Page 150 - Textos de Matemática Vol. 44

140 T. MORAN AND I. SPITKOVSKY
Since wT w is the (n, n)-entry of R2, from here and (6.3) we conclude:
n−1 n−1 β2λ2  n−1 β2 2 2 222 jj j
βj =c μ  On the other hand,
(μ−λj)2+μ+ μ−λj . j=1
 n−1 β2
(6.5) μ +
j=1
j=1

.
 n−1  n−1 β2  n−1 β2λ2  n−1 β2 2

(6.6) 1 = ⟨en,en⟩ = ⟨Ren,R−1en⟩ = ⟨w,y⟩ = c2μ2 1+ j
j=1
(μ − λj )2  Plugging in the value of c2μ2 from (6.6) into (6.5), we arrive at
 j  
μ2+ β21+ j = jj+μ+ j.
(μ − λj )2 
Somewhat lengthy but straightforward algebraic manipulations show that this
is equivalent to
j=1
j=1
j=1
n−1 n−1 β2  n−1 β2 2
 j  
(6.7) β2 j= j.
(μ − λj )2  μ − λj  j=1
j=1
j=1
(μ − λj )2   μ − λj  j=1
By the Cauchy-Schwarz inequality, (6.7) holds if and only if the vectors [β1 ...βn−1]and[β1/(μ−λ1)...βn−1/(μ−λn−1)]arecollinear.But,starting with n = 3, this is not the case — a contradiction.
With Lemma 6.1 at our disposal, the rest is easy.
Theorem 6.2. Let A be an invertible 3-by-3 pure almost normal matrix. Then ∆(A) is unitarily irreducible.
Indeed, ∆(A) has no normal eigenvalues according to Lemma 6.1. For n = 3, this is equivalent to being unitarily irreducible.
Theorem 6.3. Let A be an invertible almost normal matrix. Then ∆(A) is normal if and only if either A is normal itself or its PAN-rank equals 2 and the eigenvalues μ1 , μ2 of the block Aa of its representation (2.1) satisfy μ1 μ2 ≤ 0.
In particular, ∆(A) may be normal for 2-by-2 non-normal (and thus pure almost normal) matrices A. This shows that the restriction on n in Lemma 6.1 is essential.