How to prove any $N \times N$ matrix $M$, satisfying $M^2 = \lambda^2 I_{N\times N}$, $\lambda\neq 0$ is diagonalizable. I want to prove following things. 

Any $N \times N$ matrix $M$, satisfying $M^2 = \lambda^2 I_{N\times N}$, with $\lambda\neq 0$ is diagonalizable. 

Can you give me some hint or proof of this? 

My trail starts from determinant
\begin{align}
0=det(M^2 - \lambda^2 I) = det(M - \lambda I) det(M+ \lambda I)
\end{align}
thus i conclude that either $det(M - \lambda I)=0$ or $ det(M+ \lambda I)=0$. 
But i don't know how to go further..
As $M^2$ is in diagonal form, then we can write it's diagonalization $M^2=I(\lambda^2 I)I=Q\Lambda Q^{-1}$. Then we can write $M^2 = I(\lambda I)(\lambda I)I=Q\sqrt{\Lambda}\sqrt{\Lambda}Q^{-1}$. Then we can also write $M^2=Q\sqrt{\Lambda}Q^{-1}Q\sqrt{\Lambda}Q^{-1} = I(\lambda I)II(\lambda I)I=MM$. So from this derivation we see that $M=\lambda I$ with possible eigenvalues $\pm \lambda$.
As $M^2$ is in diagonal form, then we can write it's diagonalization $M^2=I(\lambda^2 I)I=Q\Lambda Q^{-1}$. Then we can write $M^2 = I(\lambda I)(\lambda I)I=Q\sqrt{\Lambda}\sqrt{\Lambda}Q^{-1}$. Then we can also write $M^2=Q\sqrt{\Lambda}Q^{-1}Q\sqrt{\Lambda}Q^{-1} = I(\lambda I)II(\lambda I)I=MM$. So from this derivation we see that $M=\lambda I$ with possible eigenvalues $\pm \lambda$.