Transformation of cross product

[tex]
(Av)\times (Aw) = (\textrm{det}\;A) (A^{-1})^{T}(v\times w)
[/tex]

Is an identity, I believe. So for orthogonal A then what you put should be correct as it is a special case of the above.

I've spent 10 minutes trying to prove it though and I'm not getting anywhere, heh.

 

EDIT: I've since realised that this proof only works if A is a normal matrix - i.e:

[tex]AA^*=A^*A[/tex]

or

[tex]AA^T=A^TA[/tex]

if A is real.

Since I've assumed that an eigenbasis of A exists. Of course this is true for the set of orthogonal matrices that you were considering originally though, as well as eg. the symmetric and skew-symmetric matrices, so the proof is still mainly valid.

__________________

Been thinking about this a while, and I think I have a proof.

For a while I tried to prove it by using the spectral decomposition [tex]A = MDM^T[/tex], but after some fruitless algebra I ended up just resigning myself to choosing a basis. So here goes.

Please note that I study engineering though, not maths, so this may not be totally rigorous, or even correct:


Let's (hopefully without any loss of generality) choose our basis of the 3-dimensional euclidean space to be the (normalised) eigenbasis from A, i.e. we're assuming that the vectors u and w can be written as a sum of the normalised eigenvectors of A.

With this choice of basis, our matrix A becomes diagonal:

[tex]A = \begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix}[/tex]

Where the lambdas are the eigenvalues of A. Now we are trying to prove:

[tex]
(Av)\times (Aw) = |A| (A^{-1})^{T}(v\times w)
[/tex]

Note that:

[tex] |A| = \lambda_1\lambda_2\lambda_3 [/tex]

[tex] (A^{-1})^T = A^{-1} = \begin{bmatrix} 1\over\lambda_1 & 0 & 0 \\ 0 & 1\over\lambda_2 & 0 \\ 0 & 0 & 1\over\lambda_3 \end{bmatrix}[/tex]

Substituting these in to the right, and doing the matrix multiplications on the left (easy as A is diagonal) we get:

[tex]
\begin{bmatrix} \lambda_1 v_1 \\ \lambda_2 v_2 \\ \lambda_3 v_3 \end{bmatrix} \times \begin{bmatrix} \lambda_1 w_1 \\ \lambda_2 w_2 \\ \lambda_3 w_3 \end{bmatrix} = \lambda_1\lambda_2\lambda_3 \begin{bmatrix} 1 \over \lambda_1 & 0 & 0 \\ 0& 1 \over \lambda_2 & 0 \\ 0 & 0 & 1 \over \lambda_3 \end{bmatrix} \begin{bmatrix} (v\times w)_1 \\ (v\times w)_2\\ (v\times w)_3 \end{bmatrix}
[/tex]

So:

[tex]
\begin{bmatrix} \lambda_1 v_1 \\ \lambda_2 v_2 \\ \lambda_3 v_3 \end{bmatrix} \times \begin{bmatrix} \lambda_1 w_1 \\ \lambda_2 w_2 \\ \lambda_3 w_3 \end{bmatrix} = \lambda_1\lambda_2\lambda_3 \begin{bmatrix} (v\times w)_1 \over \lambda_1 \\ (v\times w)_2 \over \lambda_2\\ (v\times w)_3 \over \lambda_3 \end{bmatrix}
[/tex]

hence:

[tex]
\begin{bmatrix} \lambda_1 v_1 \\ \lambda_2 v_2 \\ \lambda_3 v_3 \end{bmatrix} \times \begin{bmatrix} \lambda_1 w_1 \\ \lambda_2 w_2 \\ \lambda_3 w_3 \end{bmatrix} = \begin{bmatrix} \lambda_2\lambda_3(v\times w)_1 \\ \lambda_1\lambda_3(v\times w)_2 \\ \lambda_1\lambda_2(v\times w)_3 \end{bmatrix}
[/tex]

NB: I'm using [tex](v \times w)_n[/tex] to mean the n'th component of the cross product.

Then on performing these cross products we get:

[tex]\begin{bmatrix} \lambda_2 \lambda_3 (v_2 w_3 - v_3 w_2) \\ \lambda_1 \lambda_3 (v_3 w_1 - v_1 w_3) \\ \lambda_1 \lambda_2 (v_1 w_2 - v_2 w_1) \end{bmatrix} = \begin{bmatrix} \lambda_2 \lambda_3 (v_2 w_3 - v_3 w_2) \\ \lambda_1 \lambda_3 (v_3 w_1 - v_1 w_3) \\ \lambda_1 \lambda_2 (v_1 w_2 - v_2 w_1) \end{bmatrix}[/tex]

Which completes the proof. Hopefully it doesn't matter that I've chosen a specific basis? I think the result should still apply generally. :)

 

Haha, brilliant. Can't wait. :tongue:

 

Hm, yeah. I did try proving it somewhat like that at first and got bogged down in the algebra. I like it more than mine (using the epsilon symbol (tensor?) is inspired, I didn't think of that) because mine has the assumption that you _have_ an eigenbasis, which I'm not sure is entirely justified.

I can't see where your transpose has gone though. :)

EDIT: Ahh I see, well spotted :)