Page 92 - Textos de Matemática Vol. 39

82 CHARLES R. JOHNSON AND YULIN ZHANG
are all nonzero, k = 0, ..., n − 2, or equivalently by Sylvester’s identity,
k+2 i=1
i̸=k+1
k = 0, ..., n − 2. Of course, such conditions are generic, and it may happen that A (with the desired leading principal minors) fails them while a similarity of A (with the same leading principal minors) does not.
It is clear that if a similarity of A has a special LU factorization (for β1,...,βn, γ1,...,γn) and if any equal βi′s (resp. equal γj′s) are consecutively labelled, then A has a nonderogatory factorization for β1, ..., βn and γ1, ..., γn, as the L and U are nonderogatory (and even a bit more). Note that whether a similarity has a special LU factorization is a property both of the similarity class of A (i.e; its Jordan form) and also of the β1 , ..., βn , γ1 , ..., γn (including their ordering). We next exhibit a general constraint, on A when there is such a similarity.
Let λ ∈ σ(A) and suppose that A has the special LU factorization A = LU. Then, A−λI = LU −λI = L(U −λL−1). Since the subdiagonal of L is totally nonzero, the same is true of L−1. Since β1, ..., βn (γ1, ..., γn) are the diagonal entries of L (U ), the diagonal entries of U − λL−1 are
γ1− λ,γ2− λ,...,γn− λ. β1 β2 βn
det A[1, ..., k, k + 2] ̸=
βiγi,
As L−1 is also lower triangular, the sub-and super-diagonal of U − λL−1 are both totally nonzero. If one of the diagonal entries γi − λ of U − λL−1 were
βi
0, then rank(U −λL−1) would be at least 2. But, γi − λ = 0 if and only if
βi
λ = βiγi. Since rank(A−λI) =rank(U −λL−1), and rank(A−λI) is a similarity
invariant, we may conclude the following.
Lemma 1.2. Suppose that the nonsingular matrix A ∈ Mn and β1, ..., βn, γ1, ..., γn ∈ C are given so that β1 ···βnγ1 ···γn = detA. If there is an i such that βiγi =λ∈σ(A)andrank(A−λI)≤1,thennosimilarityofAhasaspecial LU factorization relative to β1, ..., βn, γ1, ..., γn.
Because of the lemma, we call a pair, consisting of a matrix A ∈ Mn and ordered numbers β1, ..., βn, γ1, ..., γn, such that rank(A − βiγiI) = 1 for some i, exceptional. For an exceptional pair, there is no similarity with a special LU factorization. However, we will show that in other relevant cases there will be a similarity with a special LU factorization, and it is relatively easy to nonderogatorially factor in another way in exceptional cases. Recall that a scalar matrix is one for which rank(A − λI) = 0 for some λ and that these are exceptional in the case of Sourour’s theorem. Our exceptional matrices