Grinshpan
The cross product transformation
1 Choose a nonzero (force) vector in R3, say v = 2 . For x in R3 (position vector), 3 consider the linear transformation T (x) = x × v, where × is the cross product operation. Geometrically, x × v is a vector orthogonal to both x and v and such that x, v, x × v form a right-hand triple:
The length of x × v is ∥x∥ ∥v∥ |sin α|, where α is the angle between x and v. In particular, x × v is the zero vector whenever x is a scalar multiple of v. Also note that x × v = −v × x.
For the standard basis vectors, we have: e1 × e2 = e3, e2 × e3 = e1, e3 × e1 = e2. In terms of coordinates, [ ] [ ] [ ] x1 v1 x2v3 − x3v2 x2 × v2 = −x1v3 + x3v1 . x3 v3 x1v2 − x2v1 Using geometry, we see that ker T is the line through the origin spanned by v and im T is the plane through the origin orthogonal to v. The standard matrix of T is [ ] [ ] 0 3 −2 A = e1 × v e2 × v e3 × v = −3 0 1 . 2 −1 0 ([ ]) ([ ] [ ]) 1 −2 −3 We have ker T = ker A = span 2 and im T = im A = span 1 , 0 . 3 0 1
What are the eigenvalues of A? Since A is not invertible, one of its eigenvalues is λ = 0. The corresponding eigenspace is one-dimensional: ([ ]) 1 E0 = ker(A − 0 · I) = ker A = span 2 . 3
For every x in E0, we have Ax = 0x = 0. Does A have other eigenvalues?
−λ 3 −2 −λ 1 −3 1 −3 −λ det(A − λI) = −3 −λ 1 = (−λ) −3 + (−2) −1 −λ 2 −λ 2 −1 2 −1 −λ = −λ3 − λ − 9λ + 6 − 6 − 4λ = −λ(λ2 + 14). Since λ2 + 14 = 0 has no real roots, A has no real eigenvalues other than λ = 0. 3 Furthermore, because E0 is one-dimensional, it is not possible to find a basis for R consisting of the eigenvectors of A; not even two linearly independent real eigenvectors.
However, we may still be able to find a basis for R3 more suitable than standard for the purposes of representing our transformation T. Let us exploit the fact that ker T and im T are orthogonal subspaces. Consider, for instance, the following choice: [ ] [ ] [ ] 1 −2 −3 B : w1 = 2 , w2 = 1 , w3 = 0 . 3 0 1
3 It is easy to check that B is indeed a basis for R . Note that w1 = v is in ker A, w1 is orthogonal to both w2 and w3, and w2, w3 is a basis for im A. We have [ ] 0 T (w1) = w1 × w1 = 0 = 0w1 + 0w2 + 0w3 0 [ ] 3 T (w2) = w2 × w1 = 6 = 0w1 + 6w2 − 5w3 −5 [ ] −2 T (w3) = w3 × w1 = 10 = 0w1 + 10w2 − 6w3. −6 [ ] 0 0 0 So that T in the B-basis is represented by B = 0 6 10 . 0 −5 −6 In many respects, B is simpler than A, but it lacks skew-symmetry. Perhaps, a better choice would be to make w2, w3 an orthogonal basis for im A. Consider [ ] [ ] [ ] 1 −2 −3 ˜ B : w1 = 2 , w2 = 1 , w˜3 = w1 × w2 = −6 , 3 0 5 an orthogonal basis for R3. A quick check shows that, in the B˜-basis, T is represented by [ ] 0 0 0 B˜ = 0 0 14 . 0 −1 0 Let us finally adjust the lengths to obtain an orthonormal basis for R3 : 1 −2 −3 1 1 1 Bo : u1 = √ 2 , u2 = √ 1 , u3 = √ −6 . 14 3 5 0 70 5 Do you see what is then the matrix of T ?
0 0√ 0 Answer: B0 = 0√ 0 14 . The skew-symmetry is restored. 0 − 14 0
The action of T can be read off this matrix. Given any x = c1u1 + c2u2 + c3u3, T eliminates its u1-component (orthogonal√ projection onto the u2u3-plane), stretches the u2u3-component by a factor of ∥v∥ = 14 (force magnitude) and rotates the result by an ◦ angle of 90 about the u1-axis (clockwise). In the language of physics/mechanics, T gives the torque.
Note that since the basis Bo is orthonormal,[ the] coordinates with respect to Bo are easy −1 ⊤ to find, the similarity matrix S = u1 u2 u3 is orthogonal, S = S , and we have ⊤ A = SB0S .