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Unformatted text preview: Week 8: 1. 6.1 1st order systems of DE (briefly!) 2. 5.4 Eigenvalues and Eigenvectors 3. 5.5 Eigenspaces and Diagonalization 2 A vector v = 0 in R n (or in C n ) is an eigenvector with eigenvalue λ of an nby n matrix A if Av = λv. We rewrite the vector equation as ( A λI n ) v = 0 , which is a homogeneous system with coef matrix ( A λI ) , and we want λ so that the system has a nontrivial solution. We see that the eigenvalues are the roots of the characteristic polynomial P ( λ ) = 0 , where P ( λ ) = det( A λI ) . To find the eigenvectors we find the distinct roots λ = λ i , and for each i solve ( A λ i I ) v = 0 . Problem 1. Find the eigenvalues and eigenvectors of A = 3 1 5 1 ¶ Solution. P ( λ ) = det ( A λI 2 ) = fl fl fl fl 3 λ 1 5 1 λ fl fl fl fl = ( 1 λ )(3 λ ) 5 = λ 2 2 λ 3 5 = λ 2 2 λ 8 = ( λ 4)( λ + 2) , so λ 1 = 4 , λ 2 = 2 . For λ 1 = 4 , we reduce the coef matrix of the system ( A 4 I ) x = 0 , A 4 I = 3 4 1 5 1 4 ¶ = 1 1 5 5 ¶ → 1 1 ¶ , so for x 2 = a we get x 1 = a, and the eigenvectors for λ 1 = 4 are the vectors ( x 1 , x 2 ) = a ( 1 , 1) with a = 0 , from the space with basis ( 1 , 1) . 4 For λ 2 = 2 , we start over with coef A + 2 I = 3 + 2 1 5 1 + 2 ¶ → 5 1 ¶ . To find a spanning set without fractions, we anticipate that finding x 1 will use division by 5, and take x 2 = 5 b. Then 5 x 1 x 2 = 0 gives 5 x 1 = 5 b, so x 1 = b, and ( x 1 , x 2 ) = b (1 , 5) , spanned by (1 , 5) [or, if you must, ( 1 5 , 1)]. Problem 2. Find the eigenvalues and eigenvectors of A = 10 12 8 2 8 12 6 . Solution. Since A is 3by3 , the characteristic polynomial P ( λ ) is cubic, and we can make this problem difficult by neglecting to think strategically. How should P ( λ ) be given; for example, do we need the coef of λ 2 ?...
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This note was uploaded on 02/29/2008 for the course MATH 205 taught by Professor Zhang during the Spring '08 term at Lehigh University .
 Spring '08
 zhang
 Linear Algebra, Algebra, Eigenvectors, Vectors

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