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

Define a sequence x_{0}, x_{1}, ... by the initial conditions x_{0} = a, x_{1} = b and x_{2} = c together with the recursion relation for n = 0, 1, 2, ... (b) If the matrix A from part (a) is defined in MATLAB/Octave, we can do the following calculations >eig(A) ans = 1.83929 + 0.00000i 0.41964 + 0.60629i 0.41964  0.60629i >abs(eig(A)) ans = 1.83929 0.73735 0.73735 Describe how you could make further use of the eig command and other MATLAB/Octave commands to determine all (possibly complex) initial values a, b and c for which as . 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

First, we notice that the recursion relation X_{n+1} = AX_{n} implies that 
Hint 2 

Write the vector of initial conditions X_{0} as a linear combination of the eigenvectors, then apply A^{n+1} to each of those. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. First, we notice that the recursion relation X_{n+1} = AX_{n} implies that Since A is a 3×3 matrix with three distinct eigenvalues λ_{1} = 1.83929, λ_{2} = 0.41964 + 0.60629i, λ_{3} = 0.41964 + 0.60629i there is a basis of corresponding eigenvectors v_{1}, v_{2}, v_{3}. Let us rewrite X_{0} with respect to this basis: Then Now we see that the last two terms will vanish as n goes to infinity, because and . However, since 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, lies in the plane that is spanned by v_{2} and v_{3} but has no component in direction of v_{1}, then the sequence will converge to zero: x_{n} → 0 as n → ∞. Eigenvectors v_{2} and v_{3} 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 