Page 135 - Textos de Matemática Vol. 44

LARGEST EIGENVALUE 125
where 2 ≤ k ≤ n and P is a permutation matrix. Then λ = (k − 1)m. Proof. Apply Lemma 2.1 to A/m.
3. Remarks
Remark 3.1. Let 0 < c < m. Because the positive entries of A/c are greater
than 1, we can apply (2.1) to A/c. We obtain
λ ≤ 1(−c + c2 + 4∥A∥2) =: 1f(c).
Since
22
′c f(c)=−1+ c2+4∥A∥2 <0,
we have f(c) > f(m), and (2.7) does not improve.
Remark 3.2. The bound (2.7) improves the well-known bound
λ ≤ ∥A∥
strictly. To show it, apply (2.4) (clearly λ > 0 there) or note that
∥A∥2 − 1(−m + m2 + 4∥A∥2)2 = 1(m m2 + 4∥A∥2 − m2) > 0. 42
Remark 3.3. We can drop the assumption concerning zero diagonal, but then we need one more estimate and lose precision. So let A now be nonnegative and symmetric. Let M be its largest diagonal entry, and let A0 denote the matrix obtained from A by replacing all its diagonal entries with zero. Assuming A0 ̸= O, let m be its smallest positive entry. We have
(3.1) λ = λ(A) ≤ λ(MI + A0) = M + λ(A0)
≤ M + u(A0) = M + 1(−m + m2 + 4∥A0∥2) =: u′.
2
For the first inequality, see, e.g., [4, Corollary 8.1.19].
Remark 3.4. Yuan [9, Theorem 1] improved (1.1), assuming additionally that A is irreducible (actually it is enough that A has no zero rows), to
√
λ≤ 2e−n+1
and gave equality conditions. It seems that his method cannot be directly modified to work if A is not a (0, 1)-matrix.