Un théorème sur les matrices inversibles

A∈Mn(R)
A∈Mn(R)B∈Mn(R)
AB =BA =In
A∈Mn(R)
B∈Mn(R)AB =In
4
1ere
3eme 4eme
1ere
AB =Indet(A).det(B) = det(In) = 1 det(A)6= 0 A
2eme
f:M7→ MA Mn(R)
MKer f MA = 0 ⇒M=M(AB) = (MA)B= 0.
Ker f={0}f Mn(R)
C CA =In
C=C(AB) = (CA)B=B AB =BA =In
3eme
AB =InCA =InB=C
A, B, C ∈Mn(R)
1≤i≤n BiCiieme B C
AB =C⇒ ∀i∈ {1, . . . , n}, ABi=Ci.

AB =C i ∈ {1, . . . , n}
∀k∈ {1, . . . , n},[ABi]k=
n
X
s=0
aksbsi = [AB]ki =cki ABi=Ci
3
EiBiieme InB
AB =In⇒ ∀i, ABi=Ei
A, B In
Pn
i=1 λiBi= 0 APn i=1 λiBi=Pn i=1 λiABi=Pn i=1 λiEi= 0
λi= 0 i(Ei)1≤i≤nMn,1(R)
(Bi)1≤i≤nn Mn,1(R)
n
(Bi)1≤i≤nMn,1(R)
X∈Mn,1(R)x1, . . . , xnX
(Bi)1≤i≤nX=Pn
i=1 xiBi
(BA)X=Pxi(BA)Bi=PxiB(ABi) = PxiBEi=PxiBi=X
∀X∈Mn,1(R),(BA)X=X
X=EiBAEi=Ei
BA = (BA)In= (BA)(XEi) = XBAEi=XEi=In
4eme 1ere
(∃B, AB =In)⇒(A)⇒(A)