Linear Algebra

Topic: Linear Recurrences 421So we can describe the set of solutions of our linear homogeneous recurrencerelation of degree k (again, without any initial conditions) by taking linearcombinations of a set having only k-many linearly independent functions.To produce those equations we give the recurrence f(n + 1) = a n f(n) + · · · +a n−k f(n − k) a matrix formulation.⎛⎞a n a n−1 a n−2 . . . a n−k+1 a n−k1 0 0 . . . 0 00 1 00 0 1⎜.⎝ . ... .0 0 0 . . . 1 0⎛ ⎞ ⎛⎞f(n) f(n + 1)f(n − 1)⎜ ⎟⎝ . ⎠ = f(n)⎜⎟⎝ . ⎠⎟⎠ f(n − k) f(n − k + 1)We want the characteristic function of the matrix, the determinant of A − λIwhere the above matrix is A. The pattern in the 2×2 case()a n − λ a n−1= λ 2 − a n λ − a n−11 −λand the 3×3 case⎛⎞a n − λ a n−1 a n−2⎜⎝1 −λ 00 1 −λ⎟⎠ = −λ 3 + a n λ 2 + a n−1 λ + a n−2leads us to expect (and Exercise 4 verifies) that this is the characteristic equation.a n − λ a n−1 a n−2 . . . a n−k+1 a n−k1 −λ 0 . . . 0 00 1 −λ0 0 1. . ... .∣ 0 0 0 . . . 1 −λ ∣= ±(−λ k + a n λ k−1 + a n−1 λ k−2 + · · · + a n−k+1 λ + a n−k )The ± is irrelevant to find the roots so we will drop it. We say that thepolynomial is ‘associated’ with the recurrence relation.If −λ k + a n λ k−1 + a n−1 λ k−2 + · · · + a n−k+1 λ + a n−k has no repeated rootsthen the matrix is diagonalizable and we can, in theory, get a formula for f(n)as in the Fibonacci case. But because we know that the subspace of solutionshas dimension k, we do not need to do the diagonalization calculation providedthat we can exhibit k linearly independent functions satisfying the relation.Where r 1 , r 2 , . . . , r k are the distinct roots, consider the functions f r1 (n) = r n 1through f rk (n) = r n kof powers of those roots. Exercise 5 shows that each is asolution of the recurrence and that they form a linearly independent set. So ifthe roots r 1 , . . . , r k of the associated polynomial are distinct then any solution