# The projection operator on a finite dimensional vector space is diagonalizable

This question has been answered before, but I want to check if my solution using minimal polynomials is good.

A projection matrix satisfies $M^2 = M$, so it satisfies the polynomial equation $M(M-1) = 0$. Thus the minimal polynomial must be either $M$, $M-1$, or $M(M-1)$. Since in all cases the polynomial factors into distinct linear factors, the matrix is diagonaliazable.

• Perfectly fine. Just saying $X(X-1)$ is a polynomial with simple roots is enough, though. – Gabriel Romon Mar 12 '16 at 22:40