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1 This article appeared in a journal published by Elsevier. The attached copy is furnished to the author for internal non-commercial research and education use, including for instruction at the authors institution and sharing with colleagues. Other uses, including reproduction and distribution, or selling or licensing copies, or posting to personal, institutional or third party websites are prohibited. In most cases authors are permitted to post their version of the article e.g. in Word or Tex form) to their personal website or institutional repository. Authors requiring further information regarding Elsevier s archiving and manuscript policies are encouraged to visit:

Linear Algebra and its Applications 431 2009) 1616 1619 Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: www.elsevier.

Université Paul Sabatier, 118 route de Narbonne, 31062 Toulouse, France A R T I C L E I N F O A B S T R A C T Article history: Received 26 May 2008 Accepted 25 May 2009 Available online 9 July 2009

2 Linear Algebra and its Applications ) Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: One more simple proof of the Craig Sakamoto theorem Hervé Carrieu a, Patrice Lassère b, a Lycée Camille Guérin, Poitiers, France b Institut de Mathématiques, UMR CNRS 5580, Université Paul Sabatier, 118 route de Narbonne, Toulouse, France A R T I C L E I N F O A B S T R A C T Article history: Received 26 May 2008 Accepted 25 May 2009 Available online 9 July 2009 We give one more elementary proof of the Craig Sakamoto s theorem: given A, B S n R) such that deti n xa yb) = deti n xa) deti n yb), x, y R; then AB = Elsevier Inc. All rights reserved. Submitted by C.K. Li AMS classification: 15A15 15A18 15A24 15A48 Keywords: Craig Sakamoto s theorem Linear and bilinear algebra 1. Introduction The Craig Sakamoto theorem asserts that Given A, B S n R) then deti n xa yb) = deti n xa) deti n yb), x, y R if and only if AB = 0. For an historical viewpoint of this result coming from statistical-probabilities, the interested reader can look at [1,2,5,6]; since his first statement it has inspired many proofs see [3,4,8 10], and the references listed in the previous papers). The purpose of this note is to give one more new let us hope ) proof of this theorem using the elementary machinery of linear and bilinear algebra. Corresponding author. addresses: herve.carrieu@ac-poitiers.fr H. Carrieu), lassere@math.ups-tlse.fr P. Lassère) /$ - see front matter 2009 Elsevier Inc. All rights reserved. doi: /j.laa

3 H. Carrieu, P. Lassère / Linear Algebra and its Applications ) The proof The key of the proof is the following: Property ). Let A, B S n R) \{O Mn R)} such that deti n xa yb) = deti n xa) deti n yb), x, y R. Then B or A, the same by symmetry) admits a nonzero eigenvalue λ such that kera) ker I n λ B) 1 /={0}. First, we are going to show how this property implies the Craig Sakamoto s theorem Proof of the Craig Sakamoto s theorem We do it by induction on the size n of the matrices. For n = 1, the Craig Sakamoto s theorem is clear, so let n 2 and suppose it true up to rank n 1; let A, B S n R) \{O Mn R)} we exclude the trivial case where one of the two matrices is zero) such that deti n xa yb) = deti n xa) deti n yb), x, y R. We have to prove that AB = 0. Because of property ) there exists a nonzero eigenvalue of B, sayλ, such that kera) keri λb) /={0}; so choose e λ kera) keri λb) \{0} and consider an orthogonal basis B of R n with first term e λ. Because of the choice ) of e λ, the ) symmetrics matrices of A and B in the basis B have 0 0 λ 0 the respective shapes 0 A and 0 B where the matrices A and B belong to S n 1 R). An elementary computation gives deti n xa yb) = 1 λy) det deti n xa) deti n yb) = 1 λy) det I n 1 xa yb ), I n 1 xa ) det I n 1 yb ), so det I n 1 xa yb ) = det I n 1 xa ) det I n 1 yb ), x, y R. Then, by the induction hypothesis A B = 0, and we have ) ) ) AB = P λ 0 0 A 0 B P = P A B P = 0 and we are done Proof of property ) For it, we first need two lemmas. Lemma 1. Let U = u ij )) a symmetric positive semi-definite matrix; if a diagonal coefficient u ii 1 i n) is equal to zero, then u ij = u ji = 0 for all 1 j n. Proof of the lemma 1. This is classical and elementary see [7, problem 20.1]) but for this demonstration to be self-contained we include the proof: let U such a matrix with u ii = 0 and, by contradiction, suppose that there exists a coefficient u ji /= 0. Let X t = x k ) n 1 the vector where x j = 1, x i = tu ji, t R and where the other components are zero, then t X t UX t = u jj + 2tu 2 ji change sign when t runs through R which is impossible.

4 1618 H. Carrieu, P. Lassère / Linear Algebra and its Applications ) Lemma 2. Let U, V S n R). Suppose U 0 and Then t R : detu tv) = 0. kerv) keru) /={0}. Remark. Note that this lemma is obviously false without some symmetry hypothesis; for example, consider U = ) I n , V = ) 0 1 O n 1 0. Proof of the lemma 2. V is diagonalisable in an orthonormal basis: ) P O n R) : PV t Dr 0 P = Diagλ 1,..., λ r,0,...,0) =, λ 0 0 i /= 0 note that r < n because detu) = 0) then we have: ) ) ) PU t P tpv t U1 U P = 2 Dr 0 U1 td t t = r U 2 U 2 U t. U 2 U 3 And in fact, it is possible to choose P at the expense of changing the n r lasts basis vectors) to have also U 3 diagonal: precisely, let Q O n r so that QU t 3 Q is diagonal, then, in the new basis associated ) Ir 0 to the orthogonal matrix 0 Q our matrix PU t P tpv t P is U 1 td r U ) t 2 U 2 QU t 3 Q and is in the required shape. So let us consider such a P, because of the hypothesis, the polynomial R t detpu t P td) = 0 is nul; the coefficient of t r being up to a sign) λ 1...λ r detu 3 ),wehavedetu 3 ) = 0. Now, detu 3 ) = 0 and U 3 diagonal implies that the positive symmetric matrix PU t P admits a diagonal element equal to zero, say the i-th i {r + 1,..., n}): then by lemma 1 the i-th column in PU t P is also null, e.g. PU t Pe i = 0; but, because i {r + 1,..., n} we have also ) PV t Dr 0 Pe i = e 0 0 i = 0. Consequently U t Pe i = V t Pe i = 0, and finally t Pe i keru) kerv). What we had to prove. Now the proof of the property ) is easy: Proof of property ). Let λ be a non zero eigenvalue of B,wehave det I n λ 1 B xa) = 0, x R and we will be in the case of the lemma 2 with V = A and U = I n λ 1 B who will be positive semi-definite if we choose λ as the greatest nonzero eigenvalue of B. Acknowledgments It is our pleasure to thank our colleague Jean Baptiste Hiriart-Urruty University of Toulouse) who first told us about the Craig Sakamoto s theorem. References [1] M.F. Driscoll,W.R. Gundberg, A history of the development of Craig s theorem, Am. Stat ) [2] J.B. Hiriart-Urruty, Le théorème déterminental en bref, RMS )

5 H. Carrieu, P. Lassère / Linear Algebra and its Applications ) [3] Chi-Kwong Li, A simple proof of the Craig Sakamoto s theorem, Linear Algebra Appl., ) [4] M. Matsuura, On the Craig Sakamoto s theorem and Olkin s determinantal result, Linear Algebra Appl ) [5] J. Ogawa, A history of the development of Craig Sakamoto s theorem from Japanese standpoint, Proc. Ann. Inst. Stat. Math ) [6] J. Ogawa, I. Olkin, A tale of two countries: The Craig Sakamoto Matusita theorem, J. Stat. Plann. Inference, in press. [7] V.V. Prasolov, Problems and Theorems in Linear Algebra, Translation of Mathematical Monographs, vol. 134, AMS, [8] T. Ogosawara, M. Takahashi, Independence of quadratics quantities in a normal system, J. Sci. Hiroshima Univ ) 1 9. [9] I. Olkin, A determinantal proof of the Craig Sakamoto s theorem, Linear Algebra Appl ) [10] O. Taussky, On a matrix theorem of A.T. Craig and H. Hotelling, Indagationes Math )

This article appeared in a journal published by Elsevier. The attached copy is furnished to the author for internal non-commercial research and

This article appeared in a journal published by Elsevier. The attached copy is furnished to the author for internal non-commercial research and education use, including for instruction at the authors institution