A=
1 1 ··· 1
1 2 ··· 2
1 2 ··· n
= (min(i, j))1i,jn∈ Mn(R)
A
H=1
i+j11i,jn∈ Mn(R)
H
A= (ai,j )∈ S++
n(R)
ϕ(X, Y ) =tXAY Mn,1(R)
i6=j
a2
i,j < ai,iaj,j
A= (ai,j )1i,jn∈ Sn(R)Ap= (ai,j )1i,jp
p∈ {1, . . . , n}A
det A > 0
A
p∈ {1, . . . , n}det Ap>0
nN
A∈ S++
n(R)B∈ S+
n(R)
In+AB
AGLn(R)
tAA ∈ S++
n(R)
S∈ S++
n(R)
S2=tAA
AGLn(R),(O, S)∈ On(R)× S++
n(R), A =OS
A∈ Mn(R)A
P∈ Mn(R)A=tP P
A
PGLn(R)A=tP P
Mn(R)S++
n(R)
A1, . . . , AkS++
n(R)λ1, . . . , λk
A=
k
X
i=1
λiAiB=
k
X
i=1 |λi|Ai
XRn
tXAXtXBX
|det A| ≤ det B
A∈ S++
n(R)B∈ S+
n(R)
C∈ S++
n(R)C2=A1
D=CBC
(det(I+D))1/n 1 + (det D)1/n
(det(A+B))1/n (det A)1/n + (det B)1/n
A, B ∈ S++
n(R)
(A+B)16=A1+B1
S∈ Sn(R)S
S∈ S++
n(R)S∈ S+
n(R)
A
T=
1··· 1
(0) 1
GLn(R)
A=tT T
X∈ Mn,1(R)
tXAX =t(T X)T X =kT Xk20
T X = 0 X= 0
H
H
(P, Q)7→ Z1
0
P(t)Q(t) dt
Rn1[X]H
ϕ A
A∈ S++
n(R)
E1, . . . , EnMn,1(R)
ϕ(Ei, Ej) = ai,j ϕ(Ei, Ei) = ai,i ϕ(Ej, Ej) = aj,j
a2
i,j ai,iaj,j
EiEj
Ei=Ej
ASp AR
+A
det A A
det A > 0
Ap∈ Sp(R)X∈ Mp,1(R)tXApX=tX0AX0
X0∈ Mn,1(R)X
A∈ S++
n(R)Ap∈ S++
p(R) det Ap>0
n= 1
n1
A∈ Sn+1(R)p∈ {1, . . . , n + 1}det Ap>0
A
A=AnCn
tCnλ
An∈ S++
n(R)
Pn∈ On(R)tP AP =Dn= diag(λ1, . . . , λn)λi>0
Pn+1 =PnXn
0 1 GLn+1(R)
tPn+1APn+1 =DnYn
tYn
Yn=tPn(AXn+Cn)
Xn=A1
nCn
tPn+1APn+1 =Dn0
0
λn+1 >0 det A > 0λ1. . . λn+1 >0
A A
In+AB =AA1+B
A1+B
X∈ Mn,1(R)\ {0},tX(A1+B)X=tXA1X+tXBX > 0
A1+BS++
n(R)In+AB
X
tXtAAX =t(AX)AX 0
tXtAAX = 0 =AX = 0 =X= 0
P∈ On(R)
tPtAAP = diag(λ1, . . . , λn)λi>0
S=tPdiag(pλ1,...,pλn)P
O=AS1A=OS tOO =tS1tAAS1=In
O∈ On(R)A=OS
A=OS S2=tAA
λSp(tAA)
Ker(tAA λIn) = Ker(S2λIn)
Ker(S2λIn) = Ker(SλIn)Ker(S+λIn)
λ > 0
Ker(S+λIn) = {0}
Sp SR
+λSp(tAA)
Ker(tAA λIn) = Ker(SλIn)
S
Mn,1(R) =
λSp(tAA)Ker(tAA λIn)
A=tP P A
P A
A=tQDQ Q ∈ On(R)
D= diag(λ1, . . . , λn)λi0λi>0A
P= ∆Q∆ = diag(λ1,...,λn)
Vect S++
n(R) = Sn(R)
tXAX =Pk
i=1 λitXAiXtXAiX0
|tXAX| ≤ Pk
i=1 |λi|tXAiX=tXBX
B=In
A X |tXAX| ≤ tXX
λ|λ| ≤ 1|det A| ≤ 1 = det B
λiB∈ S++
n(R)
B=C2C∈ S++
n(R)A0=C1AC1∈ Sn(R)
XRn
|tXA0X|=t(C1X)A(C1X)t(C1X)B(C1X) = tXX
|det A0| ≤ 1|det A| ≤ (det C)2= det B
A
A=tP DP P ∈ On(R), D = diag(λ1, . . . , λn)λi>0
C=tPP∆ = diag(1/λ1,...,1/λn)
tD=DtXDX =t(CX)B(CX)0D∈ S+
n(R)
D=tQD0Q Q ∈ On(R), D0= diag(µ1, . . . , µn)µi0
n
Y
i=1
(1 + µi)1/n 1 +
n
Y
i=1
µ1/n
i
µiµi
x7→ ln(1 + ex)
a1, . . . , anR,ln 1+e1
nPn
i=1 ai1
n
n
X
i=1
ln (1 + eai)
ai= ln µi
ln 1 +
n
Y
i=1
µ1/n
i!ln
n
Y
i=1
(1 + µi)1/n
(det C)2det(A+B) = det(CAC +CBC) = det(I+D)
det A= 1/(det C)2det B= det D/(det C)2
(det(I+D))1/n 1 + (det D)1/n
(A+B)1=A1+B1
(A+B)(A1+B1) = In
BA1+AB1+In=On
A
B+AB1A+A=On
X∈ Mn,1(R)
tXBX +t(AX)B(AX) + tXAX = 0
tXBX, t(AX)B(AX),tXAX > 0
(i, j)S
(1)i+ji,j
i,j (i, j)S
i j S
S
(i, j) (j, i)
S
S∈ S++
n(R)
Com S=t(Com S) = det(S)S1
S S1det S > 0 Com S
S∈ S+
n(R)t > 0
St=S+tIn∈ S++
n(R)
Com(St)∈ S++
n(R)
Com(S) = lim
t0±Com(St)∈ S++
n(R) = S+
n(R)
1 / 5 100%
La catégorie de ce document est-elle correcte?
Merci pour votre participation!

Faire une suggestion

Avez-vous trouvé des erreurs dans linterface ou les textes ? Ou savez-vous comment améliorer linterface utilisateur de StudyLib ? Nhésitez pas à envoyer vos suggestions. Cest très important pour nous !