Engineering Analysis

Engineering Analysis Chapter 6 Eigenvalues

we will be concerned with the equation Ax = λx. This equation occurs in many applications of linear algebra. If the equation has a nonzero solution x, then λ is said to be an eigenvalue of A and x is said to be an eigenvector belonging to λ. • Wherever there are vibrations, there are eigenvalues, the natural frequencies of the vibrations. If you have ever tuned a guitar, you have solved an eigenvalue problem. When engineers design structures, they are concerned with the frequencies of vibration of the structure. • The eigenvalues of a boundary value problem can be used to determine the energy states of an atom or critical loads that cause buckling in a beam.

6.1 Eigenvalues and Eigenvectors Many application problems involve applying a linear transformation repeatedly to a given vector. The key to solving these problems is to choose a coordinate system or basis that is in some sense natural for the operator and for which it will be simpler to do calculations involving the operator. With respect to these new basis vectors (eigenvectors) we associate scaling factors (eigenvalues) that represent the natural frequencies of the operator.

6.1 Eigenvalues and Eigenvectors Definition EXAMPLE 2 Let Solution Let A be an n×nmatrix. A scalar λ is said to be an eigenvalue or a characteristic Value of A if there exists a nonzero vector x such that A x = λ x. The vector x is saidto be an eigenvector or a characteristic vector belonging to λ. it follows that λ = 3 is an eigenvalue of A and x = (2, 1 is an eigenvector belonging to λ. Actually, any nonzero multiple of x will be an eigenvector, because

6.1 Eigenvalues and Eigenvectors For example, (4, 2 is also an eigenvector belonging to λ = 3 The equation Ax = λx can be written in the form Thus, λ is an eigenvalue of A if and only if (1) has a nontrivial solution.

6.1 Eigenvalues and Eigenvectors Equation (1) will have a nontrivial solution if and only if A − λI is singular, or, equivalently det(A − λI) =0 (2) If the determinant in (2) is expanded, we obtain an nth-degree polynomial in the variable λ: • This polynomial is called the characteristic polynomial, and equation (2) is called the characteristic equation, for the matrix A. • The roots of the characteristic polynomial are the eigenvalues of A. • If roots are counted according to multiplicity, then the characteristic polynomial will have exactly n roots • Thus, A will have n eigenvalues, some of which may be repeated and some of which may be complex numbers.

6.1 Eigenvalues and Eigenvectors Let A be an n ×n matrix and λ be a scalar. The following statements are equivalent: (a) λ is an eigenvalue of A. (b) (A − λI) x = 0 has a nontrivial solution. (c) N (A − λI) {0} (d) A − λI is singular. (e) det(A − λI) = 0 EXAMPLE 3 Find the eigenvalues and the corresponding eigenvectors of the matrix

6.1 Eigenvalues and Eigenvectors Solution The characteristic equation is Thus, the eigenvalues of A are λ1 = 4 and λ2 = −3. To find the eigenvectors belonging to λ1 = 4, we must determine the null space of A − 4I. Solving (A − 4I)x = 0, we get Hence, any nonzero multiple of (2, 1 is an eigenvector belonging to λ1, and {(2, 1 } is a basis for the eigenspace corresponding to λ1. Similarly, to find the eigenvectors for λ2, we must solve

6.1 Eigenvalues and Eigenvectors (A + 3I)x = 0 In this case, {(−1, 3 } is a basis for N(A + 3I) and any nonzero multiple of (−1, 3 is an eigenvector belonging to λ2. Example The linear system can be written in matrix form as Find the eigenvalues and corresponding eigenvectors of the matrix A

6.1 Eigenvalues and Eigenvectors Solution The characteristic equation of A is The factored form of this equation is , so the eigenvalues of A are . eigenvector of A (λI – A) x =0

6.1 Eigenvalues and Eigenvectors If λ = 2 then Solving this system yields =-t, = t the eigenvectors of A corresponding to λ = -5 solve this system

6.1 Eigenvalues and Eigenvectors Example Find the eigenvalues of Solution The characteristic polynomial of A is The eigenvalues of A must therefore satisfy the cubic equation Thus the eigenvalues of A are

6.1 Eigenvalues and Eigenvectors EXAMPLE 4 Let Find the eigenvalues and the corresponding eigenspaces Solution Thus, the characteristic polynomial has roots λ1 = 0, λ2 = λ3 = 1. The eigenspacecorresponding to λ1 = 0 is N(A), which we determine in the usual manner:

6.1 Eigenvalues and Eigenvectors Setting x3 = , we find that x1 = x2 = x3 = . Consequently, the eigenspace corresponding to λ1 = 0 consists of all vectors of the form (1, 1, 1)T . To find the eigenspacecorresponding to λ = 1, we solve the system (A − I)x = 0: Setting x2= and x3=β, we get x1=3−β. Thus, the eigenspace corresponding to λ = 1 consists of all vectors of the form

6.1 Eigenvalues and Eigenvectors Example Find bases for the eigenspacesof Solution The characteristic equation of matrix A is , or, in factored form • thus the eigenvalues of A are =1 and =2 , so there are two eigenspaces of A. is an eigenvector of A corresponding to if and only if x is a nontrivial solution of —that is, of (3)

6.1 Eigenvalues and Eigenvectors If , λ = 2 then 3 becomes Solving this system using Gaussian elimination yields Since are linearly independent, these vectors form a basis for the eigenspace corresponding to λ = 2 .

6.1 Eigenvalues and Eigenvectors If λ = 1 , then 3 becomes Solving this system yields Thus the eigenvectors corresponding to λ = 1are the nonzero vectors of the form is a basis for the eigenspace corresponding to λ = 1. .

6.1 Eigenvalues and Eigenvectors • EXAMPLE Let Find the eigenvalues and the corresponding eigenvector Solution

6.1 Eigenvalues and Eigenvectors

6.1 Eigenvalues and Eigenvectors EXAMPLE 5 Let Compute the eigenvalues of A and find bases for the corresponding eigenspaces. Solution The roots of the characteristic polynomial are λ1 = 1 + 2i, λ2 = 1 − 2i. It follows that {(1, i } is a basis for the eigenspace corresponding to λ1 = 1 + 2i. Similarly, and {(1,−i } is a basis for N(A − λ2I).

6.1 Eigenvalues and Eigenvectors A square matrix A is invertible if and only if λ= 0 is not an eigenvalue of A.

6.1 Eigenvalues and Eigenvectors Complex Eigenvalues If A is an n × n matrix with real entries, then the characteristic polynomial of A will have real coefficients, and hence all its complex roots must occur in conjugate pairs. Thus, if λ = a + bi (b 0) is an eigenvalue of A, then = a − bi must also be an eigenvalue of A. The sum of the diagonal elements of A is called the trace of A and is denoted by tr(A).

6.1 Eigenvalues and Eigenvectors EXAMPLE 6 If Then det(A) = −5 + 18 = 13 and tr(A) = 5 − 1 = 4 The characteristic polynomial of A is given by and hence the eigenvalues of A are λ1 = 2 + 3i and λ2 = 2 − 3i. Note that

6.1 Eigenvalues and Eigenvectors If A is an n x n triangular matrix (upper triangular, lower triangular, or diagonal), then the eigenvalues of A are the entries on the main diagonal of A Example By inspection, the eigenvalues of the lower triangular matrix

6.1 Eigenvalues and Eigenvectors Similar Matrices We close this section with an important result about the eigenvalues of similar matrices. Recall that a matrix B is said to be similar to a matrix A if there exists a nonsingular matrix S such that B = AS. Theorem 6.1.1 Let A and B be n×n matrices. If B is similar to A, then the two matrices have the same characteristic polynomial and, consequently, the same eigenvalues. EXAMPLE 7 Given

6.1 Eigenvalues and Eigenvectors It is easily seen that the eigenvalues of T are λ1 = 2 and λ2 = 3. If we set A = TS, then the eigenvalues of A should be the same as those of T. We leave it to the reader to verify that the eigenvalues of this matrix are λ1 = 2 and λ2 = 3

6.2 Systems of Linear Differential Equations A differential equation is an equation that involves an unknown function and its derivatives. An important, simple example of a differential equation is where r is a constant. The idea here is to find a function x(t) that will satisfy the given differential equation. This differential equation is discussed further subsequently. linear algebra is helpful in the formulation and solution of differential equations. Homogeneous Linear Systems We consider the first-order homogeneous linear system of differential equations,

6.2 Systems of Linear Differential Equations We can write above eq. in matrix form by letting

6.2 Systems of Linear Differential Equations Eigenvalues play an important role in the solution of systems of linear differential equations. In this section, we see how they are used in the solution of systems of linear differential equations with constant coefficients. We begin by considering systems of first-order equations of the form

6.2 Systems of Linear Differential Equations then the system can be written in the form Let us consider the simplest case first. When n = 1, the system is simply y Clearly, any function of the form

6.2 Systems of Linear Differential Equations satisfies equation (1). A natural generalization of this solution for the case n > 1 is to take where x = (x1, x2, . . . , xn. To verify that a vector function of this form does work, we compute the derivative Now, if we choose λ to be an eigenvalue of A and x to be an eigenvector belonging to λ, then

6.2 Systems of Linear Differential Equations Hence, Y is a solution of the system. Thus, if λ is an eigenvalue of A and x is an eigenvector belonging to λ, then x is a solution of the system = AY. This will be true whether λ is real or complex. EXAMPLE 1 Solve the system Solution is a solution of the system.

6.2 Systems of Linear Differential Equations ExampleFor the system the matrix has eigenvalues λ1 = 2 and λ2 = 3 with respective associated eigenvectors These eigenvectors are automatically linearly independent, since they are associated with distinct eigenvalues Hence the general solution to the given system is

6.2 Systems of Linear Differential Equations In terms of components, this can be written as Example Consider the following homogeneous linear system of differential equations: Solution The characteristic polynomial of A is

6.2 Systems of Linear Differential Equations so the eigenvalues of A are λ1 =1, λ2=2, and λ3=4. Associated eigenvectors are respectively. The general solution is then given by where b1, b2, and b3 are arbitrary constants.

6.2 Systems of Linear Differential Equations EXAMPLE For the linear system of previous Example, solve the initial value problem determined by the initial conditions x1(0)=4, x2(0)= 6, and x3(0) = 8. Solution We write our general solution in the form x = Pu as or Therefore, the solution to the initial value problem is

6.2 Systems of Linear Differential Equations Complex Eigenvalues Let A be a real n × n matrix with a complex eigenvalue λ = a + bi, and let x be an eigenvector belonging to λ. The vector x can be split up into its real and imaginary parts Since the entries of A are all real, it follows that = a − bi is also an eigenvalue of A with eigenvector

6.2 Systems of Linear Differential Equations and hence x and are both solutions of the first-order system = AY. Any linear combination of these two solutions will also be a solution. Thus, if we set then the vector functions Y1 and Y2 are real-valued solutions of = AY. Taking the real and imaginary parts of we see that

6.2 Systems of Linear Differential Equations EXAMPLE 2 Solve the system Solution Let

6.2 Systems of Linear Differential Equations

6.2 Systems of Linear Differential Equations Higher Order Systems Given a second-order system of the form we may translate it into a first-order system by setting

6.2 Systems of Linear Differential Equations and • then • and The equations can be combined to give the 2n × 2n first-order system

6.2 Systems of Linear Differential Equations EXAMPLE 3 Solve the initial value problem Solution

6.2 Systems of Linear Differential Equations The coefficient matrix for this system, has eigenvalues Corresponding to these eigenvalues are the eigenvectors

6.2 Systems of Linear Differential Equations Thus, the solution will be of the form We can use the initial conditions to find c1, c2, c3, and c4. For t = 0, we have or, equivalently,

6.2 Systems of Linear Differential Equations The solution of this system is c = (2, 1, 1, 0, and hence the solution to the initial value problem is Therefore,

6.2 Systems of Linear Differential Equations In general, if we have an mth-order system of the form where each Ai is an n×n matrix, we can transform it into a first-order system by setting We will end up with a system of the form

Engineering Analysis