Consider the length 3 recurrence xk+3 = axk + bxk+1 + cxk+2. If is any eigenvalue
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xk+3 = axk + bxk+1 + cxk+2.
If λ is any eigenvalue of A, show that X = [1 λ λ2]T is a λ-eigenvector.
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We have so c A x x 3 a bx cx ...View the full answer
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