Introduction to the Modeling and Analysis of Complex Systems

82 CHAPTER 5. DISCRETE-TIME MODELS II: ANALYSISdynamics can be described asx t = Ax t−1 , (5.33)where x is the state vector of the system and A is the coefficient matrix. Technically, youcould also add a constant vector to the right hand side, such asx t = Ax t−1 + a, (5.34)but this can always be converted into a constant-free form by adding one more dimension,i.e.,⎛ ⎞ ⎛ ⎞ ⎛ ⎞y t = ⎝ x t⎠ = ⎝ A a⎠ ⎝ x t−1⎠ = By t−1 . (5.35)10 1 1Therefore, we can be assured that the constant-free form of Eq. (5.33) covers all possiblebehaviors of linear difference equations.Obviously, Eq. (5.33) has the following closed-form solution:x t = A t x 0 (5.36)This is simply because A is multiplied to the state vector x from the left at every time step.Now the key question is this: How will Eq. (5.36) behave when t → ∞? In studyingthis, the exponential function of the matrix, A t , is a nuisance. We need to turn it intoa more tractable form in order to understand what will happen to this system as t getsbigger. And this is where eigenvalues and eigenvectors of the matrix A come to play avery important role. Just to recap, eigenvalues λ i and eigenvectors v i of A are the scalarsand vectors that satisfyAv i = λ i v i . (5.37)In other words, throwing at a matrix one of its eigenvectors will “destroy the matrix” andturn it into a mere scalar number, which is the eigenvalue that corresponds to the eigenvectorused. If we repeatedly apply this “matrix neutralization” technique, we getA t v i = A t−1 λ i v i = A t−2 λ 2 i v i = . . . = λ t iv i . (5.38)This looks promising. Now, we just need to apply the above simplification to Eq. (5.36).To do so, we need to represent the initial state x 0 by using A’s eigenvectors as the basisvectors, i.e.,x 0 = b 1 v 1 + b 2 v 2 + . . . + b n v n , (5.39)