directly decompose the regular representation of $C_n$

Question

I want to decompose the regular representation (over $\mathbb{C}$) of cyclic group $G=\langle g \rangle$ of order $n$ without using character theory. So far, I learned that irreducible representations of $G$ are $1$-dimensional representations given by $$\rho_j(g)=\omega^j, 1 \leq j \leq n$$ where $\omega=e^{2\pi i/n}$ is the $n^{\text{th}}$ root of unity. If we write $G=\left\{ g_i=g^i : 1\leq i \leq n\right\}$ then the regular representation $\rho(g)$ is given by the matrix with $(i,j)$-th entry being coefficient of $g_j$ in $g_ig=g_{i+1}$, i.e. $$\rho(g)= \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0\\ 0 &0 & 1 & 0 & \cdots & 0\\ & & & \vdots & & \\ & & & \vdots & & \\ 0 & 0& 0 & \cdots & 0 & 1\\ 1 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix}$$

Now I don't know how to write the regular representation as the direct sum of irreducible representations I found. I think when we say decompose the regular representation means decompose the corresponding vector space $V$ as a direct sum of subspaces $W_i$. Here I know that $W_i$ are one dimensional so $V=\bigoplus W_i$. But what is the natural way to define $W_i$?

Answer 1

The $W_i$ are just the eigenspaces of the matrix $\rho(g)$, with eigenvalue $\omega^i$ on $W_i$. In this case, there is a neat way to explicitly write down the eigenspaces: the vector $$v_i=(1,\omega^i,\omega^{2i},\dots,\omega^{(n-1)i})$$ is easily seen to be an eigenvector for $\rho(g)$ with eigenvalue $\omega^i$. So, $W_i$ is just the span of $v_i$ for each $i$.

(In case the formula for $v_i$ seems miraculous, note that it is easy to derive: if $v=(a_0,a_1,\dots,v_{n-1})$ then $\rho(g)v=(a_1,a_2,\dots,a_0)$ so to be an eigenvector with eigenvalue $\omega^i$ we must have $a_1=\omega^ia_0$, $a_2=\omega^ia_1=\omega^{2i}a_0$, and so on.)