= Solution
Write $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. The two off-diagonal entries of $M^2=M$ give
$$
b(a+d)=b,\qquad c(a+d)=c,
$$
or $b(\operatorname{tr}M-1)=c(\operatorname{tr}M-1)=0$. 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> is
$$
\boxed{M^2=M\quad\Longrightarrow\quad M\text{ is diagonal or }\operatorname{tr}M=1.}
$$
For the diagonal case, idempotence further forces $a,d\in\{0,1\}$. The stated alternative is only necessary: $\begin{pmatrix}1&1\\1&0\end{pmatrix}$ has <trace> one but its square is $\begin{pmatrix}2&1\\1&1\end{pmatrix}$, 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 $M^2-M+(\det M)I=0$.
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