# directly decompose the regular representation of $C_n$

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 $$W_i$$?

• Eigenvectors${}$? Commented Nov 18, 2019 at 7:28

## 1 Answer

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.)