[Rotations in R^3 ] Consider R∶ R^3 → R^3 the linear transformation that rotates π/3 around the z-axis

A vector $(x,y,z)$ is mapped to $(\cos(\pi/3) x+ \sin(\pi/3) y, -\sin(\pi/3)x +\cos(\pi/3)y, z)$. Mapping has matrix representation:
$$\begin{pmatrix}\cos(\pi/3) & \sin(\pi/3) &0 \\ -\sin(\pi/3) & \cos(\pi/3) &0 \\ 0 &0&1\end{pmatrix}$$

The eigenvalues of $R$ are $1$, with $(0,0,1)$ being a basis of the eigenspace corresponding to this eigenvalue. To see this note that the characteristic polynomial of $R$ is:

$$\det(\lambda\Bbb1 -R) = (\lambda-1)\cdot ( (\lambda-\cos(\pi/3))^2+\sin(\pi/3)^2) = (\lambda-1)\cdot (\lambda^2-2\lambda\cos(\pi/3) +1)  $$
Using $\cos(\pi/3)=\frac12$ this may be further simplified to:
$$(\lambda-1)(\lambda^2-\lambda+1)$$
This polynomial admits only one real root, namely $\lambda+1$ (use the $pq$ formula on the second factor if you have to), and this root has order $1$. So $R$ has at one real eigenvalue and this eigenvalue has a eigenspace of dimension $1$ (being $?1$ but $?$ the mutliplicity of the corresponding root, which is $1$). 

Since $R$ does not admit a basis of eigenvectors it cannot be diagonalisable.