Page 71 - Matrices theory and applications

3. Matrices with Real or Complex Entries
54
and γ ≺ β =(β 1 ,... ,β n−1 ).

We apply the lemma to the sequences α = λ, β = a.Since γ ≺ a ,the
induction hypothesis tells us that there exists a real symmetric matrix S
of size (n − 1) × (n − 1) with diagonal a and spectrum γ. From Corollary
n
3.3.2, there exist a vector y ∈ IR and b ∈ IR such that the matrix
T
S y
Σ=
y b

has spectrum λ.Since s n (a)= s n (λ) = Tr Σ = Tr S + b = s n−1 (a )+ b,we
have b = a n . Hence, a is the diagonal of Σ.
We prove now Lemma 3.4.1. Let ∆ be the set of sequences δ of n−1real
numbers satisfying
(3.4)
α 1 ≤ δ 1 ≤ α 2 ≤ ··· ≤ δ n−1 ≤ α n
together with
k k

δ j ≤ β j , ∀k ≤ n − 2. (3.5)
j=1 j=1
We must show that there exists δ ∈ ∆ such that s n−1 (δ)= s n−1 (β ). Since

n
∆ is convex and compact (it is closed and bounded in IR ), it is enough to
show that

inf s n−1 (δ) ≤ s n−1 (β ) ≤ sup s n−1 (δ). (3.6)
δ∈∆ δ∈∆

On the one hand, α =(α 1 ,... ,α n−1 ) belongs to ∆ and s n−1 (α ) ≤

s n−1 (β ) from the hypothesis, which proves the first inequality in (3.6).
Let us now choose a δ that achieves the supremum of s n−1 over ∆. Let r

be the largest index less than or equal to n−2 such that s r (δ)= s r (β ), with
r = 0 if all the inequalities are strict. From s j (δ) <s j (β )for r< j < n−1,

one has δ j = α j+1 , since otherwise, there would exist > 0such that
ˆ j ˆ
δ := δ + e belong to ∆, and one would have s n−1 (δ)= s n−1 (δ)+ ,
contrary to the maximality of δ. Now let us compute

s n−1 (δ) − s n−1 (β ) = s r (β) − s n−1 (β)+ α r+2 + ··· + α n
= s r (β) − s n−1 (β)+ s n (α) − s r+1 (α)
≥ s r (β) − s n−1 (β)+ s n (β) − s r+1 (β)
= β n − β r+1 ≥ 0.

This proves (3.6) and completes the proof of the lemma.