Some inequalities on the spectral radius of nonnegative tensors

Lemma 2.1

[22] Let A ∈ T m , n be nonnegative. Then, there exists λ 0 ≥ 0 and x 0 ∈ ℝ + n such that A x 0 m − 1 = λ 0 x 0 [ m − 1 ] .

Lemma 2.2

[22] Let A ∈ T m , n be irreducible nonnegative. Then, there exists λ ₀ > 0 and x 0 ∈ ℝ + + n such that A x 0 m − 1 = λ 0 x 0 [ m − 1 ] . Moreover, if λ is an eigenvalue with a nonnegative eigenvector, then λ = λ ₀. If λ is an eigenvalue of A , then |λ| ≤ λ ₀.

Lemma 2.3

[23] Let A , ℬ ∈ T m , n be nonnegative. Then,

1. ρ ( A ) is an eigenvalue of A with a nonnegative eigenvector corresponding to it;

2. If λ is an eigenvalue of A with a positive eigenvector, then λ = ρ ( A ) ;

3. If A ≤ ℬ , then ρ ( A ) ≤ ρ ( ℬ ) ;

4. ρ ( A ) = max x ∈ ℝ + n \ { 0 } min x i ≠ 0 ( A x m − 1 ) i x i m − 1 ;

5. If x ∈ ℝ + + n , then ρ ( A ) ≤ max 1 ≤ i ≤ n ( A x m − 1 ) i x i m − 1 ;

6. Let A k = A + 1 k J , k = 1, 2,… Then, lim k → ∞ ρ ( A k ) = ρ ( A ) .

Lemma 3.1

Let A ∈ T m , n be nonnegative, and let { ℰ k } k = 1 ∞ be a set of mth order n-dimensional nonnegative tensors with lim k → ∞ ℰ k = O . If A has an eigenvalue with a positive eigenvector corresponding to it, then lim k → ∞ ρ ( A + ℰ k ) = ρ ( A ) .

Proof

By Lemma 2.3 (ii), there exists x = ( x 1 , x 2 , … , x n ) T ∈ ℝ + + n such that A x m − 1 = ρ ( A ) x [ m − 1 ] . Then, by Lemma 2.3 (iii) and (v),

Since lim k → ∞ ℰ k = O , lim k → ∞ max 1 ≤ i ≤ n ( ℰ k x m − 1 ) i x i m − 1 = 0 . Thus, lim k → ∞ ρ ( A + ℰ k ) = ρ ( A ) .□

Lemma 3.2

Let A ∈ T m , n be nonnegative, and let { ℰ k } k = 1 ∞ be a set of mth order n-dimensional nonnegative tensors such that ℰ k ≤ A for k = 1, 2,… with lim k → ∞ ℰ k = O . Then, lim k → ∞ ρ ( A − ℰ k ) = ρ ( A ) .

Proof

By Lemma 2.3 (i), there exists nonzero y = ( y 1 , y 2 , … , y n ) T ∈ ℝ + n such that A y m − 1 = ρ ( A ) y [ m − 1 ] . Then, by Lemma 2.3 (iii) and (iv),

Since lim k → ∞ ℰ k = O , lim k → ∞ max y i ≠ 0 ( ℰ k y m − 1 ) i y i m − 1 = 0 . Thus, lim k → ∞ ρ ( A − ℰ k ) = ρ ( A ) .□

Theorem 3.3

Let A ∈ T m , n be nonnegative, and suppose A has an eigenvalue with a positive eigenvector corresponding to it. Let { A k } k = 1 ∞ be a set of mth order n-dimensional nonnegative tensors with lim k → ∞ A k = A . Then, lim k → ∞ ρ ( A k ) = ρ ( A ) .

Proof

Let A − A k = ℰ k + − ℰ k − , k = 1, 2,…, where

Then, ℰ k + ≥ 0 ,   ℰ k − ≥ 0 ,   A − ℰ k + ≥ 0 ,   lim k → ∞ ℰ k + = 0 , lim k → ∞ ℰ k − = 0 . Since A k = A + ℰ k − − ℰ k + ≤ A + ℰ k − , by Lemma 2.3 (iii), ρ ( A k ) ≤ ρ ( A + ℰ k − ) . Then, by Lemma 3.1, lim k → ∞ ρ ( A k ) ≤ lim k → ∞ ρ ( A + ℰ k − ) = ρ ( A ) . Since A k ≥ A k − ℰ k − = A − ℰ k + , by Lemma 2.3 (iii), ρ ( A k ) ≥ ρ ( A − ℰ k + ) . Then, by Lemma 3.2, lim k → ∞ ρ ( A k ) ≥ lim k → ∞ ρ ( A − ℰ k + ) = ρ ( A ) . This completes the proof.□

Lemma 3.4

Let A ∈ T m , n be positive, and let x = ( x 1 , x 2 , … , x n ) T ∈ ℝ + + n , u = ( u 1 , u 2 , … , u n ) T ∈ ℝ + n , v = ( v 1 , v 2 , … , v n ) T ∈ ℝ + n . Then,

Proof

Since the function f(t) = log t is concave on (0, +∞), we have

Then,

which implies

This completes the proof.□

Lemma 3.5

Let A ∈ S m , n be nonnegative, and let x = ( x 1 , x 2 , … , x n ) T ∈ ℝ + n . Then, A x m ≤ ρ ( A ) ∑ i = 1 n x i m .

Proof

By Lemma 2.3 (vi), it suffices to consider the case when A is positive. First suppose x = ( x 1 , x 2 , … , x n ) T ∈ ℝ + + n . Note that Lemma 2.2 implies that ρ ( A ) is an eigenvalue of A with a positive eigenvector corresponding to it. Let u = ( u 1 , u 2 , … , u n ) T ∈ ℝ + + n be the eigenvector corresponding to ρ ( A ) , and let v = ( v 1 , v 2 , … , v n ) T ∈ ℝ + + n with v i = x i m u i m − 1 , i = 1, 2,…,n. Since A is symmetric,

Thus,

By Lemma 3.4, A x m ≤ v T ( A u m − 1 ) = v T ρ ( A ) u [ m − 1 ] = ρ ( A ) ∑ i = 1 n v i u i m − 1 = ρ ( A ) ∑ i = 1 n x i m .

Next suppose x = ( x 1 , x 2 , … , x n ) T ∈ ℝ + n . For k = 1, 2,…, let x ( k ) = x + 1 k e , where e is the vector with each component equal to 1. Then, x ( k ) ∈ ℝ + + n and lim k → ∞ x ( k ) = x . By what we have proved above, A ( x ( k ) ) m ≤ ρ ( A ) ∑ i = 1 n ( x i + 1 k ) m for k = 1, 2,…. The conclusion holds when k → ∞.□

Theorem 3.6

Let A , ℬ ∈ S m , n be nonnegative. Then, ρ ( A + ℬ ) ≤ ρ ( A ) + ρ ( ℬ ) .

Proof

By Lemma 2.3 (i), let x = ( x 1 , x 2 , … , x n ) ∈ ℝ + n be the eigenvector of A + ℬ corresponding to ρ ( A + ℬ ) with ∑ i = 1 n x i m = 1 . Then, by Lemma 3.5, ρ ( A + ℬ ) = ρ ( A + ℬ ) ∑ i = 1 n x i m = ∑ i = 1 n x i ( ( A + ℬ ) x m − 1 ) i = ( A + ℬ ) x m = A x m + ℬ x m ≤ ρ ( A ) ∑ i = 1 n x i m + ρ ( ℬ ) ∑ i = 1 n x i m = ρ ( A ) + ρ ( ℬ ) . □

Lemma 3.7

Let x = ( x 1 , x 2 , … , x n ) T ∈ ℝ + n with ∑ i = 1 n x i m = 1 . Then, | | x ⊗ m | | F ≤ n m 2 − 1 .

Proof

Since ∑ i = 1 n x i m = 1 with m ≥ 2 by the power mean inequality [31, p. 203], we have ( x 1 2 + x 2 2 + ⋯ + x n 2 n ) 1 2 ≤ ( x 1 m + x 2 m + ⋯ + x n m n ) 1 m = ( 1 n ) 1 m . Then, | | x ⊗ m | | F 2 = ∑ i 1 , i 2 , … , i m = 1 n x i 1 2 x i 2 2 ⋯ x i m 2 = ( x 1 2 + x 2 2 + ⋯ + x n 2 ) m ≤ n m − 2 . This completes the proof.□

Lemma 3.8

Let A ∈ T m , n be nonnegative. Then, ρ ( A ) ≤ n m 2 − 1 | | A | | F .

Proof

Let x = ( x 1 , x 2 , … , x n ) T ∈ ℝ + n be the eigenvector of A corresponding to ρ ( A ) with ∑ i = 1 n x i m = 1 . By Lemma 3.7 and the Cauchy–Schwarz inequality, ρ ( A ) = ρ ( A ) ∑ i = 1 n x i m = ∑ i = 1 n x i ( A x m − 1 ) i = ∑ i , i 2 , … , i m = 1 n a i i 2 ⋯ i m x i x i 2 ⋯ x i m = 〈 A , x ⊗ m 〉 ≤ | | A | | F | | x ⊗ m | | F ≤ n m 2 − 1 | | A | | F .□

Theorem 3.9

Let A , ℬ ∈ S m , n be nonnegative. Then, | ρ ( A ) − ρ ( ℬ ) | ≤ n m 2 − 1 | | A − ℬ | | F .

Proof

Since A − ℬ ≤ | A − ℬ | , A ≤ | A − ℬ | + ℬ . By Theorem 3.6, ρ ( A ) ≤ ρ ( | A − ℬ | + ℬ ) ≤ ρ ( | A − ℬ | ) + ρ ( ℬ ) . Thus, ρ ( A ) − ρ ( ℬ ) ≤ ρ ( | A − ℬ | ) . Similarly, ρ ( ℬ ) − ρ ( A ) ≤ ρ ( | A − ℬ | ) . Then, by Lemma 3.8, | ρ ( A ) − ρ ( ℬ ) | ≤ ρ ( | A − ℬ | ) ≤ n m 2 − 1 | | A − ℬ | | F .□

Theorem 4.1

Let A ∈ T m , n be nonnegative. Then, ( ρ ( A [ r ] ) ) 1 r ≤ ( ρ ( A [ s ] ) ) 1 s , with r ≥ s > 0.

Proof

By Lemma 2.3 (vi), we may suppose A is positive. Let A [ s ] = ℬ = ( b i 1 i 2 ⋯ i m ) > O , and let t = r s ≥ 1 . It suffices to prove that ρ ( ℬ [ t ] ) ≤ ( ρ ( ℬ ) ) t . By Lemma 2.2, there exists x = ( x 1 , x 2 , … , x n ) T ∈ ℝ + + n such that ℬ x m − 1 = ρ ( ℬ ) x [ m − 1 ] . Then, for i = 1, 2,…,n,

By Lemma 2.3 (v),

This completes the proof.□

Theorem 4.2

Let A 1 , A 2 , … , A k ∈ T m , n be nonnegative. Then, ρ ( A 1 [ α 1 ] ∘ A 2 [ α 2 ] ∘ ⋯ ∘ A k [ α k ] ) ≤ ( ρ ( A 1 ) ) α 1 ( ρ ( A 2 ) ) α 2 ⋯ ( ρ ( A k ) ) α k with α i >0 and ∑ i =1 k α i ≥ 1 .

Proof

By continuity of the spectral radius, we may suppose A 1 , A 2 , … , A k are all positive. Let α = ∑ i =1 k α i ≥ 1, and let A i = ( a i 1 i 2 ⋯ i m ( i ) ) , i = 1, 2,…,k. By Theorem 4.1,

By Lemma 2.2, for i = 1, 2,…,k, there exist x ( i ) = ( x 1 ( i ) , x 2 ( i ) , … , x n ( i ) ) T ∈ ℝ + + n such that A i ( x ( i ) ) m − 1 = ρ ( A i ) ( x ( i ) ) [ m − 1 ] . Let A = A 1 [ α 1 α ] ∘ A 2 [ α 2 α ] ∘ ⋯ ∘ A k [ α k α ] , and let x = ( x ( 1 ) ) [ α 1 α ] ∘ ( x ( 2 ) ) [ α 2 α ] ∘ ⋯ ∘ ( x ( k ) ) [ α k α ] = ( x 1 , x 2 , … , x n ) T ∈ ℝ + + n . For i = 1, 2,…,n,