How does this action decompose into irreducible representations? I am considering the action of $\mathbb{Z}/3$ on $\mathbb{C}^3$ given by the formula $x \mapsto y \mapsto z \mapsto x$. I want to determine its decomposition into irreducible representations, but I'm having some trouble.
I know that any representation of $\mathbb{Z}/3$ can be expressed as the direct sum of $1$-dimensional representations, but I'm having trouble understanding what to do from here. I know the $\mathbb{R}$-vector subspaces invariant under this action on $\mathbb{R}^3$ (See What are the $\mathbb{R}$-vector subspaces invariant under this action?), but I don't know where to go next, or what to even write down.
Any insight would be appreciated!
 A: Hint: The matrix
$${\bf M} = \left[\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right]$$
represents the action if the elements x,y,z correspond to $[1,0,0]^T, [0,1,0]^T, [0,0,1]^T$. Which is the basis $\bf S$ and $\bf C$ with smallest blocks on the diagonal so that $${\bf M = SCS}^{-1}$$?

EDIT
As it seems the OP has solved the problem, some clarifications may be in order:
If we solve $\det({\bf M}-\lambda {\bf I}) = 0$ we will get the irreducible representations split in roots $\lambda$ and subspaces spanned by vectors which have no roots (in our scalar field). Over $\mathbb C$, it is easy to show that $$\{\lambda_1,\lambda_2,\lambda_3\} = \left\{1,\exp\left(\frac{2\pi i} {6}\right),\exp\left(\frac{-2\pi i} {6}\right)\right\}$$
and then the irreducible representations will correspond to the 1 dimensional eigenspaces. However if we work over $\lambda \in \mathbb R$, then we need to investigate the spaces spanned by non-real eigenvalues. If we multiply them together: $$(\lambda-\lambda_2)(\lambda-\lambda_3)=\left(\lambda - \exp\left(\frac{2\pi i} {6}\right)\right)\left(\lambda - \exp\left(\frac{-2\pi i} {6}\right)\right) = \lambda^2+\lambda+1$$
Which we see has real coefficients and for example can be put on similar form to example the factors' companion matrix
$$\left[\begin{array}{cc}0&-1\\1&-1\end{array}\right]$$
or the popular 2x2 real block representation of complex conjugates:
$$a+bi \Leftrightarrow \left[\begin{array}{cc}a&-b\\b&a\end{array}\right]$$
We can choose whichever is most convenient for us, but it is the size of the block that determines the size of the irreducible representation which is 2 dimensional. 

So for the real case it is $\rho = \rho_1 \oplus \rho_2$, 
where $\rho_1$ is 1 dimensional corresponding to the root $\lambda = 1$ (or linear factor $(\lambda-1)$), 
and $\rho_2$ is 2 dimensional corresponding to quadratic factor $\lambda^2+\lambda+1$ which can't be factored any further (over $\mathbb R$).
