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02.6.1. Oblique Projectors—Solution

Published online by Cambridge University Press:  01 December 2003

Götz Trenkler
Götz Trenkler
Affiliation:
University of Dortmund, Germany
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Extract

02.6.1. Oblique projectorsTwo solutions are published, proposed independently by Götz Trinkler (the poser of the problem) and by Hans Joachim Werner. Excellent solutions have been independently proposed by G. Dhaene, L. Lauwers, S. Lawford, and H. Neudecker.—solution.

Information

Type
PROBLEMS AND SOLUTIONS
Information
Econometric Theory , Volume 19 , Issue 6 , December 2003 , pp. 1195 - 1196
Copyright
© 2003 Cambridge University Press

It is well known that an oblique projector P can be written as

where r denotes the rank of P and U is a unitary matrix (cf. Hartwig and Loewy, 1992). Because K is not Hermitian we have K ≠ 0.

Consider the matrix

where C = (Ir + KK*)−1 and D = K*C. Then

are both Hermitian. Furthermore, RPR = R and PRP = P. Thus R satisfies the four conditions of the Moore–Penrose inverse of P, i.e., R = P+.

If P+ were a projector we would get C2 = C, which implies C = I, and consequently K = 0. This is a contradiction.

References

REFERENCE

Hartwig, R.E. & R. Loewy (1992) Maximal elements under the three partial orders. Linear Algebra and Its Applications 175, 39–61.Google Scholar
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