LECTURE 14: EXAMPLES OF CHANGE OF BASIS AND MATRIX TRANSFORMATIONS. QUADRATIC FORMS. Prof. N. Harnew University of Oxford MT 2012 1 Outline: 14. EXAMPLES OF CHANGE OF BASIS AND MATRIX TRANSFORMATIONS. QUADRATIC FORMS. 14.1 Examples of change of basis 14.1.1 Representation of a 2D vector in a rotated coordinate frame 14.1.2 Rotation of a coordinate system in 2D 14.2 Rotation of a vector in fixed 3D coord. system 14.2.1 Example 1 14.2.2 Example 2 14.3 MATRICES AND QUADRATIC FORMS 14.3.1 Example 1: a 2 × 2 quadratic form 14.3.2 Example 2: another 2 × 2 quadratic form 14.3.3 Example 3: a 3 × 3 quadratic form 2 14.1 Examples of change of basis 14.1.1 Representation of a 2D vector in a rotated coordinate frame I Transformation of vector r from Cartesian axes (x; y) into frame (x 0; y 0), rotated by angle θ 3 0 x0 = r cos α y = r sin α x = r cos(θ + α) y = r sin(θ + α) 0 x cos α ! y 0 = y sin α ! x = cos(θ+α) sin(θ+α) 0 0 x cos α = x0 cos θ cos α − x0 sin θ sin α y sin α = y sin θ cos α + y cos θ sin α 0 0 Since x0 sin α = y 0 cos α Since y cos α = x sin α 0 0 x = x0 cos θ −y 0 sin θ y = x sin θ +y cos θ I Coordinate transformation: 9 These equations 9 x cos θ − sin θ x0 = (1) > > y sin θ cos θ y 0 > relate the coordinates > => of r measured in the => I Take the inverse: (x; y) frame with those > > x0 cos θ sin θ x > measured in the rotated > 0 = (2) ;> ;> y − sin θ cos θ y (x 0; y 0) frame 4 14.1.2 Rotation of a coordinate system in 2D I Start from the familiar orthonormal basis 1 0 je i = (≡ x^); je i = (≡ y^) (3) 1 0 2 1 I Transform the basis via a rotation through an angle θ cos θ − sin θ New basis : je0 i = (≡ x^0); je0 i = (≡ y^0) 1 sin θ 2 cos θ (4) 5 0 I The transformation matrix S is determined from jeii = Sjeii S S 1 cos θ S = cos θ 11 12 = ) 11 (5) S21 S22 0 sin θ S21 = sin θ S S 0 − sin θ S = − sin θ 11 12 = ) 12 (6) S21 S22 1 cos θ S22 = cos θ cos θ − sin θ Hence S(θ) = (7) sin θ cos θ As expected, the basis transformation matrix jei ! je0i is the inverse of the transformation (x; y) ! (x 0; y 0) of the components derived in the previous sub-section. I The inverse transformation matrix rotates backwards cos θ sin θ S−1(θ) = ≡ S(−θ) (8) − sin θ cos θ I It is easy to show via substitution that two successive rotations S(θ)S(α) = S(α)S(θ) = S(θ + α) 6 14.2 Rotation of a vector in fixed 3D coord. system I In 3D, we can rotate a vector r about any one of the three axes r0 = R(θ) r A rotation about the z axis is given by 0 cos θ − sin θ 0 1 Rz (θ) = @ sin θ cos θ 0 A (9) 0 0 1 Note that rotations of a vector in a fixed coordinate system transform in the same way as rotations of the base vectors (see previous section). I For rotations about the x and y axes 0 1 0 0 1 0 cos γ 0 sin γ 1 Rx (α) = @ 0 cos α − sin α A ; Ry (γ) = @ 0 1 0 A 0 sin α cos α − sin γ 0 cos γ (10) I But note that now for successive rotations: Rz (θ)Rx (α) 6= Rx (α)Rz (θ) 7 14.2.1 Example 1 Rotate the unit vector 0 cos θ − sin θ 0 1 (1; 0; 0) by 90◦ about Rz (θ) = @ sin θ cos θ 0 A the z-axis 0 0 1 0 0 −1 0 1 0 1 1 0 0 1 ! @ 1 0 0 A @ 0 A = @ 1 A 0 0 1 0 0 ! Which is a unit vector along the y-axis (as expected). nd ◦ I Now make a 2 rotation of 90 about the x-axis: 0 1 0 0 1 0 0 −1 0 1 0 1 1 0 0 1 ◦ ◦ Rx (90 )Rz (90 ) = @ 0 0 −1 A @ 1 0 0 A @ 0 A = @ 0 A 0 1 0 0 0 1 0 1 So ... by now the procedure of matrix multiplication should be clear: the exact form of the row/column multiplication is necessary to make a linear transformation between two bases. It is also the required 8form for rotations of vectors in their associated vector space(s). 14.2.2 Example 2 ◦ I Rotate the vector r = (1; 2; 3) by 30 about the y axis. p sin 30◦ = 1=2; cos 30◦ = 3=2 I The rotation matrix is p 0 cos γ 0 sin γ 1 0 3=2 0 1=2 1 Ry (γ) = @ 0 1 0 A = @ 0 1p 0 A (11) − sin γ 0 cos γ −1=2 0 3=2 p p 0 x0 1 0 3=2 0 1=2 1 0 1 1 0 3=2 + 3=2 1 0 @ y A = @ 0 1p 0 A @ 2 A = @ 2 p A (12) z0 −1=2 0 3=2 3 −1=2 + 3 3=2 I As a check - rotate back ! use inverse matrix p p 0 x 1 0 3=2 0 −1=2 1 0 3=2 + 3=2 1 @ y A = @ 0 1p 0 A @ 2 p A (13) z 1=2 0 3=2 −1=2 + 3 3=2 p p 0 3=4 + 3 3=4 + 1=4 − 3 3=4 1 0 1 1 = @ p 2 p A = @ 2 A as required: (14) 3=4 + 3=4 − 3=4 + 9=4 3 9 14.3 MATRICES AND QUADRATIC FORMS Best illustrated by a few examples. 14.3.1 Example 1: a 2 × 2 quadratic form 2 2 T I Represent equation x + y = 1 in matrix form X AX = 1 I Matrix A is a transformation matrix which represents the conic form of the equation. x 1 0 x x2 + y 2 = (x; y) = (x; y) = 1 y 0 1 y (15) 10 14.3.2 Example 2: another 2 × 2 quadratic form I Matrix representation: 5 2 x (x; y) = 5 (16) 2 3 y I Write in equation form: 5x + 2y = (x; y) (17) 2x + 3y = 5x2 + (2xy + 2xy) + 3y 2 = 5x2 + 4xy + 3y 2 ! 5x2 + 4xy + 3y 2 = 5 11 14.3.3 Example 3: a 3 × 3 quadratic form 2 2 2 I Represent x + y − 3z + 2xy + 6xz − 6yz = 4 in the matrix form (X T AX). I Write 0 a b c 1 0 x 1 0 ax + by + cz 1 (x; y; z) @ d e f A @ y A = (x; y; z) @ dx + ey + fz A (18) g h i z gx + hy + iz = ax 2 + bxy + cxz + dxy + ey 2 + fyz + gxz + hyz + iz2 I Comparing terms: [x 2] ! a = 1;[y 2] ! e = 1;[z2] ! i = −3 [xy] ! b + d = 2;[xz] ! c + g = 6;[yz] ! f + h = −6 (underconstrained) 0 1 I In echelon form: 1 0 0 T set b = c = f = 0 X @ 2 1 0 A X = 4 (19) 6 −6 −3 I In symmetrical form: 0 1 1 3 1 T set b = d, c = g, f = h X @ 1 1 −3 A X = 4 (20) 3 −3 −3 12.