Prove Determinant of Rotation Matrix is 1 w/Continuity Argument

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Homework Help Overview

The discussion revolves around proving that the determinant of a rotation matrix is always 1, specifically through the use of a continuity argument. Participants are exploring the definitions and properties of rotation matrices and their determinants.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the meaning of a continuity argument in the context of matrices and determinants. Some suggest that a straightforward approach involves calculating the determinant of a specific rotation matrix. Others propose using Euler's rotation theorem to express a general rotation matrix as a product of simpler rotation matrices.

Discussion Status

The discussion is ongoing, with various interpretations of the continuity argument being explored. Some participants have provided definitions and properties of rotation matrices, while others are seeking clarification on the rigor of proposed proofs.

Contextual Notes

There is a mention of the need for definitions related to rotation matrices and the implications of isometric transformations. Participants are also considering the implications of orientation preservation in relation to the determinant.

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What's a continuity argument? For example, a question asks to prove that the determinant of a rotation matrix is always 1 using a continuity argument?
 
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Anyone? How would i prove in general that the det of a rotation matrix is 1?
 
I am unsure of what you mean by a continuity argument as it pertains to matrices and determinants. But you can straightforwardly show that the determinant of a rotation matrix is 1 by writing down the matrix and taking its determinant.

For instance, a clockwise rotation by an angle [itex]\theta[/itex] about the [itex]z[/itex] axis is described by the following matrix:

[tex]R_z(\theta)=\left[\begin{array}{ccc}\cos(\theta)&sin(\theta)&0 \\ -\sin(\theta)&\cos(\theta)&0 \\ 0&0&1 \end{array}\right ][/tex]

It's a piece of cake to show that [itex]det(R_z(\theta))=1[/itex].
 
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I'm not sure what the question means by continuity argument either. That's why I asked.

I know could just take the determinant of the rotation matrix. Since by euler's rotation theorem any rotation matrix M can be expressed as rotations over three perpendicular axis, det M = det R1 * det R2 * det R3 = 1, where R1, R2, and R3, are just the general rotation matrices for rotating over x,y, and z axis. But I wasn't sure if this 'proof' is sufficiently rigorous and general.
 
When asking questions like this one it would be nice to define what you are talking about. In this case a rotation matrix. The definition
Rotation matrix: A matrix for which det(A)=1
would be very helpful

Another definition is
Rotation matrix: An isometric orientation preserving linear transform
by isometric I mean the inner product defined by A matches the one defined by I (idenity matrix)
(Ax)'Ay=x'y
where ' is the adjoint (conjugate transpose or if A is real transpose)
immediately we have
A'A=I
This only insures abs(det(A))=1
so we turn to the orientation preserving bit
That is fancy talk, but it means that I is our prototype rotation
that is
T(x,y,z)=(-x,y,z)
would be a "bad" or improper rotation
we want for A a rotation and B any matrix
det(AB)=det(B)
which again does not teach us much
I said we want I to be a prototype
det(I)=1 but again we want to go further
say h is a small number and we wish to construct an almost rotation
I+Ah
we have
(I+Ah)'=I+A'h
and
(I+Ah)^-1~I-Ah
thus we require
A'=-A
we could also make better almost rotations
I+Ah+A^2h^2/2+A^3h^3/6+...
The ultimate result being the actual rotation
exp(Ah)
and
of course
A'=-A->tr(A)=0
det(exp(At))=exp(tr(At))=exp(0)=1
Thus rotation matricies are of the form exp(A) where A'=-A thus have det=1
 

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