 Solution

Applying $A^2=A$ to an eigenvector gives ${\lambda }_j^2={\lambda }_j$, so each eigenvalue is either zero or one. If all are one, the spectral theorem gives $A=I$, hence $\det A=1$ and $\mathrm{tr}A=n$. Otherwise, if $m$ eigenvalues are one, with $0\le m<n$, thenThus the trace of this orthogonal projection equals its rank, including $m=0$ for the zero matrix.