Show that every $u\in V$ is a sum of eigenvectors of P, i.e. $u=v+w$ where $u$ and $v$ are eigenvectors of P Let $V$ be a vector space and $P:V\rightarrow V$ is a linear transformation such that $P \circ P=P. $
Show that every $u\in V$ is a sum of eigenvectors of P, i.e. $u=v+w$ where $v$ and $w$ are eigenvectors of P.
Here's what I think. If $u$ and $v$ are eigenvectors of $P$, then shouldn't they span $V$ itself, which makes any $u\in V$ be a linear combination of $u$ and $v$?
 A: Take $u=Pv$ and $w=v-Pv$. Check that both two vectors are eigenvectors of $P$ associated to eigenvalue $1$ and $0$ respectively.
A: I'm not sure if i can assume that $V$ is of finite dimension. I will assume this, of dimension $n < \infty$. 
We can consider $P$ as $n\times n$ matrix. $P^2 = P$, i.e. $P^2 - P = 0$, so the polynomial $p(x) = x^2 - x = x(x-1)$ satisfies $p(P) = 0$. Since the minimal polynomial of $P$ is factor of $p(x)$, and each case its minimal polynomial is product of distinct linear factors, i.e. $P$ is diagonalizable. 
Since $P$ is diagonalizable, the set of eigenvectors span $V$. Let $\mathbf{v}, \lambda$ be an eigenvector and eigenvalues respectively, we have $P \mathbf{v} = \lambda \mathbf{v}$ so $P^2\mathbf{v} = \lambda^2 \mathbf{v} = \lambda \mathbf{v}$, i.e. eigenvalue is 0 or 1. 
Let the eigenbasis of $V$ be $\{v_1, \cdots, v_k, w_1, \cdots, w_\ell\}$, where $v_*$ has eigenvalue 0 and $w_*$ has eigenvalue 1. For all $u$ there exists coefficient $\alpha_i, \beta_j$ satisfying $$u = \sum_{i = 1}^k\alpha_i v_i + \sum_{j = 1}^{\ell} \beta_j w_j$$. Let $v = \sum_{i = 1}^k\alpha_i v_i$, $w = \sum_{j = 1}^{\ell} \beta_j w_j$, check that $v$ and $w$ are eigenvectors of eigenvalue 0 and 1 respectively.
