New aspects on square roots of a real 2 2 matrix and their geometric applications

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1 MATHEMATICAL SCIENCES AND APPLICATIONS E-NOTES X (X 1-6 (018 c MSAEN New aspects on square roots of a real matrix and their geometric applications Mircea Crasmareanu*, Andrei Plugariu (Communicated by Cihan ÖZGÜR Abstract We present a new study on the square roots of real matrices with a special view towards examples, some of them inspired by geometry. Keywords: real matrix; square root. AMS Subject Classification (010: Primary: 15A4 ; Secondary: 15A18. *Corresponding author a c We begin with the following general matrix A = M b d (R and ask: is there a matrix B M (R such that B = A? Such a matrix B is called square root of A. We point out that the more complicated case of a real matrix of order 3 is discussed in [4]. Although the case we consider is also well studied (according to the bibliography of [3] we add several examples and facts concerning this notion as well as a series of geometrical applications. The Euclidean example We recall the n-orthogonal group: O(n = {A M n (R : A t A = I n }; is the invariant group of the Euclidean inner product <, > (yielding the usual Euclidean norm. If A O(n then (det A t (det A = deti n = 1 implies that det A = ±1. Hence, the orthogonal group splits into two components: O(n = SO(n O (n where SO(n contains the matrices from O(n having deta = 1 and O (n those with deta = 1; represents the disjoint reunion of sets. SO(n is a subgroup in O(n and is called n-special orthogonal group. O (n is not closed under product: A 1, A O (n implies that A 1 A SO(n. Since M 1 (R = R we have that O(1 = {±1} with SO(1 = {1} and O (1 = { 1}; we remark that O(1 contains the integer unit roots! We know O( as well: cos t sin t cos t sin t R(t = SO(, S(t = O (, t R. sin t cos t sin t cos t Hence, we have that: S(t cos t sin t cos t sin t = = I sin t cos t sin t cos t which means that any S(t is a root of the unit matrix I. We recall that from a geometrical point of view a square root of the unit matrix is called almost product structure, see for example [6]. Geometrical significance: R(t is the matrix of rotation of angle t in trigonometrical sense (i.e anticlockwise around the origin and S(t is the matrix of axial symmetry with respect to d t/ =line from plane R which contains the origin O and makes the oriented angle t/ with Ox. We have that S(t S(t 1 = A(t t 1 S(t 1 S(t. Received : , Accepted : dd month yyyy

2 M. Crâşmăreanu, A. Plugariu We return to the general case of matrix A. We remind that A has two invariants: T ra := a + d, det A := ad bc. Properties: i T r : M (R R is a linear operator: T r(αa 1 + βa = αt ra 1 + βt ra, ii det : M (R R is a multiplicative function: det(a 1 A = det A 1 det A, iii The characteristic equation of A: A T ra A + det A I = O. The multiplicative property of the determinant yields: The necessary condition for existence of square roots: Hence we assume from now on that det A 0. B : B = A det A 0. Revised Euclidean example 1: T rs(t = 0, dets(t = 1 which says that S(t does not admit roots. A root of order 4 of unit matrix is called structure of electromagnetic type according to [7, p. 3807]. We also have relationships between the invariants of A and B: T ra = (T rb det B, det A = (det B. (0 Proof. It is enough to proof the first identity. We write the characteristic equation of B: which gives: We square this relation: or: From (1 we have that: which is replaced in square brackets from (3: A T rb B + det B I = O (1 A = T rb B det B I. ( A = (T rb A T rb det B B + (det B I A (T rb A + det B[T rb B] (det B I = O. (3 T rb B = A + det B I (4 A (T rb A + det B[A + det B I ] (det B I = A [(T rb det B] A + (det B I = O and by comparing with the characteristic equation of A we obtain the conclusion. The relation (4 is fundamental for finding B and we have two cases: Case I: T rb = 0 implies that: A = det B I. Case II T rb 0 implies that: B = 1 T rb [A + det B I ]. (5 From the first relation (0 we have that: hence, if A ai, we obtain that: II1 T ra + det A 0 implies that A does not have roots, II T ra + det A > 0 but T ra det A 0 implies that A has two roots: (T rb = T ra + det A (6 1 B ± = ± T ra + det A [A + det AI ] (7

3 New aspects on square roots of a real matrix and their geometric applications 3 II3 T ra det A > 0 (which implies that T ra + det A > 0 implies that A has four roots: 1 B ± (ε = ± T ra + ε det A [A + ε det AI ], ε = ±1. (8 Revised Euclidean example For A(t we have: T rr(t = cos t, det R(t = 1, T rr(t + det R(t = 4 cos t, T rr(t det R(t = (cos t 1 0. (9 From Case II, we get that R(t has two roots: B ± (t = ± 1 ( cos t + 1 sin t cos t sin t cos t + 1 = ±R t. (10 The relation (10 can be considered the matrix version of the well-known Moivre s relation from complex algebra (C, +, : (cos t + i sin t = cos(t + i sin(t. (11 The group law of SO( is: R(t 1 R(t = R(t 1 + t = R(t R(t 1 which gives: R(t = R(t and the fact that SO( is a group isomorphic to the multiplicative group (S 1, of all unit complex numbers. Inspired by characteristic equation of A we introduce the characteristic polynomial of A, namely p A R[X]: p A (X = X T ra X + det A. (1 We know that the possible real roots of p A are called eigenvalues of A and are useful in studying the diagonalisation of A. So, if the eigenvalues exist and are different, we denote them by λ 1 < λ and it follows that A admits a diagonal form: A = S 1 diag(λ 1, λ S (13 with S GL(, R=-general linear group i.e. the group of all real invertible matrices of order. Obviously, the condition of existence and inequality for λ 1, holds when the discriminant (p A is strictly positive: The relationship between (p A and (p B is given by: Proposition Let B be a square root of A. Then: (p A := (T ra 4 det A. (14 (p A = (T rb (p B. (15 Thus, if T rb 0 then A has different eigenvalues if and only if B has different eigenvalues. Proof. The relation (15 is a direct consequence of (0. Corollary Suppose the matrix A with det A > 0 has the root B with T rb 0. Assume that A is diagonalisable with S GL(, R and different eigenvalues λ 1 < λ. Then, 0 < λ 1 < λ and B is diagonalisable with the same matrix S having the eigenvalues { λ 1, λ } or { λ 1, λ } or { λ 1, λ } or { λ 1, λ }. Equivalently, we are in case II3 with: 1 B ± (ε = ± λ + ε [A + ε λ 1 λ I ] = S diag(± λ 1, ± λ S 1. (16 λ 1 Proof. Because det A > 0 we have that λ 1 and λ have the same sign. We suppose that λ 1 < λ < 0. From (6 we have that (T rb = λ 1 + λ ± λ 1 λ > 0. It follows only the case with + i.e. λ 1 λ + λ 1 λ > 0 but this is impossible because of the AM-GM inequality. Recall that the AM GM inequality states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list.

4 4 M. Crâşmăreanu, A. Plugariu The golden example It is known that the golden proportion (or the golden number is the positive root, φ = 5+1, of the equation [6]: x x 1 = 0. (17 The negative root is φ 1 = 1 5. Let us consider the matrix: 3 A =, T ra = 6, det A = 5. (18 3 A is diagonalisable, being symmetric, with 0 < λ 1 = 1 < λ = 5. We have that: T ra + ε det A = 6 + ε 5 = ( 5 + ε (19 We are in Case II3 and for example: B ± (1 = ± ( = ± 1 ( φ φ φ 1 φ φ = ± φ 1 φ. (0 By analogy with the problem studied here, we call the matrix A M n (R satisfying A A I n = O n, as being an almost golden structure. In [6] we study the relationship between almost golden structures and almost product structures. We return to the case I given by A = ai and we present the solution from [3, p. 491]. We have, irrespective of a s sign, an infinity of roots: c s B ± (c, s := ±, c R, s R. (1 c a c s If a = 0 then we add the infinite set of almost tangent structures: ( 0 0 B ± (u := ± u 0, u R. ( If a > 0 then we add the infinite set: ( ± a 0 B ± (u := u a ( ± a 0, B ± := 0 ± a. (3 Revised Euclidean example 3 For a = 1 the family B + (c, s becomes: c s B(c, s = c 1 c s (4 which gives: B(cos t, sin t = S(t. (5 This way we obtain the matrices from O (. We consider now a right triangle with sides x, y and hypotenuse z. We have that: S(t = 1 x y. (6 z y x If (x, y, z (N 3 then (x, y, z is a Pythagorean triple. This example of almost product structures provided by Pythagorean triples appears on the Web page [1]. In [5] we gave a method for finding matrices A M 3 (R which transforms a Pythagorean triple into another Pythagorean triple. Open problem Do the matrices A which preserve Pythagorean triples admit roots? We return now to the given corollary: a symmetric matrix A with different and strictly positive eigenvalues is positive definite, []. Thus, A defines a new inner product on R n : < x, ȳ > A :=< x, Aȳ >. (7

5 New aspects on square roots of a real matrix and their geometric applications 5 If A admits B as a root then: If B is also symmetric, which happens when n =, then: hence: < x, ȳ > A :=< x, B ȳ >=< B t x, Bȳ >. (8 < x, ȳ > A :=< B x, Bȳ > (9 x A = B x. (30 Thus, for nonzero vectors x, ȳ R n, the angle ϕ A ( x, ȳ between them with respect to <, > A is given by: cos ϕ A ( x, ȳ = cos ϕ(b x, Bȳ. (31 Generalized golden example The matrix A M (R is called bi-symmetric if it has the form: a b A =. (3 b a Then det A = a b and, for having roots, we assume that a > b. We are in case II3 and obtain that: B ± (ε = ± 1 ( a + b + ε a b a + b ε a b a + b ε a b a + b + ε a b (33 which gives us the result that any of its root is also bi-symmetric. The conversely: If B is bi-symmetric then B is bi-symmetric is obvious from calculus. Hyperbolic example We consider: ( cosh t sinh t A(t := sinh t cosh t A is a bi-symmetric matrix with a > b and with formula (33 we obtain that: ( cosh t B ± (1 = ± sinh t sinh t sinh t cosh t, B ± ( 1 = ± cosh t cosh t sinh t Fibonacci example In [8, p. 4] is introduced the Q-Fibonacci matrix: 1 1 Q = 1 0 that has the natural powers: ( Q n Fn+1 F = n F n F n 1. (34. (35 (36. (37 Because of the golden example, we consider the matrix: Fn+1 F Q(n = n. (38 F n F n+1 With relation (33 we have the roots: Q ± (n, ε = ± 1 ( Fn+ + ε F n Fn+ ε Fn+ F n ε F n Fn+ + ε. (39 F n Almost complex example A root of the matrix I is called almost complex structure. According to (1 we have: ( B ± (s, c := ± s 1 s c c s, s R, c R. (40

6 6 M. Crâşmăreanu, A. Plugariu An interesting particular case is: ( sinh t cosh t B ± (sinh t, cosh t := B(t = ± cosh t sinh t. (41 Acknowledgements The authors are extremely indebted to two anonymous referees for their extremely useful remarks and improvements. References [1] [] [3] Anghel, N., Square roots of real matrices, Gaz. Mat. Ser. B 118 (013, no. 11, [4] Anghel, N., Square roots of real 3x3 matrices vs. quartic polynomials with real zero, An. Ştiinţ. Univ. Ovidius Constanţa, Ser. Mat. 5 (017, no. 3, [5] Crasmareanu, M., A new method to obtain Pythagorean triple preserving matrices, Missouri J. Math. Sci. 14 (00, no. 3, MR (003h:15041, Zbl [6] M. Crasmareanu, M. and Hreţcanu, Cristina-Elena, Golden differential geometry, Chaos, Solitons & Fractals 38 (008, no. 5, MR 45653(009k:53059 [7] Reyes, E., Cruceanu, V. and Gadea, P. M., Structures of electromagnetic type on vector bundle, J. Phys. A, Math. Gen. 3 (1999, no. 0, Zbl [8] Stakhov A. and Aranson S., The golden non-euclidean geometry. Hilbert s fourth problem, golden dynamical systems, and the fine-structure constant. With the assistance of Scott Olsen. Series on Analysis, Applications and Computation 7. Hackensack, NJ: World Scientific, 017. Zbl Affiliations MIRCEA CRASMAREANU ADDRESS: University "Al. I.Cuza", Faculty of Mathematics, , Iasi-Romania. mcrasm@uaic.ro ANDREI PLUGARIU ADDRESS: University "Al. I.Cuza", Faculty of Mathematics, , Iasi-Romania. plugariu.andrei@yahoo.ro