证明
Proposition 4. Let K=Λ_+(p,e) . For real symmetric X,Z and y∈^3 , X0: (X,Z,y)∈K…Proposition 3. The polynomial p in is homogeneous of degree 20 on ^4×^4×^3 , a…观点
Write the hypothetical pencil as L(X,Z,y)=L_1(X)+L_2(Z)+L_3(y). We first remove the pencil's common kernel and normalize two positive blocks. We then rescale the pencil and prove that its limit represents the same inequality. Compression and normalization. We use the elementary fact that A+H0, A-H0 A⊆ H. Indeed, if w∈ A , the two nonnegative quadratic forms on w sum to zero. A positive semidefinite matrix annihilates every vector on which its quadratic form vanishes, so both (A+H)w and (A-H)w are zero. Since e is interior to K , for every ambient vector x there is an ε>0 with L(e)±εL(x)0 . The preceding fact shows that L(e) is a common kernel of the whole pencil. Compressing to its orthogonal complement preserves the represented cone and makes L(e)0 . This complement is nonzero: otherwise the pencil would vanish identically, whereas (-I_4,I_4,0)∉K by. The same slice shows that L_1 and L_2 are positive maps. Set U= L_1(I_4) . Positivity at I_4±εX , for small ε>0 , and the same kernel argument show that every L_1(X) annihilates U . Thus, in U^⊥⊕U , L_1(X)= D(X), 0; 0, 0 . Both summands are nonzero. If U^⊥=0 , then L_1=0 and the pencil accepts (-I_4,I_4,0) , contradicting. If U=0 , then L_1(I_4)0 , so L_1(I_4)-εL_2(I_4)0 for sufficiently small ε>0 . This would accept (I_4,-εI_4,0) , again contradicting that slice. Let a= U^⊥ , c= U , and let E(Z) be the lower diagonal block of L_2(Z) . Both D and E are positive maps. Moreover, D(I_4)0 by the definition of U , and E(I_4)0 because it is the compression of L(e)0 to U . Apply the fixed block diagonal congruence diag(D(I_4)^ -1/2 ,E(I_4)^ -1/2 ) and retain the notation for the resulting blocks. We now have D(I_4)=I_a and E(I_4)=I_c . Rescaling the pencil. Write the other blocks as L_2(Z)= F(Z), C(Z); C(Z)^, E(Z) , L_3(y)= A(y), B(y); B(y)^, G(y) . Fix y and put Z_0=Φ_y(I_4) . By and Φ_ sy =s^2Φ_y , the point (I_4,s^2Z_0,sy) lies in K for every real s . Its lower pencil block therefore satisfies s^2E(Z_0)+sG(y)0 (s∈). Dividing by |s| and taking limits through positive and negative s gives G(y)0 and -G(y)0 . Hence G(y)=0 . For fixed X0 , Z and y , and any real s≠0 , congruence of L(X,s^2Z,sy) by diag(I_a,s^ -1 I_c) gives M_s(X,Z,y)= D(X)+sA(y)+s^2F(Z), B(y)+sC(Z); B(y)^+sC(Z)^, E(Z) . Congruence preserves positive semidefiniteness for either sign of s . The exact slice gives M_s(X,Z,y)0 ZΦ_y(X^ -1 ), s≠0. As s→0 , these matrices converge to M(X,Z,y)= D(X), B(y); B(y)^, E(Z) . It follows immediately that ZΦ_y(X^ -1 ) implies M(X,Z,y)0 . Recovering the inequality from the limit. For the converse, suppose that M(X,Z,y)0 . Positivity and unitality give D(X)λ_ (X)I_a0. For every δ>0 , M(X,Z+δI_4,y) =M(X,Z,y)+diag(0,δI_c) 0. To see strict positivity, a vector with nonzero lower component has strictly positive contribution from the added term. A nonzero vector with lower component zero has strictly positive quadratic form under D(X) . Consequently, for each fixed δ>0 , the matrices M_s(X,Z+δI_4,y) are positive definite for sufficiently small nonzero s . Equation yields Z+δI_4Φ_y(X^ -1 ) . Letting δ0 proves the converse. We have therefore shown ZΦ_y(X^ -1 ) M(X,Z,y)0. Taking the Schur complement of the positive definite block D(X) gives.