Page 134 - Textos de Matemática Vol. 44

124 J. K. MERIKOSKI AND A. KOVACˇEC
dependent to give equality. Equivalently, the first row vector of B and the vec- tor y(1) = (0 1(k−1))T /√k must be linearly dependent. Hence b12 = · · · = b1k. Similar reasoning applies to other rows of B. Therefore all off-diagonal entries of B are equal and in fact, due to the above paragraph, equal to one.
We have now shown that a necessary condition for equality in (2.1) is that A is of the form (1.2). To show sufficiency, note first that since
∥A∥2 = ∥Jk∥2 = k2 − k,
the left-hand side of (2.1) is k − 1. Because λ(Jk) is between the largest and smallest row sums of Jk (e.g., [4, Theorem 8.1.22]) and these sums are k − 1, also the right-hand side
(2.6) λ = λ(Jk) = k − 1.
We present also another proof of (2.6), because it applies a determinan- tal equation which is interesting in itself. Given real numbers a,x,t1,...,tk, let H = H(a,x,t1,...,tk) be the k × k matrix with diagonal (t1,...,tk), all upper diagonal entries x and lower diagonal entries a. If x ̸= a, then [6, Exer- cise XXII.3]
detH= xφ(a)−aφ(x), x−a
where
Denoting by Ik the k × k identity matrix, we therefore have
φ(s)=(t1 −s)···(tk −s).
det(tIk −Jk)= lim detH(−1,x,t,...,t)= lim xφ(−1)+φ(x)
x→−1 x→−1 x+1
x(t+1)k +(t−x)k
= lim = (t+1)k −k(t+1)k−1
x→−1 x+1
= (t + 1)k−1(t + 1 − k).
Hence the distinct eigenvalues of Jk are −1 and k − 1, and so (2.6) follows. The proof of the equality condition and of the whole lemma is now complete.
Theorem 2.1. Let m be the smallest positive entry of A. Then (2.7) λ≤ 1(−m+ m2 +4∥A∥2)=:u.
2
Equality holds if and only if
T mJk O A=P O OP,