 Solution

If $Av=\lambda v$ for a nonzero eigenvector, idempotence gives ${\lambda }^{2}v=A^{2}v=Av=\lambda v$. HenceThese are the only possible eigenvalues; either may be absent. In fact $v=Av+(I-A)v$ decomposes every vector into the image and kernel of $A$, so every idempotent matrix is diagonalizable even when it is not an orthogonal projection.