# USA MA 237 - Linear Algebra MA237 (4 pages)

Previewing page*1*of 4 page document

**View the full content.**## Linear Algebra MA237

Previewing page
*1*
of
actual document.

**View the full content.**View Full Document

## Linear Algebra MA237

0 0 59 views

- Pages:
- 4
- School:
- University of South Alabama
- Course:
- Ma 237 - Linear Algebra I

**Unformatted text preview:**

Linear Algebra MA237 Fall 2009 Dr Byrne Homework for Section 5 1 Due 11 16 1 Determine whether the given vectors v1 v2 v3 v4 are eigenvectors of the given matrix If they are state the corresponding eigenvalue 1 A 2 2 4 1 2 1 0 v1 v2 v3 v4 2 1 1 0 1 2 1 5 1 2 2 0 1 2 1 3 1 2 0 0 v1 A v1 Eigenvector 5 2 4 2 10 v2 A v2 Eigenvector 0 2 4 1 0 v3 A v3 2 4 1 6 v4 A v 4 2 4 0 0 Not an eigenvector Not an eigenvector For the matrices below find a the characteristic polynomial b the real eigenvalues c the eigenvector and d eigenspace corresponding to each eigenvalue e Describe the eigenspace geometrically i e as two lines or something else 5 4 2 A 8 1 a characteristic polynomial 5 A I 4 8 1 1 0 0 5 1 4 5 1 8 0 1 32 0 2 6 27 0 b real eigenvalues 2 6 27 3 9 0 so 3 9 1 2 c eigenvectors 3 eigenvector is 9 eigenvector is 1 1 d eigenspaces 1 2 s and s 1 1 e eigenspaces geometrically are 2 lines 4 3 A 1 5 0 6 0 1 2 Hint 5 is an eigenvalue 0 a characteristic polynomial 3 2 2 29 30 0 b real eigenvalues 5 6 1 c eigenvectors 5 eigenvector 1 3 11 1 0 6 eigenvector 1 0 1 5 1 eigenvector 9 25 1 Hint use row reduction to solve the linear systems the eigenvectors are the spanning vectors of the solution the next problem is shown in more detail d eigenspaces s 1 3 and 11 1 0 s 1 and 0 e eigenspaces geometrically are 3 lines 1 5 s 9 25 1 0 4 A 1 1 1 0 1 1 1 0 a characteristic polynomial det A I det 1 1 1 1 1 0 1 2 1 1 1 1 1 3 1 1 3 3 2 3 3 2 b eigenvalues To factor polynomials of degree 3 first check possible rational roots 1 2 try 1 1 3 3 2 doesn t work try 1 1 3 3 2 works yields 2 2 2 1 roots are 1 2 and 1 c eigenvectors 1 Solve A I x 0 1 A I x 1 1 x t s y t z s 1 1 1 1 1 1 1 x 0 0 0 1 1 1 t 1 s 0 0 1 0 0 1 0 x y z 0 x y z but y and z are free 0 Two eigenvectors 2 Solve A I x 0 2 A I x 1 1 1 2 1 1 1 x 0 2 1 0 0 x y z 0 1 0 1 1 x z 0 y z 0 only z is free 0 s 1 s s 1 s 1 1 1 1 d two distinct eigenspaces s 1 and t 1 s 0 1 0 1 e the geometry of the eigenspaces is the union of a line and a plane

View Full Document