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Published online by Cambridge University Press: 01 December 2003
02.6.1. Oblique projectors—solution.
The claimed result is an immediate consequence of the following more informative characterization.
THEOREM 1. Let P be an idempotent matrix; i.e., let P2 = P. Then P+ is idempotent if and only if P = P*, with P+ and P* denoting the Moore-Penrose inverse and the conjugate transpose of P, respectively.
Proof. Let P2 = P = P*. Then P = P3 = (P2)*, which in turn directly implies that the Penrose defining equations of P+ are satisfied for P+ = P, and so our proof of sufficiency is complete. For proving necessity, let P be such that P2 = P and (P+)2 = P+. If P = 0, then trivially P+ = 0 = P*. Next, let P ≠ 0 and let r denote the rank of P. According to the well-known singular value decomposition theorem, P can then be written as P = UDrV*, where U and V are (column-unitary matrices) such that U*U = V*V = Ir and where Dr is an r × r diagonal matrix with the r positive singular values of P along its main diagonal. In view of this decomposition, clearly

Because P+ = V(Dr)−1U*, which again is easily seen by checking the Penrose defining equations of P+, we likewise obtain

Combining these observations necessarily results in Dr = I = U*V = V*U. Consequently, (V − U)*(V − U) = 0, or, equivalently, V = U. As claimed, we thus arrive at P = UU* = P*, and so our proof is complete. █
Further equivalent conditions for an idempotent matrix P to be Hermitian are given in the following theorem.
THEOREM 2. Let P be an idempotent matrix. The following conditions are then equivalent:
(a) P = P*; i.e., P is Hermitian.
(b) P = PP*P; i.e., P is a partial isometry.
(c)

; i.e., P is an EP matrix.
(d) (P+)2 = P+.
(e) P+ = P.
(f) P+ = P*.
That (a) [hArr ] (b) [hArr ] (c) has been shown recently in Werner (2002). That (a) [hArr ] (e) [hArr ] (f) can be shown on similar lines and is therefore left to the reader.