{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw8 - Week 8 1 6.1 1st order systems of DE(briefly 2 5.4...

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Week 8: 1. 6.1 1st order systems of DE (briefly!) 2. 5.4 Eigenvalues and Eigenvectors 3. 5.5 Eigenspaces and Diagonalization
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 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 .
Image of page 2
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 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) .
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4 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
Image of page 4
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 .
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern