# Rotational invariance of cross product matrix operator

• I
• Filip Larsen
In summary: Read moreIn summary, the normal vector cross product is rotational invariant, meaning that the equation ##\mathbf R(a\times b) = (\mathbf R a)\times(\mathbf R b)## holds for any two arbitrary vectors ##a, b \in \mathbb{R}^3## and a 3x3 rotation matrix ##\mathbf R##. This also applies to the cross product matrix operator ##[a]_\times = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix}##, with ##a \times
Filip Larsen
Gold Member
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.

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) .##

fresh_42 said:
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?

Filip Larsen said:
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.

fresh_42 said:
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.

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
fresh_42 said:
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.

fresh_42 said:
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
Filip Larsen said:
OK, assuming this is a real math rated "of course" then I guess I was just worried about nothing
Yep.

## 1. What is rotational invariance of cross product matrix operator?

Rotational invariance of cross product matrix operator refers to the property of the cross product matrix operator to remain unchanged under rotations. This means that no matter how the coordinate system is rotated, the cross product matrix operator will still produce the same result.

## 2. Why is rotational invariance important in science?

Rotational invariance is important in science because it allows us to analyze and understand physical phenomena without being affected by the orientation of the coordinate system. This helps to simplify calculations and make them more accurate.

## 3. How is rotational invariance of cross product matrix operator related to vector operations?

The cross product matrix operator is used to perform vector operations such as calculating the cross product of two vectors. The rotational invariance of this operator ensures that the result of the cross product remains the same, regardless of the orientation of the coordinate system.

## 4. Can you give an example of rotational invariance in real life?

An example of rotational invariance in real life is the rotation of a bicycle wheel. No matter how the bicycle is turned or tilted, the wheel will continue to rotate in the same direction. This is because the rotational invariance of the wheel allows it to maintain its angular momentum.

## 5. How is rotational invariance tested in scientific experiments?

Rotational invariance can be tested in scientific experiments by performing the same experiment in different orientations and comparing the results. If the results are consistent, then the experiment exhibits rotational invariance. Additionally, mathematical calculations can also be used to test for rotational invariance.

• Linear and Abstract Algebra
Replies
1
Views
979
• Calculus
Replies
6
Views
695
• Linear and Abstract Algebra
Replies
2
Views
2K
Replies
1
Views
974
• Linear and Abstract Algebra
Replies
2
Views
1K
• Calculus
Replies
4
Views
544
• Classical Physics
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
2
Views
1K
• Mechanics
Replies
2
Views
696
• Linear and Abstract Algebra
Replies
1
Views
1K