Some bounds for the spectral radius of the Hadamard product of matrices Tin-Yau Tam Mathematics & Statistics Auburn University Georgia State University, May 28, 05 in honor of Prof. Jean H. Bevis
Some bounds for the spectral radius of the Hadamard product of two nonnegative matrices are given. Some results involve M- matrices. A joint work with Guang-Hui Cheng, Xiao-Yu Cheng, Ting-Zhu Huang, University of Electronic Science and Technology of China
I. Introduction Given A, B C n n, the Hadamard product of A and B is A B = (a ij b ij ). Example 1. 1 2 A =, B = 3 4 2 1, A B = 1 0 2 2 3 0 If A, B 0, then ρ(a B) ρ(a)ρ(b), (1) where ρ(a) is the spectral radius of A. Proof: The Kronecker product A B 0 since A, B 0. ρ(a B) = ρ(a)ρ(b).
A B is a principal submatrix of A B. Apply monotonicity of the Perron root. By monotonicity of the Perron root of the B 0, max b ii ρ(b). (2) i=1,...,n The lower bound is attained when B 0 is diagonal. Is it possible to have a better bound like the following for ρ(a B), where A, B 0? ρ(a B) ρ(a) max b ii (3) i=1,...,n
Example 2. 0 1 A =, B = 1 0 Evidently 1 2, A B = 2 1 0 2. 2 0 and ρ(a) = 1, ρ(b) = 3, ρ(a B) = 2, So max b ii = 1. i=1,2 ρ(a B) = 2 1 = ρ(a) = max 1 i 2 b ii.
Goal: provide a necessary condition for (3) to be valid. It turns out the condition is satisfied by an important class of matrices called inverse M- matrices.
II. Some bounds and diagonal dominance A matrix A is said to be diagonally dominant of its row entries (respectively, of its columns entries) if a ii a ij (respectively a ii a ji ) for each i = 1,..., n and all j i. Example 3. 4 1 A = 3 2 is diagonally dominant of its column entries but not of its row entries. Similarly we define diagonally subdominant of its row entries (respectively, of its columns entries) by reversing the inequalities. Strict diagonal dominance is defined similarly.
Theorem 1. Let A 0, B 0 be n n nonnegative matrices. If there exists a positive diagonal D such that 1. DBD 1 is diagonally dominant of its column (or row) entries, then and ρ(b) tr B, (4) ρ(a B) ρ(a) max b ii, (5) i=1,...,n 2. DBD 1 is diagonally subdominant of its column (or row) entries, then and ρ(a) ρ(b) tr B min b ii ρ(a B). i=1,...,n
Proof: A (DBD 1 ) = D(A B)D 1 and hence ρ(a B) = ρ(a (DBD 1 )). diag B = diag DBD 1. So we may assume that B is diagonally dominant of its column (or row) entries. A B A diag (b 11,..., b nn ) A max i=1,...,n b ii. By the monotonicity of the Perron root ρ(a B) ρ(a diag (b 11,..., b nn )) ρ(a) max i=1,...,n b ii, (6) which yields (5) immediately. To obtain (4), set A = J n in the first inequality of (6). Then ρ(b) ρ(j n diag (b 11,..., b nn )) = tr B,
since rank (J n diag (b 11,..., b nn )) 1. Remark 1. 1. The upper bound tr B in (4) is attained by B = J n. 2. Though max i=1,...,n b ii ρ(b) is true for B 0, ρ(b) tr B in (4) may not hold if the assumption in the theorem is dropped, for example, the irreducible 0 1 B = 1 0 3. It is not true that if A 0 and B 0 are both diagonally dominant of its (column) row entries, then ρ(a B) For example, 2 1 A =, B = 1 1.5 max i=1,...,n a ii max i=1,...,n b ii. 2 1, A B = 1 2 4 1, 1 3
with ρ(a B) 4.6180 > 4 = max i=1,2 a ii max i=1,2 b ii. Corollary 1. Let B 0 be an n n nonnegative matrix. If there exists a positive diagonal D such that DBD 1 is diagonally dominant of its column (or row) entries, then max{ρ(a B) : A 0, ρ(a) = 1} = max i=1,...,n b ii. M-matrices: Z n := {A R n n : a ij 0, i j}. A matrix A Z n is called an M-matrix if there exists an P 0 and s > 0 such that A = si n P and s > ρ(p ),
M n = the set of all n n nonsingular M- matrices. The matrices in M 1 n := {A 1 : A M n } are called inverse M-matrices. Known: Given A Z n. A M 1 n A is nonsingular and A 0. Corollary 2. Let A 0, B 0 be n n nonnegative matrices. If B M 1 n, then and ρ(b) tr B, ρ(a B) ρ(a) Hence if B M 1 n, then max i=1,...,n b ii. max{ρ(a B) : A 0, ρ(a) = 1} = max i=1,...,n b ii.
Proof: Since B 1 M n, there exists a positive diagonal D such that DB 1 D 1 is strictly row diagonally dominant. The inverse DBD 1 of DB 1 D 1 is strictly diagonally dominant of its column entries. Then apply Theorem 1 (1).
III. A sharper upper bound when B 1 is an M-matrix The inequality in Corollary 2 has an resemblance of a known result which asserts that if A, B M n, then where τ(a B 1 ) τ(a) min β ii, (7) i=1,...,n τ(a) = min{re λ : λ σ(a)}, and σ(a) is the spectrum of A M n. Known: τ(a) = 1 ρ(a 1 ) and is a positive eigenvalue of A M n.
The number τ(a) is often called the minimum eigenvalue of the M-matrix A. Indeed τ(a) = s ρ(p ), if A = si n P where s > ρ(p ), P 0. So τ(a) is a measure of how close A M n to be singular. Known: A B 1 M n if A, B M n. Chen (2004) provided a sharper lower bound for τ(a B 1 ) which improves (7): τ(a B 1 ) τ(a)τ(b) min i=1,...,n τ(a) min i=1,...,n β ii [( aii τ(a) + Since a ii > τ(a) for all i = 1,..., n. b ii τ(b) 1 ) ] βii b ii. (8) (9)
Inequality (8) may be rewritten in the following form: ρ((a B 1 ) 1 ) ρ(a 1 )ρ(b 1 ) min i=1,...,n [(a ii ρ(a 1 ) + b ii ρ(b 1 ) 1) β ii b ]. ii where A 0 and B M 1 n. However Chen s result is not an upper bound for ρ(a B 1 ). In view of Corollary 2 and motivated by Chen s bound and its proof, we provide a sharper upper bound for ρ(a B), where A 0 and B M 1 n.
Theorem 2. Suppose A 0, B M 1 n. 1. If A is nilpotent, i.e., ρ(a) = 0, then ρ(a B) = 0. 2. If A is not nilpotent, then ρ(a B) ρ(a) ρ(b) max i=1,...,n [ ] aii bii ρ(a) + β iiρ(b) 1 β ii ρ(a) max i=1,...,n b ii.
Remark 2. The first inequality in Theorem 2 is no long true if we merely assume that B is nonsingular nonnegative. For example, if and A B = then but A = 0 1 0 0 0 1, B = 1 0 0 0 2 0 0 0 2, B 1 = 1 2 0 0 9 1 2 0 0 1 2 2 0 1, 1 2 4 4 1 2 2 4 1 ρ(a) = 1, ρ(b) = 3, ρ(a B) = 2, ρ(a) ρ(b) = 2, max i=1,...,n not even nonnegative., [ ] aii bii ρ(a) + β iiρ(b) 1 β ii