Question

38. Generalized Eigenvectors Suppose that we wish to ex- tend the method described for finding one generalized eigenvector to finding two (or more) generalized eigen- vectors. Let's look at the case where A has multiplicity 3 but has only one linearly independent eigenvector . First, we find ui by the method described in this section. Then we find u2 such that

Generalized Eigenvectors problem, Differential Equations and Linear Algebra section 6.2, problem 38

Please solve all 3 parts, thanks!

Answer 1

Given that two generalized eigen vectors at the case where lambda has multiplicity y 3 only one linearly independent eigen vector v a) AV = lambda v (A - lambda I) = v_2 = v (A - lambda U) (A - lambda I) u_2 = v rearrange the above equation Av = lambda v A v_1 = lambda u_1 + v A v_2 = lambda v_2 + v_1 it is choin of generalized eigen vectors of length x = A x x_1 = A_1 = lambda x_1 X_1 = v e lambda t 'v' an arbitrary constant \r\n x_2' = lambda x_2 + x_1 x_2' - lambda x_2 = v e lambda t using non - homogeneous out method for solving the above equation x_2 = v_1 e^lambda t + t v e^lambda t ,where v_1 is an arbitrary constant x_3' = lambda x_3 + x_2 x_3' = lambda x_3 + v_1 e^lambda t + c e^lambda t solve again using this method for solving non - homogeneous O D E equation x_3 = v_2 e^lambda t + v_1 t e e^lambda t