 Trace of an idempotent equals its rank

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 $\mathrm{diag}(I_r,0)$. 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.