= Trace criterion for a two-by-two idempotent
{title2=$P^2=P\implies P\text{ diagonal or }\operatorname{tr}P=1$}
For $P=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, the off-diagonal equations in $P^2=P$ are $b(a+d-1)=c(a+d-1)=0$. Thus either $b=c=0$ or the <trace> is one. The condition alone is not sufficient: $\begin{pmatrix}1&1\\1&0\end{pmatrix}$ has <trace> one but is not an <idempotent matrix>. For a two-by-two <matrix> of <trace> one, idempotence is equivalent to zero <determinant>, by the <Cayley-Hamilton theorem>.
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