Page 90 - Textos de Matemática Vol. 39

80 CHARLES R. JOHNSON AND YULIN ZHANG
eigenvalue) for B and C is the more likely, but it is not obvious that this is al- ways possible. Our purpose here is to show that each of B and C may be taken to be nonderogatory, for any allowed spectra specified for them. In the process, we show that a certain, more special, sort of factorization usually exists for a similarity of A, and this may be of independent interest.
We first note that the possible Jordan forms for B and C, given their eigen- v a l u e s , a r e a s i m i l a r i t y i n v a r i a n t o f A , a s S − 1A S = S − 1B C S = ( S − 1B S ) ( S − 1C S ) . Thus for purposes of proof, A may be placed in any form allowed for it by sim- ilarity. Our general approach is to show that A is similar to a matrix with a special kind of LU factorization. Although this approach suffices to re-prove the original result (in a possibly cleaner way), it does not work in some cases for our more precise purpose. In particular, we will characterize the infrequent exceptions. However, these exceptions may be treated in another way. For this purpose (and others) we begin with a careful treatment of the 2-by-2 case, in a way different from our general approach.
Lemma 1.1. Let A ∈ M2(C) be nonsingular and nonscalar and let β1,β2,γ1 and γ2 ∈ C be such that β1β2γ1γ2 = detA. Then, there exist nonderogatory matrices B and C such that B has eigenvalues β1, β2, C has eigenvalues γ1 and γ2 and A = BC. In case A is nonsingular and scalar, the conclusion remains validincaseγi =β−1,i=1,2.
0 But such an A may be factored as
i
Proof. Via multiplication by scalars as needed, we may assume without loss
of generality that detA = detB = detC = 1. Thus, A has eigenvalues α, 1. α
We first assume that α ̸= ±1,0. If β1 ̸= β2 and γ1 ̸= γ2, there is nothing to do, as the desired statement follows from the classical result of [2]. Thus we assume, also without loss of generality, that γ1 = γ2 = 1. Since, the eigenvalues of A are distinct, we may assume
α0
1. α
⎡αβα−2 αd⎤⎡α2α−β −d⎤ ⎢ α2 − 1 ⎥ ⎢ α2 − 1 ⎥
⎢⎣ −(α2 −βα+1)2 2α−β ⎥⎦⎢⎣ (α2 −βα+1)2 αβ−2 ⎥⎦. αd(α2 −1)2 α2 −1 d(α2 −1)2 α2 −1
In this case, d is a free nonzero parameter and β = β1 + β2. Since d, α ̸= 0, both factors are nonderogatory, completing the proof in this case.