Skew-Symmetric Matrix

A square matrix $\mathbf{A}$ is skew-symmetric (or antisymmetric) if its transpose is its negative:

$${\mathbf{A}}^{\top }=-\mathbf{A}$$

In terms of its entries $a_{ij}$, this means $a_{ji}=-a_{ij}$ for all $i,j$.

Properties

• Zero Diagonal: The diagonal elements of a skew-symmetric matrix must be zero, as $a_{ii}=-a_{ii}⟹a_{ii}=0$.
• Quadratic Form: For any vector $\mathbf{x}$, the quadratic form associated with a skew-symmetric matrix is zero: $${\mathbf{x}}^{\top }\mathbf{Ax}=0$$
• Eigenvalues: The eigenvalues of a real skew-symmetric matrix are either zero or purely imaginary.
• Trace: The trace of a skew-symmetric matrix is always zero.
• Lie Algebra: The set of all $n\times n$ skew-symmetric matrices forms the Lie algebra $\mathfrak{so}(n)$ of the orthogonal group $O(n)$ and the special orthogonal group $\mathrm{SO}(n)$.

3D Case: The Hat Map

In 3D Euclidean Space, there is a one-to-one correspondence between vectors in ${\mathbb{R}}^3$ and $3\times 3$ skew-symmetric matrices. For a vector $\mathbf{\omega }=({\omega }_1,{\omega }_2,{\omega }_3{)}^{\top }$, we define the hat map $[\cdot {]}_{\times }:{\mathbb{R}}^3\to \mathrm{SO}(3)$ as:

$$[\mathbf{\omega }{]}_{\times }=\hat{\mathbf{\omega }}=\left[\begin{array}{ccc}0& -{\omega }_3& {\omega }_2\\ {\omega }_3& 0& -{\omega }_1\\ -{\omega }_2& {\omega }_1& 0\end{array}\right]$$

It can also be denoted as $[\omega \times ]$.

Relation to Cross Product

The skew-symmetric matrix $[\mathbf{\omega }{]}_{\times }$ represents the cross product operation with $\mathbf{\omega }$. For any vector $\mathbf{v}\in {\mathbb{R}}^3$:

$$[\mathbf{\omega }{]}_{\times }\mathbf{v}=\mathbf{\omega }\times \mathbf{v}$$

Application in Rigid Body Dynamics

In rigid body dynamics, skew-symmetric matrices are used to represent angular velocity. If $\mathbf{R}(t)$ is a rotation matrix, then its derivative $\dot{\mathbf{R}}(t)$ satisfies:

$$\dot{\mathbf{R}}=[{\mathbf{\omega }}_s{]}_{\times }\mathbf{R}=\mathbf{R}[{\mathbf{\omega }}_b{]}_{\times }$$

where ${\mathbf{\omega }}_s$ is the spatial angular velocity and ${\mathbf{\omega }}_b$ is the body angular velocity. The matrix ${\mathbf{R}}^{\top }\dot{\mathbf{R}}$ is always skew-symmetric.

Theorem (Skew-Symmetric Transpose Multiplication)

Let $a$ be a vector in ${\mathbb{R}}^3$ and $[a{]}_{\times }$ be the corresponding skew-symmetric matrix. Then:

$$[a{]}_{\times }^{\top }[a{]}_{\times }=|a{|}^{2}I-a{a}^{\top }$$

Proof:

First, recall that $[a{]}_{\times }^{\top }=-[a{]}_{\times }$. Thus, $[a{]}_{\times }^{\top }[a{]}_{\times }=-[a{]}_{\times }^2$.

We compute $[a{]}_{\times }^2$ by direct matrix multiplication:

$$\begin{array}{rl}[a{]}_{\times }^2& =\left[\begin{array}{ccc}0& -a_3& a_2\\ a_3& 0& -a_1\\ -a_2& a_1& 0\end{array}\right]\left[\begin{array}{ccc}0& -a_3& a_2\\ a_3& 0& -a_1\\ -a_2& a_1& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}-(a_2^2+a_3^2)& a_1{a}_2& a_1{a}_3\\ a_1{a}_2& -(a_1^2+a_3^2)& a_2{a}_3\\ a_1{a}_3& a_2{a}_3& -(a_1^2+a_2^2)\end{array}\right]\end{array}$$

Note that $|a{|}^2=a_1^2+a_2^2+a_3^2$. We can rewrite the diagonal elements as

$$-(a_2^2+a_3^2)=a_1^2-|a{|}^2$$

and so on:

$$\begin{array}{rl}[a{]}_{\times }^2& =\left[\begin{array}{ccc}{a}_1^2-|a{|}^2& a_1{a}_2& a_1{a}_3\\ a_1{a}_2& a_2^2-|a{|}^2& a_2{a}_3\\ a_1{a}_3& a_2{a}_3& a_3^2-|a{|}^2\end{array}\right]\\ & =\left[\begin{array}{ccc}{a}_1^2& a_1{a}_2& a_1{a}_3\\ a_1{a}_2& a_2^2& a_2{a}_3\\ a_1{a}_3& a_2{a}_3& a_3^2\end{array}\right]-|a{|}^2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ & =a{a}^{\top }-|a{|}^{2}I\end{array}$$

Therefore, $[a{]}_{\times }^{\top }[a{]}_{\times }=-[a{]}_{\times }^2=|a{|}^{2}I-a{a}^{\top }$.