Linear Algebra, 2020a

Topic: Linear Recurrences 477
We can find the dimension of S. Where k is the order of the recurrence, consider
this map from the set of functions S to the set of k-tall vectors.
⎛ ⎞
f(0)
f(1)
f ↦→
⎜ ⎟
⎝ . ⎠
f(k − 1)
Exercise 4 shows that this is linear. Any solution of the recurrence is uniquely
determined by the k-many initial conditions so this map is one-to-one and onto.
Thus it is an isomorphism, and S has dimension k.
So we can describe the set of solutions of our linear homogeneous recurrence
relation of order k by finding a basis consisting of k-many linearly independent
functions. To produce those we give the recurrence a matrix formulation.
⎞
a n−1 a n−2 a n−3 ... a n−k+1 a n−k
⎛
⎛
⎞
f(n)
f(n − 1)
⎜
⎝
⎟
. ⎠ = ⎜
f(n − k + 1) ⎝
1 0 0 ... 0 0
0 1 0
0 0 1
.
.
. ..
. ..
0 0 0 ... 1 0
⎛ ⎞
f(n − 1)
f(n − 2)
⎜
⎝
⎟
. ⎠
⎟
⎠ f(n − k)
Call the matrix A. We want its characteristic function, the determinant of
A − λI. The pattern in the 2×2 case
(
)
a n−1 − λ a n−2
= λ 2 − a n−1 λ − a n−2
1 −λ
and the 3×3 case
⎛
⎞
a n−1 − λ a n−2 a n−3
⎜
⎝
1 −λ 0
0 1 −λ
⎟
⎠ =−λ 3 + a n−1 λ 2 + a n−2 λ + a n−3
leads us to expect, and Exercise 5 verifies, that this is the characteristic equation.
a n−1 − λ a n−2 a n−3 ... a n−k+1 a n−k
1 −λ 0 ... 0 0
0 1 −λ
0 =
0 0 1
.
.
.
.. . ..
∣ 0 0 0 ... 1 −λ ∣
= ±(−λ k + a n−1 λ k−1 + a n−2 λ k−2 + ···+ a n−k+1 λ + a n−k )