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I am aware that an arbitrary $3D$ Rotational Matrix can be constructed by successive rotations about the $x, y$ and $z$ axes. How can I use this to show that all $3D$ rotation matrices are othogonal and have $det=1$.

Would I just need to individually prove each matrix is orthogonal, i.e show that the transpose of each matrix is the same as its inverse and then show that the product of the matrices is also orthogonal?

For showing that the $det=1$ is this just a trivial calculation of the determinant of each rotation matrix? And then using the property that the determinant of the product of matrices is just the product of each individual determinant?

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Feb 22 at 16:43
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You obtain your rotation matrix $R$ from the product of three matrices

$$R=R_x(\alpha)\,R_y(\beta)\,R_z(\gamma).$$

If $R $is orthogonal thus $R^T\,R=I_3$ where $I_3$ $3\times 3$ identity matrix

$$R^T\,R=(R_x\,R_y\,R_z)^T\,(R_x\,R_y\,R_z)=R_z^T\,R_y^T\,R_x^T\,R_x\,R_y\,R_z$$

with:

$R_x^T\,R_x=I_3\quad,R_y^T\,R_y=I_3$ and $\quad R_z^T\,R_z=I_3.$

Thus $$R^T\,R=I_3\, ,\quad R^{-1}=R^T\, ,$$

and the determinant of $R$ is:

$$\det(R)=\det(R_x)\,\det(R_y)\,\det(R_z)=1\times 1\times 1=1$$


$$S_x(\alpha)=\left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( \alpha \right) &-\sin \left( \alpha \right) \\ 0& \sin \left( \alpha \right) &\cos \left( \alpha \right) \end {array} \right] $$

$$S_y(\beta)= \left[ \begin {array}{ccc} \cos \left( \beta \right) &0&\sin \left( \beta \right) \\ 0&1&0\\ -\sin \left( \beta \right) &0&\cos \left( \beta \right) \end {array} \right] $$

$$S_z(\gamma)= \left[ \begin {array}{ccc} \cos \left( \gamma \right) &-\sin \left( \gamma \right) &0\\ \sin \left( \gamma \right) &\cos \left( \gamma \right) &0\\ 0&0&1\end {array} \right] $$

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By definition orthogonal matrices are real and satisfy $R\cdot R^\top = \hat 1$ where $\hat 1$ is the unit matrix. Taking the determinant shows that $(\hbox{Det}(R))^2=1$. The special orthogonal matrices, which form the group SO(n) rather than just O(n), are those orthogonal matrices with determinant=+1 and correspond to rotations. The simplest way to show this is to consider rotation by an arbitrary angle $\theta$. Then for $\theta=0$ one must recover the identity, which has determinant +1. Since the rotation angle is continuous, it follows that the rotation matrices must also have determinant=+1.

The orthogonal matrices not only include rotations but also reflections, which are not continuous transformation and thus cannot be reached by taking some parameter to $0$ as for the rotation matrices.

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Yes one can do it in the way you’ve described. But one must note that the definition of rotation is a continuous linear transformation that preserves length. And the property of matrices that preserve length is that they have determinant squared equals one. Plus one determinant corresponds to rotations and minus one corresponds to reflection.

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  • $\begingroup$ Ok thanks for your help. $\endgroup$ – GavinK14 Feb 22 at 16:31
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    $\begingroup$ Or minus one, reflections do preserve length as well $\endgroup$ – Cryo Feb 22 at 23:58
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    $\begingroup$ -1. Having unit determinant does not guarantee that the matrix will preserve lengths, as this answer suggests. The converse is almost true (up to a nontrivial sign) but it's a different property. $\endgroup$ – E.P. Feb 23 at 1:50
  • $\begingroup$ @EmilioPisanty right! I goofed up. Thank you. $\endgroup$ – Fellow Traveller Feb 23 at 3:12
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IMHO proving that operator that does rotation in 3D space, is relatively simple - write out the matrix:

$\left(\begin{array} \ \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{array}\right)$

And do the calculation. More interesting is doing the opposite.You can actually prove that any real-valued 3-by-3 orhogonal matrix with determinant $+1$ can be written as rotation about matrix in some basis, i.e. you can prove that for any such matrix there exists a basis where the matrix looks like this

$\left(\begin{array} \ a & -b & 0 \\ b & a & 0 \\ 0 & 0 & 1 \end{array}\right)$

where $a,b\in\mathbb{R}$ and $a^2+b^2=1$. The proof can be found in Axler's Linear algrbra done right (Operators on Inner Product Spaces).

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