Page 93 - Textos de Matemática Vol. 39

SPECTRALLY ARBITRARY FACTORIZATION: THE NONDEROGATORY CASE 83
are, somehow, the next closest thing to scalar matrices, as rank(A − λI) = 1. Fortunately, there is still a nonderogatory factorization.
It is informative to revisit the 2-by-2 case with the notion of exceptional matrices in mind. If A ∈ M2 is not scalar, it is an exceptional matrix when paired with β1,β2,γ1,γ2 (β1β2γ1γ2 = detA ̸= 0) if and only if β1γ1 ∈ σ(A) (equivalently β2γ2 ∈ σ(A)). In this event A is similar to
with t ̸= 0 and
β1γ1 t 0 β2γ2
β1 r γ1 s 0β2 0γ2
with β1s+γ2r = t provides a nonderogatory factorization in which both factors are upper triangular. If β1γ1 ̸∈ σ(A) and A is not scalar, then a similarity of A in which β1γ1 is the 1,1 entry (which always exists, see [2] or [1]) has both off-diagonal entries nonzero and, thus, has a special LU factorization. This completes an alternative proof of the 2-by-2 case and verifies
Lemma 1.3. If A ∈ M2 is nonsingular and nonscalar and β1, β2, γ1, γ2 ∈ C are such that β1β2γ1γ2 = det A, then A is similar to a matrix with special LU factorization for β1,β2,γ1,γ2 if and only if A, along with β1,β2,γ1,γ2 is not an exceptional pair.
We will need some machinery to demonstrate a converse to Lemma 1.2, including the case n = 3. We begin with a lengthy technical lemma that is more general than the case n = 3.
Lemma 1.4. Let A ∈ M3(C) be a nonsingular, nonscalar matrix and let α1, α2, ..., αk ∈ C be nonzero scalars such that
rank(A−αiI)>1, i=1,...,k.
Then, for any nonzero α ∈ C (that, in case A is not nonderogatory, is neither
among α1,α2,...,αk nor detA, in which λ is the repeated eigenvalue), there is λ2
(i) detB[1,2]=α;
(ii) rank(B[1,2]−αiI)>1,i=1,...,k;
(iii)(iiia) detB[1,2;1,3]̸=0 (iiib) detB[1,3;1,2]̸=0.
Proof. Our strategy is as follows. We write B as LU, in which lower (upper) triangular L (U ) are chosen so as to ensure that B has desired features (i), (ii) and (iii). Then, the remaining freedom in L and U is used to ensure that B lies in the similarity class of A. For this, we distinguish two cases: the one in which A is nonderogatory and the one in which A is not nonderogatory. In the latter
a matrix B, similar to A, such that