College of Engineering and Computer ScienceMechanical Engineering DepartmentMechanical Engineering 501ASeminar in Engineering AnalysisFall 2004 Number 17472 Instructor: Larry Caretto September 14 Homework SolutionsPage 375, problem 11 – Find the spectrum (all the eigenvalues) and eigenvectors for the matrix, A, at the right.We use the basic equation that the eigenvalues,  of a matrix, A, can be found from the equation Det(A – I) = 0. Writing A – I and using the expression for the 3 by 3 determinant gives100640353A0)1)(4)(3()6)(0)(3()1)(5)(0()3)(4)(0()6)(5)(0()3)(0)(0()1)(4)(3(100640353)( IADetHere we see a case of two general results: (1) the determinant of a triangular matrix is the product of the terms on the diagonal; and (2) the eigenvalues of a triangular matrix are equal to the elements on the diagonal. In this case we have three unique eigenvalues, 1 = 3, 2 = 4, and3 = 1. (Note that the ordering of the eigenvalues is arbitrary; often they are numbered in rank order. Regardless of how the eigenvalues are numbered, the eigenvectors must be numbered in a consistent manner.)In solving for the eigenvectors, we recognize that it will not be possible to obtain a unique solution. The eigenvectors are determined only to within a multiplicative constant. If we denote the components of each eigenvector as x1, x2, and x3, the solution for each eigenvector is found by substituting the corresponding eigenvalues into the equation (A – I)x = 0, and solving for the corresponding xi as the components of the eigenvector. This is shown for each eigenvector in thetable below.5x2 + 3x3 = 0x2 + 6x3 = 0 -2x3 = 0For 1 = 3, we see that the solution to the third equation is x3 = 0. With this solution, the second equation gives x2 = -6x3 = 0. We are left with no equation for x1, so we conclude that this can be any value and we denote it is a. Thus gives the first eigenvector as x(1) = [a 0 0]T.-x1 + 5x2 + 3x3 = 0 6x3 = 0 -2x3 = 0For 2 = 4, the second and third equation both give x3 = 0. The first equation gives x1 = 5x2 + 3x3 = 5x2. If we pick x2 to be an arbitrary quantity, say b, then x1 = 5b and the second eigenvector becomes x(2) = [5b b 0]T.2x1 + 5x2 + 3x3 = 0 3x2 + 6x3 = 0 0 = 0For 3 = 1, ere the third equation gives us no information, but the second equation tells us that 3x2 = -6x3. If we pick x3 = c (an arbitrary quantity), then x2 = 2c and the first equation gives x1 = (-5x2 – 3x3)/2 = (-5(-2c) – 3c)/2 = -7c/2. x(3) = [7c/2 -2c c]T.Page 379, problem 5 – Find the principal axes and the corresponding expansion (or compression) factors for the elastic deformation matrix, A, shown at the right.As shown in the text, this matrix describes a stretching where every old coordinate, x, moves to a new coordinate, y., such that y = Ax.1212123AEngineering Building Room 1333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.7062The principal directions for this deformation are the directions for which the position vector changes only in magnitude. Thus, the new position vector, y, has the same direction as the old position vector, x. This means that the two vectors are related by a multiplicative constant. E. g., y = x. This means that we have two equations: (1) the general equation y = Ax, for any coordinate direction and (2) y = x, in the principal coordinate direction. We can only satisfy these two equations if Ax = x for the principal coordinates. This gives an eigenvalues/eigenvector problem. We first solve Det(A – I) = 0, as shown below, to get the eigenvalues. 0125211231212123)(22IADetThe eigenvalues are found by the usual formula for the solution of a quadratic equation.21,2435)1(24162525)1(2)1)(1(42525212If we denote the components of each eigenvectors as x1, and x2, the solution for each eigenvectoris found by substituting the corresponding eigenvalues into the equation (A – I)x = 0, and solving for the corresponding xi as the components of the eigenvector. Here we have only two equations to solve, and one of the unknowns will be arbitrary. For 1 = 2, the two equations are02/2/21 xxand 02/21 xx. Both these equations give us 212xx . Thus, our first eigenvector is x(1) = [2a a]T, where a is arbitrary. For 2 = ½, the two equations are02/21 xxand 02/2/21 xx. Both these equations give us 122xx so that we can write our second eigenvector as x(2) = [b -2b]T.If we regard the x1 coordinate as lying along a horizontal axis and the x2 coordinate as lying alonga vertical axis, we can find the principal coordinate directions corresponding to the two eigenvalues as follows.oaaxx26.3561548.02tantan11)1(2)1(11 for 1 = 2obbxx74.54955317.02tantan11)2(2)2(12 for 2 = 1/2Here we find that the first principal direction is 35.26o with an expansion by a factor of 2; the second principal direction is –54.74o with a contraction by a factor of ½. Note that the numbering of “first” and “second” are arbitrary. Also, we see that when we are interested in only the direction of a vector, having a common factor in all components does not affect the result.Page 384, problem 4 – Is the matrix at the right symmetric, skew-symmetric or orthogonal? Find the eigenvalues of thematrix and show how this illustrates Theorems 1 on page 382 or Theorem 5 on page 384.)cos()sin(0)sin()cos(0001AThe matrix is not skew-symmetric since it has nonzero components on its principal diagonal. For a skew-symmetric matrix (A = –AT), all terms on the principal diagonal must be zero.Engineering Building Room 1333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.7062The matrix is almost symmetric, but a23 = -sin()  a32 = sin(). Thus the matrix is not symmetric.If the matrix is orthogonal, the inner product of each pair of rows must vanish.