Math 304-504Linear algebraLecture 37:Rotations in space.Orthogonal polynomials.Orthogonal matricesTheorem Given an n×n matrix A, the 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.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→ Rnisrepresented 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 = reflectio n 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, ei φ, e−i φ. B has eigenvalues−1, 1, 1. C has eigenvalues −1, ei φ, e−i φ.Rotations in spaceIf the axis of rotation is oriented, we can say abo utclockw ise or counterclockwise rotations (withrespect to the vi ew from the positi ve 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 li ne spanned by the vector a = (1, 2, 2).The rotation is assumed to be counterclockwisewhen looking from the tip of a.B =0 −1 01 0 00 0 1is the matrix of (counterclockwise)rotation by 90oabout the z-axis.We need to fi nd an orthonormal basis v1, v2, v3suchthat v3has the same direction as a. Also, the basi sv1, 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 transitio n 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 bas is,the matrix U wil l 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 = (1, 2, 2), b = (−2, −1, 2), andc = (2, −2, 1) are orthogonal.We have |a| = |b| = |c| = 3, hence v1=13b,v2=13c, v3=13a 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, ei φ, and e−i φ. Thereforedet(U − λI ) = (1 − λ)(ei φ− λ)(e−i φ− λ).Besides, det(U − λI ) = −λ3+ c1λ2+ c2λ + c3,where c1= tr U (the sum of diagonal entries).It follows thattr U = 1 + ei φ+ e−i φ= 1 + 2 cos φ.tr U = −2/3 =⇒ cos φ = −5 /6 =⇒ φ ≈ 1 46.44oOrthogonal polynomialsP: the vector space of all polynomial s with r ealcoefficients: p(x) = a0+ a1x + a2x2+ · · · + anxn.Basis for P: 1, x, x2, . . . , xn, . . .Suppose that P is endowed with an inner product.Definition. Orthogonal polynomials (relative tothe inner product) are polynomials p0, p1, p2, . . .such that deg pn= n (p0is a nonzero constant)and hpn, pmi = 0 for n 6= m.Orthogonal po lynomials can be obtained by applyingthe Gram-Schmidt orthogo nalization processto the basis 1, x, x2, . . . :p0(x) = 1,p1(x) = x −hx, p0ihp0, p0ip0(x) ,p2(x) = x2−hx2, p0ihp0, p0ip0(x) −hx2, p1ihp1, p1ip1(x) ,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .pn(x) = xn−hxn, p0ihp0, p0ip0(x) − · · · −hxn, pn−1ihpn−1, pn−1ipn−1(x) ,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Then p0, p1, p2, . . . are orthogonal
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