MATH 304Linear algebraLecture 37:Rotations in space.Orthogonal matricesDefinition. An n×n matrix A is called orthogonalif AAT= ATA = I , i.e., AT= A−1.Example. A =cos φ − sin φsin φ cos φ.• Aφis orthogonal.• Eigenvalues: λ1= cos φ + i sin φ = eiφ,λ2= cos φ − i sin φ = e−iφ.• |λ1| = |λ2| = 1, λ1= λ2.• Associated eigenvectors: v1= (1, −i),v2= (1, i).• Vectors v1and v2form a basis for C2.Consider a linear operator L : Rn→ Rn, L(x) = Ax,where A is an n×n orthogonal matrix.Theorem There exists an orthonormal basis for Rnsuch that the matrix of L relative to this basis has adiagonal block structureD±1O . . . OO R1. . . O............O O . . . Rk,where D±1is a diagonal matrix whose diagonalentries are equal to 1 or −1, andRj=cos φj− sin φjsin φjcos φj, φj∈ R.Why are orthogonal matrices called so?Theorem Given an n×n matrix A, t he followingconditions are equivalent:(i) A is orthogonal: AT= A−1;(ii) columns of A form an orthonormal basis for Rn;(iii) rows of A form an orthonormal basis for Rn.Proof: Entries of the matrix ATA are dot products ofcolumns of A. Entries of AATare dot products of rows of A.Thus an orthogonal matrix is the transition matrixfrom one orthonormal basis to another.Consider a linear operator L : Rn→ Rn, L(x) = Ax,where A is an n×n matrix.Theorem The following conditions are equivalent:(i) |L(x)| = |x| for all x ∈ Rn;(ii) L(x) · L(y) = x · y for all x, y ∈ Rn;(iii) the matrix A is orthogonal.Definition. A transformation f : Rn→ Rnis calledan isometry if it preserves distances betweenpoints: |f (x) − f (y)| = |x − y|.Theorem Any isometry f : Rn→ Rncan berepresented as f (x) = Ax + x0, where x0∈ RnandA is an orthogonal matrix.Classification of 3 ×3 orthogonal matrices:A =1 0 00 cos φ − sin φ0 sin φ cos φ, B =−1 0 00 1 00 0 1,C =−1 0 00 cos φ − sin φ0 sin φ cos φ.A = rotation about a line; B = reflection in aplane; C = rotation about a line combined withreflection in the orthogonal plane.det A = 1, det B = det C = −1.A has eigenvalues 1, eiφ, e−iφ. B has eigenvalues−1, 1, 1. C has eigenvalues −1, eiφ, e−iφ.Rotations in spaceIf the axis of rotation is oriented, we can say aboutclockwise or counterclockwise rotations (withrespect to the view from the positive semi-axis).Clockwise rotations about coordinate axescos θ sin θ 0− sin θ cos θ 00 0 1cos θ 0 − sin θ0 1 0sin θ 0 cos θ1 0 00 cos θ sin θ0 − sin θ cos θProblem. Find the matrix of the rotation by 90oabout the line spanned by the vector c = (1, 2, 2).The rotation is assumed to be counterclockwisewhen looking from the tip of c.B =0 −1 01 0 00 0 1is the matrix of (counterclockwise)rotation by 90oabout the z-axis.We need to find an orthonormal basis v1, v2, v3suchthat v3has the same direction as c. Also, the basisv1, v2, v3should obey the same hand rule as thestandard basis. Then B is the matrix of the givenrotation relative to the basis v1, v2, v3.Let U denote the transition matrix from the basisv1, v2, v3to the standard basis (columns of U arevectors v1, v2, v3). Then the desired matrix isA = UBU−1.Since v1, v2, v3is going to be an orthonormal basis,the matrix U will be orthogonal. Then U−1= UTand A = UBUT.Remark. The basis v1, v2, v3obeys the same handrule as the standard basis if and only if det U > 0.Hint. Vectors a = (−2, −1, 2), b = (2, −2, 1),and c = (1, 2, 2) are orthogonal.We have |a| = |b| = |c| = 3, hence v1=13a,v2=13b, v3=13c is an orthonormal basis.Transition matrix: U =13−2 2 1−1 −2 22 1 2.det U =127−2 2 1−1 −2 22 1 2=127· 27 = 1.(In the case det U = −1, we should interchangevectors v1and v2.)A = UBUT=13−2 2 1−1 −2 22 1 20 −1 01 0 00 0 1·13−2 −1 22 −2 11 2 2=192 2 1−2 1 21 −2 2−2 −1 22 −2 11 2 2=191 −4 88 4 1−4 7 4.U =13−2 2 1−1 −2 22 1 2is an orthogonal matrix.det U = 1 =⇒ U is a rotation matrix.Problem. (a) Find the axis of the rotation.(b) Find the angle of the rotation.The axis is the set of points x ∈ Rnsuch thatUx = x ⇐⇒ (U − I )x = 0. To find the axis, weapply row reduction to the matrix 3(U − I ):3U − 3I =−5 2 1−1 −5 22 1 −1→−3 3 0−1 −5 22 1 −1→1 −1 0−1 −5 22 1 −1→1 −1 00 −6 22 1 −1→1 −1 00 −6 20 3 −1→1 −1 00 0 00 3 −1→1 −1 00 3 −10 0 0→1 −1 00 1 −1/30 0 0→1 0 −1/30 1 −1/30 0 0Thus Ux = x ⇐⇒x − z/3 = 0y − z/3 = 0The general solution is x = y = t/3, z = t, t ∈ R.=⇒ d = (1, 1, 3) is the direction of the axis.U =13−2 2 1−1 −2 22 1 2Let φ be the angle of rotation. Then theeigenvalues of U are 1, eiφ, and e−iφ. Thereforedet(U − λI ) = (1 − λ)(eiφ− λ)(e−iφ− λ).Besides, det(U − λI ) = −λ3+ c1λ2+ c2λ + c3,where c1= tr U (the sum of diagonal entries).It follows thattr U = 1 + eiφ+ e−iφ= 1 + 2 cos φ.tr U = −2/3 =⇒ cos φ = −5/6 =⇒ φ ≈
View Full Document
Unlocking...