 Solution

If $Mv=\lambda v$ for a nonzero eigenvector, then $M^2=M$ implies $({\lambda }^2-\lambda )v=0$, hence $\lambda \in \{0,1\}$. Conversely, the real-symmetric spectral theorem gives $M=Q\Lambda Q^T$ with orthogonal $Q$. If every diagonal entry of $\Lambda$ is zero or one, then ${\Lambda }^2=\Lambda$ and therefore $M^2=M$. Thus the necessary and sufficient condition isSuch a symmetric idempotent matrix is the orthogonal projection onto its eigenvalue-one eigenspace, with its eigenvalue-zero eigenspace as the kernel. Diagonalizability is essential to the sufficiency argument from eigenvalues alone; symmetry supplies it here.