rotational invariance of cross product

Let R be a rotational 3×3 matrix, i.e., a real matrix with det𝐑=1 and 𝐑-1=𝐑T. Then for all vectors 𝐮,𝐯 in 3,


Proof. Let us first fix some right hand oriented orthonormal basis in 3. Further, let {u1,u2,u3} and {v1,v2,v3} be the componentsMathworldPlanetmathPlanetmathPlanetmath of u and v in that basis. Also, in the chosen basis, we denote the entries of R by Rij. Since R is rotational, we have RijRkj=δik where δik is the Kronecker delta symbol. Here we use the Einstein summation convention. Thus, in the previous expression, on the left hand side, j should be summed over 1,2,3. We shall use the Levi-Civita permutation symbol ε to write the cross productMathworldPlanetmath. Then the i:th coordinate of 𝐮×𝐯 equals (𝐮×𝐯)i=εijkujvk. For the kth component of (𝐑𝐮)×(𝐑𝐯) we then have

((𝐑𝐮)×(𝐑𝐯))k = εimkRijRmnujvn
= εimlδklRijRmnujvn
= εimlRkrRlrRijRmnujvn
= εjnrdet𝐑Rkrujvn.

The last line follows since εijkRimRjnRkr=εmnrεijkRi1Rj2Rk3=εmnrdet𝐑. Since det𝐑=1, it follows that

((𝐑𝐮)×(𝐑𝐯))k = Rkrεjnrujvn
= Rkr(𝐮×𝐯)r
= (𝐑𝐮×𝐯)k

as claimed.

Title rotational invariance of cross product
Canonical name RotationalInvarianceOfCrossProduct
Date of creation 2013-03-22 13:33:53
Last modified on 2013-03-22 13:33:53
Owner matte (1858)
Last modified by matte (1858)
Numerical id 6
Author matte (1858)
Entry type Theorem
Classification msc 15A72
Classification msc 15A90