{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

la_m12_handouts - MAC 2103 Module 12 Eigenvalues and...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 1 MAC 2103 Module 12 Eigenvalues and Eigenvectors 2 Rev.F09 Learning Objectives Upon completing this module, you should be able to: 1. Solve the eigenvalue problem by finding the eigenvalues and the corresponding eigenvectors of an n x n matrix. Find the algebraic multiplicity and the geometric multiplicity of an eigenvalue. 2. Find a basis for each eigenspace of an eigenvalue. 3. Determine whether a matrix A is diagonalizable. 4. Find a matrix P, P -1 , and D that diagonalize A if A is diagonalizable. 5. Find an orthogonal matrix P with P -1 = P T and D that diagonalize A if A is symmetric and diagonalizable. 6. Determine the power and the eigenvalues of a matrix, A k . http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.
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 3 Rev.09 Eigenvalues and Eigenvectors http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Eigenvalues Eigenvalues , Eigenvectors, , Eigenvectors, Eigenspace Eigenspace, Diagonalization Diagonalization and and Orthogonal Orthogonal Diagonalization Diagonalization The major topics in this module: 4 Rev.F09 What are Eigenvalues and Eigenvectors? http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. If A is an n x n matrix and λ is a scalar for which A x = λ x has a nontrivial solution x ℜⁿ then λ is an eigenvalue of A and x is a corresponding eigenvector of A. A x = λ x is called the eigenvalue problem for A. Note that we can rewrite the equation A x = λ x = λ I n x as follows: λ I n x - A x = 0 or ( λ I n - A) x = 0 . x = 0 is the trivial solution. But our solutions must be nonzero vectors called eigenvectors that correspond to each of the distinct eigenvalues.
Image of page 2
3 5 Rev.F09 What are Eigenvalues and Eigenvectors? http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Since we seek a nontrivial solution to ( λ I n - A) x = ( λ I - A) x = 0 , λ I - A must be singular to have solutions x 0 . This means that the det( λ I - A) = 0. The det( λ I - A) = p( λ ) = 0 is the characteristic equation, where det( λ I - A) = p( λ ) is the characteristic polynomial. The deg(p( λ )) = n and the n roots of p( λ ), λ 1 , λ 2 , …, λ n , are the eigenvalues of A. The polynomial p( λ ) always has n roots, so the zeros always exist; but some may be complex and some may be repeated. In our examples, all of the roots will be real. For each λ i we solve for x i = p i the corresponding eigenvector, and A p i = λ i p i for each distinct eigenvalue. 6 Rev.F09 How to Solve the Eigenvalue Problem, A x = λ x ? http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Example 1: Find the eigenvalues and the corresponding eigenvectors of A. Step 1: Find the characteristic equation of A and solve for its eigenvalues. Each eigenvalue has algebraic multiplicity 1. A = ! 3 ! 2 ! 5 0 " # $ % & ' p ( ! ) = ! I " A = ! + 3 2 5 ! = 0 = ( ! + 3) ! " (2)(5) = ! 2 + 3 ! " 10 = ( ! + 5) 1 ( ! " 2) 1 = 0 Thus , the eigenvalues are !
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
Image of page 4
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