New PDF release: Linear Complementarity, Linear and Nonlinear Programming

# New PDF release: Linear Complementarity, Linear and Nonlinear Programming

By Murty K.G.

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1 xxxii Notation Quadratic forms in matrix notations Consider the quadraticnform in nn variables x= n P P P (x : : : xn )T , f (x) = gii x2i + gij xi xj . i i j i An example for n = 3 is h(x) = 81x2 ; 7x2 + 5x x ; 6x x + 18x x . Let F = (fij ) be a square matrix of order n satisfying 1 =1 =1 = +1 1 1 2 1 3 2 2 3 fii = gii i = 1 to n fij + fji = gij for i 6= j and j > i: Then it can be veri ed that f (x) = xT Fx. In particular, if we de ne the symmetric matrix D = (dij ) of order n, where dii = gii i = 1 to n dij = dji = 12 gij for i 6= j and j > i then f (x) = xT Dx.

3 If F is PD, fii > 0 for all i. 2. 4 If F is a PSD matrix, all principal submatrices of F are also PSD. 2. 5 If F is PSD matrix, fii >= 0 for all i. 4. 6 Suppose F is a PSD matrix. If fii = 0, then fij + fji = 0 for all j . Proof. To be speci c let f11 be 0 and suppose that f12 + f21 6= 0. 4 the principal submatrix 8 f11 f12 9 8 0 f12 9 > :f f > = > :f f > 21 22 21 22 must be PSD. Hence f22y22 + (f12 + f21)y1 y2 > = 0 for all y1 y2. Since f12 + f21 6= 0, take y1 = (;f22 ; 1)=(f12 + f21) and y2 = 1.

So from the above argument f ( ) 6= 0 for any satisfying 0 < = < = 1. Clearly, f (0) = 1, and f (1) = determinant of F . 1 there exists a satisfying 0 < < 1 and f ( ) = 0, a contradiction. Hence f (1) 6< 0. Hence the determinant of F cannot be negative. Also it is nonzero. Hence the determinant of F is strictly positive. 3 If F is a PD matrix, whether it is symmetric or not, all principal subdeterminants of F are strictly positive. Proof. 2. 4 If F is a PSD matrix, whether it is symmetric or not, its determinant is nonnegative.

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