 Solution

Write $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. The two off-diagonal entries of $M^2=M$ giveor $b(\mathrm{tr}M-1)=c(\mathrm{tr}M-1)=0$. If both $b,c$ vanish, $M$ is diagonal; if either is nonzero, the trace must be one. Thus the trace criterion for a two-by-two idempotent isFor the diagonal case, idempotence further forces $a,d\in \{0,1\}$. The stated alternative is only necessary: $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has trace one but its square is $\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$, so it is not idempotent. Among two-by-two matrices of trace one, the additional condition $\det M=0$ is sufficient by the Cayley-Hamilton theorem, which gives $M^2-M+(\det M)I=0$.