Let u and v be eigenvectors of a matrix A, with corresponding eigenvalue λ and μ , and let c 1 and c 2 be scalars. Define:

x k = c 1 λ k u + c 2 μ k v   k = 1,2 , 3 …

  1. What is x k + 1 , by definition?
  2. Compute A x k from the formula for x k , and show that A x k = x k + 1 . This calculation will prove that the sequence x k defined above satisfies the difference equation x k + 1 = A x k   k = 1,2 , 3 …

Passo 1

Faaaaaala, galera... Tranquileba??

Temos resposta dada no estado, isto é:

x k + 1 = c 1 λ k + 1 u + c 2 μ k + 1 v   k = 1,2 , 3 …

Passo 2

Para b)

A x k = A x k = c 1 λ k u + c 2 μ k v

c 1 λ k A u + c 2 μ k A v

Sabendo que:

A u = λ u

A v = μ v

Podemos fazer:

c 1 λ k A u + c 2 μ k A v =

c 1 λ k λ u + c 2 μ k μ v

Assim podemos fazer:

c 1 λ k λ u + c 2 μ k μ v = c 1 λ k + 1 + c 2 μ k + 1 v

E desde que u e v   sejam autovetores:

c 1 λ k + 1 + c 2 μ k + 1 v = x k + 1

Resposta

a)

x k + 1 = c 1 λ k + 1 u + c 2 μ k + 1 v   k = 1,2 , 3 …