*This preview shows
pages
1–2. Sign up
to
view the full content.*

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **M346 Second Midterm Exam, October 23, 2009 1) Diagonalization: a)(10 pts) Find the eigenvalues and eigenvectors of A = parenleftbigg 3 8 2- 3 parenrightbigg Since the trace is zero and the determinant is -25, the eigenvalues are 5 . The eigenvectors are parenleftbigg 4 1 parenrightbigg and parenleftbigg 1- 1 parenrightbigg . b) (10 pts) Compute e Bt , where B = parenleftbigg 1 2 2 4 parenrightbigg . Since the trace is 5 and the determinant is zero, the eigenvalues are and 5 . Computing the eigenvectors we get P = parenleftbigg 2 1- 1 2 parenrightbigg , P- 1 = 1 5 parenleftbigg 2- 1 1 2 parenrightbigg , e Bt = Pe Dt P- 1 = 1 5 parenleftbigg 4 + e 5 t- 2 + 2 e 5 t- 2 + 2 e 5 t 1 + 4 e 5 t parenrightbigg . c) (10 pts) Find the eigenvalues of C = 2 2 3 7 2 5 2 8 6 16 4- 6 . You do not need to find the eigenvectors. This is block triangular. The upper left block is just B + I , with eigenvalues 1 and 6. The lower right block is 2 A , with eigenvalues 10 and -10, so C has eigenvalues 1, 6, 10, and -10. 2. Consider the system of equations x ( n +1) = A x ( n ), where A = parenleftbigg 1 2- 1 parenrightbigg , with initial condition x (0) = parenleftbigg- 1 5 parenrightbigg . a) (15 pts) Find x ( n ) for all n . Be as explicit as possible. The eigenvalues of A are 1 and- 2 , with eigenvectors b 1 = parenleftbigg 1 1 parenrightbigg and b 2 = parenleftbigg 1- 2 parenrightbigg . Since x (0) = b 1-...

View
Full
Document