Page 70 - Matrices theory and applications

3.4. The Spectrum and the Diagonal of Hermitian Matrices
from which one obtains y k−1 ≤ x k ≤ y k+1 . Therefore, there are two cases,
depending on the position of y x with respect to x k . For example, if y k ≤ x k ,
we compute
n
k

(x j − x k )+
|x j − x k | =
1
k+1
j
From the assumption, it follows that (x k − x j ). 53
n k

|x j − x k |≥ (y j − x k )+ (x k − y j )= |y j − x k |,
j k+1 1 j =k
which means that φ(x k ) ≥ 0, which contradicts the hypothesis. Hence, φ is
a nonnegative function.
Our first statement expresses an order between the diagonal and the
spectrum of a Hermitian matrix.
Theorem 3.4.1 (Schur) Let H be a Hermitian matrix with diagonal a
and spectrum λ.Then a λ.
Proof
Let n be the size of H. We argue by induction on n. We may assume that
a n is the largest component of a.Since s n (λ)=Tr A,one has s n (λ)= s n (a).
In particular, the theorem holds true for order 1. Let us assume that it holds
for order n − 1. Let A be the matrix obtained from H by deleting the nth
row and the nth column. Let µ =(µ 1 ,... ,µ n−1 ) be the spectrum of A.
Let us arrange λ and µ in increasing order. From Theorem 3.3.3, one has
λ 1 ≤ µ 1 ≤ λ 2 ≤ ··· ≤ µ n−1 ≤ λ n . It follows that s k (µ) ≥ s k (λ)for
every k< n. The induction hypothesis tells us that s k (µ) ≤ s k (a ), where

a =(a 1 ,... ,a n−1 ). Finally, we have s k (a )= s k (a), and s k (λ) ≤ s k (a)for

every k< n, which ends the induction.
.
Here is the converse.
Theorem 3.4.2 Let a and λ be two sequences of n real numbers such that
a λ. Then there exists a real symmetric matrix of size n × n whose
diagonal is a and spectrum is λ.
Proof
We proceed by induction on n. The statement is trivial if n =1. If n ≥ 2,
we use the following lemma, which will be proved afterwards.
Lemma 3.4.1 Let n ≥ 2 and α, β two nondecreasing sequences of n real
numbers, satisfying α ≺ β. Then there exists a sequence γ of n − 1 real
numbers such that

α 1 ≤ γ 1 ≤ α 2 ≤ ··· ≤ γ n−1 ≤ α n