Linear Algebra, Theory And Applications, 2012a

172 SPECTRAL THEORY
for all k ≥ 1. If A is nondefective so that it has a basis of eigenvectors, {v 1 , ··· , v n }
where
Av j = λ j v j
you can write the initial condition x 0 in a unique way as a linear combination of these
eigenvectors. Thus
n∑
x 0 = a j v j
Now explain why
x (k) =
j=1
n∑
a j A k v j =
j=1
n∑
a j λ k j v j
which gives a formula for x (k) , the solution of the dynamical system.
45. Suppose A is an n × n matrix and let v be an eigenvector such that Av = λv. Also
suppose the characteristic polynomial of A is
j=1
det (λI − A) =λ n + a n−1 λ n−1 + ···+ a 1 λ + a 0
Explain why (
A n + a n−1 A n−1 + ···+ a 1 A + a 0 I ) v = 0
If A is nondefective, give a very easy proof of the Cayley Hamilton theorem based on
this. Recall this theorem says A satisfies its characteristic equation,
A n + a n−1 A n−1 + ···+ a 1 A + a 0 I =0.
46. Suppose an n × n nondefective matrix A has only 1 and −1 as eigenvalues. Find A 12 .
47. Suppose the characteristic polynomial of an n×n matrix A is 1−λ n . Find A mn where
m is an integer. Hint: Note first that A is nondefective. Why?
48. Sometimes sequences come in terms of a recursion formula. An example is the Fibonacci
sequence.
x 0 =1=x 1 ,x n+1 = x n + x n−1
Show this can be considered as a discreet dynamical system as follows.
( ) ( )( ) ( ) ( )
xn+1 1 1 xn x1 1
=
, =
x n 1 0 x n−1 x 0 1
Now use the technique of Problem 44 to find a formula for x n .
49. Let A be an n × n matrix having characteristic polynomial
Show that a 0 =(−1) n det (A).
det (λI − A) =λ n + a n−1 λ n−1 + ···+ a 1 λ + a 0