An Explicit SOS Decomposition of A Fourth Order Four Dimensional Hankel Tensor with A Symmetric Generating Vector

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arxiv:1503.03383v1 [math.oc] 8 Mar 2015 An Explicit SOS Decomposition of A Fourth Order Four Dimensional Hankel Tensor with A Symmetric Generating Vector Yannan Chen Liqun Qi Qun Wang September 25, 2018 Abstract In this note, we construct explicit SOS decomposition of A Fourth Order Four Dimensional Hankel Tensor with A Symmetric Generating Vector, at the critical value. This is a supplementary note to Paper [3]. Key words: Hankel tensors, generating vectors, sum of squares, positive semidefiniteness. AMS subject classifications (2010): 15A18; 15A69 We construct an explicit SOS decomposition for the homogeneous polynomial f(x). Let u := ( x 2 1,x2 2,x2 3,x2 4,x ) 1x 3,x 2 x 4,x 1 x 2,x 3 x 4,x 2 x 3,x 1 x 4, be a basis of quadratics. Then, f(x) is SOS if and only if there exists a PSD matrix C such that f(x) = u Cu. Inspired by the Cholecky decomposition of C, we define the parameterized SOS decomposition of f(x) as follows f(x) = 10 q 2 k (x), School of Mathematics and Statistics, Zhengzhou University, Zhengzhou, China. E-mail: ynchen@zzu.edu.cn (Y. Chen) This author s work was supported by the National Natural Science Foundation of China (Grant No. 11401539) and the Development Foundation for Excellent Youth Scholars of Zhengzhou University (Grant No. 1421315070). Department of Applied Mathematics, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong. E-mail: maqilq@polyu.edu.hk (L. Qi). This author s work was partially supported by the Hong Kong Research Grant Council (Grant No. PolyU 502111, 501212, 501913 and 15302114). Department of Applied Mathematics, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong. Email: wangqun876@gmail.com (Q. Wang). 1

where q 1 (x) = α 11 x 2 1 +α 12 x 2 2 +α 13 x 2 3 +α 14 x 2 4 +α 15 x 1 x 3 +α 16 x 2 x 4 +α 17 x 1 x 2 +α 18 x 3 x 4 +α 19 x 2 x 3 +α 1,10 x 1 x 4, q 2 (x) = α 22 x 2 2 +α 23x 2 3 +α 24x 2 4 +α 25x 1 x 3 +α 26 x 2 x 4 +α 27 x 1 x 2 +α 28 x 3 x 4 +α 29 x 2 x 3 +α 2,10 x 1 x 4, q 3 (x) = α 33 x 2 3 +α 34x 2 4 +α 35x 1 x 3 +α 36 x 2 x 4 +α 37 x 1 x 2 +α 38 x 3 x 4 +α 39 x 2 x 3 +α 3,10 x 1 x 4, q 4 (x) = α 44 x 2 4 +α 45x 1 x 3 +α 46 x 2 x 4 +α 47 x 1 x 2 +α 48 x 3 x 4 +α 49 x 2 x 3 +α 4,10 x 1 x 4, q 5 (x) = α 55 x 1 x 3 +α 56 x 2 x 4 +α 57 x 1 x 2 +α 58 x 3 x 4 +α 59 x 2 x 3 +α 5,10 x 1 x 4, q 6 (x) = α 66 x 2 x 4 +α 67 x 1 x 2 +α 68 x 3 x 4 +α 69 x 2 x 3 +α 6,10 x 1 x 4, q 7 (x) = α 77 x 1 x 2 +α 78 x 3 x 4 +α 79 x 2 x 3 +α 7,10 x 1 x 4, q 8 (x) = α 88 x 3 x 4 +α 89 x 2 x 3 +α 8,10 x 1 x 4, q 9 (x) = α 99 x 2 x 3 +α 9,10 x 1 x 4, q 10 (x) = α 10,10 x 1 x 4. 2

The involved 55 parameters α i,j,(i j), satisfies the following 35 equality constraints v 0 = α11, 2 (1) 4v 1 = 2α 11 α 17, (2) 4v 2 = 2α 11 α 15, (3) 7 6v 2 = 2α 11 α 12 + αk7, 2 (4) 4v 3 = 2α 11 α 1,10, (5) 12v 3 = 2α 11 α 19 +2 α k5 α k7, (6) 4v 3 = 2 α k2 α k7, (7) 12 = 2α 11 α 16 +2 6 = 2α 11 α 13 + 12 = 2 1 = 7 α k7 α k10, (8) αk5 2, (9) α k2 α k5 +2 7 α k7 α k9, (10) αk2 2, (11) α 11 α 18 +2 4v 5 = 2 α k5 α k,10, (12) α k2 α k10 +2 α k3 α k7 +2 α k6 α k7, (13) α k5 α k9, (14) α k2 α k9, (15) 6v 6 = 2α 11 α 14 + 24v 6 = 2 4v 6 = 2 10 α k5 α k6 +2 α 2 k10, (16) 7 9 α k7 α k8 +2 α k9 α k10, (17) α k3 α k5, (18) 3

4v 6 = 2 6v 6 = 2 4v 5 = 2 12 = 2 6 = 2 12 = 2 1 = 4v 3 = 2 12v 3 = 2 4v 3 = 2 4v 2 = 2 6v 2 = 2 4v 1 = 2 v 0 = α k2 α k6, (19) α k2 α k3 + α k4 α k7 +2 9 αk9 2, (20) α k3 α k10 +2 α k2 α k8 +2 α k6 α k10, (21) α k5 α k8, (22) α k6 α k9, (23) α k3 α k9, (24) α k4 α k5 +2 α k2 α k4 +2 α k3 α k6 +2 8 α k8 α k10, (25) αk6 2, (26) 8 α k8 α k9, (27) αk3 2, (28) α k4 α k10, (29) α k4 α k9 +2 α k6 α k8, (30) α k3 α k8, (31) α k4 α k6, (32) α k3 α k4 +2 8 αk8 2, (33) α k4 α k8, (34) α 2 k4. (35) 4

Giving fixed v 2,v 6,v 1,v 3,v 5, we denote M 1 (v 2,v 6,v 1,v 3,v 5 ) as the minimum of the following optimization problems: M 1 (v 2,v 6,v 1,v 3,v 5 ) := min v 0 s.t. equality constraints (1)-(35). (36) By simple algebraic derivation, we have the following theorem. Theorem 1 Supposethatthe assumptions(2) and(3)in [3]hold. Thenif v 0 M 1 (v 2,v 6,v 1,v 3,v 5 ), A is SOS and 10 f(x) = q k (x) 2. Thus, M 1 (v 2,v 6,v 1,v 3,v 5 ) M 0 (v 2,v 6,v 1,v 3,v 5 ). References [1] A.A. Ahmadi and P.A. Parrilo, A convex polynomial that is not sos-convex, Mathematical Programming 135 (2012) 275-292. [2] R. Badeau and R. Boyer, Fast multilinear singular value decomposition for structured tensors, SIAM Journal on Matrix Analysis and Applications 30 (2008) 1008-1021. [3] Y. Chen, L. Qi and Q. Wang, Positive semi-definiteness and sum-of-squares property of fourth order four dimensional Hankel tensors, arxiv:1502.04566, March 2015. [4] G. Chesi, On the gap between positive polynomials and SOS of polynomials, IEEE Transactions on Automatic Control 52 (2007) 1066-1072. [5] M.D. Choi and T.Y. Lam, Extremal positive semidefinite forms, Mathematische Annalen 231 (1977) 1-18. [6] W. Ding, L. Qi and Y. Wei, Fast Hankel tensor-vector products and application to exponential data fitting, Numerical Linear Algebra with Applications, (2015), DOI: 10.1002/nla.1970. [7] C. Fidalgo and A. Kovacec, Positive semidefinite diagonal minus tail forms are sums of squares, Mathematische Zeitschrift 269 (2011) 629-645. [8] D. Hilbert, Über die Darstellung definiter Formen als Summe von Formenquadraten, Mathematical Annals, 32 (1888) 342-350. 5

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