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 ?
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$\begingroup$ 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 WillseNov 15, 2014 at 11:04
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2$\begingroup$ How do you define rotation matrix? $\endgroup$– Git GudNov 15, 2014 at 11:16
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$\begingroup$ 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). $\endgroup$– DreamerNov 15, 2014 at 17:08
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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. $\endgroup$– DreamerNov 15, 2014 at 17:11
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2 Answers
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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 $\det(M) = 1$.
answered Nov 15, 2014 at 10:57
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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). $\endgroup$– user88319Nov 16, 2014 at 1:33
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$\begingroup$ 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, 2018 at 11:55
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$\begingroup$ A rotation matrix $M$ does not need to satisfy $\det(M)=1$. This is only true if $M$ describes a proper rotation; otherwise it describes an improper rotation, and $\det(M)=-1$. $\endgroup$ Feb 17, 2021 at 22:09
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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.
answered Nov 15, 2014 at 11:36
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$\begingroup$ correct. actually, the lower left entry has to be negative $\endgroup$ Nov 15, 2014 at 12:39
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1$\begingroup$ Then it would be a rotation of $-\pi/3$.. just use that $\cos$ is an even function, and $\sin $ is odd. $\endgroup$ Nov 15, 2014 at 13:19