= Trace of an idempotent equals its rank
{title2=$P^2=P\implies\operatorname{tr}P=\operatorname{rank}P$}
For a finite-dimensional real or complex <linear projection> $P$, every vector splits as $Px+(I-P)x$ in the <image of a linear map> and the <kernel of a linear map>, respectively. Their intersection is zero, so a basis adapted to this <direct sum> represents $P$ as $\operatorname{diag}(I_r,0)$. Its <trace> and <rank> both equal $r$. In particular, a projection's <determinant> is zero unless it is the identity. Hermiticity is unnecessary for this algebraic identity; a <Hermitian matrix> projection is additionally orthogonal.
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