Page 150 - Textos de Matemática Vol. 39

140 ANA PAULA SANTANA AND HELENA ALBUQUERQUE
Lemma 2.4.
(...((Ei1 j1 φ (α1 )Ei2 j2 φ (α2 ))Ei3 j3 φ (α3 ))...)Eir jr φ (αr ) =
Iφ + α1Ei1j1 + α2Ei2j2 + ... + αrEirjr .
Proof. We use induction on r to prove this assertion. If r = 2, simple calcu- lations prove that
Ei1j1φ(α1)Ei2j2φ(α2)=Iφ+α1Ei1j1 +α2Ei2j2. Suppose that the assertion is valid for a product of k matrices. Then
(...((Ei1j1φ(α1)Ei2j2φ(α2))...Eikjkφ(αk))Eik+1jk+1φ(αk+1) = (Iφ+α1Ei1j1 +α2Ei2,j2 +...+αkEikjk)(Iφ+αk+1Eik+1jk+1)=
Iφ +α1Ei1j1 +α2Ei2j2 +...+αkEikjk +αk+1Eik+1jk+1. Lemma 2.5.
Ei1j1φ(α1)(Ei2j2φ(α2)(...(Eikjkφ(αk)X)...)) = ((Ei1j1φ(α1)Ei2j2φ(α2))...Eikjkφ(αk))X.

Proof. Once more, we use induction on r. If r = 2 we have
(Iφ + α1Ei1j1 )[(Iφ + α2Ei2j2 )X] = (Iφ + α1Ei1j1 )[X + α2Ei2j2 X] =
X+α1Ei1j1X+α2Ei2j2X=(Iφ+α1Ei1j1 +α2Ei2j2)X= (Ei1j1φ(α1)Ei2j2φ(α2))X.
Suppose that the statement holds for the product of k matrices. Then Ei1j1φ(α1)(Ei2j2φ(α2)(...(Eikjkφ(αk)(Eik+1jk+1φ(αk+1)X))...)) =
((Ei1j1φ(α1)Ei2j2φ(α2))...Eikjkφ(αk))[Eik+1jk+1φ(αk+1)X] = (Iφ +α1Ei1j1 +α2Ei2j2 +...+αkEikjk)[X+αk+1Eik+1jk+1X]= X + α1Ei1j1 X + α2Ei2j2 X + ... + αkEikjk X + αk+1Eik+1jk+1 X = ((Ei1j1φ(α1)Ei2j2φ(α2))...Eik+1jk+1φ(αk+1))X.

Theorem 2.6. Let X ∈ Mn,φ. Then there are deformed permutation matrices Pi1j1φ,Pi2j2φ,...,Pikjkφ, a matrix U in row echelon form and a lower triangular matrix L with lii = φ(i−1, i, i−1) such that
Pi1j1φ(Pi2j2φ(...(PikjkφX))...) = L.U.