Determinantal and Eigenvalue Inequalities for Matrices with Numerical Ranges in a Sector

Transcription

1 Determinantal and Eigenvalue Inequalities for Matrices with Numerical Ranges in a Sector Raymond Nung-Sing Sze The Hong Kong Polytechnic University July 29, 2014 Based on a joint work with Chi-Kwong Li 2014 Workshop on Numerical Ranges and Numerical Radii Tsinghua Sanya International Mathematics Forum (TSIMF) Sanya, China Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 1 / 25

2 Introduction - Fischer Inequality Fischer inequality A11 A 12 Suppose A = M A 21 A n, where A 11 M m, is positive semi-definite. 22 Then det(a) det(a 11) det(a 22). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 2 / 25

3 Introduction - Fischer Inequality Fischer inequality A11 A 12 Suppose A = M A 21 A n, where A 11 M m, is positive semi-definite. 22 Then det(a) det(a 11) det(a 22). Outline of Proof. Suppose A 22 is invertible. Then A11 A 12 Im A A = = 12A 1 22 A11 A 12A 1 22 A21 0. A 21 A 22 0 I n m A 21 A 22 Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 2 / 25

4 Introduction - Fischer Inequality Fischer inequality A11 A 12 Suppose A = M A 21 A n, where A 11 M m, is positive semi-definite. 22 Then det(a) det(a 11) det(a 22). Outline of Proof. Suppose A 22 is invertible. Then A11 A 12 Im A A = = 12A 1 22 A11 A 12A 1 22 A21 0. A 21 A 22 0 I n m A 21 A 22 Because A 11 A 12A 1 22 A21 is positive semi-definite, det(a) = det(a 11 A 12A 1 22 A 21) det(a 22) det(a 11) det(a 22). By continuity, the general result holds. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 2 / 25

5 Introduction - Fischer Inequality Fischer inequality A11 A 12 Suppose A = M A 21 A n, where A 11 M m, is positive semi-definite. 22 Then det(a) det(a 11) det(a 22). Outline of Proof. Suppose A 22 is invertible. Then A11 A 12 Im A A = = 12A 1 22 A11 A 12A 1 22 A21 0. A 21 A 22 0 I n m A 21 A 22 Because A 11 A 12A 1 22 A21 is positive semi-definite, det(a) = det(a 11 A 12A 1 22 A 21) det(a 22) det(a 11) det(a 22). By continuity, the general result holds. Remark: This proof reveals that estimating the determinant and eigenvalues of the Schur complement of A w.r.t. A 22, i.e., A 11 A 12A 1 22 A 21. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 2 / 25

6 3-7 June, ILAS Conference In 2013 ILAS Conference, Providence, RI, USA, 3-7 June, 2013, Minghua Lin of University of Victoria gave a talk with title Fischer type determinantal inequalities for accretive-dissipative matrices In his[ talk, he considered ] the following problem. Given an n n matrix A11 A 12 A = M A 21 A n, where A 11 M m with m n/2. Find optimal 22 (smallest) γ > 0 such that det(a) γ det(a 11) det(a 22), which is connected to the study of growth factor in Gaussian elimination. By Fischer inequality, γ = 1 if A is positive semi-definite. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 3 / 25

7 3-7 June, ILAS Conference Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 4 / 25

8 Some early results before 2013 A matrix A is called accretive-dissipative if A + A 0 and i(a A) 0. Ikramov showed that if A is accretive-dissipative, then det(a) γ det(a 11) det(a 22) with γ = 3 m. [JMS 121: (2004)] Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 5 / 25

9 Some early results before 2013 A matrix A is called accretive-dissipative if A + A 0 and i(a A) 0. Ikramov showed that if A is accretive-dissipative, then det(a) γ det(a 11) det(a 22) with γ = 3 m. Lin improved the bound to γ = { 2 3m/2 if m n/3, 2 n/2 if n/3 < m n/2. [JMS 121: (2004)] [LAA 438: (2013)] (Submitted 15 October 2012) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 5 / 25

10 Some early results before 2013 A matrix A is called accretive-dissipative if A + A 0 and i(a A) 0. Ikramov showed that if A is accretive-dissipative, then det(a) γ det(a 11) det(a 22) with γ = 3 m. Lin improved the bound to γ = { 2 3m/2 if m n/3, 2 n/2 if n/3 < m n/2. [JMS 121: (2004)] Lin further proposed the following. Conjecture 1 Suppose A is accretive-dissipative. Then [LAA 438: (2013)] (Submitted 15 October 2012) det(a) 2 m det(a 11) det(a 22). Im I m Note that the above equality holds if A = I I m I n 2m. m Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 5 / 25

11 Numerical range The numerical range of a matrix A M n is defined by W (A) = {x Ax : x C n, x x = 1}. W (A) is always nonempty and convex. W (A) = {λ} if and only if A = λi. W (A) [0, ) if and only if A 0. W (A) R if and only if A is Hermitian. W (A) = conv σ(a) if A is normal. W (A) = {µ C : t [0,2π) eit µ + e it µ λ 1(e it A + e it A )}. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 6 / 25

12 Numerical range in a Sector For any α [0, π/2), let S α = {z C : Im z (Re z) tan α}. A subset of C is a sector of half angle α if it is of the form {e iϕ z : z S α} for some ϕ [0, 2π). 2α For any matrix A, one can always find α and θ in [0, 2π) such that W (A) e iθ S α W (e iθ A) S α. W (A) (WLOG, we may assume θ = 0.) If A is positive semi-definite, then W (A) S α with α = 0. 2α If A is accretive-dissipative, then W (e iπ/4 A) S α with α = π 4. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 7 / 25

13 Results and conjectures of Drury Drury proved that if W (A) is a subset of a sector of half angle α such that 0 α < π/(2m), then det(a) sec 2 (mα) det(a 11) det(a 22). [LAA 439: (2013)] (Submitted 6 June 2013) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 8 / 25

14 Results and conjectures of Drury Drury proved that if W (A) is a subset of a sector of half angle α such that 0 α < π/(2m), then det(a) sec 2 (mα) det(a 11) det(a 22). Drury further proposed the following. Conjecture 2 [LAA 439: (2013)] (Submitted 6 June 2013) If W (A) is a subset of a sector of half angle α such that 0 α < π/2, then det(a) sec 2m (α) det(a 11) det(a 22). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 8 / 25

15 Results and conjectures of Drury Drury proved that if W (A) is a subset of a sector of half angle α such that 0 α < π/(2m), then det(a) sec 2 (mα) det(a 11) det(a 22). Drury further proposed the following. Conjecture 2 [LAA 439: (2013)] (Submitted 6 June 2013) If W (A) is a subset of a sector of half angle α such that 0 α < π/2, then det(a) sec 2m (α) det(a 11) det(a 22). Moreover, if A 11 is 1 1 nonzero and A 22 is (n 1) (n 1) invertible, then lies in the set { R = det(a) det(a 11) det(a = det(a) 22) A 11 det(a 22) re i2φ C : 0 r } 2(cos(2φ) cos(2α)), α φ α. sin 2 (2α) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 8 / 25

16 Results and conjectures of Drury y R with α = π 12 x Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 9 / 25

17 Results and conjectures of Drury y y R with α = π 12 R with α = π 6 x x Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 9 / 25

18 Results and conjectures of Drury y y R with α = π 12 R with α = π 6 x x y R with α = π 4 x Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 9 / 25

19 Results and conjectures of Drury y y R with α = π 12 R with α = π 6 x x y y R with α = π 4 R with α = π 3 x x Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 9 / 25

20 8-14 Maryland, USA After the ILAS meeting, I visited Professor Li at Maryland for a week (8-14 June) with Zejun Huang. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 10 / 25

21 8-14 Maryland, USA After the ILAS meeting, I visited Professor Li at Maryland for a week (8-14 June) with Zejun Huang. We discussed these two conjectures. After trying several different approaches, finally we were able to obtain some results. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 10 / 25

22 8-14 Maryland, USA After the ILAS meeting, I visited Professor Li at Maryland for a week (8-14 June) with Zejun Huang. We discussed these two conjectures. After trying several different approaches, finally we were able to obtain some results. At the same time, we were also exploring how to make Zongzi for celebrating Dragon Boat Festival (12th June). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 10 / 25

23 Our approach - A generalized eigenvalue problem Let A = A11 A 12 M A 21 A n, where A 11 M m with m n/2. 22 Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 11 / 25

24 Our approach - A generalized eigenvalue problem Let A = A11 A 12 M A 21 A n, where A 11 M m with m n/2. 22 We consider the following equality for determinant det(a) = det(a 22) det(a 11 A 12A 1 22 A 21) = det(a 22) det(a 11) det(i m A 1 11 A 12A 1 22 A 21). and affirm (and improve) the conjectures of Drury and Lin. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 11 / 25

25 Our approach - A generalized eigenvalue problem Let A = A11 A 12 M A 21 A n, where A 11 M m with m n/2. 22 We consider the following equality for determinant det(a) = det(a 22) det(a 11 A 12A 1 22 A 21) = det(a 22) det(a 11) det(i m A 1 11 A 12A 1 22 A 21). and affirm (and improve) the conjectures of Drury and Lin. Determine the optimal containment region of the generalized eigenvalue λ satisfying A A12 λ x = x for some nonzero x C n. 0 A 22 A 21 0 Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 11 / 25

26 Our approach - A generalized eigenvalue problem Let A = A11 A 12 M A 21 A n, where A 11 M m with m n/2. 22 We consider the following equality for determinant det(a) = det(a 22) det(a 11 A 12A 1 22 A 21) = det(a 22) det(a 11) det(i m A 1 11 A 12A 1 22 A 21). and affirm (and improve) the conjectures of Drury and Lin. Determine the optimal containment region of the generalized eigenvalue λ satisfying A A12 λ x = x for some nonzero x C n. 0 A 22 A 21 0 We can then obtain the optimal eigenvalue containment region of the matrix I m A 1 11 A 12A 1 22 A 21 in case A 11 and A 22 are invertible. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 11 / 25

27 Connection to the conjectures of Drury and Lin Let B 1 = invertible. A11 0 and B 0 A 2 = 22 0 A12 and assume that B A is Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 12 / 25

28 Connection to the conjectures of Drury and Lin A11 0 Let B 1 = and B 0 A 2 = 22 invertible. Then λ satisfies λb 1x = B 2x if and only if λx = B 1 1 B2x, 0 A12 and assume that B A is for some nonzero x C n Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 12 / 25

29 Connection to the conjectures of Drury and Lin A11 0 Let B 1 = and B 0 A 2 = 22 invertible. Then λ satisfies λb 1x = B 2x if and only if λx = B A12 and assume that B A is for some nonzero x C n B2x, i.e., λ is an eigenvalue of the matrix [ ] O A 1 11 A12, B = B 1 1 B 2 = A 1 22 A21 O Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 12 / 25

30 Connection to the conjectures of Drury and Lin A11 0 Let B 1 = and B 0 A 2 = 22 invertible. Then λ satisfies λb 1x = B 2x if and only if λx = B A12 and assume that B A is for some nonzero x C n B2x, i.e., λ is an eigenvalue of the matrix [ ] O A 1 11 A12, B = B 1 1 B 2 = A 1 22 A21 O whose eigenvalues have the form ±λ 1,..., ±λ m, 0,..., 0. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 12 / 25

31 Connection to the conjectures of Drury and Lin A11 0 Let B 1 = and B 0 A 2 = 22 invertible. Then λ satisfies λb 1x = B 2x if and only if λx = B A12 and assume that B A is for some nonzero x C n B2x, i.e., λ is an eigenvalue of the matrix [ ] O A 1 11 A12, B = B 1 1 B 2 = A 1 22 A21 O whose eigenvalues have the form ±λ 1,..., ±λ m, 0,..., 0. The matrix B 2 = [ A 1 11 A12A 1 O 22 A21 O A 1 22 A21A 1 11 A12 has eigenvalues: λ 2 1,..., λ 2 m, and λ 2 1,..., λ 2 m, 0,..., 0. ] Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 12 / 25

32 Connection to the conjectures of Drury and Lin A11 0 Let B 1 = and B 0 A 2 = 22 invertible. Then λ satisfies λb 1x = B 2x if and only if λx = B A12 and assume that B A is for some nonzero x C n B2x, i.e., λ is an eigenvalue of the matrix [ ] O A 1 11 A12, B = B 1 1 B 2 = A 1 22 A21 O whose eigenvalues have the form ±λ 1,..., ±λ m, 0,..., 0. The matrix B 2 = [ A 1 11 A12A 1 O 22 A21 O A 1 22 A21A 1 11 A12 has eigenvalues: λ 2 1,..., λ 2 m, and λ 2 1,..., λ 2 m, 0,..., 0. Thus, we need only estimate the eigenvalues of B to get information about those of B 2, and det(a) = det(a 11) det(a 22) det(i m A 1 11 A 12A 1 22 A 21). ] Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 12 / 25

33 Connection to the conjectures of Drury and Lin A11 0 Let B 1 = and B 0 A 2 = 22 invertible. Then λ satisfies λb 1x = B 2x if and only if λx = B A12 and assume that B A is for some nonzero x C n B2x, i.e., λ is an eigenvalue of the matrix [ ] O A 1 11 A12, B = B 1 1 B 2 = A 1 22 A21 O whose eigenvalues have the form ±λ 1,..., ±λ m, 0,..., 0. The matrix B 2 = [ A 1 11 A12A 1 O 22 A21 O A 1 22 A21A 1 11 A12 has eigenvalues: λ 2 1,..., λ 2 m, and λ 2 1,..., λ 2 m, 0,..., 0. Thus, we need only estimate the eigenvalues of B to get information about those of B 2, and det(a) = det(a 11) det(a 22) det(i m A 1 11 A 12A 1 22 A 21). The rest is hammering out the details. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 12 / 25 ]

34 Main result Consider the generalized eigenvalue problem A A12 λ x = x for some nonzero x C 0 A 22 A 21 0 n. (1) Li and Sze [JMAA 410: (2014)] (Submitted 5 July 2013) A11 A 12 Let A = M A 21 A n with A 11 M m be such that m n/2, and 22 W (A) be a subset of a sector of half angle α [0, π/2). (a) Suppose A 11 A 22 is singular, and x C n is a nonzero vector in the kernel. Then Ax = 0 and (1) holds for every λ C with this nonzero vector x. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 13 / 25

35 Main result Consider the generalized eigenvalue problem A A12 λ x = x for some nonzero x C 0 A 22 A 21 0 n. (1) Li and Sze [JMAA 410: (2014)] (Submitted 5 July 2013) A11 A 12 Let A = M A 21 A n with A 11 M m be such that m n/2, and 22 W (A) be a subset of a sector of half angle α [0, π/2). (a) Suppose A 11 A 22 is singular, and x C n is a nonzero vector in the kernel. Then Ax = 0 and (1) holds for every λ C with this nonzero vector x. (b) Suppose A 11 and A 22 are invertible. If λ C satisfies (1), then { } 1 λ 2 R = re i2φ 2(cos(2φ) cos(2α)) : 0 r, α φ α. sin 2 (2α) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 13 / 25

36 Optimality of the containment region Let λ C be such that λ 2 = 1 re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). y re i2φ x

37 Optimality of the containment region Let λ C be such that λ 2 = 1 re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). Suppose θ ( φ, α] satisfies r = 2(cos(2φ) cos(2θ))/ sin 2 (2θ). y re i2φ x

38 Optimality of the containment region Let λ C be such that λ 2 = 1 re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). Suppose θ ( φ, α] satisfies r = 2(cos(2φ) cos(2θ))/ sin 2 (2θ). Define a = cot θ sin φ and b = tan θ cos φ and y e iφ a + ib A = a + ib e iφ. re i2φ x

39 Optimality of the containment region Let λ C be such that λ 2 = 1 re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). Suppose θ ( φ, α] satisfies r = 2(cos(2φ) cos(2θ))/ sin 2 (2θ). Define a = cot θ sin φ and b = tan θ cos φ and y e iφ a + ib re i2φ A = a + ib e iφ. r 1e iθ Notice that A is normal and has eigenvalues e iφ + (a + ib) = r 1e iθ, r 1 0, e iφ (a + ib) = r 2e iθ, r 2 0. r 2e iθ x

40 Optimality of the containment region Let λ C be such that λ 2 = 1 re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). Suppose θ ( φ, α] satisfies r = 2(cos(2φ) cos(2θ))/ sin 2 (2θ). Define a = cot θ sin φ and b = tan θ cos φ and y e iφ a + ib re i2φ A = a + ib e iφ. Notice that A is normal and has eigenvalues Therefore, e iφ + (a + ib) = r 1e iθ, r 1 0, e iφ (a + ib) = r 2e iθ, r 2 0. r 2e iθ r 1e iθ W (A) x W (A) = {αr 1e iθ + (1 α)r 2e iθ : α [0, 1]} S θ. Finally, λ = ±(a + ib)e iθ satisfy e iφ 0 0 a + ib λ 0 e iφ x = x. a + ib 0

41 Optimality of the containment region Let λ C be such that λ 2 = 1 re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). Suppose θ ( φ, α] satisfies r = 2(cos(2φ) cos(2θ))/ sin 2 (2θ). Define a = cot θ sin φ and b = tan θ cos φ and y e iφ a + ib re i2φ A = a + ib e iφ. Notice that A is normal and has eigenvalues Therefore, e iφ + (a + ib) = r 1e iθ, r 1 0, e iφ (a + ib) = r 2e iθ, r θ r 2e iθ r 1e iθ W (A) x W (A) = {αr 1e iθ + (1 α)r 2e iθ : α [0, 1]} S θ. Finally, λ = ±(a + ib)e iθ satisfy e iφ 0 0 a + ib λ 0 e iφ x = x. a + ib 0 Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 14 / 25

42 Optimality of the containment region Example Let λ C be such that 1 λ 2 = re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 15 / 25

43 Optimality of the containment region Example Let λ C be such that 1 λ 2 = re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). Suppose θ ( φ, α] satisfies r = 2(cos(2φ) cos(2θ))/ sin 2 (2θ). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 15 / 25

44 Optimality of the containment region Example Let λ C be such that 1 λ 2 = re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). Suppose θ ( φ, α] satisfies r = 2(cos(2φ) cos(2θ))/ sin 2 (2θ). Let ( ) e iφ a + ib A = I m a + ib e iφ (e iφ I n 2m) with a = cot θ sin φ and b = tan θ cos φ. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 15 / 25

45 Optimality of the containment region Example Let λ C be such that 1 λ 2 = re i2φ with r 2(cos 2φ cos 2α)/ sin 2 (2α). Suppose θ ( φ, α] satisfies r = 2(cos(2φ) cos(2θ))/ sin 2 (2θ). Let ( ) e iφ a + ib A = I m a + ib e iφ (e iφ I n 2m) with a = cot θ sin φ and b = tan θ cos φ. Then W (A) lying in a sector of half angle θ α, and A has eigenvalues: e iφ + (a + ib) = r 1e iθ, r 1 0, with multiplicity m, e iφ (a + ib) = r 2e iθ, r 2 0, with multiplicity m, and e iφ with multiplicity n 2m, all in S α. Moreover, λ = ±(a + ib)e iφ satisfy the generalized eigenvalue problem (1), and 1 λ 2 = re 2iφ. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 15 / 25

46 Optimality of the containment region All eigenvalues of the matrix C = I m A 1 11 A12A 1 22 A21 lie in the set { } re i2φ 2(cos(2φ) cos(2α)) : 0 r, φ α. sin 2 (2α) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 16 / 25

47 Optimality of the containment region All eigenvalues of the matrix C = I m A 1 11 A12A 1 22 A21 lie in the set { } re i2φ 2(cos(2φ) cos(2α)) : 0 r, φ α. sin 2 (2α) Thus, the spectral radius of m m matrix C is bounded by 2(cos(2φ) cos(2α)) 2(1 cos(2α)) max = = sec 2 (α), φ α sin 2 (2α) sin 2 (2α) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 16 / 25

48 Optimality of the containment region All eigenvalues of the matrix C = I m A 1 11 A12A 1 22 A21 lie in the set { } re i2φ 2(cos(2φ) cos(2α)) : 0 r, φ α. sin 2 (2α) Thus, the spectral radius of m m matrix C is bounded by 2(cos(2φ) cos(2α)) 2(1 cos(2α)) max = = sec 2 (α), φ α sin 2 (2α) sin 2 (2α) and hence det(c) sec 2m (α). So det(a) = det(a 11) det(a 22)det(C) sec 2m (α) det(a 11) det(a 22). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 16 / 25

49 Optimality of the containment region All eigenvalues of the matrix C = I m A 1 11 A12A 1 22 A21 lie in the set { } re i2φ 2(cos(2φ) cos(2α)) : 0 r, φ α. sin 2 (2α) Thus, the spectral radius of m m matrix C is bounded by 2(cos(2φ) cos(2α)) 2(1 cos(2α)) max = = sec 2 (α), φ α sin 2 (2α) sin 2 (2α) and hence det(c) sec 2m (α). So det(a) = det(a 11) det(a 22)det(C) sec 2m (α) det(a 11) det(a 22). By continuity, one can remove the invertibility assumption on A 11 A 22. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 16 / 25

50 Consequences - confirmation of Conjectures 1 and 2 Corollary 1 [Conjecture 2] A11 A 12 Let A = M A 21 A n with A 11 M m such that m n/2. 22 Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 17 / 25

51 Consequences - confirmation of Conjectures 1 and 2 Corollary 1 [Conjecture 2] A11 A 12 Let A = M A 21 A n with A 11 M m such that m n/2. 22 Suppose W (A) is a subset of a sector of half angle α [0, π/2). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 17 / 25

52 Consequences - confirmation of Conjectures 1 and 2 Corollary 1 [Conjecture 2] A11 A 12 Let A = M A 21 A n with A 11 M m such that m n/2. 22 Suppose W (A) is a subset of a sector of half angle α [0, π/2). Then det(a) sec 2m (α) det(a 11) det(a 22). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 17 / 25

53 Consequences - confirmation of Conjectures 1 and 2 Corollary 1 [Conjecture 2] A11 A 12 Let A = M A 21 A n with A 11 M m such that m n/2. 22 Suppose W (A) is a subset of a sector of half angle α [0, π/2). Then det(a) sec 2m (α) det(a 11) det(a 22). If A 11 and A 22 are invertible, then the eigenvalues of the matrix lies in the region { R = C = I m A 1 11 A 12A 1 22 A 21 M m re i2φ : 0 r } 2(cos(2φ) cos(2α)), α φ α. sin 2 (2α) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 17 / 25

54 Consequences - confirmation of Conjectures 1 and 2 Corollary 1 [Conjecture 2] A11 A 12 Let A = M A 21 A n with A 11 M m such that m n/2. 22 Suppose W (A) is a subset of a sector of half angle α [0, π/2). Then det(a) sec 2m (α) det(a 11) det(a 22). If A 11 and A 22 are invertible, then the eigenvalues of the matrix lies in the region { R = C = I m A 1 11 A 12A 1 22 A 21 M m re i2φ : 0 r } 2(cos(2φ) cos(2α)), α φ α. sin 2 (2α) In particular, if A 11 M 1 and A 22 M n 1 are invertible, then det(a)/[det(a 11) det(a 22)] = det(c) R. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 17 / 25

55 Consequences - confirmation of Conjectures 1 and 2 Corollary 2 [Conjecture 1] Suppose A satisfies the hypothesis of Corollary 1 with α = π/4. Then det(a) 2 m det(a 11) det(a 22). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 18 / 25

56 Consequences - confirmation of Conjectures 1 and 2 Corollary 2 [Conjecture 1] Suppose A satisfies the hypothesis of Corollary 1 with α = π/4. Then det(a) 2 m det(a 11) det(a 22). If A 11 and A 22 are invertible, then the eigenvalues of the matrix C = I m A 1 11 A 12A 1 22 A 21 M m lies in the set {z C : z 1 1}. We thank Dr. Zejun Huang for some discussion. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 18 / 25

57 17-23 Tokyo, Japan When we were preparing the final version of our paper, we learned that Drury had independently affirmed Conjecture 2 for α [0, π/4]. [LAMA, accepted] (Submitted 18 June, 2013) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 19 / 25

58 17-23 Tokyo, Japan When we were preparing the final version of our paper, we learned that Drury had independently affirmed Conjecture 2 for α [0, π/4]. [LAMA, accepted] (Submitted 18 June, 2013) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 19 / 25

59 17-23 Tokyo, Japan When we were preparing the final version of our paper, we learned that Drury had independently affirmed Conjecture 2 for α [0, π/4]. [LAMA, accepted] (Submitted 18 June, 2013) Finally, we finished our manuscript and uploaded to ArXiv on 22th June, Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 19 / 25

60 17-23 Tokyo, Japan When we were preparing the final version of our paper, we learned that Drury had independently affirmed Conjecture 2 for α [0, π/4]. [LAMA, accepted] (Submitted 18 June, 2013) Finally, we finished our manuscript and uploaded to ArXiv on 22th June, Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 19 / 25

61 Singular value inequalities Let σ 1(X) σ n(x) 0 be the singular values of an n n matrix X. Drury and Lin [OAM, accepted] (Submitted 10 July 2013) A11 A 12 Let A = M A 21 A n with A 11 M m. Suppose W (A) is a subset of a 22 sector of half angle α [0, π/2). Then σ j ( A11 A 12A 1 22 A 21 ) sec 2 (α) σ j(a 11) j = 1,..., m. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 20 / 25

62 Singular value inequalities Let σ 1(X) σ n(x) 0 be the singular values of an n n matrix X. Drury and Lin [OAM, accepted] (Submitted 10 July 2013) A11 A 12 Let A = M A 21 A n with A 11 M m. Suppose W (A) is a subset of a 22 sector of half angle α [0, π/2). Then σ j ( A11 A 12A 1 22 A 21 ) sec 2 (α) σ j(a 11) j = 1,..., m. Notice that det ( A11 A 12A 1 22 A 21 ) = m j=1 σ j ( A11 A 12A 1 22 A 21 ) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 20 / 25

63 Singular value inequalities Let σ 1(X) σ n(x) 0 be the singular values of an n n matrix X. Drury and Lin [OAM, accepted] (Submitted 10 July 2013) A11 A 12 Let A = M A 21 A n with A 11 M m. Suppose W (A) is a subset of a 22 sector of half angle α [0, π/2). Then σ j ( A11 A 12A 1 22 A 21 ) sec 2 (α) σ j(a 11) j = 1,..., m. Notice that det ( A11 A 12A 1 22 A 21 ) = m j=1 σ j ( A11 A 12A 1 22 A 21 ) m sec 2 (α) σ j(a 11) = sec 2m (α) det(a 11). j=1 Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 20 / 25

64 Singular value inequalities Let σ 1(X) σ n(x) 0 be the singular values of an n n matrix X. Drury and Lin [OAM, accepted] (Submitted 10 July 2013) A11 A 12 Let A = M A 21 A n with A 11 M m. Suppose W (A) is a subset of a 22 sector of half angle α [0, π/2). Then σ j ( A11 A 12A 1 22 A 21 ) sec 2 (α) σ j(a 11) j = 1,..., m. Notice that det ( A11 A 12A 1 22 A 21 ) = Then m j=1 σ j ( A11 A 12A 1 22 A 21 ) m sec 2 (α) σ j(a 11) = sec 2m (α) det(a 11). j=1 det A = det ( A11 A 12A 1 22 A 21 ) det(a22) sec 2m (α) det(a 11) det(a 22), which also confirmed Conjecture 2. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 20 / 25

65 More inequalities A norm is unitarily invariant on M n if A = UAV for any A M n and any unitaries U and V M n. Corollary - Norm Inequality A11 A 12 Let A = M A 21 A n with A 11 M m. Suppose W (A) is a subset of a 22 sector of half angle α [0, π/2). Then A/A 11 sec 2 (α) A 22 for any unitarily invariant norm. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 21 / 25

66 More inequalities A norm is unitarily invariant on M n if A = UAV for any A M n and any unitaries U and V M n. Corollary - Norm Inequality A11 A 12 Let A = M A 21 A n with A 11 M m. Suppose W (A) is a subset of a 22 sector of half angle α [0, π/2). Then A/A 11 sec 2 (α) A 22 for any unitarily invariant norm. One of the major steps: If W (A) S α, then A = Xdiag ( e iθ 1,..., e iθn) X with θ j α, j = 1,..., n for some invertible X M n. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 21 / 25

67 More inequalities Fan and Hoffman Inequality For any A M n, λ j(re A) σ j(a) j = 1,..., n. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 22 / 25

68 More inequalities Fan and Hoffman Inequality For any A M n, λ j(re A) σ j(a) j = 1,..., n. Reversed Inequality Let A M n be such that W (A) S α. Then λ j(a) sec 2 (α)(re A) j = 1,..., n. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 22 / 25

69 More inequalities Fan and Hoffman Inequality For any A M n, λ j(re A) σ j(a) j = 1,..., n. Reversed Inequality Let A M n be such that W (A) S α. Then λ j(a) sec 2 (α)(re A) j = 1,..., n. F. Zhang [LAMA, accepted] (submitted 28 October 2013) Let A M n be such that W (A) S α. Then λ j(a) w sec(α)(re A). Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 22 / 25

70 More norm inequalities Lin and Zhou [JMAA 407: (2013)] (submitted 27 February 2013) Let A M n be accretive-dissipative. Then for any unitarily invariant norm, max{ A 12 2, A 21 2 } 4 A 11 A 22. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 23 / 25

71 More norm inequalities Lin and Zhou [JMAA 407: (2013)] (submitted 27 February 2013) Let A M n be accretive-dissipative. Then for any unitarily invariant norm, max{ A 12 2, A 21 2 } 4 A 11 A 22. Conjecture Let A M n be accretive-dissipative. Then for any unitarily invariant norm, max{ A 12 2, A 21 2 } 2 A 11 A 22. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 23 / 25

72 More norm inequalities Lin and Zhou [JMAA 407: (2013)] (submitted 27 February 2013) Let A M n be accretive-dissipative. Then for any unitarily invariant norm, max{ A 12 2, A 21 2 } 4 A 11 A 22. Conjecture Let A M n be accretive-dissipative. Then for any unitarily invariant norm, max{ A 12 2, A 21 2 } 2 A 11 A 22. The conjecture is confirmed by Y. Zhang. [JMAA: 412: (2014)] (Submitted 26 August 2013) Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 23 / 25

73 More norm inequalities Lin and Zhou [JMAA 407: (2013)] (submitted 27 February 2013) Let A M n be accretive-dissipative. Then for any unitarily invariant norm, max{ A 12 2, A 21 2 } 4 A 11 A 22. Conjecture Let A M n be accretive-dissipative. Then for any unitarily invariant norm, max{ A 12 2, A 21 2 } 2 A 11 A 22. The conjecture is confirmed by Y. Zhang. [JMAA: 412: (2014)] Let Then A = [ ] max{ A 12 2, A 21 2 } = 2 = 2 A 11 A 22. y (Submitted 26 August 2013) W (A) x Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 23 / 25

74 More norm inequalities F. Zhang [LAMA, accepted] (submitted 28 October 2013) Let A M n be such that W (A) S α. Then for any unitarily invariant norm, max{ A 12 2, A 21 2 } sec 2 (α) A 11 A 22. Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 24 / 25

75 Raymond Nung-Sing Sze WONRA2014, Sanya, China P. 25 / 25