Minimum Angle of Rotation Required for Jordan Product to be Indefinite

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1 International Journal of Algebra, Vol. 6, 2012, no. 1, Minimum Angle of Rotation Required for Jordan Product to be Indefinite Neeraj Bajracharya Department of Mathematics and Computer Science Southern Arkansas University Magnolia, AR 71754, USA Abstract Suppose A and B be positive definite symmetric (PDS n n matrices. Their Jordan product {A, B} := AB + BA is also symmetric but not necessarily positive definite. In our recent paper Level curves of Angle Function of a Positive Definite Symmetric Matrix, we proposed following problem for commuting PDS matrices A and B: Compute Φ(A, B := inf{φ S : S SO n and {A, B S } is indefinite}, where {A, B S } := ASBS 1 +SBS 1 A, is the Jordan product of A and SBS, 1 and φ S denotes the maximum angle by which S rotates any unit vector. We studied level curves of the angle function of a 3 3 PDS matrix A and used it to give a geometric proof for the partial solution to the problem when PDS matrices A and B have eigenvalues in an increasing order. In this paper we will provide a more elaborate proof for the same and extend the result to any pair of commuting 3 3 PDS matrices A and B. Mathematics Subject Classification: 15A18, 15A99 Keywords: Jordan product, Jordan algebra, Orthogonal matrix, Angle of an operator, Positive Definite Symmetric Matrix, Level curves 1 Introduction Given a real n n matrix A, write φ A for the maximum angle by which A rotates any unit vector. Suppose A and B be commuting positive definite symmetric (PDS n n matrices. Their Jordan product {A, B} := AB + BA is also symmetric but not necessarily positive definite. Taussky-Todd first posed the question on eigenvalues of the Jordan product of two hermitian matrices as a research problem in 1960 [6]. As an answer to it, Strang derived the bounds on

2 42 N. Bajracharya the extremal eigenvalues [1]. His results depend solely on the largest and the smallest eigenvalues of A and B. Likewise, Nicholson gave a sufficient condition for the Jordan product to be positive definite [3]. In [2], Conley, Pucci and Serrin presented necessary and sufficient conditions for the Jordan product of PDS matrices to be a positive definite and used their results to prove the validity of the strong maximum principle. It is known that if φ A +φ B π/2 for PDS matrices A and B, then the Jordan product may not be positive definite. In fact, it is possible to find special orthogonal matrix S SO n such that {A, B S } is indefinite, where B S = SBS 1. Let Φ(A, B denote the infimum of the set {φ S : S SO n and {A, B S } is indefinite}. Recently, we studied level curves of the angle function of a 3 3 PDS matrix to compute Φ(A, B, when A and B are commuting PDS matrices [7]. Our method of finding Φ(A, B was based on a geometrical argument, and was conclusive only for the case when A and B have eigenvalues in an increasing order. In this paper, we propose a method of computing Φ(A, B for any commuting 3 3 PDS matrices A and B. Through out this paper, suppose A and B are two 3 3 diagonal PDS matrices such that the Jordan product {A, B S } is indefinite for some S SO 3. Without loss of generality, let A and B be the matrices ( B B B 3 ( A A 2 0 and 0 0 A 3 respectively, where A i,b i > 0 for all i. Since {A, B S } is indefinite, there exist unit vectors a and b such that Sb = a and (Aa, SBb π/2. Define α := (a, Aa and β := (b, Bb. Since (Aa, SBb α + β, we must have α + β π/2. Let L(A, α ={a S 2 : (a, Aa =α} and L(B,β={b S 2 : (b, Bb =β} denote the level curves of the angle function of A and B for the angles α and β respectively. The proof of the Main Theorem stated below provides a basis for the computation of Φ(A, B. Main Theorem The minimum of φ S over all (a, b with α + β π/2 occurs a Aa for some (a, b L(A, α L(B,β with α + β = π/2 if and only if = a Aa b Bb. b Bb In sections 2 and 3 we first state our results from [7] for the cases when α + β = π/2 and α + β>π/2respectively and establish some preliminary results to pave our way to prove the main theorem. The problem of computing Φ(A, B, roughly speaking, is maximization of Tr(S formula from [7] with constraints α<φ A,β < φ B, and (a, b L(A, α L(B,β such that Sb = a and SBb Aa = 0. Since a and b depends on α and β, respectively, we consider the problem as a maximization problem primarily over α and β. Ifα and a are fixed, then the problem can be thought of as one of finding vector SBb for some appropriate β such that Sb = a and SBb Aa = 0. Although identification of vector SBb is not crucial in our proof of the main theorem, it does help in

3 Minimum angle required for indefiniteness 43 computing φ S in certain case. In this section, we first give a result pertaining to it. Let GC(x, y denote the great circle passing through the points x, y S 2. Suppose {e 1,e 2,e 3 } be the standard basis for R 3. From our study of level curves, we know that for all α φ A there exists at least a pair (e i,e j such that L(A, α GC(e i,e j. Thus without loss of generality fix a 0 =(a 1,a 2,a 3 L(A, α GC(e i,e j for some i, j. Lemma 1.1 If α + β>π/2, then for all b L(B,β there exists exactly two unit vectors C β := SBb such that Sb = a Bb 0 and (Aa 0, C β = π. 2 Proof Suppose there exists a unit vector c =(c 1,c 2,c 3 such that (Aa 0,c= π/2 and (a 0,c=β. Then Aa 0 c = 0 and a 0 c = cos β. Thus A i a i c i +A j a j c j = 0 and a i c i + a j c j = cos β. Solve these equations to get c i = A j cos β a i (A j A i and c j = A i cos β a j (A j A i. Since c is a unit vector we have c 2 k = a2 i a2 j (A j A i 2 cos 2 β(a 2 j a2 j + A2 i a2 i a 2 i a2 j (A. j A i 2 Note that α, β < π such that α + β π, we have cos β sin α, with 2 2 equality holding when α + β = π. Thus if α + β = π/2, the numerator of 2 the right side of the above equation is zero, and c is unique. However, if α + β>π/2, then the numerator of the above equation is positive, and a 2 i a2 j (A j A i 2 cos β(a 2 j a2 j + A2 i a2 i c k = ±. a i a j (A j A i Hence, corresponding to the positive and negative sign of c k we obtain two distinct unit vectors. Note that the vector C β discussed above was obtained for a fixed a 0, and thus it depends on β but not on b. In general, we are interested in the vector SBb for arbitrary a and b. For the rest of the paper, fix arbitrary a, b S 2 and Bb denote vector SBb by C. Define Bb α := (a, Aa, β := (b, Bb. 2 Case α + β = π 2 Suppose α + β = π/2. Thus C (a Aa = 0. In three dimension, the trace of S SO 3 is given by Tr(S = cos φ S, where φ S is the maximum angle by which S rotates any non-zero vector. Thus we seek to compute Φ(A, B by computing max{tr(s :{A, B S } is indefinite}. Following proposition was proved in [7]:

4 44 N. Bajracharya Proposition 2.1 There exists a unique matrix S SO 3 such that Sb = a and (Aa, SBb =α + β. Its trace is given by Tr(S = a Bb b Bb + b Aa Aa Bb +(b Bb (a Aa. ((b Bb Proposition 2.2 Unit vectors (a, b L(A, α L(B,β maximizes Tr(S in Proposition 2.1 if and only if it satisfies a Aa a Aa = b Bb b Bb. Proof Suppose unit vectors a, b maximizes Tr(S. If a = b, since (Aa, SBb = α + β,s must be the identity matrix and a Aa = b Bb. Suppose a b. Consider the orthonormal bases {v 1,v 2,v 3 } and {w 1,w 2,w 3 } of R 3 defined by v 1 = b, v 3 = b Bb b Bb, v 2 = v 3 v 1, w 1 = a, w 3 = a Aa a Aa, w 2 = w 3 w 1. Since S must satisfy Sb = a, Sv 1 = w 1. Tr(S is maximum if S is a rotation about b Bb. ThusSv 3 = v 3. However, SBb Span{w 1,w 2 }, and hence Sv 2 Span{w 1,w 2 }. Since v 2 v 1, Sv 2 Sv 1. Thus Sv 2 = ±w 2. By definition of the orthonormal basis {w 1,w 2,w 3 },Aalies on the first quadrant of the (w 1,w 2 plane. Since (Aa, SBb = π/2, SBb must lie on the fourth quadrant of the (w 1,w 2 plane, so Sv 2 = w 2 and Sv 3 = w 3. Hence, w 3 = v 3. For the other implication, first note that the orthogonal matrix S in general can be factored as a product of two orthogonal matrices: S 1 whose axis of rotation is along a b, and S 2 whose axis of rotation is along a such that S = S 2 S 1. From Lemma 5.7 [7] we know cos 2 ( φ S 2 = cos2 ( φ S 1 2 cos2 ( φ S 2. (1 2 Thus minimum of φ S occurs only when one of φ S1 or φ S2 is zero. In other words, Tr(S is maximum if S is either S 1 or S 2. Since a Aa = b Bb we must have S = S 1 and thus Tr(S is maximum. Note that for (a, b in Proposition 2.2 we have (b Bb (a Aa (a Aa(b Bb we have following: = 1. Hence, Theorem 2.3 The maximum of Tr(S occurs at some (a, b L(A, α a Aa L(B,β if and only if = b Bb and it is given by max Tr(S =1+ a Bb b Bb + b Aa (Aa Bb ((b Bb.

5 Minimum angle required for indefiniteness 45 3 Case α + β> π 2 Suppose α + β> π Aa. Then C (a Aa 0. The trio of unit vectors a, and 2 Aa C forms a spherical triangle on the unit sphere, whose sides along GC(a, Aa, Aa GC(a, C and GC( Aa, C are respectively α, β and π. Let θ denotes the interior Aa 2 angle between GC(a, Aa and GC(a, C. Clearly, θ ( π,π. For all values of Aa 2 α and β, θ is given by the following formula involving fundamental relations between the parts of a spherical triangle [5]: Lemma 3.1 cos θ = cot α cot β. Proof (a Aa (a C ((a C cos θ = = = cot α cot β. a Aa a C Aa sin α sin β Proposition 3.2 There exist exactly two matrices S SO 3 such that Sb = a and (Aa, SBb = π. Their traces are give by 2 Tr(S± = {( a Bb a b + cos 2 θ b Bb + b Aa Aa Bb +(b Bb (a Aa a b ((b Bb ± tan θ ( a Proof Proved in [7]. Aa (b Bb b Bb ( b Bb b Bb (a Aa }. Since both α and β are acute angles, tan θ is positive. Thus Tr(S Tr(S +. We want to maximize Tr(S + over all θ and (a, b L(A, α L(B,β. Denote Tr(S + bytr(s. As was pointed out earlier, we can factor S as a product of two orthogonal matrices: S 1 whose axis of rotation is along a b, and S 2 whose axis of rotation is along a such that S = S 2 S 1. And, the minimum of φ S occurs only when one of φ S1 or φ S2 is zero. Thus maximum of Tr(S occurs only if S is either S 1 or S 2. Thus we have following proposition: Proposition 3.3 The maximum of Tr(S in Proposition 3.2 is { (a Aa (b Bb 1 2 cos 2 Aa (b Bb } θ tan θ, ((b Bb ((b Bb if S is a rotation about a or { (a b sin 2 θ + cos 2 θ 1+ a Bb b Bb + b Aa Aa Bb ((b Bb Bb (a Aa Aa (b Bb b (a Aa } + tan θ, ((b Bb if S is a rotation about a b.

6 46 N. Bajracharya Proof For the maximum of trace of S to occur, it must either be a rotation about a or a rotation about a b. If it is a rotation about a then vectors a and b must coincide. To exclude the trivial case, suppose Aa Bb 0. Then, the formula in Proposition 3.2 reduces to 1 2 cos 2 θ { (a Aa (b Bb (a Aa(b Bb tan θ (a Aa(b Bb} Aa (b Bb. On the other hand, if S is a rotation about a b, then a (b Bb =0, (b Bb (a Aa = 1 and the formula for Tr(S (a Aa(b Bb reduces to the second formula given above. Proof of Main Theorem: Since α + β π/2, θ (π/2,π]. When θ = π, α + β = π/2 and the second formula in Proposition 3.3 reduces to the formula given in Theorem 2.3, whose maximum occurs for vectors (a, b L(A, α L(B,β if and only if a Aa = b Bb. Similarly, when α + β = π/2 and a = b, vectors a, b, Aa, Bb all lie on the same great circle. Thus we Aa Bb have Aa (b Bb = 0 and the first formula in Proposition 3.3 reduces to 1 2 (a Aa (b Bb. Furthermore, we have either a Aa = b Bb or a Aa = (a Aa(b Bb a Aa b Bb. If a Aa = b Bb, then Tr(S = 1and φ a Aa b Bb S = π. If = b Bb, then Tr(S = 3 and φ S =0. 4 Some examples Example 1: Suppose A = ( and B = ( Since φ A =48.59 and φ B =53.13,φ A + φ B > π. Hence, it is possible to find S SO 2 3 such that {A, B S } is indefinite. A short numerical calculation gives Φ(A, B = If we chose α = φ A and β = π/2 α, then for this value of α and β we obtain vectors a =(0.9354, 0, and b =(0.9922, 0, satisfying Theorem 2.3. However, φ S =27.88 Φ(A, B. From Theorem 2.3, it is clear that for trace of S to be the maximum, we must have the vectors a and (Aa Bb (a Aa(b Bb b as close as possible so that is as small as possible. For above α (Aa Bb and β the ratio is not the smallest, and as a result φ (a Aa(b Bb S is not the smallest even though a and b satisfy the theorem. We obtainφ(a, B for α = φ A φ B +π/2 =42.73,β = φ B φ A +π/2 =47.27 and a =(0.9815, 0, ,b = 2 2 (0.9857, 0, If we choose any values of α and β except and respectively, φ S > Φ(A, B. ( ( Example 2: Suppose A = 020 and B = Note that for α = φ A φ B +π/2 = 42.73,β = φ B φ A +π/2 = 47.27, there does not exist 2 2 b L(B, β GC(a, Aa. Thus Theorem 2.3 does not apply with these values of α and β. However, if β < φ ( =45.58, we can find vectors a 0 9 and b satisfying the theorem. Taking α =(φ A φ ( π/2/2 =46.50, 0 9

7 Minimum angle required for indefiniteness 47 β =(φ ( φa +π/2/2 =43.50 obtain vectors a =(0.9686, 0, ,b= 0 9 (0.9636, 0, With these vectors, we have Φ(A, B =29.9. Example 3: Suppose A = and B =. Here φ A =48.59,φ B = ( ( In this case, b / L(B, β GC(a, Aa for all α and β. Take α = 46.50,β =43.50 as before and fix a =(0.9684, 0, L(A, α. Here, S can be factored as a product of S 1 whose axis of rotation is along a b, and S 2 whose axis of rotation is along a such that S = S 2 S 1. We want to find b L(B,β such that φ S is minimum. When b =(0, , ,φ S1 = (a, b =93.81 and φ S2 = (S 1 Bb,C β = Thus using (1, we obtain φ S = Next, take b v L(B,β, the vertex of the parabola of the level curve nearest to a along GC(a, b v. For detail on such a vector we refer to [7]. For b v, we have φ S1 =80.22 and φ S2 =45.08, and φ S = References [1] G Strang, Eigenvalues of Jordan Products, The American Mathematical Monthly, 69, Jan 1962, no. 1. [2] C. H. Conley, P. Pucci, and J. Serrin, Elliptic equations and products of positive definite matrices, Mathematische Nachrichten, 278, 2005, no. 21. [3] D. W. Nicholson, Eigenvalue bounds for AB + BA, with A, B positive definite matrices, Linear Algebra and its Application, 24, 1979, [4] R. A. Horn and C. A. Johnson, Matrix Theory, Cambridge University Press [5] W. J. M Clelland and T. Preston, A Treatise on Spherical Trigonometry, Mac Millan and Co [6] O. Taussky Todd, Research Problem 2, Bull. Amer. Math. Soc., 66, 1960, 275. [7] N. Bajracharya, Level Curves of the angle function of a Positive Definite Symmetric matrix, International Journal of Algebra, Vol. 5, 2011, no. 9, Received: August, 2011