week08 - Week 8: 1. 6.1 1st order systems of DE (briefly!)...

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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 A vector v = 0 in R n (or in C n ) is an eigenvector with eigenvalue of an n-by- n matrix A if Av = v. We re-write the vector equation as ( A- I n ) v = 0 , which is a homogeneous system with coef matrix ( A- I ) , and we want so that the system has a non-trivial solution. We see that the eigenvalues are the roots of the characteristic polynomial P ( ) = 0 , where P ( ) = det( A- I ) . To find the eigenvectors we find the distinct roots = i , and for each i solve ( A- i I ) v = 0 . Problem 1. Find the eigenvalues and eigenvectors of A = 3- 1- 5- 1 Solution. P ( ) = det ( A- I 2 ) = fl fl fl fl 3- - 1- 5- 1- fl fl fl fl = (- 1- )(3- )- 5 = 2- 2 - 3- 5 = 2- 2 - 8 = ( - 4)( + 2) , so 1 = 4 , 2 =- 2 . For 1 = 4 , we reduce the coef matrix of the system ( A- 4 I ) x = 0 , 2 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 1 = 4 are the vectors ( x 1 , x 2 ) = a (- 1 , 1) with a = 0 , from the space with basis (- 1 , 1) . For 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 3-by-3 , the characteristic polynomial P ( ) is cubic, and we can make this problem difficult by neglecting to think strategically. How should P ( ) be given; for example, do we need the coef of 2 ? Were looking for the roots, so we need the factors ( - i ). For that objective there is only one good method of computing the determinant; expansion on the 2nd row. Look why the 2nd row exp is best: we get P ( ) = (2- ) fl fl fl fl 10- 8- 8- 6- fl fl fl fl = (2- )[(10- )(- 6- ) + 64] = (2- )[ 2- 4...
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week08 - Week 8: 1. 6.1 1st order systems of DE (briefly!)...

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