Suppose $π:V\to V$ is a projection matrix, does it follows that its eigenvalues are $0, 1$? Is $π$ diagonalisable?
One of the following answers is true and the other is false, but they both seem true to me, so can anyone clarify this?
1)Yes, because if π has an eigenvalue λ and an eigenvector v, it follows that $π^2(v)=π(λv)=λ^2v=π(v)=λv$, so $λ(λ-1)=0$ and $λ=0$ or $λ=1$
2)Yes, because $V=U \oplus W$, where $U=\operatorname{im}π, W=\ker π$. Then $π(v)=v$ and $π(w)=0$, for $v \in U, w \in W$. So $U=E(1,π), W=E(0,π)$ and we then take a basis of $U$ and a basis of $W$ as a basis of eigenvectors of $V$.
The first proof is wrong. All that it proves is that if $\pi$ has eigenvalues, then they can only be $0$ or $1$. But it doesn't follow from this that there must be a basis of $V$ that consists only of eigenvectors of $\pi$ (that is, that $\pi$ is diagonalizable).