TITLE: Are idempotent matrices always diagonalizable?
QUESTION [31 upvotes]: How to prove that any idempotent matrix is diagonalizable?

REPLY [9 votes]: This is quite easy. For every vector $v$ one has $A^2v=Av$, so $Av$ is in the eigenspace for $\lambda=1$. Also $A(v-Av)=0$, so $v-Av$ is in the eigenspace for $\lambda=0$. (Define the eigenspace for$~\lambda$ as $\ker(A-\lambda I)$, even in case $\lambda$ should be no eigenvalue.) So $v=Av+(v-Av)$ is in the sum of those two eigenspaces; since $v$ was arbitrary, that sum is the whole space. This means $A$ is diagonalisable (eigenvalues $0,1$ only).