Matrix representation for linear transformation on $\mathbb{R}^{3}$ I am trying to figure out how to solve this problem:
Find a matrix representation for the following linear transformation on $\mathbb{R}^{3}$: A clockwise rotation by $60^{\circ}$ around the $x$-axis.
The answer is:
$$
\begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 0 & \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}
$$
The only thing I can deduce from my book is that the lower right $2 \times 2$ matrix is
$$
\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}
$$
but I don't know where they got the first row or first column values from.
 A: Note first that your matrix must be $3\times 3$ since otherwise it wouldn't be a transformation $\mathbb{R}^3 \to \mathbb{R}^3$.  Now, consider a point on the $x$-axis.  If you're looking for a rotation about the $x$-axis then your $x$-coordinates should remain the same.  Another way of going about this is to argue that if your rotation should be of the form
$$
R \;\; =\;\; \left [ \begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h  &k \\
\end{array} \right ]
$$
then it should leave $x$-coordinates fixed the way they are.  Thus apply
$$
\left [ \begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h  &k \\
\end{array} \right ] \left [ \begin{array}{c}
x\\
0\\
0\\
\end{array} \right] \;\; =\;\; \left [ \begin{array}{c}
ax \\
dx\\
gx\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{c}
x\\
0\\
0\\
\end{array} \right ]
$$
which forces $d=g=0$ and $a = 1$.  Now for any vector of the form $\left [ \begin{array}{ccc}
0 & y & z \\
\end{array} \right ]^T$, it should also leave the $x$-component unchanged, but it will certainly alter the $y$ and $z$ components.  Apply the rotation $R$ to this vector in the same way should yield the equation $by + cz = 0$, and the only way this can be true for all $y,z \in \mathbb{R}$ is if $b = c = 0$.
A: Well your rotation around $x$-axis, fixes the x coordinate, that's why.
A: Thinking about the mechanical process of rotating a physical object in three dimensions, it is clear that this involves going around an axis. Then focussing on the plane perpendicular to the axis it is the usual 2d-rotation.
So a 3d-rotation means a choice of axis (line) and then quantum of rotation (when measured in radians a  number between $0$ and $2\pi$.)
