Page 91 - Textos de Matemática Vol. 39

SPECTRALLY ARBITRARY FACTORIZATION: THE NONDEROGATORY CASE 81
Of course α = 0 cannot occur, and the case α = −1 is the same as α = 1, again via scalar multiplication. This leaves two possibilities for A:
A=1 0 and A=1 1. 01 01
The former is straightforwardly treated, as claimed, leaving the latter.
If
A=11 01
and B and C both have repeated eigenvalues we may complete the proof by taking a nonderogatory square root of A, A1/2 and passing a scalar between the two factors as necessary. Otherwise, without loss of generality, we may assume that only C has repeated eigenvalues. Then, B has distinct eigenvalues and the proof is simply completed by writing B = AC−1 and letting B play the role of A,AtheroleofBandC−1 theroleofCinthefirstpartofthisproof.
The general approach that we take is to find a similarity of A with nonzero leading principal minors: β1γ1, β1β2γ1γ2, ..., β1 · · · βnγ1 · · · γn such that its LU factorization with L lower, and U upper triangular with diagonal entries β1, ..., βn and γ1, ..., γn respectively have totally nonzero subdiagonal in L and superdiagonal in U. This cannot always be done. For example, if
A=α1 0 0 α2
and β1γ1 = α1, then any similarity of A whose 1,1 entry is α1 must be (lower or upper) triangular because, by the trace condition, the 2,2 entry must be α2(= β2γ2) and then, as the determinant is α1α2, the product of the two off-diagonal entries must be 0. Since the similarity of A is triangular, in no LU factorization can the 2,1 entry of L and the 1,2 entry of U both be nonzero. This problem might be solved by re-labelling either the β′s or the γ′s so that β1γ1 is no longer α1. However, this cannot be done if β1 = β2 and γ1 = γ2 (in which case α2 = α1), but then, as in the proof of Lemma 1.1, there is an easy alternative.
For A ∈ Mn with leading principal minors β1γ1, β1β2γ1γ2, ..., β1 ···βnγ1 ···γn ̸= 0, we say that A has a special LU factorization if the unique lower and upper triangular factors L, with diagonal entries β1,...,βn, and U, with diagonal entries γ1,...,γn, such that A = LU, also have totally nonzero subdiagonal and superdiagonal, respectively. (Of course, if some other LU fac- torization is chosen, with diagonal entries βi′ and γi′ , i = 1, ..., n, βi′ γi′ = βi γi , the factorization will be special as well.) It is easy to see, using e.g. Cauchy-Binet, that the LU factorization of A will be special if and only if the minors
det A[1, ..., k, k + 2; 1, ..., k + 1] det A[1, ..., k + 1; 1, ..., k, k + 2]