Science:Math Exam Resources/Courses/MATH307/December 2010/Question 05 (b)/Solution 1

First, we notice that the recursion relation X_n+1 = AX_n implies that

$X_{n+1}=A{X}_n=A(A{X}_{n-1})=A(A(A{X}_{-2}))=\dots =A^{n+1}{X}_0$

Since A is a 3×3 matrix with three distinct eigenvalues λ₁ = 1.83929, λ₂ = -0.41964 + 0.60629i, λ₃ = -0.41964 + 0.60629i there is a basis of corresponding eigenvectors v₁, v₂, v₃. Let us rewrite X₀ with respect to this basis:

$X_0=\alpha v_1+\beta v_2+\gamma v_3$

Then

$\begin{array}{rl}{X}_{n+1}& =A^{n+1}{X}_0=A^{n+1}(\alpha v_1+\beta v_2+\gamma v_3)\\ & =\alpha A^{n+1}{v}_1+\beta A^{n+1}{v}_2+\gamma A^{n+1}{v}_3\\ & =\alpha {\lambda }_1^{n+1}{v}_1+\beta {\lambda }_2^{n+1}{v}_2+\gamma {\lambda }_3^{n+1}{v}_3\end{array}$

Now we see that the last two terms will vanish as n goes to infinity, because $|{\lambda }_2|<1$ and $|{\lambda }_3|<1$. However, since $|{\lambda }_1|>1$ the norm of the first vector will blow up as n approaches infinity, unless α = 0.

This is the key to this question: As long as the vector with the initial conditions, $X_0=\left[\begin{array}{c}c\\ b\\ a\end{array}\right]$ lies in the plane that is spanned by v₂ and v₃ but has no component in direction of v₁, then the sequence will converge to zero: x_n → 0 as n → ∞. Eigenvectors v₂ and v₃ can be obtained in MATLAB/Octave with

   [V,D] = eig(A);  % V is a matrix with eigenvectors in its columns
   v_2 = V[:,2];
   v_3 = V[:,3];

Then the initial values that imply x_n → 0 as n → ∞ are

$\{\left[\begin{array}{c}c\\ b\\ a\end{array}\right]\text{ such that }\left[\begin{array}{c}c\\ b\\ a\end{array}\right]=s{v}_2+t{v}_3\text{ for some }s,t\in \mathbb{ℂ}\}$