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Unformatted text preview: A]{ x} − λ [ I ]{ x} = {0} [ A] − λ [ I ] { x} = {0} At this point, we recognize the form as a null space solution problem. There are three ways in which the equation can be satisfied, two of them, [ A] − λ [ I ] = [ 0] and is singular (or rank deficient) for certain values of λ . These values of λ are the eigenvalues of the matrix [ A] . To solve for the λ ’s, recognize that a singular (or rank deficient) matrix has a determinant of zero. [ A] − λ [ I ] = 0 { x} = {0} , are trivial solutions and are rejected. The third is that the matrix [ A] − λ [ I ] Lecture Notes 14 06/16/06 12:25 PM Mechanical Vibrations I Expanding this determinant results in a polynomial, known as the characteristic polynomial, whose roots are the eigenvalues λ . After identifying the eigenvalues, go back and solve the linear equations, [ A] − λ [ I ] { x} = {0} , for each eigenvalue λ . However there is a problem, since the coefficient matrix is singular (or rank deficient) when evaluated at the eigenvalues, we cannot simply solve the equation set. In general, the matrix will be rank deficient by one, indicating that one variable is free, so the most straightforward solution approach is to simply set one of the vector { x} elements equal to one and solve as normal. (We will not get into the issues involved in the repeated root problem. In other words, where two or more of the eigenvalues λ are equal.) Lecture Notes 15 06/16/06 12:25 PM Mechanical Vibrations I Eigenvalue/Eigenvector Example
Given the following matrix, find its associated eigenvalues and eigenvectors. [ A] = First identify the eigenvalues, 4 −1 −2 5 [ A] − λ [ I ] = −1 4 −1 1 0 4 − λ −2 5 − λ 0 1 = −2 5 − λ = 0 −1 4 − λ −2 5 − λ = (4 − λ )(5 − λ ) − (−1)(−2) = 0 λ 2 − 9λ + 18 = 0 Which has the roots, λ1 = 3 and λ2 = 6 . Next identify the eigenvector associated with each eigenvalue. [ A] − λ [ I ] { x} = {0} 4 −1 1 0 x 1 4 − λ −1 x 1 0 − λ 0 1 x = −2 5 − λ x = 0 2 2 −2 5 For the first eigenvalue, λ1 = 3 ,
−1 x1 0 4 − λ1 = −2 5 − λ1 x 2 0 4 − 3 −1 x1 1 −1 x1 0 −2 5 − 3 x = − 2 2 x = 0 2 2 Notice the rank deficient nature of the resulting coefficient matrix. Set one element of the vector { x} equal to one (unity) to solve. For this example, we will use the first element. (Note that we could actually set it equal to any arbitrary value except zero and still solve. Occasionally an eigenvector will contain a zero, but we cannot arbitrarily pick that value. Finally, if after picking a value, the resulting equations cannot be solved, simply try again picking a different element.) Letting x1 = 1 , Lecture Notes 16 06/16/06 12:25 PM Mechanical Vibrations I 1 −1 1 0 −2 2 x = 0 2 We can solve either equation for x2 . Let’s pick the first for simplicity. [1 1 −1] = 1 − x2 = 0 x2 Which yields, x2 = 1 . Therefore, the first eigenvalue/eigenvector pair is: 1 λ1 = 3 , { x}1 = 1 Repeating t...
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 Fall '11
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