0

$$ A=\left[\begin{array}{lll} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] $$ $$ D=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -\left(\frac{1+\sqrt{3} i}{2}\right) & 0 \\ 0 & 0 & \left(\frac{-1+\sqrt{3} i}{2}\right) \end{array}\right] $$

$A$ is the form of the rotation matrix in $(x, y, z)$ basis (orthonormal) and $D$ is it's form in new basis $(x', y', z')$ in which it is diagonal. The rotation occurs about the axis in the direction$$\hat{n}=\frac{\hat{x}+\hat{y}+\hat{z}}{\sqrt{3}} $$

And the transformation rule between $(x, y, z)$ and $(x', y', z')$ is given as

\begin{aligned} x &=\frac{1}{\sqrt{3}}\left(x^{\prime}+\left(\frac{-1+\sqrt{3} i}{2}\right) y^{\prime}+\left(\frac{-1-\sqrt{3} i}{2}\right) z^{\prime}\right) \\ y &=\frac{1}{\sqrt{3}}\left(x^{\prime}+\left(\frac{\left.-1- \sqrt{3} i) y^{\prime}+(-1+\sqrt{3} i) z^{\prime}\right)}{2}\right)\right.\\ z &=\frac{1}{\sqrt{3}}\left(x^{\prime}+y^{\prime}+z^{\prime}\right) \end{aligned} So my question is how to find angle of rotation? Help!

ConfusedHuman
  • 105
  • One simple way would be to take any vector that's orthogonal to the axis of rotation, for example $v = (1, 1, -2)$, and compute where $A$ sends this vector, namely $Av$. Then you simply want the angle $\theta$ between $v$ and $Av$, which you can get from $\cos(\theta) = \langle v, Av \rangle / (|v| |Av|)$. –  Dec 29 '20 at 11:31

4 Answers4

1

As you know, your rotation is a rotaion around $\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)$. Take a unitary vector $v$ orthogonal to this vector; for instance, take $v=\left(\frac1{\sqrt2},-\frac1{\sqrt2},0\right)$. It happens that$$A.v=\left(0,\frac1{\sqrt2},-\frac1{\sqrt2}\right)$$and so the angle $\theta$ that you are looking after is the angle between $v$ and $A.v$. Since these are two unit vectors, $\theta$ is such that $\cos\theta=\langle v,A.v\rangle=-\frac12$, and therefore $\theta=\frac{2\pi}3$.

José Carlos Santos
  • 427,504
1

Looking at the columns, under the transformation the unit vectors $i$, $j$ and $k$ are transformed as follows:

$$i \rightarrow j$$ $$j\rightarrow k$$ $$k\rightarrow i$$

Doing the same transformation three times would be equivalent to the identity matrix. Therefore the matrix is a rotation of $120^o$ about the line $x=y=z$.

David Quinn
  • 34,121
  • This is a clever observation (+1), but it doesn't tell us whether it's +120 or -120 degrees. –  Dec 29 '20 at 11:40
  • @Bungo how do you decide what is positive and what is negative in this situation? – David Quinn Dec 29 '20 at 11:42
  • The OP defined the vector of rotation as $(1,1,1)/\sqrt{3}$, so a standard convention would be to define a right-hand coordinate system with that vector as the $z$ direction, then define positive and negative angles in accordance with that system. In any case, no matter how you define +120 and -120, only one of them is actually the correct answer, and unfortunately I don't see how your solution tells us which one. –  Dec 29 '20 at 11:49
0

Rotations in $R^3$ are the compositions of rotations in $R^2$,

A rotation in $R^2$ is written as

$\left(\begin{matrix} \cos(\alpha)&\sin(\alpha)\\ -\sin(\alpha)&\cos(\alpha) \end{matrix}\right)$

where $\alpha$ is the angle of the rotation

In $R^3$ you can write it as the product of two rotations in $R^2$ $\left(\begin{matrix} \cos(\alpha)&\sin(\alpha)&0\\ -\sin(\alpha)&\cos(\alpha)&0\\ 0&0&1 \end{matrix}\right) \left(\begin{matrix} \cos(\beta)&0&\sin(\beta)\\ 0&1&0\\ -\sin(\beta)&0&\cos(\beta) \end{matrix}\right)$

in your example you are sending the:

  • $x$-axis to the $y$-axis
  • $y$-axis to the $z$-axis
  • $z$-axis to the $x$-axis

so you are doing 2 rotations (over 2 different axis) of 90º degrees each

Bernard
  • 175,478
yugikaiba
  • 156
0

The trace (the sum of the entries on the diagonal) of a rotation matrix is $$ 1 + 2 \cos \theta $$ In this case $$ 1 + 2 \cos \theta = 0 \implies \cos \theta = -0.5 \implies \theta = 2\pi/3 $$

PM.
  • 5,249