Unformatted text preview: ) / 2 (3 × 2 n + 4 n ) / 2 ‚ . Hence x n = 1 2 (3 × 2 n4 n ) and y n = 1 2 (3 × 2 n + 4 n ). 5. Let A = • a b c d ‚ be a real or complex matrix with distinct eigenvalues λ 1 , λ 2 and corresponding eigenvectors X 1 , X 2 . Also let P = [ X 1  X 2 ]. (a) The system of recurrence relations x n +1 = ax n + by n y n +1 = cx n + dy n 67...
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 Fall '10
 Dreibelbis
 Eigenvalue, eigenvector and eigenspace, Eigenvalue algorithm, Xn

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