Page 131 - Textos de Matemática Vol. 39

PARTIAL SPECTRA OF HERMITIAN SUMS 121
Theorem 5.1. The complete restrictions in this case are those appearing in Horn’s list as explicit bounds on sums of entries in γ (both upper and lower bounds, the latter obtained using the trace condition), together with
nn (5.1) γh+1 + γk ≤ αi + βj ≤ γh−1 + γk
k̸=h i=1 j=1 k̸=h
where the first inequality is not present if h = n, and the second inequality is
not present if h = 1.
Proof. The necessity is obvious. For the sufficiency, we have to put
n n
γh = αi + βj − γk .
i=1 j=1 k̸=h
From (5.1) it follows that γh−1 ≥ γh ≥ γh+1. All ≤ inequalities in Horn’s list involving only the entries in γ (not involving γh) are present in the hypothesis. The conditions in Horn’s list involving γh are precisely those coming from the ≥ inequalities in Horn’s list involving only the entries in γ (together with the trace condition), and therefore also follow directly from the hypothesis.
The statements for n = 3 and s = 2 in the previous section are direct applications of this theorem.
6. The case s=n−2
The next natural case is s = n − 2. We end this article with just a few
remarks about it, starting with a special situation.
Theorem 6.1. Let n = 4, γ = (γ2,γ3), with γ2 ≥ γ3. There exist 4×4 Hermitian A and B, with spectra α and β respectively, such that γ2 and γ3 are the second and the third eigenvalues of A + B if and only if the entries in γ satisfy the inequalities in Horn’s list involving only γ2, γ3 and γ2 + γ3 (both upper and lower bounds, the latter obtained using the trace condition), together with
(6.1) 2γ2 +γ3 ≤ α1 +α2 +α3 +β1 +β2 +β3 ,
(6.2) γ2 +2γ3 ≥ α2 +α3 +α4 +β2 +β3 +β4 .
Proof. The necessity is obvious, with (6.1) coming from γ1+γ2+γ3 ≤α1+α2+α3+β1+β2+β3