I am trying to solve the following question:
Suppose that $V$ is a finite-dimensional vector space over $\mathbb{C}$, and that $\alpha : V → V$ is a $\mathbb{C}$-linear map such that $\alpha^n = 1$ for some $n > 1$ . Show that if $V_1$ is a subspace of $V$ such that $\alpha(V_1) \subseteq V_1$ , then there is a subspace $V_2$ of $V$ such that $V = V_1 ⊕ V_2$ and $\alpha(V_2) \subseteq V_2$ .
[Hint: Show, for example by picking bases, that there is a linear map $π : V → V_1$ with $π(x) = x$ for all $x ∈ V_1$. Then consider $ρ : V → V_1$ with $ρ(y) = \frac{1}{n}\sum\limits_{i=0}^{n-1}α^iπα^{−i}(y)$.]
I think I have a solution but I am getting an uneasy feeling about my argument as it feels convoluted, can someone please verify:
Let $v_1,v_2,\cdots,v_k$ be a basis for $V_1$. Extend this to a basis for $V$ given by $v_1,v_2,\cdots,v_k,w_1,w_2,\cdots,w_l$. Then $\pi$ is the projection map which sends $v_i$ to itself and $w_j$ to $0$ (hence it exists). We now forget about the $w_i$ as they are only used to show the projection map exists.
Define $\rho$ as in the hint. Note that $\rho(\alpha(y)) = \frac{1}{n}\sum\limits_{i=0}^{n-1} \alpha^i\pi\alpha^{-i+1}(y) = \alpha(\frac{1}{n}\sum\limits_{i=0}^{n-1} \alpha^{i-1}\pi\alpha^{-(i-1)}(y))$. This is equal to $\alpha(\rho(y))$ as $\alpha^n = 1$ so $\alpha^{-1} = \alpha^{n-1}$. Hence $\alpha\circ \rho = \rho \circ \alpha$ so they commute.
We can also note that for any $a \in V_1$ as $\alpha(V_1) \subset V_1$ we have $\rho(a) = a$. So $\text{im } \rho = V_1$. We now apply the isomorphism theorem to $\rho$. We have $\frac{V}{\ker\rho} \cong \text{im } \rho$. Taking dimensions and rearranging we get $ \dim(V) = \dim(\ker \rho)+\dim(\text{im } \rho)$. So $\dim(\ker \rho) = \dim(V)-\dim(V_1) $.
So let $x_1,x_2,\cdots ,x_l$ be a basis for $\ker \rho$. Then $v_1,v_2,\cdots,v_k,x_1,x_2,\cdots,x_l$ are all linearly independent (in V) as if $a_1v_1+\cdots+a_kv_k+b_1x_1+\cdots+b_lx_l = 0$ Then applying $\rho$ gives us $a_1v_1+\cdots+a_kv_k = 0$ so as $v_i$ are linearly independent we get $a_i = 0$ for all $i$ and as the $x_i$ are linearly independent we also get $b_i = 0$ for all $i$.
So $v_1,v_2,\cdots,v_k,x_1,x_2,\cdots,x_l$ is a linearly independent set the size of the dimension of $V$ so it is a basis. So $V_1 + \ker \rho = V$ and also $V_1 \cap \ker\rho = \{ 0\}$. So $V_1 \oplus \ker\rho = V$. Now note that if $x \in \ker \rho$ then $0 = \alpha(0) = \alpha(\rho(x)) = \rho(\alpha(x))$ so $\alpha(x) \in \ker \rho$. So $\alpha(\ker \rho) \subset \ker \rho$ and we are done by letting $V_2 = \ker \rho$.
