Page 16 - Textos de Matemática Vol. 44

6
A. KOVACˇEC
ii.
A beautiful example of a pair of 12 × 12 diagonal matrices A, C is given, showing that the region in the de Oliveira Marcus conjecture (see Section 3 below) ∆C(A) is not simply connected; i.e. can have a hole.
This example must be reproduced here: indeed, take the diagonal 2 × 2 √
√
matrices A′ = diag(1, 1+ 3 ) and C′ = diag(1,−1). Simple direct com-
−1+ 3
putation shows that as U ranges over U(2), det(U′A′U′ + iC′) is the line
segment 2 √ [√3 + i, √3 − i]. Consider then the diagonal-block 12 × 12 −1+ 3
matrices A = ⊕6i=1A′ and C = ⊕6i=1C′. Then as V ranges over U(12) det(V AV ∗ + iC) must of course contain the sixths powers of points of the
t ∈ [−1, +1] which is a closed curve containing the origin. But since the squared module of det(V AV ∗ + iC) is det A2 · det(I + (A−1V ∗CV )2), it can never be zero.
√ √√
line segment, that is, it must contain the points z(t) = ( 2 3 +i 2 t)6, −1+3 −1+3
[B86b, nondifferentiability of boundary points of Wc(A)]: Given a compact connected subset ∆ of the complex plane, a point P in the boundary ∂∆ of ∆ is a corner of ∆ if the intersection of a small enough P-centered circular disk with ∆ is contained in an P-centered sector of angle less than π. Nat´alia shows that ‘to be at a corner’ of Wc(A) means that the matrix A is of a particular form in its unitary orbit {U∗AU : U ∈ U(n)}. We have the following result.
Theorem. Assume C = diag(γ11m1 , ..., γp1mp ) with γi, pairwise distinct. So n = m1 +···+mp. If z = tr(CA) is a corner of Wc(A), then A is a direct sum of square matrices, A = A1 ⊕···⊕Ap of sizes m1,m2,··· ,mp, respectively and z = pk=1 γk trAk.
Proof. We give a sketch of the proof since the perturbation argument used is found in many others of Nat´alia’s writings. Recall by a well known fact, that if S ∈ Mn(C) is hermitian, then the matrix eiεS is unitary for any real ε. By this and a development of the left hand side we find a formula
tr(CeiεSAe−iεS) = tr(CA)−iεtr(A[S,C])+O(ε2) ∈ Wc(A).
Since by hypothesis, z = tr(CA) is a corner, no geometrically smooth curve completely inside Wc(A) can be drawn through z; hence the first order term in the formula must vanish. In other words, tr(A[S,C]) = tr(S[C,A]) = 0. This being the case for all hermitian S, one finds [C, A] = 0. This commutation rela- tion implies (γj − γl)ajl = 0 for all j, l = 1, ..., n. From this now the direct sum property of A follows which in turn implies the formula for z. (The difference to the formula given in the original article is due to a different notation.)