Properly establishing the relation between linear algebra and geometry makes it easier to obtain the three-by-three orthogonal matrix that describes a specified rotation. A Note on Rotation Matrices Jay P. Fillmore University of California at San Diego The connection between three-by-three orthogonal The rows of an orthogonal matrix A are iA, jA, kA, the matrices and the rotations of space that they describe is images of the axis vectors i = [1,0,0], etc., and these rows quite misleading when trying to describe a rotation of are orthonormal vectors, since AA T = L From AA T = I space by angles of rotation about the three coordinate we have A T = A - I and A TA = I, so the columns of A are axes. If the relation between the linear algebra and the orthonormal also. geometry is properly established, not only do obscurities The determinant of an orthogonal matrix A is + 1 or vanish but less computation is required to obtain the - 1, since AA T = I gives (det A)2 = 1. The rotation is matrix describing a specified rotation. called proper or improper (or occasionally direct or op- This article will describe all three-by-three orthogonal posite), as det A is + 1 or - 1. Proper rotations, usually matrices, what they do geometrically, and how to obtain called simply rotations, are the most useful. Let's discuss directly a matrix having prescribed geometric properties. these first. The arguments leading to the formulas are included for completeness, but the goals are the statements that con- nect these formulas with the geometry. Proper rotations-matrix known The characteristic polynomial of a three-by-three matrix Rotations A is det (XI-A) = X 3 - (trA)X 2 + .. ._det A This discussion is restricted to rigid motions of space where tr A is the trace of A and the coefficient of X is not that fix at least one point: rotations; a fixed point is chosen needed. When A is proper orthogonal, det A = + 1, and to be the origin. Such a rigid motion is then given by a this polynomial has at least one positive real root X >0. three-by-three matrix A: The row vector x is sent to xA. If f is an eigenvector accompanying X, LA =X, and Since angles are preserved, (xA) (yA) T always equals (eA ) A ) T = (Xf) (Xf) T equals gfT, so X2 =1. Thus, Xy T, where T denotes transpose, so xAA TyT=xyT. That X= + 1 and LA =Q. L is the axis of rotation of A; A is, AA T = I iS the identity; A is an orthogonal matrix. represents a rigid motion of space about a line through the Conversely, orthogonal matrices yield rotations. origin in the direction of e. 30 0272-1716/84/0200-0030$01.00 (C) 1984 IEEE IEEE CG&A Authorized licensed use limited to: IEEE Xplore. Downloaded on January 8, 2009 at 09:23 from IEEE Xplore. Restrictions apply. The angle 0 of the rotation is the angle through which Proper rotations-axis and angle known any vector in the plane through the origin and perpen- dicular to ( is moved. We can change the length of ( to A proper rotation of space is determined by its axis and length 1 and choose e andforthogonal to (so that e, f, f is angle. The problem, opposite that of the last section, is to orthonormal and right-handed. Then eA = (cos 6)e + find the proper orthogonal matrix having a prescribed vec- (sin 0)fand similarlyfA. If the rotation is referred to this tor determining the axis of the rotation and rotating by a basis, its matrix is prescribed angle about that axis. A formula for that matrix F can be obtained by elementary methods (Rogers and cos 0 sin 0 01 Adams, and Pavlidis2), but the argument to follow, -sin 0 cos0 0 which uses power series of matrices (Nobel and Daniel3), L 0 1 is chosen because it is simplest and the technique is useful in general. which we denote by R. This means that A = PRP 1, Given the vector e = [a, b, c] along the axis of rotation where P is the change-of-basis matrix; P expresses e, f, and and of length X = -/a2 + b2 + c2 not necessarily 1, in- (in terms of i, j, k-that is, e, f, and (are the rows of P. It troduce the matrix is not necessary to know P to determine the angle 0; for 0 c -b- tr A = tr (PRP- 1) = trR = 1 + 2cos 0. L = -c O a The eigenvalues of a three-by-three matrix A can be b -a O easily found by the following observations. If B is any three-by-three matrix, let Ba be its matrix of cofactors: This matrix represents the linear transformation of cross The ijth entry is (- l)i+j times the two-by-two determi- product with e; that is, xL = -xx(. Indeed, the cross nant obtained by striking thejth row and ith column from product ofx by {is obtained by projecting x onto the plane B. Ba is commonly called the adjoint or adjugate of B. perpendicular to (and rotating through a right angle there. One has BaB = (det B)IL If det B.0, this gives Cramer's The cross product rule (xxf)xm = -x(f-m) (x.m) ewith m rule B - 1 = Ba/det B; and if det B = 0, then BaB = 0 + = fgives xL 2 xX 2 + x( TQfor every shows any row e . 0 of Ba satisfies (B = 0. Eigenvectors ( x, so of A for eigenvalue X, fA = XI?or ((A - I) = 0, are thus a L2 + X2I= QT [ b [a b cI obtained as the rows of (A XXI) a. Especially, ifA is prop- er orthogonal, its axis is obtained as a row of (A - I) a. If X is not a repeated eigenvalue of A, then A - I has rank Aside: The same cross product rule also gives xx ((x m) two and all rows of (A - )J) a are in fact scalar multiples = (xx() xm- (xxm) so of one of them. x(, that the matrix corre- sponding to (x m is the "bracket" ML -LM. Note that eL = 0, so L3 + X2L = (L2 +X2I)L = Formulas: A proper orthogonal matrix A (AA T = I (T(L = 0, and we have and det A = + 1) represents a rotation, of space about a L3 = fixed axis. An eigenvector of A for the eigenvalue 1 is in X2L L6= X4L2 the direction of the axis and is obtained as any row of L4 =- _X2L2 L' = - X6L the matrix of cofactors (A - I) a. The angle 0 of rotation L5 = X4L about this axis is given by tr A = 1 + 2 cos 0. The exponential of a matrix is obtained by replacing x with a matrix in the usual series exp x = 1 + x + x2 /2! + The desired eigenvector can be obtained even more x3 /3! + . .. Every partial sum is a matrix, and the se- simply as any row of the symmetric matrix (A + A T) quences of entries converge. The argument now proceeds - (tr A - 1). This will be shown later. analogous to that of Euler's formula e t = cos t + The sign of the angle 0 changes when the eigenvector - I sin t by power series of complex numbers. ( is replaced by - (, since this reverses the direction along Using the familiar series for cosine and sine, we have the axis used for the usual right-hand rule. The proper choice of sign for 0 is that of the determinant det (x, xA, f), exp tL = I + tL + 2 LL2 + t3 L3 +t4 L4 where x is any vector not parallel to (. Because, if we make f 2! 3! 4! of length 1, choose e in the plane ofx and (, and choosefso that e, fis orthonormal and as + iL5 + L6 + ... f, right-handed before, then 5! 6! x = ue + w(, and from eA = (cos 0) e + (sin 0)fwe ob- tain det (x, xA, f) = u2 sin 0. = I+ tL + t L2 t X2L t X2L2 Example: One can use these formulas to show that the 2! 3! 4! product of rotations through angles of 10 degrees, 20 t5 * * - degrees, and 30 degrees about the x-, y-, and z- axes respec- + 4L+ t6 X4L2 - tively, done in that order, is the same as a single rotation 5! 6! through an angle of 35.8 degrees about an axis with direc- = I+ (t-X2 t + X4 t . ..)L tion cosines 0.124, 0.616, and 0.778. In general, this prod- 3! 5! uct of rotations about axes through angles ca, 3, and -y is a rotation through an angle 0 given by 1 + 2 cos 0 = + ( _ X2t + 4t_ .)L2 cos cos y + cosy cosa + cos a cos,B + sin a sinf3 sin y. 2! 4! 6! February 1984 31 Authorized licensed use limited to: IEEE Xplore. Downloaded on January 8, 2009 at 09:23 from IEEE Xplore. Restrictions apply. or This symmetric matrix is of rank one, and its rows are scalar multiples of e. The skew-symmetric part of A is a exp tL = I + sin XtL + 1 - cos XtL2 scalar multiple of L. x X 0 1 0 ( = = Aside: If [0,0,1], then L -1 000and)X= 1, Reflections 0 0 0]l i so the formula gives The transformation that reverses a vector ( and fixes 0 1 0 1 0 0 0 1 0 every vector in the plane (through the origin) perpen- exp t -1 0 0 1 0 +(sin t) -1 0 0 dicular to ( is the reflection in that plane.