Page 293 - A Course in Linear Algebra with Applications

8.2: Systems of Linear Recurrences 277

At first sight it may not seem clear how eigenvalues enter
into this problem. However, let us put the system of linear
recurrences in matrix form by writing

x a n d 4 =
(
- = -J ' (i ~'i
Then the two recurrences are equivalent to the single matrix
equation
X n+i = AX n,

while the initial conditions assert that

100'
X
° ' 10

These equations enable us to calculate successive vectors X n;
thus X\ = AX 0, X 2 = A XQ, and in general
2
n
X n = A Xo.

In principle this equation provides the solution of our prob-
lem. However the equation is difficult to use since it involves
calculating powers of A; these soon become very complicated
n
and there is no obvious formula for A .
The key observation is that powers of a diagonal matrix
are easy to compute; one simply forms the appropriate power
of each diagonal element. Fortunately the matrix A is diago-
nalizable since it has distinct eigenvalues 2 and 3. Correspond-
ing eigenvectors are found to be I J and I J; therefore the
(\ 2 \
matrix 5 = 1 1 diagonalizes A, and

1
D = S~ AS=i I °