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EIGENVECTORS, EIGENVALUES, and FINITE STRAIN I Main Topics

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EIGENVECTORS, EIGENVALUES, and FINITE STRAIN I Main Topics GG303 Lab 9 10/27/08 1 EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN I Main Topics A Motivation B Inverse and determinant of a matrix (review) C Eigenvalue problems, eigenvectors, and eigenvalues D Diagonalization of a matrix E Quadratic forms and the principal axes theorem (maxima/minima in 2D) F Strain ellipses, strain ellipsoids, and principal strains II Motivation A To find the magnitudes and directions of the principal strains from the displacement field using linear algebra rather than graphical means or calculus (e.g., Ramsay and Huber, 1984). B To introduce a powerful, broadly useful mathematical method II Inverse and determinant of a matrix (review) −1 A Inverse [A] of a matrix: AA−1 = A−1A = I, where A is nxn (square) −1 a b B Inverse [A] of a 2x2 matrix A = c d € € −1 1 d −b 1 A = [A] can not have an inverse if ad-bc = 0 € ad − bc − c a € −1 1 d −b a b 1 da − bc db − bd 1 0 2 A A = = = = I √ ad − bc − c a c d ad − bc − ca + ac −bc + ad 0 1 € C Determinant |A| of a real matrix € 1 A number that provides scaling information on a square matrix a b 2 Determinant of a 2x2 matrix: A = = ad − bc c d a b c e f d f d e 3 Determinant of a 3x3 matrix: A = d e f = a − b + c € h i g i g h g h i Stephen Martel Lab9-1 University of Hawaii € GG303 Lab 9 10/27/08 2 4 Relevance of the inverse of a matrix to a matrix equation AX = 0 a |A| ≠ 0 if the matrix has an inverse (see II.B.1). This is true for nxn matrices in general (e.g., Strang, 1998, p. 232-233). b |A| = 0 if the matrix has no inverse (see II.B.1). This is true for nxn matrices in general (e.g., Strang, 1998, p. 232-233). 5 Geometric meanings of the real matrix equation AX = 0 (See lab4!) nxn |A|≠0; A-1 exists (see II.B.1) |A|=0; A-1 does not exist (see II.B.1) 2x2 Describes two lines that Describes two lines that intersect in a intersect at one unique point line that passes through the origin; no - the origin. unique solution. The lines are co-linear. 3x3 Describes three planes that Describes three planes that intersect intersect at one unique point in a line (or a plane) that passes - the origin. through the origin; no unique solution. (See appendix for graphical examples) III Eigenvalue problems, eigenvectors and eigenvalues A Eigenvalue problems are represented by the matrix equation AX = λX, where A is a square nxn matrix, X is a non-zero vector (an nx1 column array), and λ is a number. B Eigenvectors and eigenvalues provide simple, elegant, and clear ways to represent solutions to a host of physical and mathematical problems [e.g., geometry, strain, stress, curvature (shapes of surfaces)] C Eigenvectors 1 Non-zero directional vectors that provide solutions for AX = λX 2 Vectors that maintain their orientation when multiplied by matrix A D Eigenvalues: numbers (λ) that provide solutions for AX = λX. If X is a unit vector, λ is the length of the vector produced by AX. E Eigenvectors have corresponding eigenvalues, and vice-versa F In Matlab, [v,d] = eig(A), finds eigenvectors and eigenvalues Stephen Martel Lab9-2 University of Hawaii GG303 Lab 9 10/27/08 3 G Examples 1 Identity matrix (I) 1 0 x x x = =1 0 1 y y y All vectors in the xy-plane maintain their orientation and length € when operated on by the identity matrix, so the eigenvalue for I is 1, and all vectors are eigenvectors. 2 A 2D rotation matrix for rotations in the xy plane All non-zero real vectors rotate and maintain their length. A 2D rotation matrix thus has no real eigenvectors and hence no real eigenvalues; its eigenvectors and eigenvalues are imaginary. 3 A 3D rotation matrix. The only vectors that are not rotated are along the axis of rotation, so the one real eigenvector of a 3D rotation matrix gives the orientation of the axis of rotation. 0 2 4 A symmetric matrix: A = 2 0 Matrix equation Eigenvalue Eigenvector 1 0 2 1 €2 1 2 1 A = = = 2 . 1 2 0 1 2 1 1 1 0 2 1 − 2 1 -2 1 A = = = −2 . − 1 2 0 − 1 2 − 1 − 1 € € 9 3 5 Another symmetric matrix: A = 3 1 € € Matrix equation Eigenvalue Eigenvector − 3 0.1 9 3 − 3 0.1€ − 30 0.1 − 3 0.1 10 − 3 0.1 A = = =10 . − 0.1 3 1 − 0.1 − 10 0.1 − 0.1 − 0.1 0.1 9 3 0.1 0 0.1 0 0.1 A = = = 0 . −3 0.1 3 1 −3 0.1 0 −3 0.1 −3 0.1 € € Stephen€ Martel Lab9-3 University€ of Hawaii GG303 Lab 9 10/27/08 4 H Alternative form of an eigenvalue equation 1 [A][X] = λ[X] Since λ[X] = λ[IX], subtracting λ[IX] from both sides of (H.1) yields: 2 [A-Iλ][X] = 0 (same form as [B][X] = 0; see II.C.4) I Solution conditions and connections with determinants 1 Unique trivial solution of [X] = 0 if and only if |A-Iλ|≠0 (see II.C.4). 2 Eigenvector solutions ([X] ≠ 0) if and only if |A-Iλ|=0 (see II.C.4). J Characteristic equation: |A-Iλ|=0 1 The roots of the polynomial equation produced by expanding the characteristic equation are the eigenvalues (from I.2). 2 General 2-D solution for eigenvalues a b a b Let A = , tr(A) = a + d, = ad − bc. Then c d c d a − λ b (1) A − Iλ = = 0 c d − λ € a − λ b (2) = (a − λ)(d − λ) − (b)(c) = λ2 − (a + d)λ + ad − bc = 0 c d − λ € (3) A − Iλ = λ2 − (a + d)λ + ad − bc = λ2 − tr(A)λ + A = 0 where€ tr(A) is the trace (sum of main diagonal terms) of A. The roots€ of (3), found with the quadratic equation, are the eigenvalues. 2 2 (a + d) ± (a + d) − 4(ad − bc) tr(A) ± (tr(A)) − 4 A (4) λ ,λ = = 1 2 2 2 a In general, an n x n matrix has n eigenvalues, some of which € might be identical. b Eigenvalues can be zero even though eigenvectors can not. c For both eigenvalues to be real and distinct, ((tr(A))2 > 4|A| d For both eigenvalues to be positive, ((tr(A))2 > 4|A| & |A| >0. The latter is because tr(A) must exceed the radical term in (4). Stephen Martel Lab9-4 University of Hawaii GG303 Lab 9 10/27/08 5 2 2 tr(A) + (tr(A)) − 4 A tr(A) − (tr(A)) − 4 A e λ + λ = + = tr(A) 1 2 2 2 2 2 tr(A) + (tr(A)) − 4 A tr(A) − (tr(A)) − 4 A f λ λ = = A 1 2 2 2 € a b g Eigenvalues of a real symmetric 2 x 2 matrix A = (i.e., b=c) b d € Substituting b=c in equation J.2.(4) yields 2 (a + d) ± (a + d) − 4(ad − b2)€ (1) λ ,λ = 1 2 2 (a + d) ± (a2 + 2ad + d 2) − 4ad + 4b2 (2) λ1,λ2 = € 2 (a + d) ± (a2 − 2ad + d 2) + 4b2 (3) λ1,λ2 = € 2 2 (a + d) ± (a − d) + 4b2 (4) λ ,λ = (See III.G.4 for examples) 1 2 2 € The term under the radical sign can not be negative, so the eigenvalues€ for a real symmetric 2 x 2 matrix must be real. 3 Eigenvectors for a 2x2 matrix (See III.G.4 for examples) Start with the eigenvalue equation (1) [A-Iλ][X] = 0 For a 2x2 matrix this is a − λ b x (2) = 0 c d − λ y Solving this yields the eigenvectors via their slopes (3)€ (a − λ)x + by = 0 or cx + (d − λ)y = 0 (4) by = −(a − λ)x or (d − λ)y = −cx y (λ − a) y c (5)€ = € or = x b x (λ − d) € € Stephen Martel Lab9-5 University of Hawaii € € GG303 Lab 9 10/27/08 6 4 Distinct eigenvectors of a symmetric 2x2 matrix are perpendicular The eigenvectors X1 and X2 are perpendicular if X1 • X2 = 0. Proof (1a) AX1 = λ1X1 (1b) AX2 = λ2 X 2 Equation (1a) yields a vector parallel to X1, and equation (1b) yields a€ vector parallel to X2. Dotting€ the vector of (1a) by X2 and the vector of (1b) by X1 will test whether X1 and X2 are orthogonal. (2a) X 2 • AX1 = X 2 • λ1X1 = λ1(X 2 • X1) (2b) X1 • AX2 = X1 • λ2 X 2 = λ2(X1 • X 2) If A is symmetric, then X 2 • AX1 = X1 • AX 2 , as the following lines show: x a b x x ax by € 1 2 1 2 +€ 2 (3a) • = • = ax1x2 + bx1y2 + by1x2 + dy1y2 y1 b d y2 y1 bx 2 + dy2 € x2 a b x1 x2 ax 1 + by1 (3b) • = • = ax1x2 + by1x2 + bx1y2 + dy1y2 y2 b d y1 y2 bx 1 + dy1 € Since (3a) equals (3b), the left sides of (2a) and (2b) are equal, and so€ the right sides of (2a) and (2b) must be equal too. Hence (4) λ1(X 2 • X1) = λ2(X1 • X 2) Now subtract the right side of (4) from the left side: (5) X X 0 € (λ1 − λ2)( 2 • 1) = The eigenvalues generally are different, so λ1- λ2 ≠ 0, therefore (€X 2 • X1) = 0 and hence the eigenvectors X1and X2 are orthogonal.
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