Page 98 - Textos de Matemática Vol. 39

88 CHARLES R. JOHNSON AND YULIN ZHANG
These is no loss of generality in illustrating how to do this in case n = 4 and ⎡⎢ a 0 0 0 ⎤⎥
A=⎢⎣0 a 0 0⎥⎦. 00b0
000b This matrix is clearly similar to one of the form
⎡⎢ a 0 0 0 ⎤⎥ ⎢⎣ 0 ∗ ∗ 0 ⎥⎦ ,
in which all the ∗’s are nonzero and none are equal to a or b. This matrix is in turn similar to one of the form⎡⎢ a 0 0 0 ⎤⎥
⎢⎣ 0 b 0 0 ⎥⎦ , 00∗∗
00∗∗
with the same restrictions. Now, the nonvanishing almost principal minor con- ditions are met and they are unaltered when Lemma 1.4 and Corollary 1.6 are applied to the upper left 3-by-3 block. In this way the induction step may be carried out in the diagonal cases, completing the proof.
We now turn to nonderogatory factorization in the exceptional cases. Here we take advantage of the fact that, although the β′s and γ′s are given, we may control their respective ordering (which we have not previously exploited except that equal values be consecutively labelled, which we may continue to assume ). It can happen that A, together with β1, ..., βn, γ1, ...γn is exceptional, but it is not if the β′s and γ′s are re-ordered. For example,
A=21 12
is exceptional with β1 = 1, β2 = 3 , γ1 = 1, γ2 = 2 (because the eigenvalues of 23
Aare1,and3)butnotwithβ1′ =2,β2′ =1,γ1=1,γ2=2.
Thus we first show how it can happen that A is exceptional, together with the β′s and γ′s, no matter how they are ordered. Since for n > 2, rank(A−λI)=1 for at most one value of λ, the answer is relatively simple. (For n = 2, we have nothing left to do because of Lemma 1.1.) Suppose that β1, ..., βn, γ1, ...γn and
λaregiven.Ifthereisaβi suchthatforsomej,γj = λ,callβi,γj aλ−pair. βi
By the multiplicity of βi (γj), we mean the number of distinct indices k such that βk = βi (γk = γj) and write m(βi) and m(γj). (Note that only the
0∗∗0 000b