Linear Algebra and its Applications 294 (1999) 213±215 www.elsevier.com/locate/laa
A simple proof of the product theorem for EP matrices J.J. Koliha
*
Department of Mathematics and Statistics, University of Melbourne, Parkville 3052, Australia Received 5 August 1998; accepted 16 March 1999 Submitted by B. Cain
Abstract We give a simple proof of a recent theorem of Hartwig and Katz on the product of EP matrices. Ó 1999 Elsevier Science Inc. All rights reserved. Keywords: EP matrix; Moore±Penrose inverse; Eigenprojection
The purpose of this note is to give a simple proof of a theorem recently obtained by Hartwig and Katz [3, Theorem 1] on the product of EP matrices that solved a problem open for over 25 years (see [1]). All matrices are complex and n n. A matrix A is EP if Ay A AAy , where Ay is the Moore±Penrose inverse of A (see [2]). We need the following result. Proposition 1 [2, Theorem 4.3.1]. For any matrix A the following are equivalent: (i) A is EP. (ii) N
A N
A , where N
A is the nullspace of A. (iii) R
A R
A , where R
A is the range of A. (iv) Cn N
A ? R
A. Clearly, A is EP if and only if A is EP. We observe that every EP matrix A satis®es
*
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J.J. Koliha / Linear Algebra and its Applications 294 (1999) 213±215
A UA
for some matrix U :
1
(For instance, A A2 Ay A
Ay A .) We write Ap for the eigenprojection of A at 0; if A is EP, Ap is the orthogonal projection onto N
A. Theorem 2. Let A, B be EP matrices. Then the following are equivalent: (i) AB is EP. (ii)
ABAp 0 and Bp
AB 0. (iii) N
A N
AB and R
AB R
B. (iv) N
AB N
A N
B and R
AB R
A \ R
B.
Proof. (i) ) (ii). By (1), AB U
AB for some matrix U. From Proposition 1 (iv) we deduce that Ap is Hermitian and that Ap A AAp 0. Then
ABAp U
AB Ap UB A Ap UB
Ap A 0:
Since
AB B A is also EP, the preceding argument applies to give Bp
AB
B A Bp 0. (ii) ) (iii). From
ABAp 0 and R
Ap N
A follows N
A N
AB, and from Bp
AB 0 and N
Bp R
B follows R
AB R
B. (iii) ) (iv). Suppose that N
A N
AB. The map h de®ned on the quotient space N
AB=N
B by h
x N
B Bx is a vector space isomorphism of N
AB=N
B onto N
A \ R
B. Hence dim
N
A \ R
B dim N
AB ÿ dim N
B:
2
From Cn N
B R
B we deduce that W
N
A \ R
B N
B N
A N
B N
AB: By (2), dim W dim N
AB, and W N
A N
B N
AB. Next assume that R
AB R
B. This implies N
B N
AB , and by the ®rst part of the proof, N
AB N
A N
B . This is equivalent to R
AB R
A \ R
B. (iv) ) (i). Since N
AB N
A N
B implies R
AB R
A \ R
B , we have
R
AB R
A \ R
B R
A \ R
B R
AB; and AB is EP by Proposition 1(iii). The main result of Hartwig and Katz [3] is formulated using the row space RS
A of a matrix A. Their result follows from Theorem 2 when we observe that RS
AB RS
A () N
A N
AB; RS
AB RS
A \ RS
B () N
AB N
A N
B:
J.J. Koliha / Linear Algebra and its Applications 294 (1999) 213±215
215
Corollary 3 [3, Theorem 1]. Let A, B be EP matrices. Then the following are equivalent. (i) R
AB R
A \ R
B and RS
AB RS
A \ RS
B. (ii) R
AB R
B and RS
AB RS
A. (iii) AB is EP. References [1] T.S. Baskett, I.J. Katz, Theorems on products of EPr matrices, Linear Algebra Appl. 2 (1969) 87±103. [2] S.L. Campbell, C.D. Meyer, Generalized Inverses of Linear Transformations, Pitman, London, 1979. [3] R.E. Hartwig, I.J. Katz, On products of EP matrices, Linear Algebra Appl. 252 (1997) 339±345.