ĐA THỨC MA TRẬN sự PHÂN bố GIÁ TRỊ RIÊNG, các ĐỊNH lý BIỂU DIỄN DƯƠNG và một số vấn đề LIÊN QUAN

n R R+ C N K C R n n n R Cn t t Mt(R) Mt(C) St(R) X X α C[z] R[X] R(X) t A≻0 ||A|| P Mt(R) n (X1, , Xn) Xα1 Xαn , α = (α , , α ) ∈ Nn n n n X = (X1, , Xn) R[X] Mt(R[X]) St(R[X]) T A A0 F ∈ d t×t →R f λ(F)(Y ) := Λ(fij (Y )) e r(Y ) := s ri (Y ) Xi =1 d>0 F = (fij ) ∈ St(R[X]) F ] λ(F) + cr > m1 m ∩ {y ∈ m] |λ(F)(y) λ(F) > P c m m ] λ(F) ≤ 0} m λ(F) c > −m1/m2 r m2 R U= U] r>0 U m2 > m1 m \U λ(F) U ] ≥ m1 + cm2 > 0; λ(F) + cr ] ] m R ∩ {y ∈ m] |λ(F)(y) ≤ 0} m2 m r λ(F) + cr ≥ λ(F) > d>0 F = (fij ) ∈ St(R[X]) F := (f ) ∈ S (R[Y ]) F≻0 P ij t m F + crI 1, · · · , t ^ P e f t ≻ e F m c P k λk(F), k = ^ ck λk(F) + ckr k=1,··· ,t ^ k = 1, · · · , t e λk(F) F k = 1, · · · , t m k = 1, · · · , t e Fm:= F e X F λk(F) + cr c = max ck m F + cr + crIt t m e + cr F ^ λ k( ) + cr I F t I Yi i=1 F F β BβY , Bβ ∈ St(R), F= X β|≤d | m X i =1 F h Yi m |X X β = B β Y ( Y i) β|≤d d−|β| i=1 h d h Mϕ( F h )=F F F m P P ϕ Mϕ r F F h < λIt N > d(d − 1) L − d F 2λ F= F F mλ h F d := >0 α1 Cαλ1 |α|X =N +d · · · λm αm , ( F) L := L( F h) Cα ∈ St(R) h h Mϕ Mϕ(F ) = F ϕ =1 Y! d = deg( F F h) =1 i m Xi P F F h mλ < λIt N > d(d − 1) L − d F 2λ F F h P F d := >0 (F) L := L( F h) X δi δ F= δ Cδλ1 · · · λm m , ∈{0,1} Cδ ∈ St(R[X]) T N+d Cα A A A ∈ Mt(R[X]) P λ1, · · · , λm ∈ R[X] P = {x ∈ Rn|λi(x) ≥ 0, i = 1, · · · , m} d>0 F = (fij ) ∈ St(R[X]) P F ci ∈ R m P i=1 ci ciλi(X) = Xm Xi = bij λi(X), i = 1, · · · , n, j=1 B = (bij )i=1,··· ,n;j=1,··· ,m f i, j = 1, ,t ij f ϕ ··· { r1, · · · , rs } (ϕ) e c t + cr F I h λ ≻ m F h F := F + crIt y ∈ (y) < λIt e m F m K⊆R G ∈ St(R[Y ]) c∈R y ∈ K G(y) < cIt, G(y) ≻ c>0 y∈K y ∈ K G(y) < cIt λ1(G), · · · , λt(G) K λi(G) G ∈ St(R[Y ]) c := λ (G)(y), i y∈K c := max i=1,··· ,t i = 1, · · · , t i c G−cIt i λi(G)−c i = 1, · · · , t c λi(G)(y) − c ≥ λi(G)(y) − ci ≥ y∈K i = 1, · · · , t y ∈ K G(y) < cIt, L h := L(F ) N > d(d − 1) L − d 2λ h λi(X) ( P m i=1 F N Yi) F ∈ St(R[Y ]) Yi P ′ ′ ′ ′ := {(x, y) ∈ R |λ = + x ≥ 0, λ = − x ≥ 0, λ = + y ≥ 0, λ = − y ≥ 0} c1 = c2 = c3 = c4 = P i=1 ′ ciλi (x, y) = 1 1− 1 + := + λ1 :=4 x, λ2 := 4 x, λ3 y, λ4 =1 P B = " −2 0 # i=1 λi := − y ∈ R[x, 4 y], −2 T T B · [λ1 λ2 λ3 λ4] = [x y] ϕ : R[y1, y2, y3, y4] → R[x, y] ϕ(yi) := ϕ (ϕ) λi(x, y) i = 1, 2, 3, {r1, r2} := {y1 + y2 − , y3 + y4 − 1} 2 r := r1 + r2 F := "−4 x3 + 5xy 3xx + x y + 3x 2 x y + 7x + y + − 2− 4y + x3 + 5xy # 3x − F 2 2 λ1(F) = 6x − 4x y − 4y + 6; λ2(F) = x + x y + 4x + y + (x, y) ∈ P 11 − 1− λi(F)(x, y) ≥ i = 1, 23 B − e F 34 e F(x, y) < 2I2 = (fij ) f = 4(2y 2y ) (2y 2y )(y + y2 + y3 + y4) + 7(2y1 (2y − 2y )(y + y + y3 + y4) + 3(y1 + y2 + y3 + y4) , f f f 12 f = (2y1 = f21 −2y2) (y1 + y2 + y3 (x, y) ∈ P − + y4) + 5(2y1 − 2y2)(2y3 − 2 2y2) (y1 + y2 + y3 + y4) + 2y4)(y1 + y2 + y3 + y4 ) − 3(2y1 − 2y2)(y1 + y2 + y3 + y4) , f 22 = (2y1 − 2y2) + (2y1 − 2y2) (2y3 − 2y4)(y1 + y2 + y3 + y4) + 3(2y1 − 2y2) (y1 + y2 + y3 + y4) 4(2y3 2y4)(y1 + y2 + y3 + y4) + 6(y1 + y2 + y3 + y4) f − − e F ^ λ1(F) = λ1(F) = 35y1 − 52y1 y2 2 68y y3y4 + 20y y − 52y1y e 1 3 2 2 2 2 + 6y1 y2y4 + 48y1 y3 + + 2y1y y3 + 6y1y y4 + 8y1y2y3y4 + 8y1y2y + 18y1y + 42y1y y4 + 2 4 30y1y3y4 +6y1y4 +35y2 +54y2y3 +34y2 y4 + 54y1y3 + 34y1y4 + 82y1 y2 + 2y1 y2y3 2 2 +48y2y3 +68y2 y3y4 +20y2y4 30y2y3y4 + 6y2y4 + 5y3 + 16y3 y4 + 18y3 y4 + 8y3y4 + y4 3 ^ λ2(F) = λ2(F) = 30y1 + 24y1y2 + 32y1y3 + 112y1y4 e 2 2 3 +18y2y3 +42y2y3y4 2 2 −12y1 y2 + + + 32y1y2y3 + 16y1y2y4 + 4y1y3 120y2y y + 116y2y2 + 24y y3 + 32y y2y + 16y y2y + 40y y y2 + 48y y y y + 8y y y2 + 4 1 48y y2y + 96y y y2 + 48y y3 + 30y4 + 32y3y + 112y3y + 4y2y2 + 120y2y y + 116y2y2 + 3 4 4 2 3 48y y y + 96y y y + 48y y − 2y + 8y y + 36y y + 40y y + 14y e e e 4 4 3 F := e F + cr I F 2≻ P4 i=1 h 4 4 − c > −0 125 = 16 4 4∩{λ2(F)≤0} c = 17 2 λ (F) = −2 r = 0.125 λ1(F) = 1, h F = yi (fij ) h 2 f11 = (3(y1 + y2 + y3 + y4) + (2y1 − 2y2) + (2y3 − 2y4)(y1 + y2 + y3 + y4))(y1 + y2 + y3 + y4) + 2 (6(y1 +y2 +y3 +y4) −(4(2y3 −2y4))(y1 +y2 +y3 +y4))(2y1 −2y2) + 17(0.5y1 + 0.5y2 − 2 2 0.5y3 − 0.5y4) (y1 + y2 + y3 + y4) + 17(0.5y3 + 0.5y4 − 0.5y1 − 0.5y2) (y1 + y2 + y3 + y4) f12 h h = f21 = (y1 + y2 + y3 + y4)(3(y1 + y2 + y3 + y4) + (2y1 − 2y2) + (2y3 − 2y4)(y1 + y2 + y3 + y4))(2y1 − 2y2) + (2y1 − 2y2)(6(y1 + y2 + y3 + y4) − (8y3 − 8y4)(y1 + y2 + y3 + y4))(−y1 − y2 − y3 − y4) h 2 f22 = (3(y1 + y2 + y3 + y4) + (2y1 − 2y2) + (2y3 − 2y4)(y1 + y2 + y3 + y4))(2y1 − 2y2) + (6(y1 2 + y2 + y3 + y4) −(8y3 −8y4)(y1 + y2 + y3 + y4))(−y1 −y2 −y3 −y4) + 17(0.5y1 + 0.5y2 − 2 y y y + y +y y (y + y + y − 0.5y4) + y4 ) + 17(0.5y h + 0.5y4 − h 1− λ1(F ) = 1.9706, λ2(F ) = 1.5294 λ := 1.5294 F h < 1.5294I2 L := L(F ) = (y1 + y2 + y3 + 24 167 F h y4) h 167 (y1 + y2 + y3 + y4) yi λi(x, y) F ( 1044 h N = 167 ) F = + y 4) 87 ∈ St(R[y1, y2, y3, y4]) ∗ (λ A + λB + C)x = b ... f (z) = adz + d A := max i=0, ,d−1 f (z) 2|ad|(1 + A) d−1 |a0 | (Ad + 1) ad − ≤| | z