Cross product invariant under SO(3)-matrices?

In summary, the conversation was about proving the statement M(\vec a\times\vec b)=M\vec a\times M\vec b using the fact that M is in SO(3) and preserves the orientation of the basis. The conversation includes using the epsilon tensor and the determinant of M to arrive at the desired result. A helpful source was also provided for further reference.
  • #1
Marin
193
0
Hi there!I'm trying to prove the following obvious statement, but am somehow stuck :(

Let [tex]\vec a,\ \vec b\in\mathbb{R^3}[/tex] , let M be in SO(3) and x be the cross productprove: [tex]M(\vec a\times\vec b)=M\vec a\times M\vec b[/tex]I tried using the epsilon tensor, as in physics, but it doesn't really produce an opportunity, as you can convince yourself:

[tex](M\vec a\times M\vec b)_i=\varepsilon_{ijk}(M\vec a)_j(M\vec b)_k=\varepsilon_{ijk}M_{jl} a_l M_{km}b_m...[/tex]

where as usual summation is over repeated indices

Now I want to use the fact that M is orthogonal, i.e.
[tex]M_{ij}M_{jk}^t=\delta_{ik}[/tex],
and preserves the orientation of the basis but I don't know where exactly this has to come into the proof...

What I want to end up with is

[tex]...=M_{ij}\varepsilon_{jkl}a_kb_l=(M(\vec a\times\vec b))_i[/tex]

The statement seems to me obvious and can be envisioned very quickly by the right-hand-rule; I don't know why establishing it makes real problemsIf you have any idea, I'd be glad to see it :)With regards,
marin
 
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  • #2
Hi Marlin,

I'll give you a hint (mainly 'cause I'm too lazy to latex up the answer)
The 'preserving the orientation of the basis' bit is equivalent to saying det(M)=1
Rewrite the determinant using the epsilon tensor then a little algebraic rearrangement should get you the result.
(The formula I'm thinking of can be found http://en.wikipedia.org/wiki/Cross_product#Algebraic_properties"...)
 
Last edited by a moderator:
  • #4
Thanks a lot :)

this was very helpful :)
 
  • #5
More Generally
Where A is any invertable 3x3 matrix and M(A) is the matrix of the minors of A

[tex]A\vec a\times A\vec b=M(A)(\vec a\times\vec b)[/tex]

I got this result by playing with the algebra. Hope I didn't make a mistake.:rolleyes:
 

1. What is the cross product invariant under SO(3)-matrices?

The cross product invariant under SO(3)-matrices is a mathematical concept that describes the preservation of the magnitude and direction of the cross product of two vectors when transformed by a rotation matrix in the special orthogonal group SO(3). In other words, the cross product of two vectors will remain the same regardless of the orientation of the coordinate system.

2. Why is the cross product invariant under SO(3)-matrices important?

This concept is important in various fields such as physics, engineering, and computer graphics. It allows for the accurate representation of rotations and transformations in three-dimensional space, which is essential in many real-world applications.

3. How is the cross product invariant under SO(3)-matrices calculated?

The cross product invariant can be calculated using the right-hand rule, where the magnitude of the cross product is equal to the product of the magnitudes of the two vectors multiplied by the sine of the angle between them. The direction of the cross product is perpendicular to the two vectors and follows the right-hand rule.

4. Can the cross product invariant under SO(3)-matrices be used for non-orthogonal matrices?

No, the cross product invariant is only applicable for matrices in the special orthogonal group SO(3), which consists of 3x3 rotation matrices with a determinant of 1. Non-orthogonal matrices may result in a change in the magnitude and direction of the cross product.

5. Are there any real-world examples of the cross product invariant under SO(3)-matrices?

Yes, the cross product invariant can be observed in various everyday objects such as gyroscope, bicycle wheels, and planetary motion. It is also used in computer graphics for 3D transformations and in robotics for calculating the orientation of objects in space.

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