8
$\begingroup$

I have to prove that this matrix is a rotation-matrix $$\begin{pmatrix} \frac12 & 0 & \frac{\sqrt{3}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{3}}{2} & 0 & \frac12 \end{pmatrix}$$ How do I do this? My idea is to multiplicate it with $\begin{pmatrix} x \\ y \\ z\end{pmatrix}$ and show that one component will remain unchanged . Is this enough? Do non-rotational transformations exist, which leave one component unchanged ?

$\endgroup$
  • $\begingroup$ Certainly projection onto one component leaves that component unchanged, but (in dimensions greater than $1$) projections onto a component are singular and so are not orthogonal. $\endgroup$ – Travis Nov 15 '14 at 11:04
  • 2
    $\begingroup$ How do you define rotation matrix? $\endgroup$ – Git Gud Nov 15 '14 at 11:16
  • $\begingroup$ No, simply multiplying by (x, y ,z) and showing that one component does not change is not enough. This property would be true for several non-rotational matrices as well. One examples are scaling (diagonal) matrices for the other two components (i.e. all non main-diagonal entries are zero, one main diagonal entry is one, the other ones are arbitrary). $\endgroup$ – Dreamer Nov 15 '14 at 17:08
  • 1
    $\begingroup$ Also, only matrices for the rotation around a coordinate axis leave one vector component unchanged. But matrices for the rotation around an arbitrary axis (e.g. the axis y=x) do not. $\endgroup$ – Dreamer Nov 15 '14 at 17:11
23
$\begingroup$

The following characterization of rotational matrices can be helpful, especially for matrix size $n > 2$.

$M$ is a rotational matrix if and only if $M$ is orthogonal, i.e. $MM^T = M^TM = I$, and $\det(M) = 1$.

$\endgroup$
  • 4
    $\begingroup$ Actually, if you define rotation as 'rotation about an axis,' this is false for $n>3$. The matrix $$\left[\begin{array}{cccc} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0 & 0\\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 & 0\\ 0 & 0 & \frac{\sqrt{3}}{2} & -\frac{1}{2}\\ 0 & 0 & \frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$$ satisfies all of your conditions, but has no real eigenvalues (and hence, no axis of rotation). $\endgroup$ – Strants Nov 16 '14 at 1:33
  • $\begingroup$ I don't think that the check det(M) = 1 is relevant to test if a matrix has rotation unless it is to test that it is a pure rotation. A det of 1 means, in 3 dimensions, that the cube formed by the axes given by the matrix as an area of 1 cubic unit. Consequently, this also means that the matrix does not contain scale. It is possible to have a rotation matrix with a det of 1 (eg. 2 flipped axis). $\endgroup$ – brita_ Apr 6 '18 at 11:55
8
$\begingroup$

I think there is a minus sign missing. As it is, the determinant is not $1$. After fixing, this specific case is easy.

$$\begin{pmatrix} \frac12 & 0 & -\frac{\sqrt{3}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{3}}{2} & 0 & \frac12 \end{pmatrix} = \begin{pmatrix} \cos \frac{\pi}{3} & 0 & -\sin \frac{\pi}{3} \\ 0 & 1 & 0 \\ \sin \frac{\pi}{3} & 0 & \cos \frac{\pi}{3} \end{pmatrix}$$

It is a rotation of $\pi/3$ around the $y$-axis.

$\endgroup$
  • $\begingroup$ correct. actually, the lower left entry has to be negative $\endgroup$ – Christian Nov 15 '14 at 12:39
  • 1
    $\begingroup$ Then it would be a rotation of $-\pi/3$.. just use that $\cos$ is an even function, and $\sin $ is odd. $\endgroup$ – Ivo Terek Nov 15 '14 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.