Rotational invariance of cross product matrix operator

[Filip Larsen]

TL;DR Summary

Do rotational invariance of the vector cross product also carry over to the cross product matrix operator?

Given that the normal vector cross product is rotational invariant, that is $$\mathbf R(a\times b) = (\mathbf R a)\times(\mathbf R b),$$ where $a, b \in \mathbb{R}^3$ are two arbitrary (column) vectors and $\mathbf R$ is a 3x3 rotation matrix, and given the cross product matrix operator defined by $$ \left[a\right]_\times = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix} ,$$ such that $a \times b = \left[a\right]_\times b$, my question now is if rotational invariance also applies for this operator, that is if it in general holds that $$\mathbf R \left[ a \right]_\times \stackrel{?}{=} \left[ \mathbf R a \right]_\times \mathbf R$$ Specifically for my current use, with $\mathbf C$ being 3x3 (positive semi-definite) matrix and utilizing $\mathbf R^{-1} = \mathbf R^T$ holds for a rotation matrix can I then conclude that $\left[\mathbf R a \right]_\times \mathbf C \left[ \mathbf R a \right]_\times^T = \left( \mathbf R \left[ a \right]_\times \mathbf R^T \right) \mathbf C \left(\mathbf R \left[ a \right]_\times \mathbf R^T \right)^T = \left( \mathbf R \left[ a \right]_\times \right) \left( \mathbf R^T \mathbf C \mathbf R \right) \left( \mathbf R \left[ a \right]_\times \right)^T$ always holds, as I am inclined to believe?

My (engineering) intuition tries to tell me that since the relation with the question marks holds when applied to a column vector, due to $\left( \mathbf R \left[ a \right]_\times \right) b = \mathbf R \left( a\times b \right) = \left(\mathbf R a\right) \times \left(\mathbf R b\right) = \left( \left[ \mathbf R a \right]_\times\right) \left( \mathbf R b \right) = \left( \left[ \mathbf R a \right]_\times \mathbf R \right) b$, and since an equation involving multiplication of a 3x3 matrix can be separated into 3 equations with multiplication of each column vector of the matrix, then the relation must also hold when combined back into a general 3x3 matrix, but I worry if such math hand waving has math holes in it my engineering intuition can't see.

 

[fresh_42]

If we had the conjugation as operation: $a\longmapsto \mathbf{R}a\mathbf{R}^{-1}$ then invariance would follow directly by the Lie group $SO(3)$ operation on its Lie algebra.

If we only have a rotation once on the left, I think the quickest way is simply to check it. The situation is symmetric in all coordinates, so it is sufficient to check a rotation with one given angle around the z−axis. Thus you get either a proof or a counterexample: $\mathbf{R}_z(\varphi)([a]_{\times} b) \stackrel{?}{=} (\mathbf{R}_z)(\varphi)a)_\times (\mathbf{R}_z\varphi)b) .$

 

[Filip Larsen]

Thus you get either a proof or a counterexample: $\mathbf{R}_z(\varphi)([a]_{\times} b) \stackrel{?}{=} (\mathbf{R}_z)(\varphi)a)_\times (\mathbf{R}_z\varphi)b)$.

In my answer below assume you mean $\mathbf{R}_z(\varphi)([a]_{\times} b) \stackrel{?}{=} [\mathbf{R}_z(\varphi)a]_\times (\mathbf{R}_z(\varphi)b)$, with $b$ being a 3x3 matrix.

I was rather hoping not having to write out the full equations in scalars, even if it only involves rotation around a single axis. I know I am more or less repeating my question from before, but to prove it when $b$ is a matrix would it then be sufficient to decompose $b$ into column vectors, apply the relation known to be true when $b$ is a vector, and the assemble it back again to show you end up with the same final vector?

 

[fresh_42]

In my answer below assume you mean $\mathbf{R}_z(\varphi)([a]_{\times} b) \stackrel{?}{=} [\mathbf{R}_z(\varphi)a]_\times (\mathbf{R}_z(\varphi)b)$, with $b$ being a 3x3 matrix.

No, $b$ remains a vector.

I was rather hoping not having to write out the full equations in scalars, even if it only involves rotation around a single axis.

It makes two matrix multiplications and two matrix times vector multiplications. That's not too many to do. As I said, the natural operation would be a conjugation in which case there is nothing left to prove. But in that case we would have (in Lie algebra notation) $R[a,b]R^{-1}=RabR^{-1}-RbaR^{-1}$, i.e. the rotations in the middle cancel. If they don't as in your case, then symmetry will be lost: $Rab-Rba\stackrel{?}{=}RaRb-RbRa$. This means $ab-ba=a\times b \stackrel{?}{=}aRb-bRa$. This doesn't look true, although intuition say it is. I would look for a counterexample, that is, you have to do the calculation anyway.

I know I am more or less repeating my question from before, but to prove it when $b$ is a matrix would it then be sufficient to decompose $b$ into column vectors, apply the relation known to be true when $b$ is a vector, and the assemble it back again to show you end up with the same final vector?

I don't see how $b$ is a matrix.

 

[Filip Larsen]

I don't see how $b$ is a matrix.

Let me go with that first as I suspect we are perhaps misunderstanding each other here.

As I tried to indicate in my first post, I know from the properties of the cross product matrix operator $\left[\cdot\right]_\times$ and from the rotational invariance of the vector cross product that $$ \begin{align}
\mathbf R \left[a\right]_\times b & = \mathbf R \left( \left[a\right]_\times b \right) \nonumber \\
& = \mathbf R(a\times b) \nonumber \\
& = (\mathbf R a)\times(\mathbf R b) \nonumber \\
& = \left[ \mathbf R a \right]_\times (\mathbf R b) \nonumber \\
& = \left( \left[ \mathbf R a \right]_\times \mathbf R \right) b \nonumber
\end{align} ,$$ where as before $a, b$ are vectors and $\mathbf R$ is a rotation matrix. This implies (as far as I can see) that $$\mathbf R \left[a\right]_\times = \left[ \mathbf R a \right]_\times \mathbf R$$ is a valid matrix equality. So, if you are suggesting that I prove $\mathbf{R}_z(\varphi)([a]_{\times} b) \stackrel{?}{=} [\mathbf{R}_z(\varphi)a]_\times (\mathbf{R}_z(\varphi)b)$ with $b$ being a vector then I would claim I already known this to be true.

What I am asking, to repeat, is if there are any reason why the same shouldn't also hold as I expect when applied to a 3x3 matrix $\mathbf B$, that is if there is any reason why $$ \mathbf{R}([a]_{\times}\mathbf B) = [\mathbf{R}a]_\times \mathbf{R}\mathbf B$$ or, even more to the point regarding my intended use of this, if $$ \left[\mathbf R a \right]_\times \mathbf C \left[ \mathbf R a \right]_\times^T = \left( \mathbf R \left[ a \right]_\times \right) \left( \mathbf R^T \mathbf C \mathbf R \right) \left( \mathbf R \left[ a \right]_\times \right)^T $$ holds.

 

[fresh_42]

How is the third equation true? I thought that that was the problem. If it is, then you can of course multiply it by any matrix that you want.

 

[Filip Larsen]

How is the third equation true?

If you mean $\mathbf R(a \times b) = (\mathbf Ra)\times(\mathbf Rb)$ then that is true due to the rotational invariance of the cross product.

If it is, then you can of course multiply it by any matrix that you want.

OK, assuming this is a real math rated "of course" then I guess I was just worried about nothing

 

[fresh_42]

OK, assuming this is a real math rated "of course" then I guess I was just worried about nothing

Yep.