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

n

AA

P (z) = Adzd + · · · + A1z + A0 mM

f sos,G = sup{λI|f − λI ∈ MG}.

MG

f > 0KG f ∈ MG

KG TG

KG

Rn

f (z)d

f (z) = adzd + ad−1zd−1 + · · · + a1z + a0, ai ∈ R, ∀i = 0, , d

a d i=0, ,d−1

a 0 i=1, ,d

f (z)

2

|z| ≥

1 + √1 + 4δ”.

KG

f − λI f − λI ≥ 0 K

G m

f ∈ R[X] µK

f (1, X1, , Xn) = f (X1, , Xn)

f ∈ R[X1, , Xn]f

f(0, X1, , Xn) = fd(X1

˜ ∈ R[X0, X1, , Xn]f

≤ |λI| ≤ 1, 2λI max (A d )

min λImin(Ai) λI max λImax(Ai)

≤ | | ≤ λImin(Ai+1)

i=0, ,d−1 λImax(Ai+1) i=0, ,d−1

λI ∈ C P (z)u ∈ Ct, kuk = 1

λI ∈ C P (z) x ∈ CnλI

1

≥ k(A0 + A1λI + · · · + Ad−2λId−2)xk.

|λI|d − kAd−1k λId−1 ≤ k(It · λId + Ad−1λId−1)xk.

0 <|λI|d − kAd−1kλId−1 −α|λI|d−1|λI|

− 1

< k(It · λId + Ad−1λId−1)xk − k(A0 + A1λI + · · · + Ad−2λId−2)xk

≤ k(A0 + A1λI + · · · + Ad−2λId−2)x + (Ad−1λId−1 + It · λId)xk = kP (λI)xk.

λ P (z)

2(1 + ) ( + 1) ≤ | | ≤

−1 −1 A

kP (λI)xk ≥ |λI|d − dA |λI|d−1 ≥ |λI|d − |λI|d−1 > 0.

|λI| ≤ 1 ≤ 1 + λI0A λI0 ∈ (0, 1)

g˜i(1, X1, · · · , Xn) = gi(X1, · · · , Xn) i = 1, · · · , m,(3.1)

G = {G1, · · · , Gm} ⊆ St(R[X]) F ∈ St(R[X]) deg(F) =2d, deg(GI) = 2di, i = 1, , m

(1 + X12 + · · · + Xn2)ri fi ∈ MG

r = max{ri, i = 1, · · · , t} i = 1, · · · , t

(1 + X12 + · · · + Xn2)rfi ∈ MG

X

D ≻ 0 R+n ∩ KG Ds ≻ 0 R+n ∩ (KG )2d′ \ {0}

L(F) = max ||A1||,2! 2! 1 ||A4||

||A2||, ||A3 ||,

||A1|| = 6.9646, ||A2|| = 7.6713, ||A3|| = 4.7581, ||A4|| = 12.2341.

iλIi

P R2 v0, v1, v2 {λI0, λI1, λI2}

v0 = (0, 1), v1 = (1, 0), v2 = (1, 1) λI0 = 1 − X, λI1 = 1 − Y, λI2 = X + Y − 1

P = {(x, y) ∈ R2|λI0 ≥ 0, λI1 ≥ 0, λI2 ≥ 0}.

11 = y 0 + 23y 0 y 1 + 30y 0 y 2 + 36y 0 y 1 + 79y 0 y 1 y 2 + 49y 0 y 2 + 27y 0 y 1 + 81y 0 y 1 y 2 + 89y 0 y 1 y 2

36y0y23 + 8y14 + 30y13y2 + 45y12y22 + 33y1y23 + 10y24

BP

Y = 9λI0 2λI1 λI2

+ 7λI1 + λI2 + λI3

λI(F) + cr ≥ λI(F) > 0.

∈ St(R[Y ]) Yi

i=1 Y i )N F

F

P := {(x, y) ∈ R2|λI′1 = 1 + x ≥ 0, λI′2 = 1 − x ≥ 0, λI′3 = 1 + y ≥ 0, λI′4 = 1 − y ≥ 0}.

c1 = c2 = c3 = c4 = P

i=1 ciλIi′(x, y) = 14

^ 4 3 3 3 2 2 2 2 2 2+

λI1(F) = λI1(F) = 35y1 − 52y1 y2 + 54y1y3 + 34y1y4+ 82y1 y2 + 2y1 y2y3 + 6y1 y2y4 + 48y1 y3

68y2y3y4+ 20y2y2 − 52y1y3 + 2y1y2y3 + 6y1y2y4 + 8y1y2y3y4 + 8y1y2y2 + 18y1y3+ 42y1y2y4 +

30y1y3y4 +6y1y4 +35y2 +54y2y3 +34y2 y4 +48y2y3 +68y2 y3y4 +20y2y4 +18y2y3 +42y2y3y4

30y2y3y42 + 6y2y43 + 5y34 + 16y33y4 + 18y32y42 + 8y3y43 + y44

0.5y3 − 0.5y4)2(y1 + y2 + y3 + y4)2 + 17(0.5y3 + 0.5y4 − 0.5y1 − 0.5y2)2(y1 + y2 + y3 + y4)2 f12h

= f 21h = (y 1 + y 2 + y 3 + y 4 )(3(y 1 + y 2 + y 3 + y 4 )2 + (2y 1 − 2y2 )2 + (2y 3 − 2y4 )(y 1 +

y2 + y3 + y4))(2y1 − 2y2) + (2y1 − 2y2)(6(y1 + y2 + y3 + y4)2 − (8y3 − 8y4)(y1 + y2 + y3 +

∗

(λI2A + λIB + C)x = b