week08 - 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 Rn(or

# week08 - Week 8 1 6.1 1st order systems of DE(briefly 2 5.4...

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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 λ of an n -by- n matrix A if Av = λv. We re-write 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 non-trivial 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 ,
2 A - 4 I = 3 - 4 - 1 - 5 - 1 - 4 = - 1 - 1 - 5 - 5 1 1 0 0 , 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) . ————— For λ 2 = - 2 , we start over with coef A + 2 I = 3 + 2 - 1 - 5 - 1 + 2 5 - 1 0 0 . 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 0 2 0 - 8 12 - 6 . Solution. Since A is 3-by-3 , 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 ? We’re looking for the roots, so we need the factors ( λ - λ 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 ( λ ) = (2 - λ ) fl fl fl fl 10 - λ 8 - 8 - 6 - λ fl fl fl fl = (2 - λ )[(10 - λ )( - 6 - λ ) + 64] = (2 - λ )[ λ 2 - 4 λ - 60 + 64] = - ( λ - 2) 3 .
3 Can we agree that P ( λ ) = - λ 3 + 12 λ 2 - 24 λ + 8 , [which we would have gotten from the other methods] is not as useful as having a factor [since the other factors only need a quadratic; not solving a cubic]?
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