$M^2=I_n$ implies $M$ diagonalizable Let $M$ be an  $n \times n$ matrix  such that
$M^{2}=I_n$; does this imply that $M$ is similar to a diagonal matrix $D$? How can we prove this?

Yes it is diagonalisable. Let $W_1$ be the space of generalized eigenvalues for $1$, that is vectors such that $(M-I)^nv=0$ for some $n$. Now note that 
$(M+I)W_1=W_1$ since otherwise there is an element in $W_1$ with eigenvalue $-1$. This implies that $$(M-I)W_1=(M-I)(M+I)W_1=0$$ this all generalized eigenvectors are actual eigenvectors. This implies that algebraic and geometric multiplicities are the same. Hence diagonalizable.
