quiz3sol - Course 18.06, Fall 2002: Quiz 3, Solutions 1 (a)...

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Course 18.06, Fall 2002: Quiz 3, Solutions 1 (a) One eigenvalue of A = ones (5) is λ 1 = 5, corresponding to the eigenvector x 1 = (1 , 1 , 1 , 1 , 1). Since the rank of A is 1, all the other eigenvalues λ 2 , . . . , λ 5 are zero. Check: The trace of A is 5. (b) The initial condition u (0) can be written as a sum of the two eigenvectors x 1 = (1 , 1 , 1 , 1 , 1) and x 2 = ( - 1 , 0 , 0 , 0 , 1), corresponding to the eigenvalues λ 1 = 5 and λ 2 = 0: u (0) = (0 , 1 , 1 , 1 , 2) = (1 , 1 , 1 , 1 , 1) + ( - 1 , 0 , 0 , 0 , 1) = x 1 + x 2 . The solution to d u dt = A u is then u ( t ) = c 1 e λ 1 t x 1 + c 2 e λ 2 t x 2 = (1 , 1 , 1 , 1 , 1) e 5 t + ( - 1 , 0 , 0 , 0 , 1) . (c) The eigenvectors of B = A - I are the same as for A , and the eigenvalues are smaller by 1: B x = ( A - I ) x = A x - x = λ x - x = ( λ - 1) x , where x , λ are an eigenvector and an eigenvalue of A . The eigenvalues of B are then 4 , - 1 , - 1 , - 1 , - 1, the trace is i λ i = 0, and the determinant is Q i λ i = 4. 2
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quiz3sol - Course 18.06, Fall 2002: Quiz 3, Solutions 1 (a)...

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