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Let $T$ be a linear Transformation from $\mathbb{R}^3$ to itself such that $T$ is $60^{\circ}$ clockwise rotation with fixed $z$-axis (i.e, rotate the space according to the $z$-axis) where $\mathbb{R}^3$ has $xyz$ coordinate system

Find the matrix representation of $T$ with respect to the standard matrix.

I don't know how to do this problem.

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  • $\begingroup$ The x-y rotation is independent of z, so the last value in the first to columns will be 0. The z coordinate is unchanged and so the last column will be [0,0,1]. Can you guess the form of the matrix now? $\endgroup$ – Mark Nov 16 '14 at 6:59
  • $\begingroup$ See what the transformation would do to the vectors $(1,0,0)^T, (0,1,0)^T, (0,0,1)^T$. $\endgroup$ – copper.hat Nov 16 '14 at 7:27
  • $\begingroup$ do I do anything with the 60 degrees? $\endgroup$ – user192898 Nov 16 '14 at 17:57
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Since the rotation is around the $z$-axis, we can pretend that we're making a rotation of $60^\circ$ clockwise, or better, $300^\circ = 5\pi/3$ counterclockwise in $\Bbb R^2$. A linear mapping is known if we have its values on a basis. Rotating by $\theta$ counterclockwise, $(1,0)$ goes to $(\cos \theta, \sin \theta)$ and $(0,1)$ goes to $(-\sin \theta, \cos \theta)$. So the matrix in $\Bbb R^2$ would be: $$\begin{bmatrix} \cos 5\pi/3 & -\sin 5\pi/3 \\ \sin 5\pi/3 & \cos 5\pi/3\end{bmatrix} = \begin{bmatrix} 1/2 & \sqrt{3}/2 \\ -\sqrt{3}/2 & 1/2\end{bmatrix}.$$

Bearing this in mind: $$[T] = \begin{bmatrix} 1/2 & \sqrt{3}/2 & 0 \\ -\sqrt{3}/2 & 1/2 & 0 \\ 0 & 0 & 1\end{bmatrix}.$$

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