Unformatted text preview: EE441 HW#8 Solution Spring 2007
Instructor : Dr. Jonckheere TA: Ben Raskob Due Date 3/29/2007 5.1.2
A= A  I = 1 1 2 4 1 1 2 4 = 2  5 + 6 = (  3)(  2). the eigenvalues are = 3, 2. Now find the eigenvectors: = 3 : [A  I]x1 = choose x1 = 1 2 = 2 : [A  I]x2 = choose x2 = 1 1 1 1 . Then S 1 = 2 1 = 1 1 . 2 1 1 1 2 2 x2 = 0 2 1 2 1 x1 = 0 Let S = [x1 x2 ] = Let D = 1 0 0 2 3 0 . Then A = SDS 1 and eAt = SeDt S 1 . 0 2 1 If we let c = S 1 u(0) = 1 1 2 1 0 6 = 6 6 1 2 then + 6e2t 1 1 u(t) = c1 e1 t x1 + c2 e2 t x2 = 6e3t . 2 ...
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 Spring '08
 Neely
 Eigenvalue, eigenvector and eigenspace, x1 x2, Dr. Jonckheere TA, Ben Raskob

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