Consider the following question:
Suppose that $V$ is a finite-dimensional vector space over $\mathbb{C}$, and that $\alpha: V \rightarrow V$ 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 $\alpha\left(V_1\right) \subset V_1$, then there is a subspace $V_2$ of $V$ such that $V=V_1 \oplus V_2$ and $\alpha\left(V_2\right) \subset V_2$. [Hint: Show, for example by picking bases, that there is a linear map $\pi: V \rightarrow V_1$ with $\pi(x)=x$ for all $x \in V_1$. Then consider $\rho: V \rightarrow V_1$ with $\rho(y)=\frac{1}{n} \sum_{i=0}^{n-1} \alpha^i \pi \alpha^{-i}(y)$.]
I have an argument that differed from the hint. Here is how it goes. Since $\alpha ^n =1$, it follows that the minimal polynomial of $\alpha$ devices $t^n-1$. So $0$ is not an eigenvalue, and hence $\alpha$ is a bijection. We are told that $\alpha(V_1) \subset V_1$. But the image is also a subspace. By bijectivity it follows that $\alpha(V_1) = V_1$. Now let $\{e_i\}_i$ be a basis for the $V_1$. Then they can be extended to be a basis for all of $V$, say $V$ has basis $$ \{ e_1, ... ,e_n , f_1, ... f_m\}. $$ Clearly the span of the $f_j$ gives a subspace. Call it $V_2$. By construction the direct sum equation holds. As $\alpha$ is a bijection, due to dimensional reasons, we require $\alpha (V_2) = V_2$. QED
I think there is something wrong with this argument as it is first too simple for a question that is meant to take 25-30 minutes. Moreover, it doesn't use the hint.
Can someone check if the argument holds? Moreover, if it does, could someone elaborate on the second part of the hint? I understand the first part but I don't see how considering $\rho$ leads to $V_2$. Any hints or thoughts towards it will be greatly appriciated.