Complex and Real Representations, their differences by decomposition 1. The problem statement, all variables and given/known data
Decompose $\mathbb{C}^{5}$, the 5 dimensional complex Euclidean space) into invariant subspaces irreducible with respect to the group $C_{5} \cong \mathbb{Z}_{5}$ of cyclic permutations of the basis vectors $e_{1}$ through $e_{5}$.
Hint: The group is Abelian, so all the irreps are one-dimensional. Therefore, you can use the simplified form of the projection operators, with characters. 
Further, try to do the same for $\mathbb{R}^{5}$, insisting that the basis vectors can only be combined with real coefficients. What is the difference between real and complex reps? 
2. Relevant equations
This may be the right projection operator, unsure:
$$P^{\alpha}=\frac{d_{\alpha}}{|G|} \sum_{g} \chi^{(\alpha)}(g)*O_{g}$$
3. The attempt at a solution
I am confused by the term decompose, so my attempts have been floundering. I tried to write out the character table for $\mathbb{Z}_5$ and I think I succeeded in that, but am unsure if it is needed. The hint about the projection operators served to confuse me more, although I readily understand the part about 1D irreps and Abelian. Is this asking me to construct reps (matrices) using cyclic permutations of $C_{5}$? If so, how am I supposed to use projection operators in this case to get them; This seems right however. 
Any help would be wonderful.
 A: The generator of your group acts via the linear transformation
$T:\mathbb{C}^5\rightarrow\mathbb{C}^5, (x_1,x_2,x_3,x_4,x_5)\mapsto(x_5,x_1,x_2,x_3,x_4)$.
Write
$\zeta=e^{2\pi i/5}$. Then the vector 
$$\vec{u}_j=(1,\zeta^{4j},\zeta^{3j},\zeta^{2j},\zeta^{j})$$
is an eigenvector belonging to the eigenvalue $\zeta^j$ for $j=0,1,2,3,4$. 
Therefore they each generate an invariant 1-dimensional (complex) subspace as described in mt_'s comments. [Edit] They can also be gotten by applying the projection operators
$$
P^{-j}=\frac15\sum_{k=0}^5\chi_j^*(g^k)T^k
$$
to the vector $(5,0,0,0,0)$. Here $\chi_j$ is the character $\chi_j(g^k)=\zeta^{jk}$ for all $k=0,1,2,3,4$ and $g$ is the generator of the group.[/Edit]
The real case requires a bit more work. Observe that
$\zeta$ and $\zeta^4$ as well as $\zeta^2$ and $\zeta^3$ are complex conjugates of each other.
Thus the complex vector spaces $U_1=\mathbb{C}\vec{u}_1\oplus\mathbb{C}\vec{u}_4$ and 
$U_1=\mathbb{C}\vec{u}_2\oplus\mathbb{C}\vec{u}_3$
are stable under componentwise complex conjugation, because componentwise complex conjugation simply swaps the eigenvectors. It follows that the real vector spaces
$$
V_1=U_1\cap \mathbb{R}^5\qquad\text{and}\qquad V_2=U_2\cap \mathbb{R}^5
$$
are 2-dimensional (they are the eigenspaces belonging to eigenvalue 1 of the componentwise
complex conjugation acting on $U_1$ and $U_2$ respectively), and also stable under the action of $T$. A calculation (see also answers to this question, in particular the link to Keith Conrad's lecture notes) shows that the matrix of $T$ with respect to the basis (over the reals) $\{\vec{u}_1+\vec{u}_4, i(\vec{u}_1-\vec{u}_4)\}$ of $V_1$ is the familiar rotation matrix. Therefore $T$ must act on $U_1$ as a rotation by 72 degrees.
Similarly the restriction of $T$ on $V_2$ is a rotation by 144 degrees.
A rotation of a real plane does not have any invariant subspaces, so we cannot refine the direct sum decomposition
$$
\mathbb{R}^5=\langle(1,1,1,1,1)\rangle\oplus V_1\oplus V_2
$$
any further. Here $(1,1,1,1,1)=\vec{u}_0$.
