Let $C = \langle c \rangle$ be a cyclic group of order $n$. Let $\rho$ be a matrix representation for $C$.
The matrix $\rho(C)$ has an eigenvalue $\lambda$ and a corresponding eigenvec $v$.
We then consider a line through $v$ ; $\mathbb{C}v$, which is a $C -$subspace of $V$, with the defining action $c^rv = \lambda^rv$.
$(1)$ What exactly is $V$ here? Is is it just the set of column vectors?
Now, as I understand, the group action is multiplication by a root of unity since if we let $ v \neq 0$ be a basis, then $cv = \lambda v$ for some $\lambda$. But then $v = ev = c^nv=\lambda^nv$, hence $\lambda^n = 1$.
Now, from my notes, it then states that there is a basis such that $\rho(c) = \operatorname{diag}(\lambda_1, \lambda_2, ..., \lambda_m)$ , where the $\lambda_i$ are $n-$th roots of unity.
$(2)$ Where has this come from? How is this the matrix representation? What exactly is $m$ here?
The point is that every representation is a direct product of $1$-dimensional sub-representations. Given any representation $V$ (of dimension $m$, say), there is an eigenbasis $\{v_1,\ldots,v_m\}$ so that $\mathbb{C}v_i$ is a $1$-dimensional $C$-invariant subspace.