Méthode du gradient pour systèmes linéaires symétriques définis positifs

n∈N∗n
Rnh· ,·i
k·k
AS++
n(R)bRn
Rnx
Ax=b.
A
Ψ : Rn×Rn→R; (x,y)7→txAy
aRn
A
Ψ : (x,y)7→ hx|a(y)i,
Ψ
Rn
a
a
BRn(u1, ...un)
a
x∈Rn
Ψ(x,x) =
n
X
i=1
λix0
i
2,
x0
iiexBi= 1, ..., n
x h Rn
Φ(x+h) =1
2kx+hk2
A− hb,x+hi
=1
2kxk2
A+hx,hiA+1
2khk2
A− hb,xi−hb,hi
=Φ(x) + hAx−b, hi+1
2khk2
A
=Φ(x) + hAx−b, hi+o
h→0(khk).
k · k | · kA
Rn

xRn
Φ(x)−Φ(¯ x) = kx−¯ xkA.
xRnΦ(x)−Φ(¯ x)≥0
x=¯x Φ¯x
x∈Rn∇Φ(x) = Ax−b∇Φ(x)¯x
∇Φ(a)a
Φa
f
k∇Φ(a)k2
k∇Φ(a)k2
A
f αkf f0(αk) = h∇Φ(xk+
αk∇Φ(xk),∇Φ(xk)i,
h∇Φ(xk+1),∇Φ(xk)i= 0.
(xk,xk+1)xk+1
Φ∇Φ(xk)
xk+1
Φk·kA
¯x
xk
xk+1
xk+2
¯x
∇Φ(xk+1)
xRn

kxk4
kxk2
A−1kxk2
A
≥4λMλm
(λM+λm)2.
xRn
kxk4
kxk2
A−1kxk2
A
≥4λMλm
(λM+λm)2.
x∈B1k·k |·|
1 = kxk=kMM−1Xk ≤ |M| kM−1Xk≤|M|||M−1|,
cond(M)≥1
x∈Rn2 (u1, ...un)
kAxk2=
n
X
i=1
x0
iAui
2
=
n
X
i=1
x0
iλiui
2
=
n
X
i=1
λ2
ix0 i
2≤λ2
M
n
X
i=1
x0
i
2=λ2
Mkxk2.
λM|A| ≤ λM
X u (u1, ...un)
λM
kAuk=kλMuk=|λM|=λM.
|A| ≥ λM
|A|=λM
A−1 t(A−1)=(tA)−1=A−1
A
A−1A
|A−1|= max 1
λ1
, ..., 1
λn=1 λm
,
AR∗
+x7→ 1
x
S++
n(R)S++ n(R)
k·kA−1k·kA
S++
n(R)S++ n(R)
k·kA−1k·kA

cond(A) = λM
λm
k∈Ngk=∇Φ(xk)
pNh· ,·i
kxp+1 −¯xk2
A=hA(xp+1 −¯x),xp+1 −xpi+hA(xp+1 −¯x),xp−¯xi
hA(xp+1 −¯x),xp+1 −xpi=−αphrp+1|rpi= 0
A
kxp+1 −¯xk2
A=hA(xp+1 −xp),xp−¯xi+hA(xp−¯x),xp−¯xi
=hxp+1 −xp, A(xp−¯x)i+kxp−¯xk2
A
=−αphrp,rpi+kxp−¯xk2
A
=−krpk4
krpk2
A
+kxp−¯xk2
A.
kxk+1 −¯ xk2
A
kxp−¯ xk2
A
= 1 −krkk4
kxp−¯ xk2
Akrkk2 A
kxp−¯xk2
A=hA(xp−¯x),xp−¯xi=hA(xp−¯x), A−1A(xp−¯x)i=krpk2
A−1,
kxp+1 −¯ xk2
A
kxp−¯ xk2
A
= 1 −krpk4
krpk2
A−1krpk2
A
.
1−krpk4
krpk2
Akrpk2
A−1≤1−4λMλm
(λM+λm)2=(λM−λm)2 (λM+λm)2,
kxp+1 −¯ xk2
A
kxp−¯ xk2
A
≤cond(A)−1
cond(A)+12
.
∀n∈N,kxn+1 −¯ xkA≤cond(A)−1
cond(A)+1k
kx0−¯ xkA.

xRn
λmkxk2≤ kxkA≤λMkxk,
kxn+1 −¯ xk ≤ rλM
λmcond(A)−1 cond(A)+1k
kx0−¯ xk.
kxp+1 −¯xkA≤λM−λm
λM+λmkxp−¯xkA.
1
(xk)k∈N
n= 2
λ1λ2AΦ
¯x
λ2> λ1
λ2
λ1
1
√λ1
1
√λ2
xk
k∇Φ(xk)kε
kxk−¯xkA≤ε.
(xk)k∈N
1k∈Nxk6=xk+1
¯x
(xk)k∈N
k≥2
xk=¯x xk−16=xkxk−26=xk−1
xk−1−¯x ∇Φ(xk−1)
A(xk−1−¯x)A(xk−1−¯x)A(xk−2−¯x)
0 = hxk−1−¯x, A(xk−2−¯x)i=hA(xk−1−¯x),xk−2−¯xi.
0 = hxk−1−¯x,xk−2−¯xi=hxk−1−¯x,xk−2−xk−1+ (xk−1−¯x)i