Cross product as a matrix-vector product

linear-algebra, math

As a quick refresher, the cross product of two vectors u,vR3u, v \in \R^3 is defined as follows:

u×v=i^j^k^u1u2u3v1v2v3=(u2v3u3v2u3v1u1v3u1v2u2v1), u \times v = \begin{vmatrix*} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix*} = \begin{pmatrix} u_2 v_3 - u_3 v_2 \\ u_3 v_1 - u_1 v_3 \\ u_1 v_2 - u_2 v_1 \end{pmatrix} ,

where u=(u1,u2,u3)u = (u_1, u_2, u_3) and v=(v1,v2,v3)v = (v_1, v_2, v_3). The expression in the center above represents the determinant of the 3×33 \times 3 matrix. It is important to note that the vector u×vu \times v is orthogonal to both uu and vv.

Now, observe that the cross product can be expressed as a product of a skew-symmetric matrix and a vector:

u×v=u×v=(0u3u2u30u1u2u10)(v1v2v3), u \times v = u^{\times} v = \begin{pmatrix} 0 & -u_3 & u_2 \\ u_3 & 0 & -u_1 \\ -u_2 & u_1 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} ,

where I’m defining u×u^{\times} to be the skew-symmetric matrix shown just above. A matrix is skew-symmetric if AT=AA^T = -A. Note that, by the anticommutative property of the cross product, i.e., u×v=v×uu \times v = -v \times u, we may also write

u×v=v×u=v×u=(0v3v2v30v1v2v10)(u1u2u3) u \times v = - v \times u = -v^{\times} u = -\begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}

© 2024 Peter Cheng