Page 100 - Textos de Matemática Vol. 39

90 CHARLES R. JOHNSON AND YULIN ZHANG
Why must there be such a pair i, j? Suppose not. Then, of the 6 possible such pairings, each must have one of the triple sums greater than n. A careful inspection of the possibilities, in view of (∗) yields that if this occurs, then either: m1 + m2 + n1, m2 + m3 + n2 and m1 + m3 + n3 > n or n1 + n2 + m2, n2 + n3 + m3 and n1 + n3 + m1 > n. However, either leads to a contradiction, as, in either case, the sum of the 3 sums is identically 3n, so that not all 3 sums can exceed n, completing the proof.
We note that both parts, then, of being exceptional for all orderings of the β′s are rather rare, and increasingly so for larger n. We must first have a λ for which rank(A−λI) = 1 and we must have a pair for which TM > n; and the pair must be a λ-pair.
Now we suppose that A ∈ Mn, rank(A−λI) = 1, and there is a βi such that TM(βi, λ ) > n. In this (”unavoidably exceptional”) event, we wish to
βi
nonderogatorily factor A (giving the correct eigenvalues for the factors) without
the benefit of a special LU factorization.
Lemma 1.10. Suppose that A ∈ Mn, n > 1, is nonsingular and β1,...,βn, γ1,...,γn such that β1 ···βnγ1 ···γn = detA are given. If rank(A−aI) = 1 and for every permutation τ there is an i, 1 ≤ i ≤ n, such that βτ(i)γi = a, then there exist B ∈ Mn with eigenvalues β1,...,βn and C ∈ Mn with eigenvalues γ1, ..., γn such that A = BC, and B and C are nonderogatory.
Proof. Without loss of generality, we may suppose that a = 1 and that the 1−pair βi, γj whose total multiplicity exceeds n is βi = 1, γj = 1. The former is by multiplication of A by a scalar, as necessary, which does not change factorizability, and the latter, then, by passing a scalar factor between B and C, as necessary.
By similarity, we may then assume that A=⎡⎣I 100⎤⎦,
in which λ = detA, and that β1 ···βr,γ1 ···γs = 1, with r + s the total multiplicity of the 1−pair 1,1. Then any remaining β’s and γ’s are not 1, and we suppose that they are ordered so that any equal β (γ) values occur consecutively. Let m = min{r, s} ≥ 1, and we suppose without loss of generality that m = r. For convenience, we distinguish three possibilities for consideration: m < n − 1, m = n − 1 and m = n.
If m < n − 1, re-partition A as
A=Im 0. 0 A′
0tλ