ u_z & 0 & -u_x \\ -u_y & u_x & 0\end{bmatrix}.$$

This $3 \times 3$ matrix is defined such that $[u]_\times v = u \times v$ for all $v \in \mathbb{R}^3$, where $\times$ indicates the cross product of two vectors.

This is a meaningful parametrization of all anti-symmetric matrices in $\mathbb{R}^{3 \times 3}$. These matrices are the Lie algebra corresponding to the set of 3D rotation matrices $SO(3)$.

Relationship to Linear Transforms

Important properties of the cross-product matrix can be derived from its relationship with the determinant: $$ \det\left(\begin{bmatrix} u & v & w \end{bmatrix}\right) = u^T [v]_\times w \qquad \forall \; u, v, w \in \mathbb{R}^3.$$ This can be easily proven using expansion by minors along the first column. Given an arbitrary matrix $M \in \mathbb{R}^{3 \times 3}$ we therefore have:

$$ \det\left(M\begin{bmatrix} u & v & w \end{bmatrix}\right) = u^T\left(M^T [Mv]_\times M\right)w \qquad \forall \; u, v, w \in \mathbb{R}^3.$$

The multiplicative property of determinant tells us: $$ \det\left(M\begin{bmatrix} u & v & w \end{bmatrix}\right) = \det(M) u^T [v]_\times w \qquad \forall \; u, v, w \in \mathbb{R}^3.$$ Since this holds for all $u$ and $w$ we can conclude that:

$$ \det(M) M^{-T} [v]_\times = [ Mv ]_\times M \qquad \forall; v \in \mathbb{R}^3,$$ so long as $M$ is invertible.

This identity is particularly useful when $M = C \in SO(3)$ (i.e. when $M$ is a rotation matrix). Since $C^T = C^{-1}$ and $\det(C) = 1$, we get the identity:

$$C [v]_\times = [ Cv ]_\times C \qquad \forall \; v \in \mathbb{R}^3, \; C \in SO(3).$$

Algebraic Properties

Anti-symmetry

It is easy to verify that: $$ [v]_\times u = -[u]_\times v. $$

Lie Bracket

One of the most useful identities when working with cross-product matrices is: $$ \begin{equation} [u]_\times [v]_\times = vu^T - u^T v I. \label{eq:product} \end{equation} $$

This is not too painful to verify explicitly. From this property we see that the commutator (or Lie bracket) of $[u]_\times$ and $[v]_\times$ can be written: $$ \begin{equation} [u]_\times [v]_\times - [v]_\times [u]_\times = vu^T - uv^T. \end{equation} $$ A final important equation is that: $$ [[u]_\times v]_\times = vu^T - uv^T $$ This can also be verified explicitly (as a hint, the anti-symmetry and bilinearity of the expression means it is sufficient to test the result for the pairs $(u, v) \in \{(e_x, e_y), (e_y, e_z), (e_x, e_z)\}$).

Power Relationships

We can also apply Equation $\eqref{eq:product}$ to simplify powers of $[u]_\times$: $$ \begin{align} [u ]_\times^3 &= (uu^T - \|u\|^2 I) [u]_\times = -\|u\|^2[u]_\times \\
[u ]_\times^4 &= -[u]_\times^3 [u]_\times = -\|u\|^2[u]_\times^2 \end{align} $$

Eigenvector Analysis

If $u = 0$ then all the eigenvalues are trivially zero.

Otherwise, choose $v$ be any vector such that $v^Tu = 0$ and let $w = \frac{1}{\|u\|}[u]_\times v$. We will show that the eigenvector, eigenvalue pairs are:

• $(u, 0)$
• $(v + jw, -j\|u\|)$
• $(v - jw, j\|u\|)$

Proof: It is easy to verify that $[u]_\times u = 0$. We next verify that $v + j w$ is an eigenvector with eigenvalue $-j\|u\|$: $$ \begin{align} [u]_x (v + jw) &= \|u\| w + j [u]_x w\\
&= \|u\|w + j \frac{1}{\|u\|}[u]_x^2 v \\
&= \|u\|w + j \frac{1}{\|u\|}(uu^T - \|u\|^2 I) v \\
&= \|u\|\left(w - j v \right) = -j\|u\|\left(v+jw\right). \end{align} $$ That $v - jw$ is an eigenvector with eigenvalue $j\|u\|$ can be shown similarly.

This line or reasoning can be made more concrete by analyzing the matrix $[e_x]_\times$ with $u = e_y$ and $w = e_z$, and then concluding that the resulting analysis can be transfered to an arbitrary $[v]_\times$ by a rotation matrix and scaling.