Question

To begin investigating the long-term behavior of the system, we will assume that A is an n×n matrix with n real linearly independent eigenvectors v 1​,…,v n​ . Furthermore, assume that the corresponding real eigenvalues of A satisfy the relationship ∣λ 1​ ∣>∣λ 2​ ∣≥⋯≥∣λ n​ ∣ Consider an initial vector x . Explain why there exist constants c 1​ ,…,c n​ such that x (0) =c 1​ v 1 +c 2​ v 2 +⋯+c n​ v n

​ and show that Ax (0) =c 1​ λ 1​ v 1 +c 2 λ 2​ v 2​ +⋯+c n​ λ n​ v n​ Furthermore, show that x (k) =A k x (0) =c 1 λ 1k​ v 1 +c 2​λ 2k​ v 2​ +⋯+c n​ λ nk​ v n
​

Answer 1

There exist constants c₁,...,cₙ such that x(0) = c₁v₁ + c₂v₂ + ⋯ + cₙvₙ and Ax(0) = c₁λ₁v₁ + c₂λ₂v₂ + ⋯ + cₙλₙvₙ. Furthermore, x(k) = Aᵏx(0) = c₁λ₁ᵏv₁ + c₂λ₂ᵏv₂ + ⋯ + cₙλₙᵏvₙ.

Step-by-step explanation:

To explain why there exist constants c₁,...,cₙ such that x(0) = c₁v₁ + c₂v₂ + ⋯ + cₙvₙ and to show that Ax(0) = c₁λ₁v₁ + c₂λ₂v₂ + ⋯ + cₙλₙvₙ, we can use the fact that eigenvectors are linearly independent. Since v₁,...,vₙ are linearly independent eigenvectors of A, any vector x can be expressed as a linear combination of v₁,...,vₙ, meaning x(0) = c₁v₁ + c₂v₂ + ⋯ + cₙvₙ for some constants c₁,...,cₙ. Furthermore, since λ₁,...,λₙ are the eigenvalues corresponding to v₁,...,vₙ, we have Ax(0) = c₁λ₁v₁ + c₂λ₂v₂ + ⋯ + cₙλₙvₙ. Therefore, x(0) can be expressed as a linear combination of eigenvectors of A and Ax(0) can be expressed as a linear combination of eigenvectors multiplied by their corresponding eigenvalues.

Now, to show that x(k) = Aᵏx(0) = c₁λ₁ᵏv₁ + c₂λ₂ᵏv₂ + ⋯ + cₙλₙᵏvₙ, we can use the fact that Aᵏv = λᵏv for any eigenvector v. Therefore, Aᵏx(0) = Aᵏ(c₁v₁ + c₂v₂ + ⋯ + cₙvₙ) = c₁Aᵏv₁ + c₂Aᵏv₂ + ⋯ + cₙAᵏvₙ = c₁λ₁ᵏv₁ + c₂λ₂ᵏv₂ + ⋯ + cₙλₙᵏvₙ, since Aᵏv₁ = λ₁ᵏv₁, Aᵏv₂ = λ₂ᵏv₂, etc.