How Can We Prove the Transformation of Cross Product Under Orthogonal Matrices?

[jostpuur]

If A\in O(3) and v,w\in\mathbb{R}^3, then

<br /> (Av)\times (Aw) = (\textrm{det}\;A) A(v\times w)<br />

is right, right?

Is there any simple way of proving this? It doesn't come easily like invariance of dot product. Isn't this equation also important for understanding the pseudovectors?

 

[Jmf]

<br /> (Av)\times (Aw) = (\textrm{det}\;A) (A^{-1})^{T}(v\times w)<br />

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.

 

[jostpuur]

Very interesting comment. That reminded me about the explicit formula for A^{-1} in terms of the cofactors. I didn't try to work with them earlier, because I wasn't thinking about inverses.

 

[Jmf]

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

AA^*=A^*A

or

AA^T=A^TA

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 A = MDM^T, 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:

A = \begin{bmatrix} \lambda_1 &amp; 0 &amp; 0 \\ 0 &amp; \lambda_2 &amp; 0 \\ 0 &amp; 0 &amp; \lambda_3 \end{bmatrix}

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

<br /> (Av)\times (Aw) = |A| (A^{-1})^{T}(v\times w)<br />

Note that:

|A| = \lambda_1\lambda_2\lambda_3

(A^{-1})^T = A^{-1} = \begin{bmatrix} 1\over\lambda_1 &amp; 0 &amp; 0 \\ 0 &amp; 1\over\lambda_2 &amp; 0 \\ 0 &amp; 0 &amp; 1\over\lambda_3 \end{bmatrix}

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

<br /> \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 &amp; 0 &amp; 0 \\ 0&amp; 1 \over \lambda_2 &amp; 0 \\ 0 &amp; 0 &amp; 1 \over \lambda_3 \end{bmatrix} \begin{bmatrix} (v\times w)_1 \\ (v\times w)_2\\ (v\times w)_3 \end{bmatrix}<br />

So:

<br /> \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}<br />

hence:

<br /> \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}<br />

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

Then on performing these cross products we get:

\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}

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

 

[jostpuur]

I just managed to complete my proof too. You are only couple of minutes ahead of me.

I'll start typing now! (Proof coming in my next post)

 

[Jmf]

I just managed to complete my proof too. You are only couple of minutes ahead of me.

I'll start typing now! (Proof coming in my next post)

Haha, brilliant. Can't wait.

 

[jostpuur]

A big remark is that the inverse matrix of a 3x3 matrix can be written in terms of cross products.

First check the formula for an inverse matrix:

<br /> A^{-1} = \frac{1}{\textrm{det}(A)} C^T<br />

http://en.wikipedia.org/wiki/Cofactor_(linear_algebra)

With finite amount of effort it is possible to write the cofactor matrix C as follows:

<br /> C = \big( A_{*2}\times A_{*3},\;A_{*3}\times A_{*1},\; A_{*1}\times A_{*2}\big)<br />

Here a notation

<br /> A = \big(A_{*1}, A_{*2}, A_{*3}\big)<br />

<br /> A_{*k} = \left(\begin{array}{c}<br /> A_{1k} \\ A_{2k} \\ A_{3k} \\<br /> \end{array}\right)<br />

is used.

Then:

<br /> (Av)\times (Aw) = \epsilon_{ijk} (Av)_i (Aw)_j e_k = \epsilon_{ijk} (A_{ii&#039;}v_{i&#039;}) (A_{jj&#039;} w_{j&#039;}) e_k<br /> = \big(\epsilon_{ijk} A_{ii&#039;} A_{jj&#039;} e_k\big) v_{i&#039;} w_{j&#039;}<br />

<br /> = (A_{*i&#039;} \times A_{*j&#039;}) v_{i&#039;} w_{j&#039;} =(A_{*1}\times A_{*2})(v_1w_2 - v_2w_1) \;+\; (A_{*2}\times A_{*3})(v_2w_3 - v_3w_2) \;+\; (A_{*3}\times A_{*1})(v_3w_1 - v_1w_3)<br />

<br /> = \big(A_{*2}\times A_{*3},\; A_{*3}\times A_{*1},\; A_{*1}\times A_{*2}\big)<br /> \left(\begin{array}{c}<br /> (v\times w)_1 \\ (v\times w)_2 \\ (v\times w)_3 \\<br /> \end{array}\right) = C(v\times w) = \textrm{det}(A) (A^{-1})^T (v\times w)<br />

 

[Jmf]

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 :)