which must exist between the functions f(x)and R(x).These equations are nindepen-
dent relations one can use to solve for the unknown coefficients βiin the polynomial
representation for R(x).If λiis a repeated root of C(λ) = 0, then the equations (10.46)
do not form a set of n-linearly independent equations. However, if the eigenvalue λi
is a repeated root of C(λ) = 0,then the derivative relation
dC
dλ λ=λi= 0must also be true. By differentiating equation (10.44) and evaluating the result at
λ=λi, the equations
df (x)
dx x=λi=dR
dx x=λican be used to determine the constants in the representation of R(A). That is, if mi
is the multiplicity of the characteristic root λi, then the equations
f(λi) = R(λi), df
dx x=λi=dR
dx x=λi,... , dmi− (^1) f
dx mi−^1 x=λi
=d
mi− (^1) R
dx mi−^1 x=λi
, (10 .47)
form a set of mi linear equations. These equations can be used to determine the
coefficients in the remainder polynomial R(x) and consequently the matrix R(A)
representing f(A)can be determined.
Example 10-30. Given the matrix
A=[
2 − 1
−3 4]find the matrix function f(A) = Ak,where kis a positive integer.
Solution: The characteristic equation of Ais
C(λ) = |A−λI |=∣∣
∣∣^2 −λ −^1
−3 4 −λ∣∣
∣∣=λ^2 − 6 λ+ 5 = (λ−5)(λ−1) = 0.Examination of the above equation one can see that λ 1 = 1 and λ 2 = 5 are the char-
acteristic roots or eigenvalues of the given matrix A. The Hamilton–Cayley theorem
requires that C(A) = [0], which implies
A^2 = 6A− 5 ISuccessive multiplications by the matrix Agives
A^3 = 6A^2 − 5 A= 6(6 A− 5 I)− 5 A= 31 A− 30 I
A^4 = 31A^2 − 30 A= 31(6A− 5 I)− 30 A= 156A− 155 I,