Bounds for the Perron Root of Irreducible Nonnegative Matrices

Some bounds for the Perron root ρ of positive matrices are proposed. We proved that


Introduction
Let A = (a i j ) be an irreducible nonnegative matrix of order n, the Perron root ρ of A is a positive real eigenvalue of A and any other eigenvalue λ is strictly smaller than ρ in absolute value(|λ| < ρ).It is also called the Perron-Frobenius eigenvalue,the greatest characteristic root of A or the greatest latent root of A .For the bounds of Perron root ,the following results are well known min i r i ≤ ρ ≤ max i r i .
(1) here r i is the i-th row sum.For nonnegative matrices with nonzero row sums r 1 , r 2 ,...,r n , the bounds in (1) were improved by Minc (Minc, 1988) as follows: This result was further generalized by Liu (Shulin Liu,1996).For positive matrices( a i j > 0, i, j = 1, 2, ..., n), formula (1) were also improved by A.Ostrowski (A. Ostrowski, 1952) and by A.Brauer (Brauer A., 1957).In the paper of Brauer A. (Brauer A., 1974), the following bounds were obtained. where ) , and P i = ∑ j i a i j .There are other bounds which are obtained by some special way (Linzhang Lu, 2002).
In this paper, some new bounds for the Perron root of positive matrices are proposed at first, and these bounds can also be used to estimate the Perron root of irreducible nonnegative matrix.

The Main Results
In this section, we will get some new bounds for positive matrices at first.In order to get our results, we propose some lemmas.
Lemma 1 (Shulin Liu,1996)Let p k > 0, q k > 0, k = 1, 2, ..., n, then Lemma 2 Let A = (a i j ) n×n be a positive matrix of order n and ρ its Perron root, Suppose x = (x 1 , x 2 , ..., x n ) is the positive eigenvector of ρ, x i and x j are two arbitrarily elements of x, then Proof.If x is the positive eigenvector of ρ, then consider the j-th and i-th elements of the equation Ax = ρx, one can get and by lemma 1 we know (6) holds and the proof is complete.Now we give some bounds for the Perron root of positive matrices.
Theorem 1 Let A = (a i j ) n×n be a positive matrix of order n and ρ its Perron root, then where Proof.Suppose x is the positive eigenvector of ρ, then consider an arbitrarily element of equation Ax = ρx, suppose the i-th element, we can get From the k-th and the i-th elements of the equation Ax = ρx, by lemma 2, we have Because the arbitrariness of i, so we get The proof is complete.
This upper and lower bounds sometimes are not good enough because of the rough estimate of x k x i .For example, consider the following matrix We just get 20 3 ≤ ρ ≤ 9 by (9), though this bound is better than (1), It is worse than many other bounds obtained by ( 2),(3) and so on, so we have to improve this result to get closer bounds.Note the fact that A and A m have the same eigen-space, so A m can also be used to estimate x k x i .The following lemma will show their relations.Lemma 3 Let A = (a i j ) n×n be a positive matrix of order n and ρ its Perron root, Suppose x = (x 1 , x 2 , ..., x n ) is the positive eigenvector of ρ, x i and x j are two arbitrarily elements of x, and Proof.If x is the positive eigenvector of ρ, then consider the j-th and i-th elements of the equation it a tk , so we have by lemma 1 we get The proof is complete.
Lemma 3 shows that use A m+1 to estimate x k x i will always work better than A m .If one wants to get a sharper bound, then we just need to choose a bigger m.We proposed the improved estimation as follows.
Theorem 2 Let A = (a i j ) n×n be a positive matrix of order n and ρ its Perron root, and where One step further, we can also have the following results.
Theorem 3 Let A = (a i j ) n×n be a positive matrix of order n and ρ its Perron root, and where m ′ ki , M ′ ki is the same as theorem 2. Proof.Just consider the i-th element of the equation A m x = ρ m x, together with lemma 3 we get (13).
Take m = 2, then we get the following corollary.
Corollary 1 Let A = (a i j ) n×n be a positive matrix of order n and ρ its Perron root, then where and α k denotes the k-th row of A, β t the t-th column of A, (α k , β t ) denotes the inner product of α k and β t .
Corollary 2 Let A = (a i j ) n×n be a positive matrix of order n and ρ its Perron root, then where ki and m ′ ki , M ′ ki is the same as theorem 2. Proof.By lemma 3, we know that m ′ ki , M ′ ki is both bounded and monotonic, So the limit exists, and by theorem 2 we get (15).
Corollary 3 Let A = (a i j ) n×n be a positive matrix of order n and ρ its Perron root, then where m ki1 , M ki1 is the same as corollary 2.
Now we focus on an irreducible nonnegative matrix A. If there are no zero entries in the matrix A, then it is a positive matrix, the bounds proposed above can be used directly.If there are zero entries in the matrix, the above results can not be used directly, one have to make a little change.Since for an irreducible nonnegative matrix A, let B = (A + I) n−1 , then B > 0. So one can use the above theorems to estimate the Perron root of matrix B, and then by the relation of matrix A and B's Perron root ρ(B) = (ρ(A) + 1) n−1 , one can get the bounds for the Perron root of matrix A.

Remark 1
The above methods can also be used on the transpose matrix of A.