TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...

Appendix D
In the last appendix, we will prove a result needed to Chapter 4 as follows:
Lemma D.1. Let (V,B) be a quadratic vector space, C be an invertible endomorphism of V
such that
(1) B(C(x),y) = B(x,C(y)), for all x,y ∈ V ,
(2) 3C − 2C 2 = Id.
Then there is an orthogonal basis {e1,...,en} of B such that C is diagonalizable with eigenvalues
1 and 1 2 .
Proof. First, one has (2) equivalent to C(C − Id) = 1 2 (C − Id). Therefore, if x is a vector in V
such that C(x)−x = 0 then C(x)−x is an eigenvector with respect to eigenvalue 1 2 . We prove the
result by induction on dim(V ). If dim(V ) = 1, let {e} be an orthogonal basis of V and assume
C(e) = λe for some λ ∈ C. Then by (2) one has λ = 1 or λ = 1 2 .
Assume that the result is true for quadratic vector spaces of dimension n, n ≥ 1. Assume
dim(V ) = n + 1. If C = Id then the result follows. If C = Id then there exists x ∈ V such that
C(x) − x = 0. Let e1 = C(x) − x then C(e1) = 1 2e1. If B(e1,e1) = 0 then there is e2 ∈ V such that B(e2,e2) = 0, B(e1,e2) = 1 and
V = span{e1,e2} ⊥
⊕ V1 where V1 = span{e1,e2} ⊥ . Since 1 2 = B(C(e1),e2) = B(e1,C(e2)) one
has C(e2) = 1 2e2 + y + βe1 with y ∈ V1,β ∈ C. Let f1 = C(e2) − e2 = − 1 2e2 + y + βe1 then
C( f1) = 1 2 f1 and B(e1, f1) = − 1 2 . If B( f1, f1) = 0 then let e1 = f1. If B( f1, f1) = 0 then let
e1 = e1 + f1. In the bold cases, we have B( e1, e1) = 0 and C( e1) = 1 2 e1.
⊥
Let V = C e1 ⊕ e1 ⊥
then e1 ⊥ is non-degenerate, C maps e1 ⊥ into itself. Therefore the result follows by induction.
139